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inysia [295]
3 years ago
12

If the average movement of the particles in a sweater increases, what will happen to the sweater's temperature?

Physics
1 answer:
stira [4]3 years ago
7 0
The Temperature will increase
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Which of the following factors affects the pressure of an enclosed gas?
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OverLord2011 [107]
I really don’t get this give me a second
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3 years ago
The cable of a crane is lifting a 750 kg girder. The girder increases its speed from 0.25 m/s to 0.75 m/s in a distance of 3.5 m
IgorC [24]

To solve this exercise it is necessary to apply the equations concerning Work, both by general definition and by conservation of energy.

In other words, the work done by an object due to gravity is the equivalent to that defined by the potential energy equations, that is

W= mg\Delta h

Where,

m=mass

g=gravitational acceleration

\Delta h = Change in height

On the other hand we have that the work done by tension is defined by the conservation of kinetic and potential energy, that is to say

W= \Delta KE + \Delta PE

Where,

\Delta KE = Change in Kinetic Energy

\Delta PE =Change in Potential Energy

PART A) As defined by the work done by gravity would be given by,

W = mg\Delta h

W = (750)(9.8)(3.5)

W = 25.725kJ

Therefore the work done by gravity is 25.725kJ

PART B) The work done by the tension applies the energy conservation equation, that is to say

W= \Delta KE + \Delta PE

W = (\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2)+(mgh_f-mgh_i)

W = \frac{1}{2}m(v_f^2-v_i^2)+mg(h_f-h_i)

Replacing with our values,

W = \frac{1}{2}(750)(0.75^2-0.25^2)+(750)(9.8)(3.5)

W = 25.912kJ

Therefore the work done by tension is 25.9kJ

5 0
3 years ago
A clock has radius of 0.5m. The outermost point on its minute hand travels along the edge. What is its tangential speed?
Andreas93 [3]
The angular speed of the minute hand is the ratio between the angle covered and the time taken. The angle covered by the minute hand is 2 \pi, while the time needed is t=60 min= 3600 s, so the angular speed is
\omega =  \frac{2 \pi}{3600 s}=0.0017 rad/s

And the tangential speed is given by
v=\omega r=(0.0017 rad/s)(0.5 m)=8.72 \cdot 10^{-4} m/s
6 0
3 years ago
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