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Slav-nsk [51]
2 years ago
11

1.A student attaches a string to a puck on a frictionless air table, and pulls with a constant force on the puck. a. Draw the fo

rce diagram for the puck during the pull. B. Describe the net force on the puck during the pull. C. Describe the motion of the puck during the pull.

Physics
1 answer:
Mrrafil [7]2 years ago
6 0

a)

for the puck :

F = force applied in the direction of pull

N = normal force on the puck in upward direction by the surface of table

W = weight of the puck in down direction due to force of gravity


b)

along the vertical direction , normal force balance the weight of the puck , hence the net force is same as the force of pull F .

so  F = ma                                    where m = mass of puck  , a = acceleration

Fnet = F


c)

since the net force acts in the direction of force of pull F , hence the puck accelerates in the same direction .

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A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from
Ulleksa [173]

Answer:

1270.64\ \text{J}

Explanation:

m = Mass of object = \dfrac{mg}{g}

mg = Weight of object = 20 N

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

\theta = Angle of slope = 30^{\circ}

f = Force of friction

fd = Work done against friction

The force balance of the system is

\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}

The work done against friction is 1270.64\ \text{J}.

8 0
3 years ago
A fixed mass of an ideal gas is heated from 50°C to 80°C (a) at constant volume and (b) at constant pressure. For which case do
soldi70 [24.7K]

Answer:

Specific heat at constant pressure is =  1.005 kJ/kg.K

Specific heat at constant volume is =  0.718 kJ/kg.K

Explanation:

given data

temperature T1 =  50°C

temperature T2 = 80°C

solution

we know energy require to heat the air is express as

for constant pressure and volume

Q  = m ×  c × ΔT     ........................1

here m is mass of the gas and c is specific heat of the gas and Δ T is change in temperature of the gas

here both Mass and temperature difference is equal and energy required is dependent on specific heat of air.

and here at constant pressure Specific heat  is greater than the specific heat at constant volume,

so the amount of heat required to raise the temperature of one unit mass by one degree at constant pressure is

Specific heat at constant pressure is =  1.005 kJ/kg.K

and

Specific heat at constant volume is =  0.718 kJ/kg.K

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3 years ago
A ball is dropped from a building of height h. Assume the ball starts from rest and that air friction can be ignored. Derive an
agasfer [191]

Answer:

t=\sqrt{h/g}

Explanation:

We use the kinematics equation to solve this question:

y(t)=y_{o}+v_{o}t+1/2*a*t^{2}

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y_{o}=h     initial height

y(t)=h/2     final height

So:

h/2=h-1/2*g*t^{2}

t=\sqrt{h/g}

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