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AysviL [449]
3 years ago
8

Use the following information for question 26 and 27. A 550-g ball traveling at 8.0 m/s undergoes a sudden head-on perfectly ela

stic collision with a 250-g ball traveling toward it also at 8.0 m/s. What is the speed of the 250-g mass just after the collision
Physics
1 answer:
Gnoma [55]3 years ago
6 0

Answer:

The speed of 250 g ball after collision is 14 m/s.

Explanation:

mass of first ball, m = 550 g = 0.55 kg

initial velocity of first ball,  u = 8 m/s

mass of second ball, m' = 250 g = 0.25 kg

initial velocity of second ball, u' = - 8 m/s

Let the speed of the balls is v and v' after the collision.

Use conservation of momentum

m u + m' u' = m v + m' v'

0.55 x 8 - 0.25 x 8 = 0.55 v + 0.25 v'

0.55 v + 0.25 v' = 2.4 ..... (1)

Use the formula  of coefficient of restitution,

For elastic collision, e = 1

e =\frac{v'-v}{u - u'}\\\\1 =\frac{v'-v}{8+8}\\\\16 =v' - v...... (2)

By solving (1) and (2)

v = - 2 m/s and v' = 14 m/s

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You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a c
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Answer:

Fnet = 0

Explanation:

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       F_{appx} = F_{app} * cos \theta = 295 N * cos 35 = 242 N  (1)

  • In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

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6 0
3 years ago
With detailed explaniation
belka [17]
  • Ø=37°
  • Initial velocity=u=20m/s
  • g=10m/s²

#A

\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}

\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}

\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}

\\ \rm\Rrightarrow H_{max}=20sin^237

\\ \rm\Rrightarrow H_{max}=7.2m

#B

\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}

\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}

\\ \rm\Rrightarrow R=40sin74

\\ \rm\Rrightarrow R=38.5m

#C

\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}

\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}

\\ \rm\Rrightarrow T=4sin37

\\ \rm\Rrightarrow T=2.4s

Now

\\ \rm\Rrightarrow v=u-gt

\\ \rm\Rrightarrow v=20-10(2.4)

\\ \rm\Rrightarrow v=20-24

\\ \rm\Rrightarrow v=-4m/s

4 0
2 years ago
Two forces are acting on a wheelbarrow. One force is pushing to the right and an equal force is pushing to the left. What can yo
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-- As far as we know, the forces on the wheelbarrow are balanced.

-- That tells us that the net force on the wheelbarrow is zero, just
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-- That tells us that the wheelbarrow's acceleration is zero ... its
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-- That tells us that the wheelbarrow is moving in a straight line
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may be zero, but we can't tell that from the given information.

6 0
3 years ago
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