On the bolt, F2 force will exert the most torque. Torquing a bolt involves using a wrench. There is no difference in the magnitude of forces F1 through F6.
Vector and scalar quantities both use magnitude as a common denominator. Scalar quantities are understood to be those that have only magnitude, according to their definition. Vector quantities, on the other hand, are those that have both magnitude and direction. Magnitude can be used in a number of different contexts.
where the force is being applied From the bolt's centre of rotation, Fi (and F) are separated by a distance of d. Moment of force equals torque (Rotational domain equivalent of Force)
t1 = (d)(Fcos(theta))=0.86dF
t2 = (2d)(f) = 2dF
t3=(3d)(F)(cos(theta) = 1.5dF
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What is the symbol for a variable resistor?
A. Horizontal wire
B. 2 parallel wires
C. Line with a zig zag pattern with a diagonal line drawn across it
D. Circle with an X through it
The symbol for a variable resistor is a line with a zig zag pattern with a diagonal line drawn across it.
What is a variable resistor ?
A resistor whose electric resistance value may be altered is known to as a variable resistor. A variable resistor, which works usually by sliding a contact (wiper) over a variable resistor, is basically an electro-mechanical transducer.
The flow of electrical current is restricted by a resistor. The resistivity of a fixed resistor is steady. By adjusting a slider's position, the resistance of this resistor can be altered. Some volume controls and dimmer switches employ variable resistors.
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Glucose solution is administered to a patient in a hospital. The density of the solution is 1.308 kg/l. If the blood pressure in the vein is 35.7 mmHg, then what is the minimum necessary height of the IV bag above the position of the needle?
The minimum necessary height of the IV bag above the position of the needle, if the density of the solution is 1.308kg/L, and the pressure in the vein is 35.7 mmHg is, 2.78m.
To find the answer, we need to know more about the pressure exerted by a liquid column.
How to find the height of IV bag above the position of needle?Consider a liquid of density ρ contained in a vessel of height h, the pressure exerted by the liquid column at the bottom of the vessel is given by ,[tex]P=[/tex] ρgh
In our question, it is given that,[tex]density=1.308 kg/L\\\\P=35.7 mmHg.\\[/tex]
Thus, the height of the bag h will be,[tex]h=\frac{P}{density*g} =\frac{35.7}{1.308*9.8}\\\\ h=2.78 m[/tex]
Thus, we can conclude that, the height of the Iv bag is 2.78 m above the position of needle.
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If the solution has a density of 1.308 kg/L and the vein pressure is 35.7 mmHg, the minimum height of the IV bag above the needle position is 2.78 m.
We need to learn more about the pressure that a liquid column exerts in order to determine the solution.
How can I determine the height of IV bag above where the needle is?The pressure exerted by the liquid column at the bottom of the vessel is given by for a liquid with density enclosed in a vessel of height h.[tex]P=[/tex] ρgh
In our inquiry, it is assumed that,[tex]Density=1.308kg/L\\P=35.7mmHg[/tex]
As a result, the bag's height will be,[tex]h=\frac{P}{density*g}\\\\ h=2.78m[/tex]
As a result, we may say that the I.V. bag is 2.78 m above the ground.
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Object A has the same mass as Object B but Object A is traveling faster. What can we say about the momentum of Object A compared to that of Object B?
Answer:
Object A's momentum is larger
Explanation:
as the formula for momentum goes:
P = M * V
where P is momentum, M is mass, V is velocity.
so where Va (Object A's velocity) is larger than Vb (Object B's velocity) we get:
( i ) Va > Vb
as the masses of both objects are equal, we mark:
( ii) Ma = Mb = M
we multiply both sides of ( i ) by M to get:
( iii ) Va × M > Vb × M
and we finally get:
( iv ) Pa > Pb
Answer: Object A's momentum of larger than compared to that of object B.
Explanation:
A 1300-kg car moving on a horizontal surface has speed v = 60 km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.3 m .
What is the spring stiffness constant of the spring?
Express your answer to two significant figures and include the appropriate units.
The stiffness constant of the spring is 68,290.3 N/m
Stiffness constant of the spring
Apply the principle of conservation of energy;
U = K.E
¹/₂kx² = ¹/₂mv²
kx² = mv²
k = mv²/x²
where;
m is massv is speed = 60 km/h = 16.67 m/sx is the distancek = (1300 x 16.67²)/(2.3²)
k = 68,290.3 N/m
Thus, the stiffness constant of the spring is 68,290.3 N/m.
