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SCORPION-xisa [38]
4 years ago
7

Two separate masses on two separate springs undergo simple harmonic motion indefinitely (the surface is frictionless). In CASE 1

, the spring constant is 2k, the mass is m, and the spring oscillates with amplitude d. In CASE 2, the spring constant is 2k, the mass is 2m, and the spring oscillates with amplitude 2d.
Physics
1 answer:
Artemon [7]4 years ago
4 0

Answer:

\frac{F_1}{F_2} =\frac{1}{2}

Explanation:

Assuming the we have to find ratio maximum forces on the mass in each case

we know that in a spring mass system

F= Kx

K= spring constant

x= spring displacement

Case 1:

F_1=2k\times d

case 2:

F_2=2k\times 2d

therefore, \frac{F_1}{F_2} = \frac{2K\times d}{2K\times 2d}

\frac{F_1}{F_2} =\frac{1}{2}

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Give an example of Electrical Force
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Answer: Sometimes people try to use there flashlight but it doesnt turn on they take out there battierise and they rub them together. It doesnt work so they get new battiries

Explanation: please mark me brainlies i have a goal im tryna reach thank you!!

4 0
3 years ago
An object has a mass of 5 kg and a velocity of 20 m/s. What is the momentum of the object?(1 point)
kotegsom [21]

Answer:

option C is correct

Explanation:

given mass=5 kg  

velocity=20 m/s

momentum,P=m v

p=5*20=100 kg m/s

3 0
3 years ago
Read 2 more answers
Why does the vertical component of velocity for a projectile change with time, whereas the horizontal component of velocity does
dlinn [17]

Answer

When a body moves in projectile motion it has two components. One of the components is a horizontal component of the velocity and another is a vertical component.

The velocity along the horizontal component does not change because there is no acceleration long horizontal component.  

Whereas velocity along vertical direction keeps on changing because the acceleration due to gravity is acting on the object. At a maximum height of the projectile velocity is equal to zero.

5 0
3 years ago
In the figure, if Q = 52 µC q =10 µC and d = 55 cm, what is the magnitude of the electrostatic force on q?​
GenaCL600 [577]

Answer:

F = 15.47 N

Explanation:

Given that,

Q = 52 µC

q = 10 µC

d = 55 cm = 0.55 m

We need to find the magnitude of the electrostatic force on q. The formula for the electrostatic force is given by :

F=k\dfrac{q_1q_2}{d^2}\\\\F=9\times 10^9\times \dfrac{52\times 10^{-6}\times 10\times 10^{-6}}{(0.55)^2}\\\\F=15.47\ N

So, the magnitude of the electrostatic force is 15.47 N.

4 0
3 years ago
Two basketballs of equal mass are rolling toward each other at constant velocities. The first basketball (B1) has a velocity of
slamgirl [31]

v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)

<u>Explanation:</u>

Velocity of B₁ = 4.3m/s

Velocity of B₂ = -4.3m/s

For perfectly elastic collision:, momentum is conserved

m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2

where,

m₁ = mass of Ball 1

m₂ = mass of Ball 2

v₁ = initial velocity of Ball 1

v₂ = initial velocity of ball 2

v'₁ = final velocity of ball 1

v'₂ = final velocity of ball 2

The final velocity of the balls after head on elastic collision would be

v'_2 = \frac{2m_1}{m_1+m_2} v_1 - \frac{m_1-m_2}{m_1+m_2} v_2\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} v_1 + \frac{2m_2}{m_1+m_2} v_2

Substituting the velocities in the equation

v'_2 = \frac{2m_1}{m_1+m_2} (4.3) - \frac{m_1-m_2}{m_1+m_2} (4.3)\\\\v'_1 = \frac{m_1-m_2}{m_1+m_2} (4.3) + \frac{2m_2}{m_1+m_2} (4.3)

If the masses of the ball is known then substitute the value in the above equation to get the final velocity of the ball.

5 0
3 years ago
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