The total energy of a 0.050 kg object travelling at 0.70 c is

1 answer:

4 0

Would presume the energy as kinetic energy.

E = (1/2)*mv²

But m = 0.05kg, velocity here = 0.70c, where c is the speed of light ≈ 3* 10⁸ m/s

Ke = (1/2)*mv² = 0.5*0.05*(0.7*3* 10⁸)² = 1.1025 * 10¹⁵ Joules

There is no exact match from the options.

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What is the internal energy (to the nearest joule) of 10 moles of Oxygen at 100 K?

kkurt [141]

Answer:

U = 12,205.5 J

Explanation:

In order to calculate the internal energy of an ideal gas, you take into account the following formula:

$U=\frac{3}{2}nRT$ (1)

U: internal energy

R: ideal gas constant = 8.135 J(mol.K)

n: number of moles = 10 mol

T: temperature of the gas = 100K

You replace the values of the parameters in the equation (1):

$U=\frac{3}{2}(10mol)(8.135\frac{J}{mol.K})(100K)=12,205.5J$

The total internal energy of 10 mol of Oxygen at 100K is 12,205.5 J

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Little Tammy lines up to tackle Jackson to (unsuccessfully) prove the law of conservation of momentum. Tammy’s mass is 34.0 kg a

Naily [24]

Answer:

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So here we can say that initial momentum of the system must be equal to the final momentum of the system

now we have

$m_1v_1 + m_2v_2 = 0$

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1 year ago

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miss Akunina [59]

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(M_m g)/ (M_w g) = [ (p^2 Man )/ (2 K_man)] / [ (p^2 Woman )/ (2 K_woman)

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nadya68 [22]

Answer:

44 N/m

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The extension, e, of the spring = 2.9 m - 1.4 m = 1.5 m

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