Since you aren't painting on top of the pool, you are painting the pool's four sides, its bottom, and its five faces. 104 square meters is the surface area. The total cost of the paint was $624.
What does surface area mean?A three-dimensional shape's surface area is the sum of all of its faces. Finding the area for every face and adding them together gives us the land area of a shape.
The water-resistant paint required for the pool, which is 8 meters long, 6 meters wide, and 2 meters deep, costs $6 per sq meter.
A) Because you aren't painting on top of the pool, you are painting the pool's four sides, its bottom, and its five faces.
B) The pool requires 104 square meters of paint.
Just use surface area equation assuming the pools is a rectangular prism:
SA = wl + 2(hl) + 2(hw)
Since we're only including pool's bottom, don't add times 2 for wl.
l = 8 m
w = 6 m
h = 2 m
SA = (6)(8) + 2((2)(8)) + 2((2)(6))
SA = 104 sq m
C) The total cost of the paint was $624. $624 is equivalent to $104 × 6 square meters.
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A 269 kg weather balloon is designed to lift a 2910 kg package. What volume should the balloon have after being inflated with helium in order that the total load can be lifted? (Assume standard temperature and pressure, at which the density of air is ρ[tex]_{air}[/tex] = 1.29 kg//[tex]m^{3}[/tex] and the density of helium is ρ[tex]_{He}[/tex] = 0.179 kg/[tex]m^{3}[/tex].)
As, per the buoyancy force,the volume that the balloon should have is 2863[tex]m^{3}[/tex]
What is buoyancy force?
Air buoyancy is the upward force exerted on an object by the air that is displaced by object. Air buoyancy is responsible for the buoyancy created by the displaced air.
[tex]F_{b}[/tex] = -Vρg , where V= volume of the object
ρ = density of the object
g = acceleration due to gravity
[tex]F_{b}[/tex] = buoyant force
The buoyancy force must be equal to the total load lifted
ρ[tex]_{He}[/tex] × V × g + 269 + 2910 = ρ[tex]_{air}[/tex] × V × g
0.179 × V + 3179 = 1.29 V
0.179V + 3179 = 1.29V
0.179V- 1.29V = - 3179
1.11V = 3179
On solving , we get
V = 2863 [tex]m^{3[/tex]
Therefore, the volume that the balloon should have is 2863[tex]m^{3}[/tex].
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What does it mean when we say two objects are in equilibrium
An object is considered to be in a condition of equilibrium when it is balanced with regard to all external forces.
Equilibrium:
An object is considered to be in equilibrium if both its angular acceleration and the acceleration of its center of mass are equal to zero. In layman's terms: The item must either be at rest or moving at a constant speed if it is not accelerating because F = ma (force = mass x acceleration). Even in motion, a body can be in equilibrium. This kind of equilibrium is referred to as a dynamic equilibrium.
A weight suspended by a spring or a brick laying on a flat surface is an example. The equilibrium is unstable if the force with the smallest deviation tends to increase the displacement. As an example, imagine a ball bearing on the edge of a razor blade.
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For an air bag to work, it has to inflate full of nitrogen incredibly fast-within to
milliseconds of the collision. For a 60-liter cylindrical air bag to work property, the
nitrogen gas has to reach a pressure of 2.37 atm. At 25°G, how many moles of
nitrogen gas are needed to pressurize the air bag? Given, 0.0821 L-atm/mol-Kl
5.8 moles of nitrogen gas are needed to pressurize the air bag.
What's the expression of Ideal gas equation?Ideal gas equation is PV=nRTP= pressure, V = volume, n= no. of moles of gas, R= universal gas constant, T = temperature of the gasWhat's the no. of moles of nitrogen present in a 60L air bag at 2.37 atm pressure and 25°C temperature?P= 2.37 atm, V = 60L, R= 0.0821 L-atm/mol-K, T = 25°C = 298Kn= PV/RT= (2.37×60)/(0.0821×298)
= 5.8 moles
Thus, we can conclude that 5.8 moles of nitrogen gas are needed to pressurize the air bag.
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Two long, straight, parallel wires, 10.0 cm apart carry equal 4.00-A currents in the same direction, as shown in (Figure 1).
a) Find the magnitude of the magnetic field at point P1 , midway between the wires.
b) What is its direction?
c) Find the magnitude of the magnetic field at point P2 , 25.0 cm to the right of P1 .
d) What is its direction?
e) Find the magnitude of the magnetic field at point P3 , 20.0 cm directly above P1 .
f) What is its direction?
