The output current for the given step down transformer is 30A.
What is the step down transformer?
A step down transformer is a passive device that converts high voltage power to low voltage power, while the output current is higher than the input current. They are used in power adaptors and rectifiers to decrease the voltage to the desired level. It works according to Faraday's law of Electromagnetic induction.
The current in the windings of a step down transformer is inversely proportional to the voltage in windings as:
Input voltage / Output voltage = Output current / Input current
220 / 110 = Outout current / 15
Output current = 30A
Hence, the output current (30A) obtained is higher than the given input current (15A).
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According to Howard Gardner's theories, who among the following would need to have good linguistic intelligence to be successful?
A psychologist will need to have good linguistic intelligence in other to be successful.
Who is a Psychologist?This is referred to as a professional who specializes in the handling of mental health challenges in individuals.
It is best for such professional to have a good linguistic intelligence as the right words being said to the patient will solve the problem thereby bringing in more success.
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Two planets X and Y travel counterclockwise in circular orbits about a star, as seen in the figure.
The radii of their orbits are in the ratio 4:3. At some time, they are aligned, as seen in (a), making a straight line with the star. Five years later, planet X has rotated through 88.0°, as seen in (b). By what angle has planet Y rotated through during this time?
According to mathematics, the planet's angle is stated as
dY=704 degrees.
What is the current rotational angle of planet Y?We may demonstrate this by using Kepler's third law, which asserts that a planet's orbit squared is a function of cubed radius.
The equation for the period is often expressed numerically as
[tex](periodX / periodY)^2 = (radius X / radius Y)^3[/tex]
Therefore
(pX / pY)^2 = 4^3
(pX / pY)^2 = 64
[tex]\sqrt{(pX / pY )^2}= \sqrt{64}[/tex]
pX / pY=8
In conclusion, planet Y travels 8 times further than planet X does in the same amount of time since one orbit on planet X takes 8 times longer to complete.
planet Y travels ;
dY=8 * 88.0
dY= 704 degrees
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A satellite is in a circular orbit very close to the surface of a spherical planet. The period of the orbit is 2.35 hours. What is density of the planet? Assume that the planet has a uniform density.
The density of the planet is determined as 1,974.26 kg/m³.
Density of the planet
√(⁴/₃πGρ) = 2π/(2.35 x 3600)
where;
ρ is density of the planetG is universal gravitation constant√(⁴/₃πGρ) = 2π/(2.35 x 3600)
√(⁴/₃πGρ) = 2π/8460
(⁴/₃πGρ) = (2π/8460)²
⁴/₃πGρ = 4π²/(8460)²
ρ = 12π/(8460² x 4G)
ρ = (12π) / (8460² x 4 x 6.67 x 10⁻¹¹)
ρ = 1,974.26 kg/m³
Thus, the density of the planet is determined as 1,974.26 kg/m³.
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A uniform beam of length L = 2.2 m and mass M = 49 kg has its lower end fixed to pivot at a point P on the floor, making an angle θ = 18° as shown in the diagram. A horizontal cable is attached at its upper end B to a point A on a wall. A box of the same mass M as the beam is suspended from a rope that is attached to the beam one-fourth L from its upper end.
What is the y-component Py of the force, in newtons, exerted by the pivot on the beam?
Write an expression for the tension T in the horizontal cable AB.
What is the x-component Px of the force, in newtons, exerted by the pivot on the beam?
I cannot figure part C
Hi there!
Part A.
To solve this part, all we need to do is a summation of vertical forces.
We have the following acting on the beam :
- Force of gravity (Fg, down)
- Force of tension from the rope holding the box (T, down)
- Force exerted by pivot (Py, up)
These sum to zero because the beam is not accelerating vertically.
[tex]\Sigma F = -F_g - T + P_y = 0[/tex]
[tex]P_y = F_g + T[/tex]
The tension force is equal to the box's weight because the forces on the box are balanced. Let's use values and solve.
[tex]P_y = 49(9.8) + 49(9.8) = \boxed{960.4N}[/tex]
Part B.
We must begin by doing a summation of torques. Placing the pivot point at the pivot, we have the following present:
- Force of gravity acting at the center of mass of the rod (CC, at L/2)
- The tension of the horizontal cable acting at the end of the rod (CCW, at L)
- The force of tension in the rope holding the box (CC, at 3L/4)
Since the rod is not rotating, these torques sum to zero.
The equation for torque is:
[tex]\tau = r \times F[/tex]
This is a cross-product, and you must find the lever arm (perpendicular distance between pivot and line of action of force). We will need to use trigonometry for this.
Now, let's find the torque from all three of these forces.
- Force of gravity at center:
The perpendicular distance between the force of gravity and the pivot point is the cosine with respect to the angle made with the floor.
[tex]\tau = Mg\frac{L}{2}cos(\theta) = 49(9.8)*\frac{2.2}{2} cos(18) = 502.367 Nm[/tex]
- Tension of horizontal cable:
The lever arm is the sine with respect to the angle. We will still have to solve for the value of 'T'.
