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devlian [24]
2 years ago
7

How much work is done by 0.070 m3 of gas, when the volume remains constant with pressure of 63 x 105 Pa?

Physics
1 answer:
ICE Princess25 [194]2 years ago
5 0

Answer:

W = 0 J

Explanation:

The amount of work done by gas at constant pressure is given by the following formula:

W = P\Delta V

where,

W = Work done by the gas

P = Pressure of the gas

ΔV = Change in the volume of the gas

Since the volume of the gas is constant. Therefore, there is no change in the volume of the gas:

W = P(0\ m^3)\\

<u>W = 0 J</u>

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A 0.320 kg ball approaches a bat horizontally with a speed of 14.0 m/s and after getting hit by the bat, the ball moves in the o
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time= 0.06s

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<u>Required</u>

Force F

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F=mΔv/t

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F=2.56/0.06

F=42.67N

The average force exerted on the bat 42.67N

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Twin space probes have a mass of 722 kg each. If the gravitational force between the two space probes is 8.61
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Answer:

x = t

y = \frac{1}{3}t

z =t

Explanation:

Given

r(t) = f(t)i + g(t)j + h(t)k at t = 0

Point: (f(t0), g(t0), h(t0))

r(t) = ln\ t_i + \frac{t-1}{t+2}j + t\ ln\ tk, t0 = 1 -- Missing Information

Required

Determine the parametric equations

r(t) = ln\ ti + \frac{t-1}{t+2}j + t\ ln\ tk

Differentiate with respect to t

r'(t) = \frac{1}{t}i +\frac{3}{(t+2)^2}j + (ln\ t + 1)k

Let t = 1 (i.e t0 = 1)

r'(1) = \frac{1}{1}i +\frac{3}{(1+2)^2}j + (ln\ 1 + 1)k

r'(1) = i +\frac{3}{3^2}j + (0 + 1)k

r'(1) = i +\frac{3}{9}j + (1)k

r'(1) = i +\frac{1}{3}j + (1)k

r'(1) = i +\frac{1}{3}j + k

To solve for x, y and z, we make use of:

r(t) = f(t)i + g(t)j + h(t)k

This implies that:

r'(1)t = xi + yj + zk

So, we have:

xi + yj + zk  = (i +\frac{1}{3}j + k)t

xi + yj + zk  = it +\frac{1}{3}jt + kt

By comparison:

xi = it

Divide by i

x = t

yj = \frac{1}{3}jt

Divide by j

y = \frac{1}{3}t

zk = kt

Divide by k

z = t

Hence, the parametric equations are:

x = t

y = \frac{1}{3}t

z =t

3 0
3 years ago
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