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lora16 [44]
3 years ago
6

a bullet of mass 4g when fired with a velocity of 50m/s can enter a wall upto a depth of 10cm how much will be the average resis

tence offered by the wall
Physics
1 answer:
jok3333 [9.3K]3 years ago
8 0
Mass, m = 4g = 0.004 kg
Velocity, =  50cm/s = 0.5m/s
Distance, 10cm = 0.1m
The wall would have to resist the energy acquired by the bullet.

Kenetic Energy of bullet = Resistance offered by the wall.

1/2 mv²                        =  Resistance Force * Distance

(1/2) * 0.04 * 0.5 * 0.5 =  F * 0.1

0.5 * 0.04 * 0.5 * 0.5 =  F * 0.1

0.5 * 0.04 * 0.5 * 0.5/0.1 = F

0.05 = F

Therefore, Resistance offered by the wall = 0.05 N
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Answer:

10 m/s

Explanation:

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(8 kg)(8 m/s) + (6 kg)(6 m/s) = (8 kg)(5 m/s) + (6 kg) v

64 kg m/s + 36 kg m/s = 40 kg m/s + (6 kg) v

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Answer:

You increase the acceleration of the car.

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3 years ago
The force of attraction between a ball is F=.........×10^-¹¹
DIA [1.3K]

Answer:

4.45×10¯¹¹ N

Explanation:

From the question given above, the following data were obtained:

Mass of ball (M₁) = 4 Kg

Mass of bowling pin (M₂) = 1.5 Kg

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Distance apart (r) = 3 m

Force of attraction (F) =?

The force of attraction between the ball and the bowling pin can be obtained as follow:

F = GM₁M₂ / r²

F = 6.67×10¯¹¹ × 4 × 1.5 / 3²

F = 4.002×10¯¹⁰ / 9

F = 4.45×10¯¹¹ N

Therefore, the force of attraction between the ball and the bowling pin is 4.45×10¯¹¹ N

8 0
3 years ago
A model rocket has an acceleration of 12 m/s2. A net force of 18 N is acting on the rocket. What is the mass of the rocket? its
larisa86 [58]

Answer:

C.) 1.5 kg

Explanation:

Start with the equation:

F_n_e_t=ma

Plug in what you know, and solve:

18=m(12)\\m=1.5kg

Find matching soluation:

C.) 1.5 kg

5 0
2 years ago
What is the length a rubberband was stretched if it has a spring constant of 5700N/m and is currently holding 8600J OF POTENTIAL
lozanna [386]

Answer:

\displaystyle \Delta x=1.74\ m

Explanation:

<u>Elastic Potential Energy </u>

Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

\displaystyle PE = \frac{1}{2}k(\Delta x)^2

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.

Solving for Δx:

\displaystyle \Delta x=\swrt{\frac{2PE}{k}}

Substituting:

\displaystyle \Delta x=\sqrt{\frac{2*8600}{5700}}

Calculating:

\displaystyle \Delta x=\sqrt{3.0175}

\boxed{\displaystyle \Delta x=1.74\ m}

6 0
3 years ago
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