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4 years ago

According to Newton's 3rd law, the force on the (smaller) moon due to the (larger) earth is?

1 answer:

8 0

According to Newton's 3rd law, the force on the (smaller) moon due to the (larger) earth is "equal in magnitude to, and in the opposite direction from, the force on the earth due to the moon".

<u>Option:</u> C

<u>Explanation:</u>

A force is a push or pull acting on an entity as a consequence of its contact with a particular object. Such two phenomena are called forces of action and reaction, and are the origin of the third law of motion by Newton.

This law states the force of the moon on earth is equal and opposite to the force of earth on the moon. The enticing force operating between a satellite orbiting the earth in a circular path allows the object to accelerate towards earth, but since it maintains its circular orbit, it never gets closer to earth.

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Cart A, with a mass of 0.20 kg, travels on a horizontal air trackat 3.0m/s and hits cart B, which has a mass of 0.40 kg and is i

inysia [295]

Answer:

$\large \boxed{\text{C. 2.3 m/s}}$

Explanation:

Data:

$m_{\text{A} } = \text{0.20 kg};\,v_{\text{Ai}} = \text{3.0 m/s}\\m_{\text{B} } = \text{0.40 kg};\,v_{\text{Bi}} = \text{2.0 m/s}\\$

Calculation:

This is a perfectly inelastic collision. The two carts stick together after the collision and move with a common final velocity.

The conservation of momentum equation is

$\begin{array}{rcl}m_{\text{A}}v_{\text{Ai}} +m_{\text{B}} v_{\text{Bi}}&=&(m_{\text{A}} + m_{\text{B}})v_{\text{f}}\\0.20\times 3.0 + 0.40\times 2.0 & = & (0.20 + 0.40)v_{\text{f}}\\0.60 + 0.80 & = & 0.60v_{\text{f}}\\1.40 & = & 0.60v_{\text{f}}\\v_{\text{f}}&=& \dfrac{1.40}{0.60}\\\\& = & \textbf{2.3 m/s}\\\end{array}\\\text{The centre of mass has a velocity of $\large \boxed{\textbf{2.3 m/s}}$}$

3 0

4 years ago

A pirate ship shoots a cannon towards an oncoming ship. The cannon ball has a mass of 25 kg and the ship has a mass of 2500 kg.

mars1129 [50]

Answer:

25 m/s in the opposite direction with the ship recoil velocity.

Explanation:

Assume the ship recoil velocity and velocity of the cannon ball aligns. By the law of momentum conservation, the momentum is conserved before and after the shooting. Before the shooting, the total momentum is 0 due to system is at rest. Therefore, the total momentum after the shooting must also be 0:

$m_sv_s + m_bv_b = 0$

where $m_s = 2500 kg, m_b = 25 kg$ are masses of the ship and ball respectively. $v_s = 0.25 m/s, v_b$ are the velocities of the ship and ball respectively, after the shooting.

$2500*0.25 + 25*v_b = 0$

$25v_b = -2500*0.25$

$v_b = -2500*0.25/25 = -25 m/s$

So the cannon ball has a velocity of 25 m/s in the opposite direction with the ship recoil velocity.

3 0

3 years ago

Which of the following is not an application of Doppler technology?

Jlenok [28]

The correct answer to the question above is the third option; ultrasound imaging of the liver. The ultrasound imaging of the liver is definitely not an application of Doppler technology. If the Doppler technology is being used in medical field, it would be for the ultrasound of the heart and blood vessels for examination.

8 0

3 years ago

Read 2 more answers

Please help don’t know how to show my work

Fudgin [204]

Answer:

64.50816

Explanation:

3 0

3 years ago

A rotating light is located 13 feet from a wall. The light completes one rotation every 3 seconds. Find the rate at which the li

saveliy_v [14]

Answer:

29.2 ft/s

Explanation:

The distance of the light's projection on the wall

y = 13 tan θ

where θ is the light's angle from perpendicular to the wall.

The light completes one rotation every 3 seconds, that is, 2π in 3 seconds,

Angular speed = w = (2π/3)

w = (θ/t)

θ = wt = (2πt/3)

(dθ/dt) = (2π/3)

y = 13 tan θ

(dy/dt) = 13 sec² θ (dθ/dt)

(dy/dt) = 13 sec² θ (2π/3)

(dy/dt) = (26π/3) sec² θ

when θ = 15°

(dy/dt) = (26π/3) sec² (15°)

(dy/dt) = 29.2 ft/s

5 0

4 years ago