(a) The magnitude of the magnetic field at point P1 , midway between the wires is 1.005 x 10⁻⁴ T and the direction will be out of the page.
(b) The magnitude of the magnetic field at point P2 , 25.0 cm to the right of P1 is 2.67 x 10⁻⁶ T and the direction is into the page.
(c) The magnitude of the magnetic field at point P3 , 20.0 cm directly above P1 is 3.88 x 10⁻⁶ T and the direction is downwards.
Magnetic field midway between the wiresB = μ/2π[I₁/0.5r + I₂/0.5r]
B = (μ/2π) x (I/0.5r + I/0.5r)
B = (μ/2π) x (2I/0.5r)
B = μI/0.5r
B = 2μI/r
where;
I is current in the wiresr is the distance between the wiresB = (2 x 4π x 10⁻⁷ x 4)/(0.1)
B = 1.005 x 10⁻⁴ T
The direction of the magnetic field is out of the page.
Magnetic field at 25 cm right of P1B = μI/2πd
d = 5 cm + 25 cm = 30 cm
B = (4π x 10⁻⁷ x 4)/(2π x 0.3)
B = 2.67 x 10⁻⁶ T
The direction of the magnetic field is into the page towards P1.
Magnetic field at 20 above P1B = μI/2πd
d = √(20² + 5²)
d = 20.62 cm
B = (4π x 10⁻⁷ x 4)/(2π x 0.2062)
B = 3.88 x 10⁻⁶ T
The direction of the magnetic field is downwards towards P1.
Thus, the magnitude of the magnetic field at point P1 , midway between the wires is 1.005 x 10⁻⁴ T and the direction will be out of the page.
The magnitude of the magnetic field at point P2 , 25.0 cm to the right of P1 is 2.67 x 10⁻⁶ T and the direction is into the page.
The magnitude of the magnetic field at point P3 , 20.0 cm directly above P1 is 3.88 x 10⁻⁶ T and the direction is downwards.
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A uniform meter stick of mass 0.20 kg is pivoted at the 40 cm mark. Where should one hang a mass of 0.50 kg to balance the stick?
Answer:
36 cm
Explanation:
Mass of stick; m1 = 0.20kg at midpoint.
Total length; L=1.0 m
Pivot at 0.40m
Atached mass m2 = 0.50kg
Applying rotational equilibrium we have;
Ʈnet = 0
(m1g) • r1 = (m2g) • r2
(0.2) (0.1m) = (0.5)(x)
x = 0.04m =4cm
measured away from 40cm mark gives a position on the stick of; 40cm - 4cm = 36 cm
A crate with a mass of 175.5 kg is suspended from the end of a uniform boom with a mass of 94.7 kg. The upper end of the boom is supported by a cable attached to the wall and the lower end by a pivot (marked X) on the same wall.
Calculate the tension in the cable.
The tension in the cable is equal to 323.5 N.
What is the tension in the cable?The tension, T in the cable is determined by taking moments about the pivot marked X.
The angles of the boom and the cable with the horizontal are first calculated.
Angle of the boom with horizontal, θ = tan⁻¹(5/10) = 26.56°
The angle of cable with horizontal, B = tan⁻¹(4/10) = 21.80
Taking moments about the pivot:
175.5 * cos 26.56 + 94.7 * cos 26.56 * 0.5 = T (sin(26.56 + 21.80) * 1
Tension = 241.68/0.747
Tension = 323.5 N
In conclusion, the tension in the cable helps to suspend the crate.
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A projectile leaves the ground at an angle of 60° the horizontal.Its kinetic energy is E.Neglecting air resistance, find in terms of E its kinetic energy at the highest point of the motion
The kinetic energy of the projectile at the highest point of its motion will be E/4.
What is Projectile Motion?When a projectile will be thrown obliquely near the surface of the earth, it travels a curved path under uniform acceleration directed toward the center of the Earth. The path of a particle is called a projectile while the motion of a projectile is projectile motion.
Given, the angle of projection with horizontal, [tex]\theta = 60 ^o[/tex]
Consider that 'E' is the initial value of the kinetic energy of the projectile.
The equation for the initial kinetic energy is : [tex]E =\frac{1}{2}mu^2[/tex]
where m is the mass of the given projectile.