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Tuning an Instrument. A musician tunes the C-string of her instrumeut to a fundamental frequency of 65.4 Hz. The vibrating portion of the string is 0.600 m long and has a mass of 14.4 g. (a) With what tension must the musician stretch it? (b) What percent increase in tension is needed to increase the frequency from 65.4 Hz to 73.4 Hz, corresponding to a rise in pitch from C to D?
(a) The tension the musician must stretch it is 147.82 N.
(b) The percent increase in tension is needed to increase the frequency is 26%.
Tension in the stringv = √T/μ
where;
v is speed of the waveT is tensionμ is mass per unit length = 0.0144 kg / 0.6 m = 0.024 kg/mv = Fλ
in fundamental mode, v = F(2L)
v = 2FL
v = 2 x 65.4 x 0.6 = 78.48 m/s
v = √T/μ
v² = T/μ
T = μv²
T = 0.024 x (78.48)²
T = 147.82 N
When the frequency is 73.4 Hz;v = 2FL = 2 x 73.4 x 0.6 = 88.08 m/s
T = μv²
T = (0.02)(88.08)²
T = 186.19 N
Increase in the tension= (186.19 - 147.82)/(147.82)
= 0.26
= 0.26 x 100%
= 26 %
Thus, the tension the musician must stretch it is 147.82 N.
The percent increase in tension is needed to increase the frequency is 26%.
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An Airbus A380-800 passanger airplane is cruising at constant altitude on a straight line with a constant speed. The total surface area of the two wings is 395 m2. The average speed of the air just below the wings is 259 m/s, and it is 288 m/s just above the surface of the wings.
What is the mass of the airplane? The average density of the air around the airplane is ρ^air = 1.21 kg/m^3.
The mass of the airplane of area of two wings 395m², and the average speed of lower and upper surface of the wings are, 259 and 288m/s is 386.8×10^3 kg.
To find the answer, we need to know about the Bernoulli's principle.
How to find the mass of the airplane?The Bernoulli's principle states that, the sum of pressure energy, kinetic energy and potential energy of an incompressible, non-viscous, fluid in streamlined flow is a constant.It can be expressed as,P+ [tex]\frac{1}{2}[/tex] ρv²+ρgh=a constant.
Instead of ρ we take d as density.
We have given that,[tex]A= 395 m^2\\v_l=295 m/s\\v_u=288m/s\\density=1.21kg/m^3[/tex]
We equate the principle for lower and upper surfaces of the wing like,[tex]P_1+\frac{1}{2}v_1^2d+dgh_1=P_2+ \frac{1}{2}v_2^2d+dgh_2\\here\\h_1=h_2\\thus\\P_1-P_2=\frac{1}{2}d(v_2^2-v_1^2)\\P_1-P_2=\frac{1}{2}*1.21(288^2-259^2)=9597.12 atm\\[/tex]
Thus, the mass of the airplane from the above equation will be,[tex]\frac{F}{A}=9597.12 atm\\\\ F=9597.12*395m^2=37.9*10^5 N\\\\mg=37.9*10^5 N\\\\m=\frac{37.9*10^5 N}{9.8}\\\\ m=386.8*10^3kg[/tex]
Thus, we can conclude that, the mass of the airplane is 386.8×10^3 kg.
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The mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387x 10^3 kg.
We need to be aware of the Bernoulli principle in order to determine the solution.
How can I determine an airplane's mass?According to the Bernoulli's principle, the total amount of pressure energy, kinetic energy, and potential energy in a streamlined flow of an incompressible, non-viscous fluid is constant.It can be stated as follows:P+ [tex]\frac{1}{2}[/tex]dv²+dgh = constant.
We substitute d for to represent density.
We've done that,[tex]v_1=259m/s\\v_2=288m/s\\A=395m^2\\d=1.21kg/m^3[/tex]
We compare the governing idea for the wing's bottom and upper surfaces to:[tex]P_1+\frac{1}{2}dv_1^2+dgh= P_2+\frac{1}{2}dv_2^2+dgh\\\\P_1+\frac{1}{2}dv_1^2=P_2+\frac{1}{2}dv_2^2\\\\P_1-P_2=\frac{1}{2}d(v_2^2-v_1^2)\\\\P_1-P_2=9597 atm[/tex]
Consequently, using the aforementioned equation, the airplane's mass will be,[tex]F/A= 9597 atm\\mg=9597*395 =38*10^5N\\m=\frac{38*10^5}{9.8} = 387*10^3kg[/tex]
Consequently, we can say that the mass of an airplane with two wings that are 395m2 in size and average wing surface speeds of 259 and 288 m/s is 387 x 10^3 kg.
