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HACTEHA [7]
3 years ago
12

A cylinder is fitted with a piston, beneath which is a spring, as in the drawing. The cylinder is open at the top. Friction is a

bsent. The spring constant of the spring is 3900 N/m. The piston has a negligible mass and a radius of 0.020 m.
a) When air beneath the piston is completely pumped out, how much does the atmospheric pressure cause the spring to compress?
b) How much work does the atmospheric pressure do in compressing the spring?
Physics
1 answer:
patriot [66]3 years ago
5 0
A) We know that:
Pressure = Force / Area

And for a spring,
F = kx, where k is the spring constant and x is the change in length.

P = kx/A; P = 101325 Pa
101325 = 3900x / (0.02²π)
x = 0.033 m or 3.3 cm

b) Work = 1/2 kx²
= 1/2 * 3900 * (0.033)²
= 2.12 Joules
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Bob is standing at rest. Wendy runs past him at a constant speed of 3.00 m/s. At the instant that Wendy passes him, Bob starts t
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Answer:

90 meters

Explanation:

To find the distance in which Bob catches up with Wendy, you first write down the motion equations of both Bob and Wendy.

Wendy has a constant speed, then you have:

x_1=v_1t           (1)

Bob has an accelerated motion, then, his equation of motion is:

x_2=v_2t+\frac{1}{2}at^2          (2)

v1: constant speed of Wendy = 3.00 m/s

v2: initial speed of Bob = 0 m/s

a: acceleration of Bob = 0.200m/s^2

When Bob catches up with Wendy x1 = x2, then you equal both equations and solve for time t:

x_1=x_2\\\\v_1t=\frac{1}{2}at^2\\\\t=\frac{2v_1}{a}

you replace the values of a and v1:

t=\frac{2(3.00m/s)}{0.200m/s^2}=30 s

Finally, you replace this value of time either the equation for x1 or the equation for x2, and you calculate the distance:

x_2=x_1=(3.00m/s)(30s)=90m

hence, Bob catches up with Wendy for a distance of 90m

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