The component of the velocity of the projectile in the horizontal direction:
uₓ = u cosθ
uₓ = u cos 60°
uₓ = u/2
From the equation of motion: v = u +at
v = (u/2) + (0) t
v = u/2
The final kinetic energy of the projectile:
[tex]E_f = \frac{1}{2}mv^2[/tex]
[tex]E_f = \frac{1}{2}m(\frac{u}{2} )^2[/tex]
[tex]E_f = \frac{1}{4} (\frac{1}{2}mu^2 )[/tex]
[tex]E_f = \frac{E}{4}[/tex]
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A stone is launched at an angle of 50 degree with initial velocity of 22m/s. Find out its initial and final velocity.
(a) The initial vertical velocity of the stone is 16.85 m/s and the initial horizontal velocity is 14.14 m/s.
(b) The final vertical velocity is 0 and the final horizontal velocity is 14.14 m/s.
Initial vertical velocity
The initial vertical velocity of the stone is calculated as follows;
Vi = Vsinθ
Vi = 22 x sin(50)
Vi = 16.85 m/s
Initial horizontal velocityVxi = V cosθ
Vxi = 22 x cos(50)
Vxi = 14.14 m/s
Final vertical velocity of the stoneVf = Vi - gt
where;
Vf is the final vertical velocity = 0 at maximum heightFinal horizontal velocity of the stoneVfx = Vxi = 14.14 m/s
Thus, the initial vertical velocity of the stone is 16.85 m/s and the initial horizontal velocity is 14.14 m/s.
The final vertical velocity is 0 and the final horizontal velocity is 14.14 m/s.
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Question 3 of 10
What is the primary means by which heat is transferred through fluids?
O A. Direct contact of particles
OB. Radiation
OC. Electromagnetic waves
OD. Convection currents
The primary means by which heat is transferred through fluids is convection currents (option D).
What is convection current?Convection is the transmission of heat in a fluid by the circulation of currents.
Heat can be transferred by different methods depending on the medium. Fluids like gases and liquids transfer heat through the process of convection.
Therefore, the primary means by which heat is transferred through fluids is convection currents.
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A bullet of mass 50g moving with an initial speed of 500m/s penetrates a wall and comes to rest at in 0.2seconds. calculate the deceleration of the bullet over the 0.2second.
The deceleration of the bullet over 0.2 second, given the data from the question is –2500 m/s²
What is acceleration?This is defined as the rate of change of velocity which time. It is expressed as
a = (v – u) / t
Where
a is the acceleration
v is the final velocity
u is the initial velocity
t is the time
NOTE: Deceleration is the opposite of acceleration
With the above equation for acceleration, we can obtain the deceleration of the bullet. Details below:
How to determine the deceleration of the bulletInitial velocity (u) = 500 m/sFinal velocity (v) = 0 m/sTime (t) = 0.2 sDecelration (a) =?a = (v – u) / t
a = (0 – 500) / 0.2
a = –500 / 0.2
a = –2500 m/s²
Thus, the deceleration of the bullet is –2500 m/s²
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A computer monitor accelerates electrons between two plates and sends them at high speed to form an image on the screen. If the elec- trons gain 4.1 * 10-15 J of kinetic energy as they go from one accelerat- ing plate to the other, what is the voltage between the plates?
The voltage between the plates is 3.9 × [tex]10^-^3[/tex] V.
The work-energy theorem states that the delta in the equation equals the change in kinetic energy plus the change in potential energy. Here, a charge's potential energy is expressed as qV, where V is the position's electric potential. The greater the change in voltage per unit distance, the greater the electric field.
The kinetic energy of the electrons = 4.1 × [tex]10^-^1^5[/tex] J
Charge of the electron = 1.602 × [tex]10^-^1^9[/tex] coulomb
Using,
ΔU = q × ΔV
4.1 × [tex]10^-^1^5[/tex] = 1.602 × [tex]10^-^1^9[/tex] × ΔV
ΔV = 3.9 × [tex]10^-^3[/tex] V
Therefore, the voltage between the plates is 3.9 × [tex]10^-^3[/tex] V.
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A rocket takes off from Earth's surface, accelerating straight up at 63.2 m/s2. Calculate the normal force (in N) acting on an astronaut of mass 84.4 kg, including her space suit. (Assume the rocket's initial motion parallel to the +y-direction. Indicate the direction with the sign of your answer.)