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Planet-X has a mass of 4.74×1024 kg and a radius of 5870 km.
1. What is the Second Cosmic Speed i.e. the minimum speed required for a satellite in order to break free permanently from the planet?
2. If the period of rotation of the planet is 16.6 hours, then what is the radius of the synchronous orbit of a satellite?
The minimum speed required for a satellite in order to break free permanently from the planet and the radius of the synchronous orbit of a satellite are 7.3 Km/s and 3.1 × 10⁴ km respectively.
To find the answer, we need to know about the escape velocity and time period of revolving satellite.
What's the expression of escape velocity of a satellite?Mathematically, escape velocity= √(2GM/R)G = gravitational constant, M = mass of planet, R= radius of the planetHere, M = 4.74×10²⁴kg, R = 5870 kmEscape velocity= √(6.67×10^(-11)×4.74×10²⁴/5.870×10⁶)= 7.3 Km/s
What's the expression of time period of a circularly orbiting satellite?T= {2π×r^(3/2)}/√(GM)r= (T/2π)⅔× (GM)^(1/3)r is the radius of the orbitWhat's the radius of the circular orbit, if the time period of the satellite is 16.6 hours?T = 16.6 hours = 16.6×3600 second = 59760sr = (59760/2π)^⅔× (6.67×10^(-11)×4.74×10²⁴)^(1/3)= 3.1 × 10⁴ km
Thus, we can conclude that the escape velocity and the radius of the synchronous orbit of a satellite are 7.3 Km/s and 3.1 × 10⁴ km respectively.
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Calculate the magnitude of the electric field at one corner of a square 1.10 m on a side if the other three corners are occupied by 4.05×10−6 C charges. What is the direction of the electric field?
The direction of magnetic field is south-east and the magnitude is
[tex]23.66[/tex] × [tex]10^{3} N/C[/tex].
Here, the magnitude of CD and BC will be cancelled, as they both are in the opposite direction and equal to each other.
the magnitude, towards the diagonal AC will result in CP, which is the direction of the electric field.
magnitude of electric field can be defined as :- The force per charge on the test charge is a straight forward way to define the size of the electric field.
To find the magnitude of the electric field use the formula
[tex]E = kq/ r^{2}[/tex]
inserting the values,
[tex]E = 9. 10^{9}[/tex] × [tex]4.05[/tex] × [tex]10^{-6} / 1.1 \sqrt{2}[/tex]
[tex]E= 36.45[/tex] × [tex]10^{3}[/tex] / [tex]1.54[/tex]
[tex]E = 23.66[/tex] × [tex]10^{3}[/tex] N/C
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A moonshiner makes the error of filling a glass jar to the brim and capping it tightly. The moonshine expands more than the glass when it warms up, in such a way that the volume increases by 0.6% (that is, ΔV/V0 = 6 ✕ 10-3) relative to the space available. Calculate the force exerted by the moonshine per square centimeter if the bulk modulus is 2.1 ✕ 109 N/m2, assuming the jar does not break.
The force exerted per square centimeter is 126 N/cm².
What is pressure?Pressure is the force acting per unit area.
Pressure = force/areaBased on the data given:
volume increase, ΔV/V0 = 6 * 10⁻³
Bulk Modulus, B = 2.1 * 10⁹ N/m²
Bulk modulus B of a material is ratio of change in pressure and change in volume as given below:
B = ΔP/ [(ΔV/V)]Solving for ΔP;
ΔP = B * [(ΔV/V)]
ΔP = (2.1 * 10⁹ N/m²) * (6 * 10⁻³)
ΔP = 1.26 * 10⁶ N/m²
Converting to per square centimeter
ΔP = (1.26 * 10⁶ N/m²)/10⁴
ΔP = 126 N/cm²
In conclusion, the force exerted per square centimeter is a measure of the pressure.
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What adaptation of a cactus protects it from predators? a round cactus with many spines Broad leaves Sharp spines Thick stems Yellow flowers
A round cactus with many spines is the adaptation of a cactus that protects it from predators.
