Why does the total amount of energy before and after any energy transformations remain the same?

2 answers:

8 0

The Law of Conservation of Energy states that, in an isolated system, energy remains constant and can not be created or destroyed, only transferred from one form to another. This law was created by Julius Robert Mayer.

STALIN [3.7K]3 years ago

3 0

Answer:

This is because energy can neither be created or destroyed but can only be transferred from one state to the other

Explanation:

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A 5,000 kg satellite is orbiting the Earth in a circular path. The height of the satellite above the surface of the Earth is 800

Arada [10]

Explanation:

The given data is as follows.

m = 5000 kg, h = 800 km = $0.8 \times 10^{6} m$

$R_{e} = 6.37 \times 10^{6} m$ , r = R + h = $7.17 \times 10^{6} m$

$M_{e} = 5.98 \times 10^{24}$ kg, G = $6.67 \times 10^{-11} Nm^{2}/kg^{2}$

As we know that,

$\frac{mv^{2}}{r} = \frac{GmM_{e}}{r^{2}}$

v = $\sqrt{\frac{GM_{e}}{r^{2}}}$

And, it is known that formula to calculate angular velocity is as follows.

$\omega = \frac{v}{r} = \sqrt{\frac{GM_{e}}{r^{3}}}$

v = $\sqrt{\frac{GM_{e}}{r^{3}}}$

= $\sqrt{\frac{6.67 \times 10^{-11} Nm^{2}/kg^{2} \times 5.98 \times 10^{-24} kg^{-2}}{(7.17 \times 10^{6} m)^{3}}}$

= $1.0402 \times 10^{-3} rad/s$

Thus, we can conclude that speed of the satellite is $1.0402 \times 10^{-3} rad/s$ .

6 0

3 years ago

What real-world examples show no work being done? Can you think of examples other than resisting the force of gravity?

GalinKa [24]

-- pushing on a brick wall

-- standing on your little brother's back so that he can't get up

-- taking a nap while on the job

-- squeezing anything that doesn't yield to your squeeze, such as a glass bottle or your girl friend

-- watching TV

-- solving math problems in your head

-- making pictures out of clouds in the sky

8 0

3 years ago

â/8.37 points scalcet8 12.2.037. ask your teacher my notes question part points submissions used a block-and-tackle pulley hoist

AnnyKZ [126]

Refer to the diagram shown below.

The hoist is in static equilibrium supported by tensions in the two ropes.

For horizontal force balance, obtain
T₃ cos 50 = T₂ cos 38
0.6428T₃ = 0.788T₂
T₃ = 1.2259T₂ (1)

For vertical force balance, obtain
T₂ sin 38 + T₃ sin 50 = 350
0.6157T₂ + 0.766T₃ = 350 (2)

Substitute (1) into (2).
0.6157T₂ + 0.766(1.2259T₂) = 350
1.5547T₂ = 350
T₂ = 225.124 N
T₃ = 1.2259(225.124) = 275.979

Answer:
T₂ = 225.12 N
T₃ = 275.98 N