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nikitadnepr [17]

4 years ago

10

An astronaut throws a wrench in interstellar space. How much force is required to keep the wrench moving continuously with const

ant velocity?

2 answers:

podryga [215]4 years ago

7 0

Answer: The force required to keep the wrench moving continuously in space is 0 N.

Explanation:

Force applied on an object is defined as the product of the mass and acceleration of the object.

Mathematically,

$F=ma$

where, F = force applied

m = mass of the wrench

a = acceleration of the wrench

Acceleration is defined as the rate of change of velocity with respect to time. We are given, a wrench is moving with a constant velocity, so the acceleration of the wrench will be 0.

$a=\frac{dv}{dt}=0$ (At constant velocity)

Putting values in above equation, we get:

$F=m\times 0=0$

Hence, the force required to keep the wrench moving continuously in space is 0 N.

Vlada [557]4 years ago

6 0

The answer is a force is equal to zero

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3 years ago

You placed 43.1 g of an unknown metal at 100 °C into a coffee cup calorimeter that contained 50.0 g of water that was initially

nika2105 [10]

Answer :

(a) The heat released by the metal is -312.48 J

(b) The specific heat of the metal is $0.0944J/g^oC$

Explanation :

<u>For part A :</u>

Heat released by the metal = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the metal

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $51.5J/^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_2$ = mass of water = 50.0 g

$\Delta T$ = change in temperature = $(T_{final}-T_{initial})=23.2-22.0=1.2^oC$

Now put all the given values in the above formula, we get:

$q=[(51.5J/^oC\times 1.2^oC)+(50.0g\times 4.184J/g^oC\times 1.2^oC)]$

$q=312.48J$

Thus, the heat released by the metal is -312.48 J

<u>For part B :</u>

$q=m\times c\times \Delta T$

q = heat released by the metal = -312.48 J

m = mass of metal = 43.1 g

c = specific heat of metal = ?

$\Delta T$ = change in temperature = $(T_{final}-T_{initial})=23.2-100=76.8^oC$

Now put all the given values in the above formula, we get:

$-312.48J=43.1g\times c\times 76.8^oC$

$c=0.0944J/g^oC$

Thus, the specific heat of the metal is $0.0944J/g^oC$

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