A cactus, unlike other plants, has unique adaptations in its roots, leaves, and stems that allow it to flourish in hot and dry surroundings.
The one adaptation that protects the cactus from predators is spines.
A cactus does not have any parts that resemble leaves if you could look at one closely.
Instead, the leaves are transformed into spines, which protrude from the plant's tiny bumps known as areoles.
Herbivores that live in the desert may be enticed to eat the cactus. The spines prevent these predators by modifying leaves into spines.
Other than protection, Spines perform many functions like
1) Since evaporation is a problem in a desert since water is scarce, the spines prevent excessive evaporation.
2) The spines also impede airflow and prevent evaporation by trapping air.
3) Collecting dew from the early-morning fog is another crucial job that the spines do.
The gathered dew turned into liquid water and ran down to the earth below. The plant then absorbs this water.
Hence, the adaptation of a cactus that protects it from predators is round cactus with many spines.
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Avogadro‘s number was calculated by determining The number of atoms in
Answer:
12 grams of the isotope carbon-12.
Explanation:
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Light is incident normally on the short face of a 30∘−60∘−90∘ prism (Figure 1). A drop of liquid is placed on the hypotenuse of the prism.
a) If the index of the prism is 1.50, find the maximum index that the liquid may have for the light to be totally reflected.
Express your answer using three significant figures.
1.06 is the maximum refractive index that the liquid may have for the light to be totally reflected.
Only when a light source passes from a denser to a rarer medium can it completely reflect.
When the angle of incidence surpasses a specific critical value, specular reflection occurs in the more highly refractive of the two mediums at their interface, and this reflection is known as total reflection.
sin [tex]i_{c}[/tex] = μ[tex]_{r}[/tex] / μ[tex]_{d[/tex]
From the diagram
Angle of incidence = 60°
sin60° ≥ sin[tex]i_{c}[/tex] = μ[tex]_{r}[/tex]/μ[tex]_{d}[/tex]
μ[tex]_{r}[/tex] ≤ μ[tex]_{d}[/tex] sin60°
μ[tex]_{r}[/tex] ≤ √1.5 × √3/2
= 1.06
Hence, the maximum index that the liquid may have for the light to be totally reflected is 1.06
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A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the cable is 90°. If the beam is inclined at an angle of θ = 31.0° with respect to horizontal.
The horizontal component of the force exerted by the hi.nge on the beam = 8.662×101 N
What is the magnitude of the force that the beam exerts on the hi.nge?
The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.
To find the answer, we need to know about the tension.
How to find the magnitude of the force that the beam exerts on the hi.nge?Let's draw the free body diagram of the system using the given data.From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.[tex]N_x=86.62N[/tex]
We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.[tex]N_y=F_V=mg-Tsin59\\[/tex]
To find Ny, we need to find the tension T.For this, we can equate the net horizontal force.[tex]F_H=N_x=Tcos59\\\\T=\frac{F_H}{cos59} =\frac{86.62}{0.51}= 169.84N[/tex]
Thus, the vertical component of normal reaction that exerts by the beam on the hi.nge become,[tex]N_y= (40*9.8)-(169.8*sin59)=246.4N[/tex]
Thus, the magnitude of the force that the beam exerts on the hi.nge will be,[tex]N=\sqrt{N_x^2+N_y^2} =\sqrt{(86.62)^2+(246.4)^2}=261.12N[/tex]
Thus, we can conclude that, the magnitude of the force that the beam exerts on the hi.nge is 261.12N.
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The hi.nge will be subjected to a force of 261.12N from the beam.
We must understand the tension in order to choose the solution.
How can the amount of force the beam applies on the height be determined?Let's use the provided information to create the system's free body diagram.We need to calculate the force the beam is exerting on the height using the diagram.For this, it is assumed that the horizontal component of force is 86.62N, the same as the horizontal component of the normal reaction that the beam exerts on the height.We need to identify the vertical component of the normal reaction the beam exerts on the height. We must equalize the total force acting in the vertical direction to achieve this.[tex]N_y=F_v=mg-Tsin59[/tex]
Finding the tension T is necessary to determine Ny. Thus, we can use the net horizontal force to equate this.[tex]F_H=N_x=Tcos59\\T=\frac{F_H}{cos59} =169.84N[/tex]
As a result, the normal reaction that the beam has on the height becomes, with a vertical component,[tex]N_y=(40*9.8)-(169.84*sin59)=246.4N[/tex]
As a result, the force the beam applies on the height will be of the order of,[tex]N=\sqrt{N_x^2+N_y^2} =261.12N[/tex]
Thus, we can infer that the force the beam applies to the height is 261.12N in size.