[tex]\tau = TLsin\theta = T(2.2)sin(18) = 0.68T[/tex]
- Tension of rope holding box:
The tension is equal to the weight of the box since the box isn't accelerating. Thus, the torque would be:
[tex]\tau = Mg(\frac{3L}{4}) = 49(9.8)*\frac{3(2.2)}{4}cos(18) = 753.55 Nm[/tex]
Summing with clockwise torques + and counterclockwise -:
[tex]\Sigma \tau = 502.367 + 753.55 - 0.68 T = 0 \\\\1255.917 = 0.68T\\T = \boxed{1847.38 N}[/tex]
Part C.
This part is a lot easier than it seems. All we need to do is a summation of horizontal forces.
We only have two:
- The horizontal tension in the cable to the left (1847.38 N)
- The horizontal force exerted by the pivot on the beam to the right
These two balance out because there is no acceleration of the beam horizontally, so:
[tex]\Sigma F = P_x - T = 0 \\\\P_x = T\\\\P_x = \boxed{1847.38 N}[/tex]
**to the right
Answer:
a) 960.4 N
b) T= 5/4 Mg CotanΘ
c) 1847. 38
Explanation:
a) Py= 2Mg
=2(49 x 9.8)
= 960.4
b) T= (Mg x 1/2 x cos Θ + Mg x 3/4 x cos Θ) / sin Θ
T= 5/4 X Mg cotanΘ
c) T= (5/4) x (49 x 9.8) cotan (18)
T= 1847.37954
= 1847.38
A blue train of mass 50 kg moves at 4 m/s toward a green train of 30 kg initially at rest. What is the initial momentum of the blue and green train combined?
A. 20 kgm/s
B. 50 kgm/s
C. 0 kgm/s
D. 200 kgm/s
The correct option is D. The initial momentum of the blue and green train combined during the collision is 200 kgm/s.
Initial momentum of the blue and green trainApply the principle of conservation of linear momentum as follows;
Pi = m1v1 + m2v2
where;
m1 is mass of blue trainm2 is mass of green trainv1 is velocity of blue trainv2 is velocity green trainPi is the initial momentum of the two trainsPi = (50 x 4) + 30(0)
Pi = 200 kgm/s
Thus, the initial momentum of the blue and green train combined is 200 kgm/s.
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A 67-kg skier grips a moving rope that is powered by an engine and is pulled at constant speed to the top of a 23∘ hill. The skier is pulled a distance x = 300 m along the incline and it takes 2.0 min to reach the top of the hill.
If the coefficient of kinetic friction between the snow and skis is μk = 0.10, what horsepower engine is required if 30 such skiers (max) are on the rope at one time?
Express your answer using two significant figures.
The required horsepower engine is 32 horsepower.
What is the force of the 30 skier?The force of the 30 skiers is calculated as follows:
Force = mass * accelerationMass of the skiers = 30 * 67kg = 2010 kg
Net force acting on the skiers along the x-axis
Fx = mgsinθ + f --- (1)where f is the frictional force
The kinetic frictional force, f = μN
where
μ = The coefficient of the kinetic friction
N = normal reaction
Net force acting on the skiers along y axis, the
Fy = ma
N = mg cos θ
Substituting for N above
f = μk mg cos θ
Substituting for f in (1)
F = mg sin θ + μk mg cos θ
F = mg(sinθ + μk cos θ)
Work done by the engine in pilling up the skiers, W = Fx
W = mg ( sinθ + μk cos θ)x
x = 300 m
W = (2010 kg) (9.81 m/s²) (sin 23° + (0.10) cos 23°) (300 m)
Work done, W = 2.86 * 10⁶ J
Time taken, t = 2.0 * 60sec = 120 s
Power = Work done/time taken1 horsepower = 746 W
Power = 2.86 * 10⁶/ 120 * 1/746
Power = 31.9 horsepower
In conclusion, the power of the engine is the ratio of the work done and time taken.
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A treasure chest full of silver and gold coins is being lifted from a pirate ship to the shore using two ropes as shown in the figure. The mass of the treasure chest is 75.6 kg.
The tension in Rope A is 7.42x10^2 N, and the tension Rope B carries is 7.52x10^2 N.
What is the tension in rope C?
Answer: A treasure chest full of silver and gold coins is being lifted from a pirate ship to the shore using two ropes as shown in the figure. The mass of the treasure chest is 75.6 kg. The tension in Rope A is 7.42x10^2 N, and the tension Rope B carries is 7.52x10^2 N. Then, the tension in rope C is 376N
Explanation: To find the correct answer, we have to know more about the Basic forces that acts upon a body.
