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3 years ago

If it takes 600 N to move a box 5 meters, how much work is done on the box?

Physics

2 answers:

erastovalidia [21]3 years ago

8 0

C it is 600 x 5 = 3000 J hope this helped!

Anika [276]3 years ago

3 0

Answer:

Option C is the correct option.

Explanation:

Given,

Force = 600 N

Distance = 5 meters

Work = ?

Now,

Work = Force $\times$ distance

$= 600 \times 5$

Calculate the product

$= 3000 \:$ Joule

Hope this helps...

Good luck on your assignment..

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A woman with mass 50 kg is standing on the rim of a large disk that is rotating at 0.80 rev/s about an axis through its center.

My name is Ann [436]

Answer:

angular momentum, $L=1.4\times 10^4\ kg-m^2/s$

Explanation:

Given that,

Mass of the woman, m = 50 kg

Angular velocity of the disk, $\omega=0.8\ rev/s=5.02\ rad/s$

Mass of the disk, m' = 2670 kg

Radius of the disk, R = 4 m

We need to find the magnitude of the total angular momentum of the woman–disk system. The moment of inertia of the system is equal to the sum of moment of inertia of women and the moment off inertia of the disk.

$I=mR^2+\dfrac{1}{2}m'R^2\\\\I=R^2(m+\dfrac{1}{2}m')\\\\I=(4)^2\times (50+\dfrac{1}{2}\times 270)\\\\I=2960\ kg-m^2$

The angular momentum is given by :

$L=I\omega\\\\L=2960\times 5.02\\\\L=14859.2\ kg-m^2/s$

$L=1.4\times 10^4\ kg-m^2/s$

So, the magnitude of the total angular momentum of the woman–disk system is $1.4\times 10^4\ kg-m^2/s$ . Hence, this is the required solution.

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A car traveling west in a straight line on a highway decreases its speed from 30 meters per second to 23 meters per second in 2

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A gas is compressed at a constant pressure of 0.800 atm from 12.00 L to 3.00 L. In the process, 390 J of energy leaves the gas b

Andru [333]

Answer:

a)W= - 720 J

b)ΔU= 330 J

Explanation:

Given that

P = 0.8 atm

We know that 1 atm = 100 KPa

P = 80 KPa

V₁ = 12 L = 0.012 m³ ( 1000 L = 1 m³)

V₂ = 3 L = 0.003 m³

Q= - 390 J ( heat is leaving from the system )

We know that work done by gas given as

W = P (V₂ -V₁ )

W= 80 x ( 0.003 - 0.012 ) KJ

W= - 0.72 KJ

W= - 720 J ( Negative sign indicates work done on the gas)

From first law of thermodynamics

Q = W + ΔU

ΔU=Change in the internal energy

Now by putting the values

- 390 = - 720 + ΔU

ΔU= 720 - 390 J

ΔU= 330 J

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