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Why should you use an iron hammer instead of a wooden
hammer?
Answer:
more durability and easier to pound a hammer with the heavy weight.
Explanation:
One should use an iron hammer instead of the wooden hammer because the force developed with the help of the iron hammer is far greater than the wooden hammer
What is Newton's second law?Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change of momentum.
The mathematical expression for Newton's second law is as follows
F = m*a
where F represents the force applied
m is the mass of the object
a is the acceleration of the object
As we know that force is a product of mass and acceleration, let us assume that both iron and wooden hammers have the same dimension , considering the acceleration applied is the same while hammering ,
As for the same dimension the mass of iron hammer would be greater as result force applied through the iron hammer would be grater
Since the iron hammer's force is so much stronger than the wooden hammer's force, one should use an iron hammer instead of a wooden hammer.
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A 69-kg base runner begins his slide into second base when he is moving at a speed of 3.2 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base.
(a) How much mechanical energy is lost due to friction acting on the runner?
J
(b) How far does he slide?
m
The mechanical energy lost is 353.28 Joules . The distance he slid off is 0.16m.
we know:-
Mass = 69 kg
Speed = 3.2 m/s
Coefficient of friction ( ratio of friction force and normal force ) = 0.70
Acceleration due to gravity, g = 9.8 m/s^2
(a) To determine the amount of mechanical energy that is lost because of friction acting on the runner, we would calculate the change in kinetic energy:
[tex]KE = \frac{1}{2} m (v-u)^{2}[/tex]
[tex]= \frac{1}{2} 69 ( 3.2-0 )^{2}[/tex]
[tex]= 353.28[/tex] Joules
Mechanical energy = 353.28 Joules
(b) To determine how far (distance) the runner slide:
acceleration = ug
[tex]= 3.2[/tex] × [tex]9.8[/tex]
[tex]= 31.36[/tex] [tex]m/s^{2}[/tex]
distance ,
[tex]V^{2} = U^{2} + 2aS[/tex]
[tex]( 3.2)^{2} = 0 + 2 ( 31.36) S[/tex]
[tex]10.24 = 62.72 S[/tex]
[tex]S = {\frac{10.24}{62.72} }[/tex]
Distance, S = 0.16 m
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Glucose solution is administered to a patient in a hospital. The density of the solution is 1.308 kg/l. If the blood pressure in the vein is 35.7 mmHg, then what is the minimum necessary height of the IV bag above the position of the needle?
The minimum necessary height of the IV bag above the position of the needle is 0.37 m.
we know that,
P = ρgh
where,
P = 35.7mmHg
= 4759.609 Pa
g = 9.8[tex]\frac{m}{s^{2} }[/tex]
ρ = 1.308 kg / m^3
now, substituting all the values, we get,
4759.609 = 1.308 × 9.8 × h
h = 0.37 m
The minimum necessary height of the IV bag above the position of the needle = 0.37 m.
what is an IV bag ?
A reagent, also called as an analytical reagent, is a substance or compound that is added to a system in chemistry to bring about a chemical reaction or examine to see if one happens. Even though the terms "reagent" and "reactant" are frequently used synonymously, "reactant" refers to a substance that is consumed during a chemical reaction.
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The decibel level of the sound from a certain hair dryer is measured at 60 dB. Find the intensity of the sound.
Based on the calculations, the sound intensity level is equal to 1.0 × 10⁻⁵ W/m².
How to determine intensity of the sound?Mathematically, sound intensity level can be calculated by using this formula:
[tex]\beta = 10 log(\frac{I}{I_{ref}} )[/tex]
Where:
I is the intensity of the sound.
Note: The reference value of sound intensity is equal to 1.0 × 10⁻¹² W/m².
Rewriting the formula, we have:
β/10 = logI - logIo
Substituting the parameters into the formula, we have;
60/10 = logI - log(1.0 × 10⁻¹²)
6 = logI + 12
logI = 6 - 12
logI = -6
I = 1.0 × 10⁻⁵ W/m².