What is force and which are the basic forces that acts upon a body?A push or a pull which changes or tends to change the state or rest, or motion of a body is called Force.Force is a polar vector as it has a point of application.Positive force represents repulsion and the negative force represented attraction.There are 3 main forces acting on a body, such as, weight mg, normal reaction N, and the Tension or pulling force.How to solve the problem?Given that,
[tex]T_A=7.42*10^2N\\T_B=7.54*10^2N[/tex]
From the free body diagram, we get Tension in the rope C as,[tex]T_C=T_B sin30\\thus,\\T_C=7.52*10^2*0.5=376N[/tex].
Thus, we can conclude that the tension in the rope c will be 376N.
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376N is the tension in rope C.
In order to determine the right response, we must have a better understanding of the fundamental forces that affect a body.
What exactly is force, and what are the fundamental forces that affect a body?Force is a push or a pull that modifies or tends to modify the condition, rest, or motion of a body.Given that it has a point of application, force is a polar vector.Repulsion is represented by positive force, and attraction by negative force.A body is subject to three main forces: weight mg, normal response N, and tension or pulling force.How can the issue be resolved?Tension in the rope C is determined from the free body diagram as,[tex]T_c=T_Bsin30\\T_c=7.52*10*2*0.5=376N[/tex]
Thus, we can infer that the rope's c tension will be 376N.
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A rigid, nonconducting tank with a volume of 4 m3 is divided into two unequal parts by a thin membrane. One side of the membrane, representing 1/3 of the tank, contains nitrogen gas at 6 bar and 100oC, and the other side, representing 2/3 of the tank, is evacuated. The membrane is ruptured and the gas fills the tank. (a) What is the final temperature of the gas? How much work is done? Is the process reversible? (b) How much work is done if the gas is returned to its original state by a reversible process? Assume nitrogen ideal gas for which Cp = (7/2) R and Cv = (5/2)R.
The final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.
To find the answer, we need to know about the thermodynamic processes.
How to find the final temperature of the gas?Any processes which produce change in the thermodynamic coordinates of a system is called thermodynamic processes.In the question, it is given that, the tank is rigid and non-conducting, thus, dQ=0.The membrane is raptured without applying any external force, thus, dW=0.We have the first law of thermodynamic expression as,[tex]dU=dQ-dW[/tex]
Here it is zero.[tex]dU=0[/tex],
As we know that,[tex]dU=C_pdT=0\\\\thus, dT=0\\\\or , T=constant\\\\i.e, T_1=T_2[/tex]
Thus, the final temperature of the system will be equal to the initial temperature,[tex]T_1=T_2=100^0C=373K[/tex]
How much work is done?We found that the process is isothermal,Thus, the work done will be,[tex]W=RT*ln(\frac{V_2}{V_1} )=373R*ln(\frac{4}{\frac{4}{3} })\\ \\W=409.8R J[/tex]
Where, R is the universal gas constant.
What is a reversible process?Any process which can be made to proceed in the reverse direction is called reversible process.During which, the system passes through exactly the same states as in the direct process.Thus, we can conclude that, the final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.
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The system's final temperature will be 373K, which is the same as its starting temperature. The system exerts 409.8R Joules of work.
We need to understand the thermodynamic processes in order to locate the solution.
How can I determine the gas's final temperature?Thermodynamic processes are any actions that result in modifications to a system's thermodynamic coordinates.Given that the tank is stiff and non-conducting, the answer to the question is that dQ=0.Without using any external force, the membrane is torn; hence, dW=0.The first law of thermodynamics is expressed as follows:[tex]dU=dQ-dW[/tex] , It is 0 here.
As we are aware,[tex]dU=C_pdT=0\\dT=0\\T=constant\\T_1=T_2=373K[/tex]
As a result, the system's final temperature will be equal to its starting temperature.
How much work is expended?The process is isothermal, as we discovered.As a result, the work will be,[tex]W=RT ln(\frac{V_2}{V_1} )=373R*ln(3 )\\W=409.8R Joules[/tex]
R is the gaseous universal constant.
A reversible process is what?Reversible processes are any operations that have the ability to be reversed.The system goes through the exact same states as it did during the direct procedure throughout this time.Thus, we can draw the conclusion that the system's end temperature will be 373K, the same as its starting temperature. The system exerts 409.8R Joules of work.
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I AM OFFERING 100 POINTS!!! I REALLY NEED HELP!!!!!!
Part C
Now prepare the cold sand and cold water samples from part A:
Fill a 100-milliliter container with 50 grams of sand. Fill a 100-milliliter container with 50 grams of cold tap water. Fill the last 100-milliliter container with 100 grams of cold tap water. Use the scale to measure the masses.
*Image should be there*
2. Pour all the ice cubes into a tub, and fill it with cool tap water to a depth of 2 inches. Place the sand and water samples in the ice water. Cover the entire tub.
*Image should be there*
3. Every 15 minutes, remove the cover and check the temperatures of the samples using the three thermometers. Wait 30 seconds before recording the thermometer reading. Once the temperatures of the three samples are no more than a degree apart, record the temperatures.