______N
From the calculation, the normal force is 6161.2 N.
What is the normal force?The normal force is given by the expression;
N - mg = ma
Then;
N = mg + ma
m = 84.4 kg
g = 9.8 m/s^2
a = 63.2 m/s2
Now we have;
N = m(g + a)
N = 84.4 (9.8 + 63.2)
N = 6161.2 N
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If the range of a projectile is and 256√3 m in the maximum height reached is 64 m. calculate the angle of projection
The angle of projection given that the range is 256√3 m and the maximum height reached is 64 m is 30°
Data obtained from the questionRange (R) = 256√3 mMaximum height (H) = 64 mAcceleration due to gravity (g) = 9.8 m/s²Angle of projection (θ) = ?How to determine the angle of projectionR = u²Sine(2θ) / g
256√3 = u²Sine(2θ) / 9.8
Cross multiply
256√3 × 9.8 = u²Sine(2θ)
Divide both sides by Sine(2θ)
u² = 256√3 × 9.8 / Sine(2θ)
H = u²Sine²θ / 2g
64 = [256√3 × 9.8 / Sine(2θ)] × [Sine²θ / 2 × 9.8]
64 = [256√3 / Sine(2θ)] × [Sine²θ / 2]
Recall
Sine²θ = SineθSineθ
Sine2θ = 2SineθCosθ
Thus,
64 = [256√3 / 2SineθCosθ] × [SineθSineθ / 2]
64 = 256√3 × Sineθ / 4Cosθ
Recall
Sineθ / Cosθ = Tanθ
Thus,
64 = 256√3 / 4 × Tanθ
Divide both side by 256√3 / 4
Tanθ = 64 ÷ 256√3 / 4
Tanθ = 0.5774
Take the inverse of Tan
θ = Tan⁻¹ 0.5774
θ = 30°
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M1 is a spherical mass (25.0 kg) at the origin. M2 is also a spherical mass (10.6 kg) and is located on the x-axis at x = 94.8 m. At what value of x would a third mass with a 19.0 kg mass experience no net gravitational force due to M1 and M2?
The point where m3 experiences a zero net gravitational force due to M1 and m2 is 57.42 m.
Position of the third mass
m1<------(x)------> m3 <-----------(94.8 m - x)-------->m2
a point, x, where m3 experiences a zero net gravitational force due to M1 and m2;
Force on m3 due to m1 = Force on m3 due to m2
Gm1m3/d² = Gm2m3/r²
m1/d² = m2/r²
where;
d is the distance between m1 and m3 = xr is the distance between m3 and m2 = 94.8 - xm1/(x²) = m2/(94.8 - x)²
m1(94.8 - x)² = m2x²
(94.8 - x)² = (m2/m1)x²
(94.8 - x)² = (10.6/25)x²
(94.8 - x)² = 0.424x²
(94.8 - x)² = (0.651)²x²
94.8 - x = 0.651x
94.8 = 1.651x
x = 94.8/1.651
x = 57.42 m
Thus, the point where m3 experiences a zero net gravitational force due to M1 and m2 is 57.42 m.
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An 87-kg football player traveling 5.2 m/s is stopped in 1.0 s by a tackler.
What is the original kinetic energy of the player?
Express your answer to two significant figures and include the appropriate units.
What average power is required to stop him?
Express your answer to two significant figures and include the appropriate units.
The original kinetic energy of the player and the average power is required to stop him are 1200 J and 1200 W respectively.
What is Kinetic Energy ?The energy possessed by a body in motion is known as Kinetic Energy. The S. I unit is Joule.
Given that an 87-kg football player traveling 5.2 m/s is stopped in 1.0 s by a tackler. The given parameters are;
Mass m = 87 KgVelocity v = 5.2 m/sTime t = 1 sThe original kinetic energy of the player can be calculated by using the formula K.E = 1/2m[tex]v^{2}[/tex]
Substitute all the parameters into the formula
K.E = 1/2 x 87 x [tex]5.2^{2}[/tex]
K.E = 1176.24
K.E = 1200 J
Power is the rate at which work is done.
Work done = energy
The average power is required to stop him can be calculated by using the formula P = E/t
Substitute all the parameters into the formula
P = 1200/1
P = 1200 W
Therefore, the original kinetic energy of the player and the average power is required to stop him are 1200 J and 1200 W respectively.