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Drag each label to the correct location on the image. Jessica is visiting a park with her mother. Jessica sits on a swing. Her mother pulls the swing to a height of 3 meters above the ground and lets it go. The image shows Jessica at three positions on the swing. Jessica‘s mass is 44 kilograms and the maximum velocity of the swing is 5 meters/second. What’s the energy she has at each position shown? Ignore friction and air resistance. Use g = 9.8 m/s2, PE = m × g × h, and KE = 0 joules KE = 550 joules PE = 862.4 joules
The energy she has at each position shown are:
Position at maximum height -- 1,294 JPosition at minimum height --0 JPosition at maximum velocity - 550 JPosition at minimum velocity - 0 JWhat does velocity implies?Velocity is known to be a term that connote the direction of any kind of a moving body or an object.
Note that the Speed is known to be a a scalar quantity and as such, Velocity is said to be a vector quantity.
Note also that from the question given, Jessica's height of the swing 3 meters above the ground, therefore:
Jessica Position at maximum height :
PE = mgh
PE = 44kg x 9.8m/s² x 3
PE = 1,294 J
Jessica Position at minimum height:
PE = 0 J
Jessica Position at maximum velocity:
KE = 1/2 x mv²
KE = 1/2 x 44kg x (5m/s²)²
KE = 550 J
Jessica Position at minimum velocity:
KE = 0 J
Therefore, The energy she has at each position shown are:
Position at maximum height -- 1,294 JPosition at minimum height --0 JPosition at maximum velocity - 550 JPosition at minimum velocity - 0 JLearn more about velocity from
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Consider the following isobaric process for air, modelled as a Calorically Perfect Ideal Gas
(CPIG), from state 1 to state 2. P1 = 200 kPa, T1 = 500 K, T2 = 600 K. Show that the
condition satisfies the second law of thermodynamics. (Show all relevant steps involved).
These given conditions satisfy the second law of thermodynamics.
As the process is isobaric
So there will be a straight line of P= 200kPa in P-v and P-T planes
P1 = P2 = 100kPa
For perfect ideal gas, v-T plane:
[tex]v = (\frac{R}{P}) T[/tex]
[tex]v_{1} = (\frac{R}{P_{1} }) T_{1}[/tex] = 287 × 500/200000 = 0.717 m³/kg
[tex]v_{2} = (\frac{R}{P_{2} }) T_{2}[/tex] = 287 × 600/200000 = 0.861m³/kg
As it is the calorically perfect gas
de = [tex]c_{v}[/tex]dT
Integration on both sides
e2 - e1 = [tex]c_{v}[/tex](T2 - T1)
= ( 716.5J/kg/K) (600-500)
= 71650 J/kg
also,
Tds = de + Pdv
Tds = [tex]c_{v}[/tex]dT +Pdv
For ideal gas
V = RT/P
dv = Rdt/P - RTdp/P²
Tds = [tex]c_{v}[/tex]dT + Rdt - RTdp/P
ds = ([tex]c_{v}[/tex] + R)dT/T - RdP/P
ds = ([tex]c_{v} + c_{p} -c_{v}[/tex])dT/T - RdP/P
ds = [tex]c_{p}[/tex]dT/T - RdP/P
Integration on both sides
s2 - s1 = [tex]c_{p}[/tex]ln (T2/T1) - R ln (P2/P1)
Since P is constant
s₂ - s₁ = [tex]c_{p}[/tex] ln (T2/T1)
= 1003.5 ln (600/500)
= 1003.5 × 0.182
= 182.95 J/kg/K
w = Pdv
[tex]w_{12}[/tex] = P(v₂ - v₁)
= 2,00,000 ( 0.861 - 0.717)
= 28,800 J/kg
de = δq -δw
δq = de + δw
q₁₂ = (e₂ - e₁) + w₁₂
= 71,650 + 28,800 = 1,00,450 J/kg
Now in this process, the gas is heated from 500 K to 600 K. We would expect at a minimum that the surroundings were at 600 K.
Let’s check for second law satisfaction.
s₂ - s₁ ≥ q₁₂ / Tₓ
182.95 ≥ 1,00,450 / 600 K
182.95 J/kg/K ≥ 167.41 J/kg/K
Hence this condition satisfies the second law of thermodynamics
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A contestant in a winter games event pulls a 36.0 kg block of ice across a frozen lake with a rope over his shoulder as shown in Figure 4.29(b). The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03.
Figure 4.29
(a) Calculate the minimum force F he must exert to get the block moving.