*Image should be there
Answer:
Every 15 minutes, remove the cover and check the temperatures of the samples using the three thermometers. Wait 30 seconds before recording the thermometer reading. Once the temperatures of the three samples are no more than a degree apart, record the temperatures.
(three 100-milliliter containers (each with a thermometer) in an ice bath inside an uncovered basin, with one container holding 50 grams of sand, one holding 50 grams of cold water, and one holding 100 grams of cold water)
Next, prepare the hot water.
-Fill each of the three 200-milliliter containers with 100 grams of hot tap water. Measure the mass using the mass scale.
(100 grams of hot water in a 200-milliliter container atop a mass scale)
-Prepare a hot-water bath by boiling water in a pot. Use the heat mitts to pour the hot water from the pot into the second tub. Fill the tub to a depth of about 2 inches. Carefully place the three containers of hot water into the bath without submerging them. Cover and wait for five minutes until the temperatures stabilize.
(three 200 mL containers (each holding 100 grams hot water) in a covered tub containing 2 inches of boiling water)
Prepare to mix the cold samples with hot water:
-Have three empty 300-milliliter mixing containers and three thermometers ready. Timing is important. Uncover the cold-water bath, and pour each cold sample into a different mixing container.
(three 300-milliliter containers (each with a thermometer), one holding 50 grams of sand, one holding 50 grams of cold water, and one holding 100 grams of cold water)
-Uncover the hot-water bath, and use a heat mitt to remove the three containers of hot water from the hot-water bath. Pour 100 grams of hot water into each mixing container.
(three 300 mL containers (each with a thermometer), one holding a 150-gram mixture of sand and hot water, one holding a 150-gram mixture of cold water and hot water, and one holding a 200-gram mixture of cold water and hot water)
Explanation:
"You can't see the forest for the trees" might seem an appropriate analogy for astronomers attempting to determine the shape of the Milky Way galaxy when we are in fact located inside the galaxy. Discuss techniques used by astronomers to determine the type of galaxy in which we live, why it is so difficult to determine the shape of our galaxy, and where our Sun is located in our galaxy?
Some of the techniques used by astronomers to determine the type of galaxy in which we live are:
radio, optical, infraredx-ray astronomyWhat is Astronomy?This refers to the study of heavenly bodies and space and other things that space is made up of.
Hence, we can see that the reason why it is so difficult to determine the shape of our galaxy is that astronomers can only infer its presence from the motions of stars in the galaxy, and a precise shape is difficult to be determined.
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The pulley is assumed massless and frictionless and rotates freely about its axle. The
blocks have masses m1 = 40 g and m2 = 20 g, and block m1 is pulled to the right by a
horizontal force of magnitude F = 0.03 N. Find the magnitude of the acceleration of block m2 and the tension in the cord if the surface is frictionless
The magnitude of the acceleration of block m2 is 0.5 m/s² and the tension in the cord if the surface is frictionless is 0.02 N.
Acceleration of block m2
The acceleration of bock m2 is calculated from Net force exerted by the pulley.
F - T = m2a
F - m1a = m2a
F = m2a + m1a
F = a(m2 + m1)
a = F/(m1 + m2)
a = (0.03) / (0.04 + 0.02)
a = 0.5 m/s²
Tension in the cordT = m1a
T = 0.04 x 0.5
T = 0.02 N
Thus, the magnitude of the acceleration of block m2 is 0.5 m/s² and the tension in the cord if the surface is frictionless is 0.02 N.
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. A bargain hunter purchases a "gold" crown at a flea market. After she gets home, she hangs it from a scale and finds its weight to be 7.84 N. She then weighs the crown while it is immersed in water, as shown below in Figure, and now the scale reads 6.86 N. Is the crown made of pure gold? Find the density of the crown and compare it to the density of gold. wag PODECID LE B water
Answer:
Average density of the crown: approximately [tex]8\; {\rm g \cdot mL^{-1}}[/tex].
Hence, if this crown contains no empty space, this crown is not made of pure gold.
Explanation:
Let [tex]m(\text{crown})[/tex] and [tex]V(\text{crown})[/tex] denote the mass and volume of this crown. Let [tex]g[/tex] denote the gravitational field strength.
Since this crown is fully immersed in water, the volume of water displaced [tex]V(\text{water, displaced})[/tex] is equal to the volume of this crown:
[tex]V(\text{water, displaced}) = V(\text{crown})[/tex].
The mass of water displaced would be:
[tex]\begin{aligned}m(\text{water, displaced}) &= \rho(\text{water}) \, V(\text{water, displaced}) \\ &= \rho(\text{water}) \, V(\text{crown})\end{aligned}[/tex].
The weight of water displaced would be [tex]m(\text{water, displaced})\, g = \rho(\text{water}) \, V(\text{crown})\, g\end{aligned}[/tex].
