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Vladimir79 [104]
3 years ago
9

Force F1 acts on a particle and does work W1. Force F2 acts simultaneously on the particle and does work W2. The speed of the pa

rticle does not change. Which one of the following must be true?a. W1 is zero, and W2 is positiveb. W1 = - W2c. W1 is positive, and W2 is positived. W1 is positive, and W2 is zero
Physics
1 answer:
egoroff_w [7]3 years ago
7 0

Answer:

b.) W1 = -W2

Explanation:

According to Newton's third law of motion, action and reaction are equal and opposite. For the particle to maintain a constant speed, it means that the two workdone on the particle are of eqaul magnitude but act in opposite direction. The two weights, instead of adding up annul each other and has no effect on the speed of the particle.

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A lawn mower is pushed with a force of 79 N. If 11,099 J of work are done on mowing the lawn, what is the total distance the law
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What is the average velocity of a car if it travels from position 25m to a position of -7m in 34 seconds?
zubka84 [21]

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Read 2 more answers
A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule
ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

p_i = 0

The final total momentum is instead:

p_f = m_a v_a + m_c v_c

where

m_a = 125 kg is the mass of the astronaut

v_a = 2.50 m/s is the velocity of the astronaut

m_c = 1900 kg is the mass of the capsule

v_c is the velocity of the capsule

Since the total momentum must be conserved, we have

p_i = p_f = 0

so

m_a v_a + m_c v_c=0

Solving the equation for v_c, we find

v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

F \Delta t = \Delta p

The change in momentum of the astronaut is

\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s

And the duration of the push is

\Delta t = 0.600 s

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

K=\frac{1}{2}mv^2

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J

3 0
3 years ago
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