40.873
(b) What is its acceleration once it starts to move, if that force is maintained?
m/s2
(a) The minimum force F he must exert to get the block moving is 38.9 N.
(b) The acceleration of the block is 0.79 m/s².
Minimum force to be applied
The minimum force F he must exert to get the block moving is calculated as follows;
Fcosθ = μ(s)Fₙ
Fcosθ = μ(s)mg
where;
μ(s) is coefficient of static frictionm is mass of the blockg is acceleration due to gravityF = [0.1(36)(9.8)] / [(cos(25)]
F = 38.9 N
Acceleration of the blockF(net) = 38.9 - (0.03 x 36 x 9.8) = 28.32
a = F(net)/m
a = 28.32/36
a = 0.79 m/s²
Thus, the minimum force F he must exert to get the block moving is 38.9 N.
The acceleration of the block is 0.79 m/s².
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Technetium -99 (half-life = 6.01) is used in the medical imaging. How many half-lives would go by in 44.0h?
The number of half-lives that would go by in 44 h is 7 half-lives
What is half life?This is the time taken for half a substance to decay
How to determine the number of half lifeThe number of half-lives that will elaspe after 44 h can be obtained as illustrated below:
Half-life (t½) = 6.0 hTime (t) = 44 hNumber of half-lives (n) =?n = t / t½
n = 44 / 6.01
n = 7
Thus, 7 half-lives will elaspe after 44 h
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A baseball player holds a 1.42 N baseball in his hand, a distance of 34.0 cm from the elbow joint as shown in the figure below. The biceps attached at a distance of 2.75 cm from the elbow exerts an upward force of 12.6 N on the forearm. Consider the forearm and hand to be a uniform rod with a mass of 1.20 kg. Calculate the net torque acting on the forearm and hand.
Answer:
90.3N
Explanation:
⊥mg = (0.170 m)(1.20 kg) 9.81 m/s
τ ball = r⊥Wball = (0.340 m)(1.42 N) = − 0.483 N ⋅m
F − 2.001− 0.483 N ⋅m = 0
F = 2.484 N ⋅m
0.0275 m = 90.3 N
The net torque acting on the forearm and hand is 90.3N
What is torque?
Torque is a measure of the pressure that can motivate an object to rotate about an axis. simply as pressure is what causes an object to accelerate in linear kinematics, torque is what reasons an item to collect angular acceleration. Torque is a vector amount.
⇒mg = (0.170 m)(1.20 kg) 9.81 m/s
⇒torque = r⊥weight of the ball
⇒ (0.340 m)(1.42 N) = − 0.483 N ⋅m
⇒F = − 2.001− 0.483 N ⋅m = 0
⇒F = 2.484 N ⋅m
⇒0.0275 m = 90.3 N
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An action force is 50 N to the left. The reaction force must be:
A. 50 N right
B. 50 N down
C. 50 N left
D. 50 N up
All moving objects have ____________.
A. Force
B. Distance
C. Momentum
D. Time
Answer:
I think it's force
Explanation:
Mark as brainliest if it is right
Answer: A
Explanation:
Force can cause a stationary object to start moving or a moving object to change its speed or direction or both
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 the force exerted by the hi.nge on the beam is 114.77 N.
Tension in the cable
Apply the principle of moment and calculate the tension in the cable;
Clockwise torque = TL sinθ
Anticlockwise torque = ¹/₂WL
TL sinθ = ¹/₂WL
T sinθ = ¹/₂W
T = (W)/(2 sinθ)
T = (29 x 9.8)/(2 x sin57)
T = 169.43 N
Vertical component of the forceT + F = W
F = W - T
F = (9.8 x 29) - 169.43
F = 114.77 N
Thus, the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.
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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.
Tension in rope A:
7.42×102 N
Tension in rope B:
7.52×102 N
What is the tension in rope C?
Tension in the rope C is 1.24× 10² N.
To find the answer, we need to know about the horizontal component of tension in the rope B.
What's the angle made by the rope B by horizontal?From the figure of the answer, in the triangle PQR, tan(θ)= PQ/QR = 6/1= 6 θ= tan inverse of 6 = 80.5°What's the horizontal component of the tension in rope B?Horizontal component= tension in rope B × cos80.5°
= 7.52×10² N × cos80.5°
= 1.24×10² N
What's the tension in the rope C?From the figure, we have found that the tension in rope C = horizontal component of the tension in rope BSo, tension in rope C= 1.24×10² NThus, we can conclude that the tension in the rope C is 1.24× 10² N.