The buoyancy force on this crown is equal to the weight of water that this crown displaced:
[tex]F(\text{buoyancy}) = \rho(\text{water}) \, V(\text{crown})\, g[/tex].
The magnitude of this buoyancy force is [tex]7.84\; {\rm N} - 6.86\; {\rm N} = 0.98\; {\rm N}[/tex]. Rearrange the equation for buoyancy to find [tex]V(\text{crown})[/tex]:
[tex]\begin{aligned} V(\text{crown}) &= \frac{F(\text{buoyancy}) }{\rho(\text{water}) \, g}\end{aligned}[/tex].
Since the weight of this crown is [tex]\text{weight}(\text{crown}) = m(\text{crown})\, g[/tex], the mass of this crown would be [tex]m(\text{crown})= \text{weight}(\text{crown}) / g[/tex].
The average density of this crown would be:
[tex]\begin{aligned}\rho(\text{crown}) &= \frac{m(\text{crown})}{V(\text{crown})} \\ &= \frac{\text{weight}(\text{crown}) / g}{F(\text{buoyancy}) / (\rho(\text{water})\, g)} \\ &= \frac{\text{weight}(\text{crown})}{F(\text{buoyancy})}\, \rho(\text{water}) \\ &= \frac{7.84\; {\rm N}}{0.98\; {\rm N}}\times 1.000 \; {\rm g\cdot mL^{-1}} \\ &= 8.0\; {\rm g \cdot mL^{-1}}\end{aligned}[/tex].
The density of pure gold is significantly higher than [tex]8.0\; {\rm g\cdot mL^{-1}}[/tex]. Hence, if this crown contains no empty space (i.e., no air bubble within the crown), the crown would not be made of pure gold.
A block of mass
m = 2.50 kg
is pushed
d = 2.10 m
along a frictionless horizontal table by a constant applied force of magnitude
F = 14.0 N
directed at an angle
= 25.0°
below is a photo of the horizontal as shown in the figure below.
(a) Determine the work done by the applied force.
_____J
(b) Determine the work done by the normal force exerted by the table.
_____J
(c) Determine the work done by the force of gravity.
_____J
(d) Determine the work done by the net force on the block.
_____J
(a) The work done by the applied force is 26.65 J.
(b) The work done by the normal force exerted by the table is 0.
(c) The work done by the force of gravity is 0.
(d) The work done by the net force on the block is 26.65 J.
Work done by the applied force
W = Fdcosθ
W = 14 x 2.1 x cos25
W = 26.65 J
Work done by the normal forceW = Fₙd
W = mg cosθ x d
W = (2.5 x 9.8) x cos(90) x 2.1
W = 0 J
Work done force of gravityThe work done by force of gravity is also zero, since the weight is at 90⁰ to the displacement.
Work done by the net force on the block∑W = 0 + 26.65 J = 26.65 J
Thus, the work done by the applied force is 26.65 J.
The work done by the normal force exerted by the table is 0.
The work done by the force of gravity is 0.
The work done by the net force on the block is 26.65 J.
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Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude of the tension in the string labeled C is 56.3 N. Calculate the magnitude of the tension in the string marked A.
The magnitude of the tension in the string marked A is 39.5 N.
What is the tension in A?The tension in A is determined thus:
The angle at A, θ = tan⁻¹(3/8) = 20.56
When extrapolated below negative x, the angle at B, α = tan⁻¹(5/4) = 51.34
When extrapolated below negative x, the angle at C, β = tan⁻¹(1/6) = 9.46
Taking the horizontal components of tension;
56.3cos(9.46) = A * cos(20.56) + B * cos(51.34)
0.6247B= 55.53 - 0.936A
B = (55.53 - 0.936A)/0.6247 ----(1)
Taking the vertical components of tension;
56.3 * sin(9.46) + A * sin(20.6) = B * sin(51.3)
9.25 + 0.35A = 0.78B ---- (2)
substitute the value (1) in (2)
9.25 + 0.35A = 0.78{(55.53 - 0.936A)/0.6247}
(9.25 + 0.35A) * 0.6247 = 43.31 - 0.73A
0.22A + 0.73A = 43.31 - 5.78
0.93A = 37.53
A = 39.5 N
In conclusion, the tension in A is determined by solving for the vertical and horizontal components of tension.
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For light of wavelength 589 nm, calculate the critical angles for the following substances when surrounded by air.
polystyrene
flint glass
(1) The critical angles for the air is 90⁰,
(2) The critical angles for the polystyrene is 39.1 ⁰ and
(3) The critical angles for the flint glass is 37.2 ⁰.
Critical angles for the different mediumθc = sin⁻¹( 1/η)
where;
η is the refractive indexRefractive index of air = 1
Refractive index of polystyrene = 1.5865
Refractive index of flint glass = 1.655
Critical angles for airθc = sin⁻¹( 1/1)
θc = 90⁰
Critical angles for polystyreneθc = sin⁻¹( 1/1.5865)
θc = 39.1 ⁰
Critical angles for flint glassθc = sin⁻¹( 1/1.655)
θc = 37.2 ⁰
Thus, the critical angles for the air is 90⁰, the critical angles for the polystyrene is 39.1 ⁰ and the critical angles for the flint glass is 37.2 ⁰.
