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e-lub [12.9K]
3 years ago
15

To demonstrate an elastic collision, a teacher places a tennis ball on the floor. She wants to hit the ball so that the followin

g conditions are satisfied.
1. After collision, the tennis ball should roll with the velocity of the ball that hit it
2. At the instant of collision, the ball that hit the tennis ball should stop.
What ball should she use?
Golf ball, tennis ball, basketball, or bowling ball?
Physics
1 answer:
kodGreya [7K]3 years ago
0 0
A tennis ball

The two balls need to have the same mass to meet this condition.
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calculate the efficiency of a light bulb that had an input of 400 j transfers 100j as a light and 300j as heat.​
beks73 [17]
Efficiency = useful energy out / total energy in x 100
= 100/400 x 100
=0.25 x 100
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25%
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2 years ago
A lawn roller in the form of a thin-walled, hollow cylinder with mass M is pulled horizontally with a constant horizontal force
olga nikolaevna [1]

Answer:

a_{cm}=\frac{F}{2M}\\\\F_{fr}=\frac{F}{2}

Explanation:

Given the mass as M, the rotational inertia of the mower is;

I_{cm}=MR^2

-The roller doesn't slip while rolling;

v_{cm}=wR, a_{cm}=\alpha R

\sum F_x=Ma_x, -F+F_{fr}=-Ma_{cm}\\\\a_{cm}=\frac{F-Fr}{M}    \ \ \ \ \ \ \ \ eqtn1\\\\\sum \tau_{cm}=I_{cm}\alpha, F_{fr}(R)}=(MR^2)(\frac{a_{cm}}{R}), ->F_{fr}=Ma_c_m\ \ \  \ \ \ \ eqtn2\\\\a_{cm}=\frac{F-Ma_{cm}}{M}, ->F=2Ma_{cm}\\\\a_{cm}=\frac{F}{2M}\\\\\\\therefore F_{fr}=M(\frac{F}{2M})\\\\F_{fr}=\frac{F}{2}

6 0
3 years ago
Determine the change in electric potential energy of a system of two charged objects when a -2.1-C charged object and a -5.0-C c
elena55 [62]

Answer:

Change in electric potential energy ∆E = 365.72 kJ

Explanation:

Electric potential energy can be defined mathematically as:

E = kq1q2/r ....1

k = coulomb's constant = 9.0×10^9 N m^2/C^2

q1 = charge 1 = -2.1C

q2 = charge 2 = -5.0C

∆r = change in distance between the charges

r1 = 420km = 420000m

r2 = 160km = 160000m

From equation 1

∆E = kq1q2 (1/r2 -1/r1) ......2

Substituting the given values

∆E = 9.0×10^9 × -2.1 ×-5.0(1/160000 - 1/420000)

∆E = 94.5 × 10^9 (3.87 × 10^-6) J

∆E = 365.72 × 10^3 J

∆E = 365.72 kJ

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