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If the velocity of an object is 9 m/s and its momentum is 72 kgm/s, what is its mass
An object with a velocity (v) of 9 m/s and a linear momentum (p) of 72 kg.m/s, has a mass (m) of 8 kg.
What is momentum?In Newtonian mechanics, linear momentum, or simply momentum, is the product of the mass and velocity of an object.
It is a vector quantity, possessing a magnitude and a direction.
The mathematical expression for momentum is:
p = m . v
where,
p is the linear momentum of the object.m is the mass of the object.v is the velocity of the object.An object has a velocity (v) of 9 m/s and its linear momentum (p) is 72 kg.m/s. We will use the definition of linear momentum to calculate the mass of the object.
p = m . v
m = p / v
m = (72 kg.m/s) / (9 m/s) = 8 kg
An object with a velocity (v) of 9 m/s and a linear momentum (p) of 72 kg.m/s, has a mass (m) of 8 kg.
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A 40.0-kg child running at 8.00 m/s suddenly jumps onto a stationary playground merry-go-round at a distance 1.50 m from the axis of rotation of the merry-go-round. The child is traveling tangential to the edge of the merry-go-round just before jumping on. The moment of inertia about its axis of rotation is 600 kg ∙ m2 and very little friction at its rotation axis. What is the angular speed of the merry-go-round just after the child has jumped onto it?
The angular speed of the merry-go-round just after the child has jumped onto it is 0.696 rad/s.
What is principle of conservation of angular momentum?
The principle of conservation of angular momentum states that the sum of initial angular momentum is equal to final angular momentum.
Li = Lf
Li = Ii ωi + Ic ωc
Li = Iiωi + MR²(V/R)
Li = Iiωi + MRV ----- (1)
Angular speed of the merry go round after the child jumps on itLf = If ωf
ωf = Lf/If
If = Im + MR²
ωf = Lf / ( Im + MR²)
Recall, Lf = Li
ωf = (Iiωi + MRV) / ( Im + MR²)
where;
Iiωi is the initial angular momentum of the merry - go round = 0M is mass of the childR radius of rotationV is tangential speed of the childIm is the moment of inertia of the merry go roundωf = (Iiωi + MRV) / ( Im + MR²)
ωf = (0 + 40 x 1.5 x 8) / (600 + 40(1.5)²)
ωf = (480) / (690)
ωf = 0.696 rad/s
Thus, the angular speed of the merry-go-round just after the child has jumped onto it is 0.696 rad/s.
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An air-filled 39.1 μH solenoid has a length 4.0 cm and a cross-sectional area 0.60 cm2. How many turns are in this solenoid?
a.21000
b.144
c.12
d.120
e.1200
No. of turns in the solenoid is an option (b) 144.
The self-inductance of a long solenoid depends only on its physical properties (such as the number of turns of wire per unit length and the volume), and not on the magnetic field or the current.
Self-inductance of solenoid = 39.1 μH
= 39.1 × [tex]10^-^6[/tex] H
Length of the solenoid = 4.0 cm
Cross-sectional area = 0.60 cm²
Expression for the self-inductance of a coil ;
L = µ₀N²A / [tex]l[/tex]
where,
L = Self- Inductance
N = No. of turns.
A = Cross-sectional area
[tex]l=[/tex] Length of the solenoid
L =( 4π × [tex]10^-^7[/tex] × N² × 0.60 ) / 4.0
39.1× 4.0 / 4π × [tex]10^-^7[/tex] × 0.60 = N²
N² = 2.07 × [tex]10^6[/tex]
N = 144
Therefore, the no. of turns of the solenoid is 144.
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Two bullets are fired at the same time with the same kinetic energy.
If one bullet has twice the mass of the other, what is the ratio of the speed of the lighter bullet to the speed of the heavier?
Which can do the most work?
The ratio of the speed of the lighter bullet to the speed of the heavier will be four times the heavier bullet.
Both can do same amount of work.
What is kinetic energy?Kinetic energy of a body is the energy due to the motion of the body.
Kinetic energy = mv²/2
where m is mass and v is velocity of the object.
Since both objects have the same kinetic energy but one bullet has twice the mass of the other, the ratio of the speed of the lighter bullet to the speed of the heavier will be four times the heavier bullet.
The kinetic energy of both bullets is the same. Hence, they can do equal amount of work.
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