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A pendulum in motion can either swing from side to side or turn in a continuous circle. The point at which it goes from one type of motion to the other is called the separatrix, and this can be calculated in most simple situations. When the pendulum is prodded at an almost constant rate though, the mathematics falls apart. Is there an equation that can describe that kind of separatrix?
The kind of equation that can be used to differentiate the kind of separatrix that shows change on motion is
H = 2g/l.
What is simple pendulum?A simple pendulum can be defined as the equipment that displays an oscillatory motion when a mass is tied on a rope and is suspended from it.
The various movements that occur using a simple pendulum is translational ( side to side) or continuous circle (oscillatory motion).
The equation that show that a change from one type of motion to another is H = 2g/l.
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Radiation from the sun hits Earth unequally and is absorbed by different materials in varying amounts. This is called
A. the Coriolis Effect
B. a thermocline
OC. El Nino Southern Oscillation
OD. differential heating
Radiation from the sun hits earth unequally and is absorbed by different materials in varying amounts. This is called differential heating.
What is differential heating?Differential heating is the property that causes different surfaces to heat up and cool down at different rates. The earth's surface receives different magnitudes of solar radiation and also the earth's surfaces absorb thermal energy in different magnitudes.
The color, shape, texture, surface and presence of constructions can influence the heating or cooling of the earth.
The earth in the equator line is heated more by solar action than that of the poles, since it receives more amount of radiation per unit area.
In general, dry surfaces heat up and cool down faster than wet ones.
Therefore, we can confirm that when radiation from the sun hits earth unequally and is absorbed by different materials in varying amounts is called differential heating.
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A train traveling initially at 16 m/s is under constant acceleration of 2 m/2. How far will it travel in 20 s? What will its final velocity be
Answer: A train traveling initially at 16 m/s is under constant acceleration of 2 m/2. At a distance of 720m it will travel in 20 s, and the final velocity will be 56m/s.
Explanation: To find the answer, we need to know about uniformly accelerated motion.
How to solve the problem?Given that,[tex]u=16m/s\\a=2 m/s^2\\t=20s[/tex]
We have to find the distance travelled by the train.As we have,[tex]S=ut+\frac{1}{2}at^2[/tex]
Substituting values, we get,[tex]S=(16*20)+\frac{2*20^2}{2} =720 m.[/tex]
We have the equation for final velocity as,[tex]v^2=u^2+2aS\\thus,\\v=\sqrt{u^2+2aS} =\sqrt{16^2+(2*2*720)} =56 m/s.[/tex]
Thus, we can conclude that, a train traveling initially at 16 m/s is under constant acceleration of 2 m/2. At a distance of 720m it will travel in 20 s, and the final velocity will be 56m/s.
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A train experiencing constant acceleration of 2 m/s^2 is moving at an initial speed of 16 m/s. It will go 720 meters in 20 seconds, with a final velocity of 56 m/s.
Understanding uniformly accelerated motion is necessary in order to determine the solution.
How can the issue be resolved?We need to determine how far the train has traveled. We have,[tex]S=ut+\frac{1}{2}at^2 \\S=720m\\where,\\u=16m/s, a=2m/s^2,t=20s[/tex]
The formula for final velocity is as follows:[tex]v^2-u^2=2aS\\v=\sqrt{u^2+2aS} \\v=56m/s[/tex]
Thus, we may say that a train moving at 16 m/s initially experiences constant acceleration of 2 m/2. It will go 720 meters in 20 seconds, with a final velocity of 56 meters per second.
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A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long as 2.20 m have been recorded. If a frog jumps 2.20 m and the launch angle is 36.5°, find the frog's launch speed and the time the frog spends in the air. Ignore air resistance.
(a)the frog's launch speed (in m/s)
(b)the time the frog spends in the air (in s)
The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
To find the answer, we need to know about the time of flight and range of projectile motion.
What's the expression of range of a projectile motion?Range = U²× sin(2θ)/gU= initial velocity, θ= angle of projectile and g= acceleration due to gravity U=√{Range×g/sin(2θ)}Here, range= 2.20m, = 36.5°U= √{2.20×9.8/sin(73)}U= √{2.20×9.8/sin(73)} = 22.5m/s
What's the expression of time of flight in projectile motion?Time of flight= (2×U×sinθ)/g So, T= (2×22.5×sin36.5°)/9.8= 2.73 s
Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
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Find the minimum diameter of an l = 19.3 m long steel wire that will stretch no more than 7.32 mm when a mass of 350 kg is hung on the lower end. (Hint: The Young's modulus of steel is 200.0 GPa.)
The minimum diameter of the steel wire is 7.58 mm.
