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4 years ago

13

If a rock is sitting on a table, the rock exerts a downward force on the table.

1 answer:

daser333 [38]4 years ago

7 0

This is true. There is a downward force on the table that stabilizes the rock to stay there.

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9. Luke is an astronaut that moves his position in the space shuttle by climbing a

BartSMP [9]

Explanation:

Remember that power is defined as how much Work is done per unit of time.

$P = \frac{dW}{dt}$

Work is defined as the amount of force applied across a certain distance.

$W = F*d$

Since in both cases of climbing the ladder (on Earth and the moon), Luke coveres the same amount of <em>distance </em>in the same amount of <em>time</em>, we are only left with one difference between the two cases - gravity.

If you were to carry your backpack on the moon with the same load of text books, it would take less force to pick it up on the moon. Therefore, Luke expends less effort on the environment with less gravity - the moon.

To find the difference factor - you would want to divid the gravitational contants between earth and the moon.

6 0

3 years ago

Read 2 more answers

Which sound is most likely the most intense

MrMuchimi

Are there answers choice or a picture I can see?

4 0

4 years ago

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.03 s la

Nookie1986 [14]

Answer:

$h=53.09m$ (2)

$v_{min}>5.05m/s$

$v_{max}$

Explanation:

<u>a)Kinematics equation for the first ball:</u>

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h$ initial position is the building height

$v_{o}=8.9m/s$

The ball reaches the ground, y=0, at t=t1:

$0=h+v_{o}t_{1}-1/2*g*t_{1}^{2}$

$h=1/2*g*t_{1}^{2}-v_{o}t_{1}$ (1)

Kinematics equation for the second ball:

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h$ initial position is the building height

$v_{o}=0$ the ball is dropped

The ball reaches the ground, y=0, at t=t2:

$0=h-1/2*g*t_{2}^{2}$

$h=1/2*g*t_{2}^{2}$ (2)

the second ball is dropped a time of 1.03s later than the first ball:

t2=t1-1.03 (3)

We solve the equations (1) (2) (3):

$1/2*g*t_{1}^{2}-v_{o}t_{1}=1/2*g*t_{2}^{2}=1/2*g*(t_{1}-1.03)^{2}$

$g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)$

$g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)$

$-2v_{o}t_{1}=g*(-2.06*t_{1}+1.06)$

$2.06*gt_{1}-2v_{o}t_{1}=g*1.06$

$t_{1}=g*1.06/(2.06*g-2v_{o})$

vo=8.9m/s

$t_{1}=9.81*1.06/(2.06*9.81-2*8.9)=4.32s$

t2=t1-1.03 (3)

t2=3.29sg

$h=1/2*g*t_{2}^{2}=1/2*9.81*3.29^{2}=53.09m$ (2)

b) $t_{1}=g*1.06/(2.06*g-2v_{o})$

t1 must : t1>1.03 and t1>0

limit case: t1>1.03:

$1.03>9.81*1.06/(2.06*g-2v_{o})$

$1.03*(2.06*9.81-2v_{o})$

$20.8-2.06v_{o}$

$(20.8-10.4)/2.06$

$v_{min}>5.05m/s$

limit case: t1>0:

$g*1.06/(2.06*g-2v_{o})>0$

$2.06*g-2v_{o}>0$

$v_{o}$

$v_{max}$

8 0

4 years ago

4. What would be the mass of #3 in kilograms?

Nezavi [6.7K]

Answer:

0.003

Explanation:

3 is in grammes so you divide by 1000

7 0

3 years ago

A physicist found that a force of 0.68 N was measured between two charged spheres. The distance between the spheres was 1.0m. Ca

amid [387]

The answer would be 0.40m. you are finding how far the distance is between 10 and 50

3 0

3 years ago