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The specific heat of copper is 0.385 J/g°C.
How much heat is needed to raise the temperature of 6.00 g of copper by 15.0°C?
35.0 J
90.0 J
234 J
34.7 J
The amount of heat needed to raise the temperature of 6.00 g of copper by 15.0°C is 34.65J (option A).
How to calculate amount of heat?The amount of heat absorbed or released by a substance can be calculated using the following formula:
Q = mc∆T
Where;
Q = quantity of heat absorbed or releasedm = mass of the substancec = specific heat capacity∆T = change in temperatureQ = 6 × 0.385 × 15
Q = 90 × 0.385
Q = 34.65J
Therefore, the amount of heat needed to raise the temperature of 6.00 g of copper by 15.0°C is 34.65J.
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Two hockey pucks are moving towards each other. (Assume no friction.) The first one is 0.13 kg and moving at a speed of 1.11 m/s, while the second puck is 0.16 kg and moving at 1.21 m/s, and they collide. (Assume elastic collision.) After collision, the second puck ends up with a speed of 1.16m/s at an angle of 42 degrees below its original path, while the first puck ends up with an unknown speed at an angle above its original path. Find the final speed and angle of the first puck.
The final speed and angle of the first puck are 1.17 m/s and 54.5° respectively.
What happened in an Elastic Collision ?In an elastic collision, both momentum and energy are conserved. But only momentum is conserved in inelastic collision.
Given that two hockey pucks are moving towards each other. The first one is 0.13 kg and moving at a speed of 1.11 m/s, while the second puck is 0.16 kg and moving at 1.21 m/s, and they collide. (Assume elastic collision.) After collision, the second puck ends up with a speed of 1.16m/s at an angle of 42 degrees below its original path, while the first puck ends up with an unknown speed at an angle above its original path.
The given parameters are;
M1 = 0.13 kgM2 = 0.16 kgU1 = 1.11 KgU2 = 1.21 KgV1 = ?V2 = 1.16 kgФ1 = ?Ф2 = 42°The mathematical representation of the above question will be in two components.
Horizontal component
M1U1 - M2U2 = M1V1cosФ - M2V2cosФ
Substitute all the parameters
0.13 x 1.11 - 0.16 x 1.21 = 0.13 x V1 cosФ - 0.16 x 1.16cos42
0.1443 - 0.1936 = 0.13V1cosФ - 0.1379
0.13V1cosФ = 0.0886
V1cosФ = 0.0886/0.13
V1cosФ = 0.6815 ........ (1)
Vertical component
0 = M1V1sinФ - M2V2sinФ
M1V1sinФ = M2V2sinФ
Substitute all the parameters
0.13 x V1 sinФ = 0.16 x 1.16sin42
V1 sinФ = 0.1242/0.13
V1 sinФ = 0.9553 ......... (2)
Divide equation 2 by 1
V1 sinФ / V1 cosФ = 0.9553/ 0.6815
Tan Ф = 1.40
Ф = [tex]Tan^{-1}[/tex](1.4)
Ф = 54.5°
Substitute Ф into equation 2
V1 sin54.5 = 0.9553
V1 = 0.9553 / 0.8141
V1 = 1.17 m/s
Therefore, the final speed and angle of the first puck are 1.17 m/s and 54.5° respectively.
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A mechanic pushes a 2.30 ✕ 103-kg car from rest to a speed of v, doing 4,800 J of work in the process. During this time, the car moves 30.0 m. Neglecting friction between car and road, find v and the horizontal force exerted on the car.
(a) the speed v
______m/s
(b) the horizontal force exerted on the car (Enter the magnitude.)
_______N
The horizontal force applied is 160 N while the velocity is 2.03 m/s.
What is the speed of the car?The work done by the car is obtained as the product of the force and the distance;
W = F x
F = ?
x = 30.0 m
W = 4,800 J
F = 4,800 J/30.0 m
F = 160 N
But F = ma
a = F/m
a = 160 N/2.30 ✕ 10^3-kg
a= 0.069 m/s
Now;
v^2 = u^2 + 2as
u = 0/ms because the car started from rest
v = √2as
v = √2 * 0.069 * 30
v = 2.03 m/s
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A yellow train of mass 100 kg is moving at 8 m/s toward an orange train of mass 200 kg traveling in the opposite direction on the same track at a speed of 1 m/s. What is the initial momentum of the yellow and orange trains combined?