Stress applied to the steel wireYoung's modulus = stress/strain
strain = e/l = (7.32 x 10⁻³ m) / (19.3 m) = 3.79 x 10⁻⁴
stress = Young's modulus x strain
stress = 200 x 10⁹ N/m² x 3.79 x 10⁻⁴ = 7.59 x 10⁷ N/m²
Area of the wirestress = Force/Area
Area = Force/stress
Area = mg/stress
Area = (350 x 9.8) / (7.59 x 10⁷)
Area = 4.519 x 10⁻⁵ m²
Minimum diameter of the wireArea = πd²/₄
πd²/₄ = 4.519 x 10⁻⁵ m²
πd² = 4(4.519 x 10⁻⁵)
d² = (4 x 4.519 x 10⁻⁵)/π
d² = 5.75 x 10⁻⁵
d = √(5.75 x 10⁻⁵)
d = 7.58 x 10⁻³ m
d = 7.58 mm
Thus, the minimum diameter of the steel wire is 7.58 mm.
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A rectangular loop of wire with dimensions 1.80 cm by 9.00 cm and resistance 0.800 Ω is being pulled to the right out of a region of uniform magnetic field. The magnetic field has magnitude 2.60 T and is directed into the plane of (Figure 1) .
a) At the instant when the speed of the loop is 3.00 m/s and it is still partially in the field region, what is the magnitude of the force that the magnetic field exerts on the loop?
b) What is the direction of the force that the magnetic field exerts on the loop?
(a) The magnitude of the force that the magnetic field exerts on the loop is 0.042 N.
(b) The direction of the force that the magnetic field exerts on the loop will be out of the plane.
Magnetic force exerted on the loop
F = BIL
where;
I is current in the loopL is length of the loopemf = BVb
where;
b is breadth of the loopV is velocityB is magnetic fieldemf = 2.6 x 3 x 0.018 = 0.1404 V
Current in the loop, I = emf/R = 0.1404/0.8 = 0.18 A
Magnetic forceF = BIL
where;
L is length of the loopF = 2.6 x 0.18 x 0.09 = 0.042 N
Direction of the forceThe direction of the force that the magnetic field exerts on the loop will be out of the plane.
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Why is air resistance friction not useful for an airplane?
A. Causes turbulence
B. Speeds it up
C. Slows it down
D. Creates heat
The correct option is C. Air resistance friction is not useful for an airplane because it slows it down.
What is air resistance?Air resistance is the opposition to motion of an object caused by air flow.
High air resistance may cause turbulence during the motion of an air plane. This can lead to engine failure or some mishap during flight.
Air resistance is also a type of friction between air and another material such as airplane.
Effect of air resistance on airplaneFriction between the air and the plane slows the airplane down. This is known as air resistance, or drag.
The faster an object travels through the air, the more it has to fight against drag.
Thus, air resistance friction is not useful for an airplane because it slows it down.
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The head of a hammer with a mass of 1.5 kg is allowed to fall onto a nail from a height of 0.70 m .
What is the maximum amount of work it could do on the nail?
Express your answer to two significant figures and include the appropriate units.
Why do people not just "let it fall" but add their own force to the hammer as it falls?
a)To make the trajectory of the fall more precise.
b)To increase the power of the hammer.
Maximum work done by the hammer on the is 10 J.
Force is added to the hammer in order to increase the power of the hammer; option B.
What is work done?Work done is defined as the product of force and the distance travelled by force.
Energy is used to do work.
The potential energy of the hammer is converted to work done.
Potential energy = mghThe maximum amount of work it could do on the nail is given below:
Maximum work = 1.5 * 9.81 * 0.7
Maximum work = 10 J
Force is added to the hammer in order to increase the power of the hammer.
In conclusion, the work done by the hammer is obtained from the potential energy of the hammer.
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what is the wavelength of a tubas A2(110hz) tone in air at standard conditions/
The wavelength of A2 with frequency (110hz) is 3.01 m.
What is the wavelength of a sound wave?Wavelength of a sound wave is the distance between successive similar points in the wave such as rarefactions or compressions.
A tuba is a musical instrument that produces sound waves.
Wavelength is related to frequency and velocity by the formula below:
Wavelength = velocity/frequencyThe wavelength of A2 with frequency (110hz) is 3.01 m.
In conclusion, the wavelength of a wave is inversely proportional to frequency.
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For the circuit shown below with a 12.0 V battery, find the total current through the battery and the current through each resistor with the following resistances: R1=5.0 Ohms, R2=10.0 Ohms, R3=15.0 Ohms, R4=33.0 Ohms. Show all your work!
The total current through the battery is 4.77 A.
The current through resistor R1 is 2.4 A.
The current through resistor R2 is 1.2 A
The current through resistor R3 is 0.8 A
The current through resistor R4 is 0.36 A.
