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1 year ago

I need help with those 2.

Physics

1 answer:

Nimfa-mama [501]1 year ago

5 0

Hi there!

Question 1:

We know that Impulse = Δp = mΔv, so:

I = 0.045(70) = 3.15 Ns

I = F · t, so:

3.15/0.1 = F

F = 31.5 N

Question 2.

We can find the impulse using:

I = Ft

I = 300 · 0.04 = 12 Ns

Find the change in velocity using:

I = mΔv

12/0.13 = Δv = 92.31 m/s

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A person holding a 15.0 kg containing one 50.0 g bullet is riding on a train that is traveling at 75.0 km/h east. If the man fir

Lana71 [14]

Answer:

The velocity of the gun relative to the ground is 19.66 m/s

Explanation:

Given data,

The mass of the gun, M = 15.0 kg

The mass of the bullet, m = 50 g

The velocity of the train, v = 75 km/h

= 20.83 m/s

The velocity of bullet relative to train, V' = 350 m/s

The velocity of bullet relative to ground, V = 350 + 20

= 370 m/s

According to the law of conservation of momentum,

Mv' + mV' = 0

$v' = -\frac{mV'}{M}$

$v' = -\frac{0.050\times 350}{15}$

= -1.17 m/s

Therefore, the velocity of the gun with,

v₀ = V + v'

= 20.83 - 1.17

= 19.66 m/s

Hence, the velocity of the gun relative to the ground is 19.66 m/s

8 0

2 years ago

Hello guys I have a quick question and please help me! Thanks!

xenn [34]

Answer: 4 herz is the answer!

Explanation:

6 0

2 years ago

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Whenever two Apollo astronauts were on the surface of the Moon, a third astronaut orbited the Moon. Assume the orbit to be circu

inysia [295]

Answer:

$v = 1,582 \ \frac{m}{s}$

Explanation:

We know that for circular motion the centripetal acceleration $a_c$ is:

$a_c = \frac{v^2}{r}$

where v is the speed and r is the radius.

The centripetal acceleration for the astronaut must be the gravitational acceleration due to the gravity, as there are no other force. So

$a_c = 1.27 \frac{m}{s^2}$ .

The radius of the orbit must be the radius of the Moon, plus the 270 km above the surface

$r = 1.7 * 10^6 \ m + 270 \ km$

$r = 1.7 * 10^6 \ m + 270 * 10^3 \ m$

$r = 1.7 * 10^6 \ m + 0.270 * 10^6 \ m$

$r = 1.97 * 10^6 \ m$

We can obtain the speed as:

$v^2 = a_c r$

$v = \sqrt{a_c r}$

$v = \sqrt{1.27 \frac{m}{s^2} * 1.97 * 10^6 \ m}$

$v = \sqrt{ 2.509 \ 10^6 \ \frac{m^2}{s^2}}$

$v = 1.582 \ 10^3 \ \frac{m}{s}$

$v = 1,582 \ \frac{m}{s}$

And this is the orbital speed.

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2 years ago

a ball is whirled on a string and then the string breaks. what causes the ball to move off in a straight line ?

katrin [286]

It would be called Inertia

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3 years ago

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A 3.0 kg object moving 8.0 m/s in the positive x direction has a one-dimensional elastic collision with an object (mass = M) ini

finlep [7]

<h2>Option 2 is the correct answer.</h2>

Explanation:

Elastic collision means kinetic energy and momentum are conserved.

Let the mass of object be m and M.

Initial velocity object 1 be u₁, object 2 be u₂

Final velocity object 1 be v₁, object 2 be v₂

Initial momentum = m x u₁ + M x u₂ = 3 x 8 + M x 0 = 24 kgm/s

Final momentum = m x v₁ + M x v₂ = 3 x v₁ + M x 6 = 3v₁ + 6M

Initial kinetic energy = 0.5 m x u₁² + 0.5 M x u₂² = 0.5 x 3 x 8² + 0.5 x M x 0² = 96 J

Final kinetic energy = 0.5 m x v₁² + 0.5 M x v₂² = 0.5 x 3 x v₁² + 0.5 x M x 6² = 1.5 v₁² + 18 M

We have

Initial momentum = Final momentum

24 = 3v₁ + 6M

v₁ + 2M = 8

v₁ = 8 - 2M

Initial kinetic energy = Final kinetic energy

96 = 1.5 v₁² + 18 M

v₁² + 12 M = 64

Substituting v₁ = 8 - 2M

(8 - 2M)² + 12 M = 64

64 - 32M + 4M² + 12 M = 64

4M² = 20 M

M = 5 kg

Option 2 is the correct answer.

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2 years ago