A. 200 kgm/s
B. 1000 kgm/s
C. 800 kgm/s
D. 600 kgm/s
The initial momentum of the yellow and the orange train is 1000kgm/s.
Momentum is the product of the mass and velocity of any object.
Momentum is denoted by P.
Momentum P = mv , where m = mass and v = velocity.
Given:Mass of the orange train = 200kg
Velocity of the orange train = 1m/s
So, the momentum of the orange train will be,
∴ P = mv
P = 200 x 1
P = 200 kgm/s
∴ The initial momentum of the orange train is 200kgm/s.
Mass of the yellow train = 100kg
Velocity of the yellow train = 8m/s
So, the momentum of the yellow train will be,
∴ P = mv
P = 100 x 8
P = 800 kgm/s
∴ The initial momentum of the yellow train is 800kgm/s.
Therefore, the initial momentum of the yellow and the orange train is 1000kgm/s.
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A 269 kg weather balloon is designed to lift a 2910 kg package. What volume should the balloon have after being inflated with helium in order that the total load can be lifted? (Assume standard temperature and pressure, at which the density of air is ρair = 1.29 kg/m3 and the density of helium is ρHe = 0.179 kg/m3.)
The volume of the helium balloon in order to lift the weight is 17,760m³.
To find the answer, we need to know about the buoyant force.
What's the buoyant force?When a lighter object is kept in a higher density medium, it experiences a force along upward by that medium. This is buoyant force.Mathematically, buoyant force= density × volume of the object×gWhat's the volume of helium required to lift the 269kg weather balloon and 2910kg package?To lift the weight, the buoyant force must equal to the weight.If V is the volume of helium, buoyant force= V×0.179×gSo, V×0.179×g = (269+2910)g=> V= 3179/0.179 = 17,760m³
Thus, we can conclude that the volume of the helium balloon in order to lift the weight is 17,760m³.
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A wire carrying a 25.0 A current bends through a right angle. Consider two 2.00 mm segments of wire, each 3.00 cm from the bend (Figure 1).
a) Find the magnitude of the magnetic field these two segments produce at point P , which is midway between them.
b) Find the direction of the magnetic field at point P
The magnitude of the magnetic field and the direction of the magnetic field at point P is mathematically given as
B=1.9*10^{-5}T
To determine the magnetic field direction, use the right-hand -rule on the page magnetic field is going.
What is the magnitude of the magnetic field these two segments produce at point P, which is midway between them.?Generally, the equation for Biot savant law is mathematically given as
[tex]B=(\frac{u}{4\pi}*{\frac{Ilsin\theta}{r^2})[/tex]
Their net field is
Bn=2B
[tex]Bn=2* B=(\frac{u}{4\pi}*{\frac{Ilsin\theta}{r^2})[/tex]
Hence
[tex]B=(\frac{4*\p *10^{-7}}{4\pi}*{\frac{(30)(2*10^{-3}sin45)}{\sqrt{(3*10^{-2})^2+((3*10^{-2})^2)}/2})[/tex]
B=1.9*10^{-5}T
In conclusion, To determine the magnetic field direction, use the right-hand -rule on the page magnetic field is going.
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Answer:
The magnitude of the magnetic field at the point P is [tex]1.57*10^{-5}T[/tex] and the field is pointing into the page.
Explanation:
The general form of a similar question to this is:
[tex]\vec{B} = \frac{\mu _{0} }{4\pi } * \oint \frac{Id\vec{l} \times \hat{r}}{r^{2} }[/tex]
where [tex]\vec{B}[/tex] is the vector of the Magnetic Field, [tex]\mu _{0}[/tex] is the Free Space Permeability Constant (equal to [tex]4\pi * 10^{-7} \frac{N}{A^2}[/tex]), [tex]I[/tex] is the current, and [tex]r[/tex] is the distance from the segment to the point P. (I will get to the [tex]d\vec{l} \times \hat{r}[/tex] term in a bit)
This equation is fairly complicated. Luckily, it can be simplified by looking at the magnitude and direction separately.
The first thing to simplify is the cross product. Due to the fact that a cross product can be simplified from [tex]\vec{x} \times \vec{y}[/tex] to [tex]xy\sin(\theta)[/tex], where [tex]\theta[/tex] is the angle between the 2 vectors, and [tex]\hat{r}[/tex] is the unit vector of [tex]r[/tex] (i.e. [tex]\hat{r} = \vec{r}/r[/tex]) we can simplify [tex]d\vec{l} \times \hat{r}[/tex] to just [tex]dl \sin(\theta)[/tex].
