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2 years ago

7

In order to move a bag of dog food across a room that measures 8 meters in length, you apply 10 Newtons of force. How much work

was done?

1 answer:

Lady bird [3.3K]2 years ago

7 0

Answer:

80 joules of work

Explanation:

Use W=F·d

W=10N(8M)= 80 Joules

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To help pick to up the metal and place it in a incinerator <span />

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Jamal plugged his radio into the wall. The radio's plug had copper wires surrounded by rubber. The rubber protects Jamal from___

Mariana [72]

The rubber protects him from being electrocuted by the flow of current going through the plug.
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3 years ago

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

If a car was moving at a speed of 25mph for 6 hours, how far would the car travel?

Anestetic [448]

Answer:

150 miles

Explanation:

25mph x 6hrs = 150miles

7 0

3 years ago

What force must the worker exert to get the box moving & what force must the worker exert to accelerate the box at 0.1 meter

photoshop1234 [79]

Since static friction is the minimum force required to just start the motion of a stationary object.

Here if we need to start an object from rest then we required F = 700 N

So for the first part of the above problem Force will be F = 700 N

Now if the box is already moving then we will have to use kinetic friction force between box and floor

now we can write the equation of net force as

$F - F_k = m*a$

here

$F_k = kinetic friction = 220 N$

$m = mass = 500 kg$

$a = acceleration = 0.1 m/s^2$

now we will have

$F - 220 = 500* 0.1$

$F = 220 + 50 = 270 N$

3 0

3 years ago