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babymother [125]
3 years ago
10

Kevin is refinishing his rusty wheelbarrow. He moves his sandpaper back and forth 45 times over a rusty area, each time moving a

total distance of 0.15 m. Kevin pushes the sandpaper against the surface with a normal force of 1.8 N. The coefficient of friction for the metal/sandpaper interface is 0.92. How much work is done by the kinetic frictional force during the sanding process
Physics
1 answer:
dmitriy555 [2]3 years ago
5 0
W = _|....F*dx*cos(a)........With F=force, x=distance over which force acts on object,
.......0.............................and a=angle between force and direction of travel.

Since the force is constant in this case we don't need the equation to be an integral expression, and since the force in question - the force of friction - is always precisely opposite the direction of travel (which makes (a) equal to 180 deg, and cos(a) equal to -1) the equation can be rewritted like so:

W = F*x*(-1) ............ or ............. W = -F*x

The force of friction is given by the equation: Ffriction = Fnormal*(coeff of friction)

Also, note that the total work is the sum of all 45 passes by the sandpaper. So our final equation, when Ffriction is substituted, is:

W = (-45)(Fnormal)(coeff of friction)(distance)
W = (-45)...(1.8N).........(0.92).........(0.15m)
W = ................-11.178 Joules
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A Carnot engine operates with an efficiency of 26.0% when the temperature of its cold reservoir is 281 K. Assuming that the temp
VLD [36.1K]

Answer:

The temperature of cold reservoir should be 246.818 K for efficiency of 35%

Explanation:

In first case we have given efficiency of Carnot engine = 26 % = 0.26

Temperature of cold reservoir T_L=281K

We know that efficiency of Carnot engine is given by

\eta =1-\frac{T_L}{T_H}

0.26 =1-\frac{281}{T_H}

T_H=379.72K

For second Carnot engine efficiency is given as 35% = 0.35

And temperature of hot reservoir is same so T_H=379.72K

So 0.35=1-\frac{T_L}{379.72}

T_L=246.818K

So the temperature of cold reservoir should be 246.818 K for efficiency of 35%

4 0
3 years ago
What do fuel cells, batteries and, solar cells have in common A.the all produce static electricity B. they are all sources of di
Vaselesa [24]
The correct answer is B :)
3 0
4 years ago
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A yacht is pushed along at a constant velocity by the wind. The wind force is 180N at an angle of 25 degrees to the direction of
damaskus [11]
The parallel component is given by
F=180cos(25)=163.14N
4 0
3 years ago
If 3600 j of work is done in 3.0 s what is the power<br>0.00083W<br>1200W<br>3600W<br>11000W
Viktor [21]

Answer:

1200 W

Explanation:

Power is given by the ratio between work done and time taken:

P=\frac{W}{t}

where W is the work done and t the time taken.

In this problem, W = 3600 J and t = 3.0 s. Therefore, the power in this exercise is

P=\frac{3600 J}{3.0 s}=1200 W

5 0
4 years ago
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A block is resting on a platform that is rotating at an angular speed of 2.4 rad/s. The coefficient of static friction between t
Sloan [31]

Answer:

r = 0m is the Minimum distance from the axis at which the block can remain in place wothout skidding.

Explanation:

From a sum of forces:

Ff = m*a   where Ff = μ * N    and a = \frac{V^2}{r}=\omega^2*r

N - m*g = 0   So, N = m*g.   Replacing everything on the original equation:

\mu*m*g = m*\omega^2*r   (eq2)

Solving for r:

r = \frac{\mu*g}{\omega^2}=1.41m

If we analyze eq2 you can conclude that as r grows, the friction has to grow (assuming that ω is constant), so the smallest distance would be 0 and the greatest 1.41m. Beyond that distance, μ has to be greater than 0.83.

4 0
3 years ago
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