Next, we will look at the integral. In this scenario, everything will function as a constant, so we can essentially ignore it.
Finally, [tex]\frac{\mu_{0}}{4\pi}[/tex] simplifies down to [tex]10^{-7}[/tex].
This gives us our new equation for the Magnetic Field produced by a single segment at a point:
[tex]B = \frac{Il\sin\theta}{r^{2}}*10^{-7}[/tex]
Now we need to find values for [tex]r[/tex] and [tex]\theta[/tex]. Luckily, we are dealing with a 45-45-90 triangle with sides of [tex]1.5 cm[/tex]. This means the distance [tex]r[/tex] is [tex](1.5\sqrt2)cm[/tex]! Similarly, because it is a 45-45-90 triangle, our [tex]\theta[/tex] is [tex]45\textdegree[/tex]!
Now we can start plugging things in:
[tex]B = \frac{(25A)(2*10^{-3}m)\sin(45\textdegree)}{(1.5\sqrt2*10^{-2}m)^2}*10^{-7}\frac{N}{A^2}[/tex]
[tex]B = 7.86^{-6} \frac{N}{A}[/tex] or [tex]T[/tex]
This is the magnitude due to only one single segment. In order to find the total field, we need to know the direction of the field due to each segment.
Finding the direction is really easy. Just use the right hand rule. Point your thumb in the direction of the current and curl the rest of your fingers around an imaginary pole. The direction your fingers point is the direction of the field. In this case, the field lines due to the segments point into the page in the 4th quadrant (the origin is the bend). This means that at point P, both segments induce the same field in the same direction. Therefore, we can take our value from before and double it, giving us our final answer:
[tex]B = 1.57*10^{-5} T[/tex]; into the page.
If you speed through a construction zone while workers are present, your fines could be:.
If you speed through a construction zone while workers are present, your fines could be as much as one thousand dollars.
What is a Fine?
This is referred to as the amount which is instructed by a court or authority to be paid as a result of it being a penalty for various types of offences. each crime has its specific fine which helps to serve as deterrent for unlawful behavior in the community.
it is always best not to speed when within a construction zone which has workers present in the area. This is ideal as it helps to prevent incidences of accidents or death.
it is therefore the reason why a fine of 1000 dollars is to be paid so that people can think of such high amount before performing certain types of activities when driving and makes it the most appropriate choice.
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light travel
3, 00,000 km/s. Is it velocity or speed?
Piston 1 in the figure has a diameter of 1.87 cm.
Piston 2 has a diameter of 9.46 cm. In the absence of friction, determine the force F, necessary to support an object with a mass of 991 kg placed on piston 2. (Neglect the height difference between the bottom of the two pistons, and assume that the pistons are massless).
The force F, necessary to support an object with a mass of 991 kg placed on piston 2. is 20.61J.
How to calculate the value?It should be noted that by Pascal Law, the pressure on piston 1 will have the same value as the pressure on piston 2.
This will be:
(991 × 10) /(π × 0.0946/2)²
= 9910/0.022
= 450454.6 Pa
F1 = A1 × 450454.6 = 3.14 × (0.0187/2)² × 450454.6
= 123.64
F = 123.64/6
F = 20.61
Therefore, the force F, necessary to support an object with a mass of 991 kg placed on piston 2. is 20.61J.
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A copper transmission cable 50.0 km long and 10.0 cm in diameter carries a current of 105 A. What is the potential drop across the cable? Let ρcopper = 1.72 × 10—8 Ω • m.
A) 5.75 V
B) 5.48 V
C) 11.5 V
D) 16.9 V
5.48 V is the potential drop across the cable for a copper transmission cable of 50.0 km long and 10.0 cm in diameter carries a current of 105 A
Ohm's Law states that the potential drop is determined by the equation: V = IR, where I is the current and R is the wire resistance.
R=PL/A
Under the assumption that all physical parameters and temperatures remain constant, Ohm's law asserts that the voltage across a conductor is directly proportional to the current flowing through it.
Only when the given temperature and the other physical variables remain constant does Ohm's law apply. Increasing the current causes the temperature to rise in some components. The filament of a light bulb serves as an illustration of this, where the temperature increases as the current increases. Ohm's law cannot be applied in this situation. The filament of the lightbulb defies Ohm's Law.
