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Setler79 [48]
3 years ago
9

A ball with a mass of 275 g is dropped from rest, hits the floor, and rebounds upward. If the ball hits the floor with a speed o

f 2.60 m/s, rebounds with a speed of 1.84 m/s, and is in contact with the floor for 1.40 ms,determine the following.
(a) magnitude of the change in the ball's momentum (Let up be in the positive j hat direction.)
_______________kg
Physics
1 answer:
Jobisdone [24]3 years ago
5 0

Answer:

-0.209 kg.m/s

Explanation:

The mass of the ball, m = 275g or 0.275 kg

Speed or velocity, v = 2.60 m/s

Momentum, P = mv

Momentum when velocity is 2.60 = 0.275 x 2.60 = 0.715 kg.m/s

Speed or velocity, v = 1.84 m/s

Momentum, P = mv

Momentum when velocity is 1.84= 0.275 x 1.84 = 0.506 kg.m/s

Change in magnitude =  0.506 - 0.715 = -0.209 kg.m/s

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What science did the study of the night sky eventually become?
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A 2.50 gram rectangular object has measurements of 22.0 mm, 13.5 mm, and 12.5 mm. what is the object's density in units of g/ml?
Zinaida [17]
G/mL is equivalent to g/cm^3, so we first convert the dimensions into cm:
2.20 cm, 1.35 cm, and 1.25 cm
Then the total volume is: V = lwh = 3.7125 cm^3
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4 0
4 years ago
A machinist is required to manufacture a circular metal disk with area 1300 cm2. (a) What radius produces such a disk
vichka [17]

Answer:

Radius r = 20.34 cm

The radius that can produces such a disk is 20.34 cm

Explanation:

Area of a circle;

A = πr^2

A = area

r = radius

Making r the subject of formula;

r = √(A/π) ........1

Given;

A = 1300 cm^2

Substituting into the equation 1;

r = √(1300/π)

r = 20.34214472564 cm

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6 0
4 years ago
A solid 0.4550 kg ball rolls without slipping down a track toward a vertical loop of radius R = 0.6750 m . What minimum translat
Talja [164]

Answer:

u = 3.35 m/s

Explanation:

given,

mass , m = 0.455 kg  

R = 0.675 m

Height of Loop = 1.021 m

the speed required at the top of loop be v

equating the force vertically

m g =\dfrac{mv^2}{r}

9.81 =\dfrac{v^2}{0.675}

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v = 2.57 m/s

Let the initial speed of ball be u

using conservation of energy

\dfrac{1}{2}mu^2 + \dfrac{1}{2}I\omega^2 + m g h = \dfrac{1}{2}I\omega^2 + \dfrac{1}{2}mv^2+ m g (2 R)

where, I =\dfrac{2}{5}mr^2

\dfrac{1}{2}mu^2 + \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{u}{r})^2 + m g h = \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{v}{r})^2 + \dfrac{1}{2}mv^2+ m g (2 R)

0.7 u^2 + g H = 0.7 v^2 + g(2R)

0.7 u^2 +9.81 \times 1.021= 0.7\times 2.57^2 + 9.81 \times 2\times 0.675)

0.7 u² = 7.85092

u² = 11.2156

u = 3.35 m/s

the initial  speed is 3.35 m/s

8 0
3 years ago
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