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3 years ago

8

The gravitational acceleration on the moon is about one sixth the size of the gravitational on earth. According to newton's seco

nd law of motion,what happends to an astronaut who goes to the moon

1 answer:

alexandr402 [8]3 years ago

7 0

Weight of a person on the earth is given by,

W= mg,

Where, g is the gravitational acceleration of the earth and m is the mass of the person.

Gravitational acceleration of the moon is one-sixth of the earth.

Therefore, the weight of a astronaut who goes to the moon will be one sixth of that on earth. Hence, the astronaut will weight less.

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Sergio039 [100]

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3 years ago

Read 2 more answers

A climber using bottled oxygen accidentally drops the oxygen bottle from an altitude of 4500 m. If the bottle fell straight down

Trava [24]

Answer:

v= 300 m/s

Explanation:

Given that

altitude ,h= 4500 m

The mass ,m = 3 kg

Lets take acceleration due to gravity , g= 10 m/s²

The speed before impact at sea level = v

Initial speed ,u = 0 m/s

We know that

v²=u²+2 g h

v=final speed

u=initial speed

h=height

Now by putting the values in the above equation

v² = 0²+ 2 x 10 x 4500

v²=90000

v= 300 m/s

Therefore the speed at sea level will be 300 m/s.

3 0

3 years ago

When you do work on an object some of your energy is ____ to that object

HACTEHA [7]

Maybe the word could be converted?

8 0

3 years ago

Two boxers are fighting. Boxer 1 throws his 5 kg fist at boxer 2 with a speed of 9 m/s.

Sladkaya [172]

Answer:

0.001 s

Explanation:

The force applied on an object is equal to the rate of change of momentum of the object:

$F=\frac{\Delta p}{\Delta t}$

where

F is the force applied

$\Delta p$ is the change in momentum

$\Delta t$ is the time interval

The change in momentum can be written as

$\Delta p=m(v-u)$

where

m is the mass

v is the final velocity

u is the initial velocity

So the original equation can be written as

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

In this problem:

m = 5 kg is the mass of the fist

u = 9 m/s is the initial velocity

v = 0 is the final velocity

F = -45,000 N is the force applied (negative because its direction is opposite to the motion)

Therefore, we can re-arrange the equation to solve for the time:

$\Delta t=\frac{m(v-u)}{F}=\frac{(5)(0-9)}{-45,000}=0.001 s$

4 0

3 years ago

A 2.64-kg copper part, initially at 400 K, is plunged into a tank containing 4 kg of liquid water, initially at 300 K. The coppe

marin [14]

Answer:

a) $T_f=305.7049\ K$

b) $\Delta S=313.51\ J.K^{-1}$

Explanation:

Given:

mass of copper, $m_c=2.64\ kg$

initial temperature of copper, $T_{ic}=400\ K$

specific heat capacity of copper, $c_c=385\ J.kg^{-1}.K^{-1}$

mass of water, $m_w=4\ kg$

initial temperature of water, $T_{iw}=300\ K$

specific heat capacity of water, $c_w=4200\ J.kg^{-1}.K^{-1}$

a)

<u>∵No heat is lost in the environment and the heat is transferred only between the two bodies:</u>

Heat rejected by the copper = heat absorbed by the water

$2.64\times 385\times (400-T_f)= 4\times 4200\times (T_f-300)$

$T_f=305.7049\ K$

b)

<u>Now the amount of heat transfer:</u>

$Q=m_c.c_c.(T_{ic}-T_{f})$

$Q=2.64\times 385\times (400-305.7049)$

$Q=95841.5841\ J$

∴Entropy change

$\Delta S=\frac{dQ}{T}$

$\Delta S=\frac{95841.5841}{305.7049}$

$\Delta S=313.51\ J.K^{-1}$

5 0

4 years ago