Total resistance of the circuitThe total resistance of the circuit is calculated as follows;
1/Rt = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄
1/Rt = 1/5 + 1/10 + 1/15 + 1/33
1/Rt = 0.397
Rt = 1/0.397
Rt = 2.518 ohms
Total current flowing in the circuitI = V/Rt
I = 12/2.518
I = 4.77 A
Current in resistor R₁I₁ = V/R₁
I₁ = 12/5
I₁ = 2.4 A
Current in resistor R₂I₂ = 12/10
I₂ = 1.2 A
Current in resistor R₃I₃ = 12/15
I₃ = 0.8 A
Current in resistor R₄I₄ = 12/33
I₄ = 0.36 A
Thus, the total current through the battery is 4.77 A.
The current through resistor R1 is 2.4 A.
The current through resistor R2 is 1.2 A
The current through resistor R3 is 0.8 A
The current through resistor R4 is 0.36 A.
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What is the speed of a giraffe that has a
mass of
75 kg and a kinetic energy of 600 J?
O 2 m/s
O
4 m/s
O 8 m/s
O 16 m/s
Name two units for measuring the diameter of nucleus atom.
The two units for measuring the diameter of nucleus atom are femtometre and metre.
How do you measure the size of the nucleus ?
Nucleus size is expressed in fermi, often known as femtometers. between a lighter and a heavier nucleus. Despite its modest size, the nucleus contains the majority of an atom's mass. The weight or mass of the atom's nucleus and neutrons are determined by neutrons.
femtometre (fm), which equals [tex]10^{-15}[/tex] metre.
A nucleus' diameter largely depends as to how many particles it contains, from about 4 fm for a light nucleus like carbon to 15 fm for a heavy nucleus as lead.
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The paths of two small satellites, M1 = 4.00 kg and M2 = 1.00 kg, are shown below, drawn to scale, with M1 corresponding to the circular orbit. They orbit around a massive star, also shown below. The orbits are in the plane of the paper.
The period of M1 is T1 = 34.0 years. Calculate the period of M2, in years.
The period of M2, in years. is mathematically given as
T2= 134.3968years
What is the period of M2, in years.?
M1 = 4.00 kg
M2 = 1.00 kg
T1 = 34.0 years.
Generally, the equation for is mathematically given as
T2 = T1 (a2/a1)3/2
T2= 34.0 * (5/2)^{3/2}
T2= 134.3968years
In conclusion, the period of M2, in years
T2= 134.3968years
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A tennis player tosses a tennis ball straight up and then catches it after 1.77 s at the same height as the point of release.
(a) What is the acceleration of the ball while it is in flight?
magnitude ______ m/s2
Which direction?
1. Upward
2. Downward
3. The magnitude is zero
(b) What is the velocity of the ball when it reaches its maximum height?
magnitude _______________ m/s
Which direction?
1. Upward
2. Downward
3. The magnitude is zero
(c) Find the initial velocity of the ball.
______m/s upward
(d) Find the maximum height it reaches.
___________m
(a) The acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.
(b) The velocity of the ball when it reaches its maximum height is zero.
(c) The initial velocity of the ball is 17.36 m/s.
(d) The maximum height it reaches is 15.36 m.
Acceleration of the ball
The acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.
Velocity of the ball at maximum heightThe velocity of the ball decreases as the ball moves upwards and eventually becomes zero at maximum height.
Initial velocity of the ballv = u - gt
at maximum height, final velocity, v = 0
0 = u - gt
u = gt
u = 9.81 x 1.77
u = 17.36 m/s
Maximum height reached by the projectileh = ut - ¹/₂gt
h = 17.36(1.77) - ¹/₂(9.81)(1.77²)
h = 15.36 m
Thus, the acceleration of the ball while it is in flight has a magnitude of 9.81 m/s2 in downward direction.
The velocity of the ball when it reaches its maximum height is zero.
The initial velocity of the ball is 17.36 m/s.
The maximum height it reaches is 15.36 m.
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The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down. Initial speed of the rock = 1.92×101 m/s. (Glob has no atmosphere, so no energy is lost to air friction. G = 6.67×10-11 Nm2/kg2.)
A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Calculate the speed of the satellite.
The orbiting speed of the satellite orbiting around the planet Glob is 60.8m/s.
To find the answer, we need to know about the orbital velocity a satellite.
What's the expression of orbital velocity of a satellite?Mathematically, orbital velocity= √(GM/r)G= gravitational constant= 6.67×10^(-11) Nm²/kg², M = mass of sun , r= radius of orbitWhat's the orbital velocity of the satellite in a circular orbit with a radius of 1.45×10⁵ m around the planet Glob of mass 7.88×10¹⁸ kg?Here, M= 7.88×10¹⁸ kg, r= 1.45×10⁵ mOrbital velocity of the orbiting satellite = √(6.67×10^(-11)×7.88×10¹⁸/1.45×10⁵)= 60.8m/s
Thus, we can conclude that the speed of the satellite orbiting the planet Glob is 60.8m/s.
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