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2 years ago

A rocket ship in space is at rest relative to a rock floating in deep space. The

1 answer:

5 0

So the equation used in this problem is ΔX=V0*T+1/2AT^2 the X is the distance, v0 is initial velocity, T is time, and a is acceleration. So when we plug these values it we get: 108= 0•T+1/2•3•T^2,the 0•t disappears, and the 1/2•3 gets us 1.5, so we have 108=1.5T^2, then we divide 108 by 1.5 which gets us 72=t^2, and we then take the square root and get 8.49=T so the answer is 8.49 seconds.

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A cannon of mass 6.43 x 103 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires a 7

Nata [24]

Answer:

The velocity of the shell when the cannon is unbolted is 500.14 m/s

Explanation:

Given;

mass of cannon, m₁ = 6430 kg

mass of shell, m₂ = 73.8-kg

initial velocity of the shell, u₂ = 503 m/s

Initial kinetic energy of the shell; when the cannon is rigidly bolted to the earth.

K.E = ¹/₂mv²

K.E = ¹/₂ (73.8)(503)²

K.E = 9336032.1 J

When the cannon is unbolted from the earth, we apply the principle of conservation of linear momentum and kinetic energy

change in initial momentum = change in momentum after

0 = m₁u₁ - m₂u₂

m₁v₁ = m₂v₂

where;

v₁ is the final velocity of cannon

v₂ is the final velocity of shell

$v_1 = \frac{m_2v_2}{m_1}$

Apply the principle of conservation kinetic energy

$K = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2\\\\K = \frac{1}{2}m_1(\frac{m_2v_2}{m_1})^2 + \frac{1}{2}m_2v_2^2\\\\K = \frac{1}{2}m_2v_2^2(\frac{m_2}{m_1}) + \frac{1}{2}m_2v_2^2 \\\\K = \frac{1}{2}m_2v_2^2 (\frac{m_2}{m_1} + 1)\\\\2K = m_2v_2^2 (\frac{m_2}{m_1} + 1)\\\\v_2^2 = \frac{2K}{M_2(\frac{m_2}{m_1} + 1)} \\\\v_2^2 = \frac{2*9336032.1}{73.8(\frac{73.8}{6430} + 1)}\\\\$

$v_2^2 = 250138.173\\\\v_2 = \sqrt{250138.173} \\\\v_2 = 500.14 \ m/s$

Therefore, the velocity of the shell when the cannon is unbolted is 500.14 m/s

3 0

3 years ago

A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying p

IRINA_888 [86]

Answer:

normal force = 10 N

Explanation:

Given data

frictional force = 0.400 N

coefficient of kinetic friction = 0.04

Solution

we get here normal force that is express as

normal force = $\frac{Frictional\ force}{coefficient\ of\ friction}$ ............1

put here value and we will get value

normal force = $\frac{0.400}{0.04}$

solve it we get

normal force = 10 N

5 0

3 years ago

How does air resistance affect forward motion

lord [1]

It creates friction on the forward moving object, causing it to loose momentum, until finally, it stops.

Hope this helps!

8 0

4 years ago

Read 2 more answers

Is your increase in gravitational potential energy the same in both cases? When Climbing a mountain on a zigzag path and on a st

baherus [9]

Answer:

The increase in gravitational potential energy is the same in both cases

Explanation:

It is easier to climb a mountain in a zigzag way rather than climbing on a straight line but since the distance is the same ( vertical height ) , mass and gravity is the same. Hence the increase in gravitational potential energy is the same in both cases.

gravitational potential energy = mgh ( same in both cases )

m = mass

g = acceleration due to gravity

h = distance ( vertical height )

7 0

3 years ago

Which of the following is a common downside of net pens, one of the most common forms of aquaculture?

Eduardwww [97]

Answer:

Answer is overcrowding aka answer choice A. I got the question and got it right. Please mark brainliest. Have a good day! :)

Explanation:

8 0

3 years ago

Read 2 more answers