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3 years ago

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19. Calculate the work done when a 30 N force pushes a rock 10 m. a) 3J b) 0.33 J c) 40 J d) 300 J

1 answer:

aksik [14]3 years ago

5 0

19. Calculate the work done when a 30 N force pushes a rock 10 m. a) 3J b) 0.33 J c) 40 J d) 300 J

20. How far did a rock move if a force of 200 N moved it doing 100 J of work? a) 2 m b) 0.5 m c) 20,000 m d) 300 m

21. An explosion of gas from a Hawaiian volcano blows a rock into the sky. the rock has 2000J of kinetic energy. What was its mass if the rock had a velocity of 20m/s. a) 200Kg b) 20,000Kg c) 10Kg d) 200Kg

22. How much potential energy does a 2 kg book have if it is 2 meters from the ground? a) 39.2J b) 12 J c) 19.6 J d) 98 J

23. 45J of work were done on the object that moved 5M. How much force was applied to the object? a) 8N b) 9N c) 225N d) ON

24. If you do 40 joules of work lifting a 10N box, how high is the shelf you move it to. a) 400m b) 50m c) 40m d) 4m

25. A force of 100 newtons was necessary to lift a rock. A total of 150 joules of work was done. How far was the rock lifted? a) 1.5 m b) 15 m c) 0.15 m d) 150 m19. Calculate the work done when a 30 N force pushes a rock 10 m. a) 3J b) 0.33 J c) 40 J d) 300 J

20. How far did a rock move if a force of 200 N moved it doing 100 J of work? a) 2 m b) 0.5 m c) 20,000 m d) 300 m

21. An explosion of gas from a Hawaiian volcano blows a rock into the sky. the rock has 2000J of kinetic energy. What was its mass if the rock had a velocity of 20m/s. a) 200Kg b) 20,000Kg c) 10Kg d) 200Kg

22. How much potential energy does a 2 kg book have if it is 2 meters from the ground? a) 39.2J b) 12 J c) 19.6 J d) 98 J

23. 45J of work were done on the object that moved 5M. How much force was applied to the object? a) 8N b) 9N c) 225N d) ON

24. If you do 40 joules of work lifting a 10N box, how high is the shelf you move it to. a) 400m b) 50m c) 40m d) 4m

25. A force of 100 newtons was necessary to lift a rock. A total of 150 joules of work was done. How far was the rock lifted? a) 1.5 m b) 15 m c) 0.15 m d) 150 m19. Calculate the work done when a 30 N force pushes a rock 10 m. a) 3J b) 0.33 J c) 40 J d) 300 J

20. How far did a rock move if a force of 200 N moved it doing 100 J of work? a) 2 m b) 0.5 m c) 20,000 m d) 300 m

21. An explosion of gas from a Hawaiian volcano blows a rock into the sky. the rock has 2000J of kinetic energy. What was its mass if the rock had a velocity of 20m/s. a) 200Kg b) 20,000Kg c) 10Kg d) 200Kg

22. How much potential energy does a 2 kg book have if it is 2 meters from the ground? a) 39.2J b) 12 J c) 19.6 J d) 98 J

23. 45J of work were done on the object that moved 5M. How much force was applied to the object? a) 8N b) 9N c) 225N d) ON

24. If you do 40 joules of work lifting a 10N box, how high is the shelf you move it to. a) 400m b) 50m c) 40m d) 4m

25. A force of 100 newtons was necessary to lift a rock. A total of 150 joules of work was done. How far was the rock lifted? a) 1.5 m b) 15 m c) 0.15 m d) 150 m

<em><u>please</u></em><em><u> </u></em><em><u>mark</u></em><em><u> </u></em><em><u>me</u></em><em><u> </u></em><em><u>as</u></em><em><u> </u></em><em><u>brainliest</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

<em><u>follow</u></em><em><u> </u></em><em><u>me</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

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What will the magnitude of the field be if the 10 nc charge is replaced by a 20 nc charge? Assume the system is big enough to co

MArishka [77]

Answer:

Same magnitude of the 10 nc charge cause the electric field is external.

Explanation:

To do a better explanation, let's go and suppose we have an electric field of, 1300 N/C with a 10 nC charge.

As the system we are talking about is really big, and the charge is small, we can assume always if the charge is sitting right in the same point where the electric field is, then, the electric field would not suffer any kind of alteration in it's value. Therefore, no matter what value of the charge is sitting here, the electric field is independent of the charge, so it would not feel any alteration. However, the force that the charge is feeling would be stronger than in the first case.

F = qE

If charge is doubled, then the force would be bigger in the second case than in the first case, but electric field remain the same value.

