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Scilla [17]
2 years ago
5

A billiards ball B rests on a horizontal surface and is struck by another billiards ball A of the same mass m = 0.2 kg. Ball A i

s initially moving at 18 m/s and is deflected 28° from its original direction. Assume the collision is elastic. a) find the final velocity vector of each ball, b) find the mechanical energy both before and after the collision.
Physics
1 answer:
SOVA2 [1]2 years ago
7 0

Answer:

v1 = 15.90 m/s

v2 = 8.46 m/s

mechanical energy before collision = 32.4 J

mechanical energy after collision = 32.433 J

Explanation:

given data

mass m = 0.2 kg

speed = 18 m/s

angle =  28°

to find out

final velocity and  mechanical energy both before and after the collision

solution

we know that conservation of momentum remain same so in x direction

mv = mv1 cosθ + mv2cosθ

put here value

0.2(18) = 0.2 v1 cos(28) + 0.2 v2 cos(90-28)

3.6 =  0.1765 V1 + 0.09389 v2    ................1

and

in y axis

mv = mv1 sinθ - mv2sinθ

0 = 0.2 v1 sin28 - 0.2 v2 sin(90-28)

0 = 0.09389 v1 - 0.1768 v2   .......................2

from equation 1 and 2

v1 = 15.90 m/s

v2 = 8.46 m/s

so

mechanical energy  before collision = 1/2 mv1² + 1/2 mv2²

mechanical energy before collision = 1/2 (0.2)(18)² + 0

mechanical energy before collision = 32.4 J

and

mechanical energy after collision = 1/2 (0.2)(15.90)² + 1/2 (0.2)(8.46)²

mechanical energy after collision = 32.433 J

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A 6.8 kg bowling ball and 7.4 kg bowling ball rest on a rack 0.74 m apart. What is the force of gravity pulling each ball toward
zimovet [89]

The gravitational force between the two balls is 6.13\cdot 10^{-9} N

Explanation:

The magnitude of the gravitational force between two objects is given by:

F=G\frac{m_1 m_2}{r^2}

where :

G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2} is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between the objects

For the balls in this problem,  we have

m_1 = 6.8 kg

m_2 = 7.4 kg

r = 0.74 m

Substituting into the equation, we find the gravitational force between the two balls:

F=(6.67\cdot 10^{-11})\frac{(6.8)(7.4)}{(0.74)^2}=6.13\cdot 10^{-9}N

Learn more about gravitational force:

brainly.com/question/1724648

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4 0
3 years ago
How much heat is needed to vaporize 10.00 grams of water at 100.0°C? The latent heat of vaporization of water is 2,259 J/g
Nata [24]

Answer:

Heat of vaporization will be 22.59 j    

Explanation:

We have given mass m = 10 gram

And heat of vaporization L = 2.259 J/gram

We have to find the heat required to vaporize 10 gram mass

We know that heat of vaporization is given by Q=mL, here m is mass and L is latent heat of vaporization.

So heat of vaporization Q will be = 10×2.259 = 22.59 J

8 0
2 years ago
Read 2 more answers
Explain why a man using a parachute falls through air slowly while a stone falls through air very fast
Korolek [52]

Answer:because the parachute is built so the wind will push up on it make the man/woman glide or fall slowly will a stone which has a lot of desentity

falls through the air faster do to is weight and shape .

Explanation:

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2 years ago
Solve for M₂
soldi70 [24.7K]

Explanation:

M₂ = Fr²/GM₁

M₂ = [(132N)(.243m)²]/[(6.67*10^-11N*m²/kg)(1.175*10^4kg)]

M₂ = (7.79N*m²)/(7.84*10^-7N*m²)

M₂ = 9.94*10^6 kg

5 0
3 years ago
tourist travels 1500 miles using two planes. The second plane averages 50 miles per hour faster than the first plane. The touris
Genrish500 [490]

Answer

given,

tourist travels = 1500 miles

second plane averages 50 miles per hour faster than the first plane.

x = y + 50

The tourist uses the slower plane for the first 500 and the faster plane for the next 1000 miles.

total flying time = 6.5 hours

\dfrac{500}{y} + \dfrac{1000}{x} = 6.5

\dfrac{500}{y} + \dfrac{1000}{y + 50} = 6.5

\dfrac{500(50 + y)+ 1000 y}{y(y + 50)} = 6.5

25000 + 600 y = 6.5 y^2 + 325 y

6.5 y^2 - 275 y - 25000 = 0

on solving equation

y = 86.67 mile/hr

x = 86.67 + 50

x = 136.67 mile/hr

8 0
3 years ago
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