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2 years ago

How fast would a(n) 85 kg man need to run in order to have the same kinetic energy as an 8.0 g bullet fired at 410 m/s?

Physics

1 answer:

kap26 [50]2 years ago

7 0

Answer:

V = 3.97 m/s

Explanation:

Mass of a man, M = 85 kg

Mass of a bullet, m = 8 g = 0.008 kg

Speed of bullet, v = 410 m/s

We need to find the speed of a man in order to have the same kinetic energy as that of the bullet. Let the kinetic energy of the bullet is k. So,

$k=\dfrac{1}{2}mv^2\\\\k=\dfrac{1}{2}\times 0.008\times 410^2\\\\k=672.4\ J$

Since, k = K (K is the kinetic energy of the man and Let V is the speed)

$K=\dfrac{1}{2}MV^2\\\\V=\sqrt{\dfrac{2K}{M}} \\\\V=\sqrt{\dfrac{2\times 672.4}{85}} \\\\V=3.97\ m/s$

So, the speed of the man is 3.97 m/s.

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If 27 J of work are needed to stretch a spring from 15 cm to 21 cm and 45 J are needed to stretch it from 21 cm to 27 cm, what i

kupik [55]

Answer:

9 cm.

Explanation:

The energy used for stretch the spring from $15 cm$ to $21 cm$ will be , $E_{1}=27J$

The energy used for stretch the spring from $21 cm$ to $27 cm$ will be , $E_{2}=45J$

using the energy of spring formula ,we find that

$27 = \frac{1}{2}K((21-L^{2})-(15-L^{2}))$

$45 = \frac{1}{2}K((27-L^{2})-(21-L^{2}))$

Dividing both the equation will get,

$\frac{3}{5}=\frac{(21-L)^{2}-(15-L)^{2}}{(27-L)^{2}-(21-L)^{2}}\\5((21-L)^{2}-(15-L)^{2})=3((27-L)^{2}-(21-L)^{2})\\3(729 - 54L + L^{2}- 441 + 42L - L^{2} ) = 5(441 - 42L + L^{2} - 225 + 30L - L^{2} )\\3(288 - 12L) = 5(216 - 12L)\\24L = 216\\L = 9 cm$

Therefore, the natural length of the spring is, 9 cm.

4 0

2 years ago

For a spacecraft or a molecule to leave the moon, it must reach the escape velocity (speed) of the moon, which is 2.37 km/s. The

guapka [62]

Answer:

Vrms = 291 m/s

Explanation:

The root mean square velocity or vrms is the square root of the average square velocity and is. vrms=√3RTM. Where M is equal to the molar mass of the molecule in kg/mol.

Temperature = 365 K

Root mean square velocity = ?

molar mass of oxygen = 16 g/mol.

But xygen gas (O2) is comprised of two oxygen atoms bonded together. Therefore:

molar mass of O2 = 2 x 16

molar mass of O2 = 32 g/mol

Convert this to kg/mol:

molar mass of O2 = 32 g/mol x 1 kg/1000 g

molar mass of O2 = 3.2 x 10-2 kg/mol

Molar mass of Oxygen = 3.2 x 10-2 kg/mol

Vrms = √[3(8.3145 (kg·m2/sec2)/K·mol)(365 K)/3.2 x 10-2 kg/mol]

Vrms = 291 m/s

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