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shutvik [7]
3 years ago
14

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]3 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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Speed is the same as the initial: 25m/s.

*if* you need vectors though:

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3 0
3 years ago
Three moles of a monatomic ideal gas are heated at a constant volume of 1.20 m3. The amount of heat added is 5.22x10^3 J.(a) Wha
k0ka [10]

Answer:

A) 140 k

b ) 5.22 *10^3 J

c) 2910 Pa

Explanation:

Volume of Monatomic ideal gas = 1.20 m^3

heat added ( Q ) = 5.22*10^3 J

number of moles  (n)  = 3

A ) calculate the change in temp of the gas

since the volume of gas is constant no work is said to be done

heat capacity of an Ideal monoatomic gas ( Q ) = n.(3/2).RΔT

make ΔT subject of the equation

ΔT = Q / n.(3/2).R

    = (5.22*10^3 ) / 3( 3/2 ) * (8.3144 J/mol.k )

    = 140 K

B) Calculate the change in its internal energy

ΔU = Q  this is because no work is done

therefore the change in internal energy = 5.22 * 10^3 J

C ) calculate the change in pressure

applying ideal gas equation

P = nRT/V

therefore ; Δ P = ( n*R*ΔT/V )

                        = ( 3 * 8.3144 * 140 ) / 1.20

                        = 2910 Pa

3 0
3 years ago
A piston-cylinder chamber contains 0.1 m3 of 10 kg R-134a in a saturated liquid-vapor mixture state at 10 °C. It is heated at co
vaieri [72.5K]

Answer:

(A) 10132.5Pa

(B)531kJ of energy

Explanation:

This is an isothermal process. Assuming ideal gas behaviour then the relation P1V1 = P2V2 holds.

Given

m = 10kg = 10000g, V1 = 0.1m³, V2 = 1.0m³

P1 = 101325Pa. M = 102.03g/mol

P2 = P1 × V1 /V2 = 101325 × 0.1 / 1 = 10132.5Pa

(B) Energy is transfered by the r134a in the form of thw work done in in expansion

W = nRTIn(V2/V1)

n = m / M = 10000/102.03 = 98.01mols

W = 98.01 × 8.314 × 283 ×ln(1.0/0.1)

= 531kJ.

6 0
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Answer:

1

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Conductivity is required for the electric current to flow.
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