4 0

3 years ago

9. Captain America is chasing Red Skull. He plans to throw his shield to knock down Red Skull but needs to know how fast Red Sku

Sedaia [141]

Red Skull's relative velocity to Captain America, towards the left front of the

truck is approximately <u>33.23 m/s</u> in a direction from the North of

approximately <u>9.18°</u>.

Reasons:

Assumptions;

Taking the north direction as positive.

The activity takes place on the trucks.

The trucks are moving towards each other.

Solution:

Vector form of net speed of Red Skull, is given as follows;

v₁ = -( $\frac{\sqrt{2} }{2}$ × 3.5)·i + ( $\frac{\sqrt{2} }{2}$ × 3.5 + 12.5)·j

Vector form of the net speed of Captain America is given as follows;

v₂ = ( $\frac{\sqrt{2} }{2}$ × 4.0)·i - ( $\frac{\sqrt{2} }{2}$ × 4.0 + 15)·j

Relative velocity, v₁₂ = v₁ - v₂

∴ v₁₂ = (-( $\frac{\sqrt{2} }{2}$ × 3.5) - ( $\frac{\sqrt{2} }{2}$ × 4.0))·i + (( $\frac{\sqrt{2} }{2}$ × 3.5 + 12.5) + ( $\frac{\sqrt{2} }{2}$ × 4.0 + 15))·j

v₁₂ = $-\frac{ 15 \cdot \sqrt{2} }{4}$ ·i + $\frac{ 110 + 15 \cdot \sqrt{2} }{4}$ ·j

Red Skull's velocity relative to Captain America, v₁₂ = $-\frac{ 15 \cdot \sqrt{2} }{4}$ ·i + $\frac{ 110 + 15 \cdot \sqrt{2} }{4}$ ·j

v₁₂ ≈ -5.3·i + 32.8·j

Therefore;

Red Skull appears to be moving West at <u>5.3 m/s</u> and North at <u>32.8 m/s</u>

The direction is $arctan \left(\frac{32.8}{-5.3} \right) \approx -80.2^{\circ}$

Therefore;

Red Skull appear to be moving at 90° - 80.2° ≈ 9.18° towards the left front end of the truck moving North

The magnitude of the velocity, |v₁₂|, is given as follows;

$|v_{12}| = \sqrt{\left(-\frac{ 15 \cdot \sqrt{2} }{4}\right)^2 + \left(\frac{ 110 + 15 \cdot \sqrt{2} }{4}\right)^2} = \dfrac{ 5 \cdot \sqrt{130+33 \cdot\sqrt{2} } }{2} \approx 33.23$ ·

The magnitude of Red Skull's velocity relative to Captain America is,

therefore;

|v₁₂| ≈ <u>33.23 m/s</u>

Learn more here:

brainly.com/question/24430414

6 0

3 years ago

A 70.0-kg skier is sliding at 4 m/s when they slide down a 2m high hill. At the bottom of the hill they run into a large 2800 N/

Katarina [22]

Answer:

$\Delta x=245\ mm$

Explanation:

Given:

mass of skier, $m=70\ kg$

initial velocity of skier, $u=4\ m.s^{-1}$

height of the hill, $h=2\ m$

spring constant, $k=2800\ N.m^{-1}$

<u>final velocity of skier before coming in contact of spring:</u>

Using eq. of motion:

$v^2=u^2+gh$

$v^2=4^2+9.8\times 2$

$v=5.9666\ m.s^{-1}$

<u>Now the time taken by the skier to reach down:</u>

$v=u+gt$

$5.9666=4+9.8\ t$

$t=0.2007\ s$

<u>Now we calculate force using Newton's second law:</u>

$F=\frac{dp}{dt}$

$F=\frac{m(v-u)}{t}$

$F=\frac{70\times(5.9666-4)}{0.2007}$

$F\approx686\ N$

<u>∴Compression in spring before the skier momentarily comes to rest:</u>

$\Delta x=\frac{F}{k}$

$\Delta x=\frac{686}{2800}$

$\Delta x=0.245\ m$

$\Delta x=245\ mm$

4 0

3 years ago

In a tug of war, two teams exerts a force of 30 N each. What is the net force acting on the rope?

Arte-miy333 [17]

The force exerted by each team is 30 N. The forces are in the direction opposite to each other.

Force exerted by team 1, $F_{1}=30 N$

Force exerted by team 2, $F_{2}=-30 N$

Net force, $F=F_{1} +F_{2}$

$=30+(-30)=0$

Therefore, net force on the rope is 0 N.

4 0

3 years ago

pochemuha

B

HOPE IT HELPS LET ME KNOW IF U NEED EXPLANATION

4 0

3 years ago