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Ganezh [65]
3 years ago
6

A golf ball traveling 3m/s to the right collides in a head-on collision with a stationary bowling ball in a friction-free enviro

nment. If the collision is almost perfectly elastic, the speed of the golf ball immediately after the collision is a) slightly greater b) much less than 3 m/s c) equal to 3 m/s d) slightly less than 3 m/s
Physics
1 answer:
Vladimir79 [104]3 years ago
3 0

Answer: equal to 3 m/s

Explanation:

Speed of golf ball will be equal to 3 m/s because in Perfect Elastic Collision Energy is conserved .

So speed of golf ball will be same in order to Satisfy

Initial Kinetic Energy =Final Kinetic Energy

Considering Bowling ball remains at rest after collision other wise some energy will also be acquired by bowling ball which automatically decreases the amount of Kinetic Energy of golf ball resulting its speed to decrease by some extent.  

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The answer is ....  C. 17,705.1 J
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What energy is calculated from an object’s mass, height, and the acceleration due to gravity? A. elastic kinetic B. elastic pote
Alex777 [14]

Answer: C. gravitational kinetic

Explanation: Gravitational potential energy is the energy calculated from an object's mass height and the acceleration due to gravity. The gravitational potential energy is the energy an object has as a result of the position of the object in a gravitational field.

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How do you find average speed on a graph
coldgirl [10]

Answer:

speed equals distance over time 50 divided by 5.

5 0
3 years ago
A ball with 100 J of PE is released from a height of 10 m. What will be the KE of the ball at 5
harkovskaia [24]

Answer:

The kinetic energy is: 50[J]

Explanation:

The ball is having a potential energy of 100 [J], therefore

PE = [J]

The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.

E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\

In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.

When the elevation is 5 [m], we have a potential energy of

P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\

This energy is equal to the kinetic energy, therefore

Ke= 50 [J]

8 0
3 years ago
Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 mm away horizontal
Harrizon [31]

Answer:

a. 8.96 m/s b. 1.81 m

Explanation:

Here is the complete question.

a) A long jumper leaves the ground at 45° above the horizontal and lands 8.2 m away.

What is her "takeoff" speed  v 0 ?

b) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 m away horizontally and 2.5 m, vertically below.  

If she long jumps from the edge of the left bank at 45° with the speed calculated in part a), how long, or short, of the opposite bank will she land?

a. Since she lands 8.2 m away and leaves at an angle of 45 above the horizontal, this is a case of projectile motion. We calculate the takeoff speed v₀ from R = v₀²sin2θ/g. where R = range = 8.2 m.

So, v₀ = √gR/sin2θ = √9.8 × 8.2/sin(2×45) = √80.36/sin90 = √80.36 = 8.96 m/s.

b. We use R = v₀²sin2θ/g to calculate how long or short of the opposite bank she will land. With v₀ = 8.96 m/s and θ = 45

R = 8.96²sin(2 × 45)/9.8 = 80.2816/9.8 = 8.192 m.

So she land 8.192 m away from her bank. The distance away from the opposite bank she lands is 10 - 8.192 m = 1.808 m ≅ 1.81 m

8 0
3 years ago
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