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A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
If the angle between the beam and the cable is θ = 57.0° what is the vertical component of the force exerted by the hi.nge on the beam?
The vertical component of force exerted by the hi.nge on the beam will be,142.10N.
To find the answer, we need to know more about the tension.
How to find the vertical component of the force exerted by the hi.nge on the beam?Let's draw the free body diagram of the system.To find the vertical component of the force exerted by the hi.nge on the beam, we have to balance the total vertical force to zero.[tex]F_V+T sin\alpha -mg=0\\F_V=mg-Tsin\alpha \\[/tex]
To find the answer, we have to find the tension,[tex]Tlsin\alpha - mg\frac{l}{2}sin\beta =0\\ \\Tlsin\alpha = mg\frac{l}{2}sin\beta\\\\Tsin57=\frac{mg}{2}sin90\\\\T=\frac{mg}{2sin57} =169.43N[/tex]
Thus, the vertical component of the force exerted by the hi.nge on the beam will be,[tex]F_V=(29*9.8)-(169.43*sin57)=142.10N[/tex]
Thus, we can conclude that, the vertical component of force exerted by the hi.nge on the beam will be,142.10N.
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The hi.nge will apply a force of 142.10N on the beam in the vertical direction.
We must learn more about the tension in order to find the solution.
How can I determine the vertical component of the force the hi.nge has on the beam?Let's create the system's free body diagram.We must balance the total vertical force to zero in order to get the vertical component of the force applied to the beam by the height.[tex]F_V=mg-Tsin\alpha[/tex]
We must identify the tension in order to find the solution.[tex]Tlsin\alpha =mg\frac{l}{2}sin\beta \\T=\frac{mgsin90}{2sin57} =169.43N[/tex]
Consequently, the force that the height exerts on the beam will have a vertical component that is,[tex]F_v=(29*9.8)-(169.43*sin57)=142.10N[/tex]
This leads us to the conclusion that the vertical component of the force the hi.nge exerts on the beam will be 142.10N.
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What are the balanced forces for someone in a parachute?
A. Gravity and air resistance
B. Centripetal and air resistance
C. Gravity and centripetal
D. Gravity and Earth
100m with a constant speed 200km/hr the pilot drops abomb from the plane. determine (neplect air resistance of friction) X Q
The horizontal distance XQ traveled by the bomb is 250 m.
Distance X Q
Let the XQ be the horizontal distance traveled by the bomb.
Time for the bomb to drop from 100 mh = vt + ¹/₂gt
Let the vertical velocity = 0
h = 0 + ¹/₂gt²
h = ¹/₂gt²
2h = gt²
t² = 2h/g
t = √(2h/g)
t = √(2 x 100 / 9.8)
t = 4.5 s
Horizontal distance traveled by the bombXQ = vx(t)
where;
vx is horizontal speed, = 200 km/hr = 55.56 m/s
XQ = 55.56 x 4.5
XQ = 250 m
Thus, the horizontal distance XQ traveled by the bomb is 250 m.
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emily is standing 150 feet from a circular target with radius 3 inches. will she hit the target if her aim is off by 0.2 degrees?
Answer:
no
Explanation:
The angle subtended by the radius of the target at Emily's distance can be found using the tangent relation.
tan(α) = opposite/adjacent = (1/4 ft)/(150 ft) = 1/600
The angle is found using the inverse relation -
α = arctan(1/600) ≈ 0.095°
If Emily's aim is off by 0.2°, she will miss the target by several inches.
Emily's projectile will miss her aiming point by ... (150 ft)tan(0.2°) ≈ 0.524 ft ≈ 6.28 in
Which of the following changes would increase the force between two
charged particles to 9 times the original force?
A. Decreasing the distance between the particles by a factor of 3
B. Decreasing the amount of charge on one of the particles by a
factor of 9
C. Increasing the distance between the particles by a factor of 3
D. Increasing the amount of charge on each particle by a factor of 9
The answer is A. Decreasing the distance between the particles by a factor of 3.
The Universal Law of Gravitation is :
F = Gm₁m₂ / r² (where 'r' is the distance between them)
Since force is inversely proportional to the square of the distance between them, distance has to be decreased by a factor of 3 to increase the force to 9 times the original force.