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Scrat [10]
3 years ago
7

Two carts undergo an inelastic collision where they stick together. Cart A has an initial velocity v0, and the second cart B is

initially at rest. After the collision, it is observed that the ratio of the final kinetic energy system to its initial kinetic energy is KfK0= 1/6. Determine the ratio of the carts' masses, mBmA. (Assume the track is frictionless.)
Physics
1 answer:
dimulka [17.4K]3 years ago
3 0

Answer:

Explanation:

Initial kinetic energy of the system = 1/2 mA v0²

If Vf be the final velocity of both the carts

applying conservation of momentum

final velocity

Vf = mAvo / ( mA +mB)

kinetic energy ( final ) =  1/2 (mA +mB)mA²vo² /  ( mA +mB)²

= mA²vo²  / 2( mA +mB)

Given 1/2 mA v0²  / mA²vo²  / 2( mA +mB) = 6

mA v0² x ( mA +mB) / mA²vo² = 6

( mA +mB) / mA = 6

mA + mB = 6 mA

5 mA = mB

mB / mA = 5 .

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You make a(n) _____ by fastening together usually perpendicular parts with the ends cut at an angle.
nadezda [96]

Answer:

Miter joint

Explanation:

Made by fastening together usually perpendicular parts with the ends cut at an angle

8 0
3 years ago
A person is standing on a level floor. His head, upper torso, arms, and hands together weigh 458 N and have a center of gravity
Step2247 [10]

Answer:

the location of the center of gravity for the entire body is 1.08 m

Explanation:

Given the data in the question;

w1 = 458 N, y1 = 1.34 m

w2 = 120 N, y2 = 0.766 m

w3 = 89.8 N, y2 = 0.204 m

The location arrangement of the body part is vertical, locate the overall centre of gravity by simply replacing the horizontal position x by the vertical position y as measured relative to the floor.

so,

Y_{centre of gravity} = (w1y1 + w2y2 + w3y3 ) / ( w1 + w2 + w3 )

so we substitute in our values

Y_{centre of gravity} = (458×1.34 + 120×0.766 + 89.8×0.204 ) / ( 458 + 120 + 89.8 )

Y_{centre of gravity} = 723.9592 / 667.8

Y_{centre of gravity} = 1.08 m

Therefore, the location of the center of gravity for the entire body is 1.08 m

5 0
3 years ago
Bonus Points. A ball was thrown straight up in the air from 1.2 meter above the grouind. After 3 seconds the ball returnes to th
nekit [7.7K]

Answer:

Explanation:

given that

Distance above the ground, s = 1.2 m

Time taken by the ball, t = 3 s

Velocity of the ball, v = 1.2/3 = 0.4 m/s

Maximum height reached by the ball is then given by the formula

H = v² / 2g

H = 0.4² / 2 * 9.8

H = 0.16 / 19.6

H = 0.0082 m or rather, 0.82 cm

5 0
3 years ago
The spring of modulus k = 200 n /m is compressed a distance of 300 mm and suddenly released with the system at rest. determine t
DiKsa [7]
I attached the missing picture.
Let's analyze the situation as spring goes from stretched to unstretched state.
When you stretch the string you have to use force against ( you are doing work) this energy is then stored in the spring in the form of potential energy. When we release the spring the energy is being used to push the two carts. When the spring reaches its unstretched length its whole initial potential energy has been used on the carts, and this is the moment when two carts have maximum velocity.
The potential energy of compressed ( stretched) spring is:
E_p=\frac{1}{2}kx^2
The kinetic energy of two carts is:
E_{k1}+E_{k2}=m_1\frac{v_1^2}{2}+m_2\frac{v_2^2}{2}
So we have:
E_p=E_{k1}+E_{k2}\\ \frac{1}{2}kx^2=m_1\frac{v_1^2}{2}+m_2\frac{v_2^2}{2}
Momentum also has to be conserved:
m_1v_1-m_2v_2=0\\ m_1v_1=m_2v_2\\ v_1=\frac{m_2}{m_1}v_2
Momentum before the release of the spring is zero so it has to stay zero. We plug this back into the expresion we got from law of conservation of energy and we get:
v_2^2=\frac{m_1^2}{m_2^2-m_1^2}kx^2=4.05\\
v_2=\sqrt{4.05}=2.012\frac{m}{s}
Now we go back to the momentum equation:
v_1=\frac{m_2}{m_1}v_2\\
v_1=4.69\frac{m}{s}

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