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Pavlova-9 [17]
3 years ago
15

Olaf is standing on a sheet of ice that covers the football stadium parking lot in Buffalo, New York; there is negligible fricti

on between his feet and the ice. A friend throws Olaf a ball of mass 0.400 kg that is traveling horizontally at 10.9 m/s . Olaf's mass is 70.2 kg. (a) If Olaf catches the ball, with what speed vf do Olaf and the ball move afterward? Express your answer numerically in meters per second. (b) If the ball hits Olaf and bounces off his chest horizontally at 8.10 m/s in the opposite direction, what is his speed vf after the collision? Express your answer numerically in meters per second.
Physics
1 answer:
Citrus2011 [14]3 years ago
6 0

Explanation:

Momentum is conserved.

a) In the first scenario, Olaf and the ball have the same final velocity.

mu = (M + m) v

(0.400 kg) (10.9 m/s) = (70.2 kg + 0.400 kg) v

v = 0.0618 m/s

b) In the second scenario, the ball has a final velocity of 8.10 m/s in the opposite direction.

mu = mv + MV

(0.400 kg) (10.9 m/s) = (0.400 kg) (-8.10 m/s) + (70.2 kg) v

v = 0.108 m/s

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olga55 [171]

Answer:

Change in momentum, \Delta p=6.3\ kg-m/s

Explanation:

It is given that,

Mass of the basketball, m = 601 g = 0.601 kg

The basketball makes an angle of 29 degrees to the vertical, it hits the floor with a speed, v = 6 m/s

It bounces up with the same speed, again moving to the right at an angle of 29 degree to the vertical. We need to find the change in momentum. It is given by :

\Delta p=mv\ cos\theta-(-mv\ cos\theta)

\Delta p=2mv\ cos\theta

\Delta p=2\times 0.601\times 6\times \ cos(29)

\Delta p=6.3\ kg-m/s

So, the change in momentum of the basketball is 6.3 kg-m/s. Hence, this is the required solution.

6 0
3 years ago
A thin, light wire 75.1 cm long having a circular cross section 0.555 mm in diameter has a 25.4 kg weight attached to it, causin
seraphim [82]

Answer:

(a) 3.23×10⁸ N/m²

(b)  1.46×10⁻³

(c) 2.21×10¹¹ N/m²

Explanation:

(a) Stress = Force/Area.

Stress = F/A................ Equation 1

But,

F = mg................. Equation 2

Where m = mass, and g = acceleration due to gravity

A = πd²/4................. Equation 3

d = diameter of the circular cross section.

Substitute equation 2 and equation 3 into equation 1

Stress = 4mg/πd²............. Equation 4

Given: m = 25.4 kg, d = 0.555 mm = 0.000555 m

Constant: g = 9.8 m/s², π = 3.142

Substitute these values into equation 4

Stress = 4(25.4)(9.8)/(3.142×0.00555²)

Stress = 995.68/(3.08×10⁻⁶)

Stress = 3.23×10⁸ N/m²

(b)

Strain = ΔL/L.............. Equation 5

Where ΔL = extension, L = Length.

Given: ΔL = 1.1 mm = 0.0011 m, L = 75.1 cm = 0.751 m

Substitute into equation 5

Strain = 0.0011/0.751

Strain = 1.46×10⁻³.

(c)

Young modulus = Stress/Strain

Young modulus = 3.23×10⁸/ 1.46×10⁻³

Young modulus = 2.21×10¹¹ N/m²

3 0
3 years ago
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4 0
3 years ago
A train moving at 5 m/sec passed a track gang and then accelerated uniformly a rate of 1.2 m/sec/sec for 5 min. How far did the
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Answer:

S = V0 t + 1/2 a t^2

S = 5 m/s * 300 s + 1/2 * 1.2 m/s * (300 s^2)

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                V2 = V0 + a t  = 5 + 1.2 * 300 = 365 m/s

Vav = (V1 + V2) / 2 = (5 + 365) / 2 = 185 m/s     (note uniform motion)

S = 185 * 300 = 55,500 m

We calculated V2 above at 365 m/s  the speed after 300 sec

3 0
3 years ago
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