A golfer hits a ball and gives it an initial velocity of 40 m/s , at an angle of 30 degrees above the horizontal.

W is the work done on the system, and K, U, and Eth are the kinetic, potential, and thermal energies of the system, respectively

MArishka [77]

Answer:

1) a block going down a slope

2) a) W = ΔU + ΔK + ΔE, b) W = ΔE, c) W = ΔK, d) ΔU = ΔK

Explanation:

In this exercise you are asked to give an example of various types of systems

1) a system where work is transformed into internal energy is a system with friction, for example a block going down a slope in this case work is done during the descent, which is transformed in part kinetic energy, in part power energy and partly internal energy that is represented by an increase in the temperature of the block.

2)

a) rolling a ball uphill

In this case we have an increase in potential energy, if there is a change in speed, the kinetic energy also increases, if the change in speed is zero, there is no change in kinetic energy and there is a change in internal energy due to the stationary rec in the point of contact

W = ΔU + ΔK + ΔE

b) in this system work is transformed into internal energy

W = ΔE

c) There is no friction here, therefore the work is transformed into kinetic energy

W = ΔK

d) if you assume that there is no friction with the air, the potential energy is transformed into kinetic energy

ΔU = ΔK

7 0

3 years ago

What type of nuclear decay releases energy but not a particle?

statuscvo [17]

I believe gamma decay but i may be wrong

7 0

3 years ago

Read 2 more answers

(15pts) A hungry 12.0 kg fish is coasting from west to east at 75 cm/s when it suddenly swallows a 1 kg fish swimming towards it

faust18 [17]

Answer:

The speed of the big fish after swallowing the small fish is 0.38 m/s.

Explanation:

Consider west to east direction as positive and the opposite direction as negative.

Given:

Mass of big fish (m₁) = 12.0 kg

Initial velocity of big fish (u₁) = 75 cm/s = 0.75 m/s

Mass of small fish (m₂) = 1 kg

Initial velocity of small fish (u₂) = -4 m/s (Direction is opposite to u₁)

After swallowing the small fish, both the fishes move together with same velocity. Let the velocity be 'v'.

So, as there are no effects of drag or any other forces, the given scenario can be considered as a case of inelastic collision where the objects move together with same velocity after collision.

The momentum is conserved in inelastic collision. Therefore,

Initial momentum of the fishes = Final momentum of the fishes

$m_1u_1+m_2u_2=(m_1+m_2)v\\\\v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}$

Now, plug in the given values and solve for 'v'. This gives,

$v=\frac{12.0\times 0.75+1\times (-4)}{12.0+1}\\\\v=\frac{9-4}{13}\\\\v=\frac{5}{13}=0.38\ m/s$

Therefore, the speed of the big fish after swallowing the small fish is 0.38 m/s

3 0

3 years ago

Chris walks 25 m in the positive direction on a number line, then turns around and walks 15 m in the opposite direction. What is

7nadin3 [17]

Answer:

$|d| = |(25 - 15)| \\ |d| = 10 \: m$

3 0

2 years ago

A pilot is upside down at the top of an inverted loop of radius 3.20 x 103 m. At the top of the loop his normal force is only on

n200080 [17]

Answer:

6858.5712 m/s

Explanation:

Given that:

Radius, r

R = 3.20 * 10^3.

Normal force = 0.5 * normal weight

Normal force = Fn ; Normal weight = Fg

Fn = 0.5Fg

Recall:

mv² / R = Fn + Fg

Fn = 0.5Fg

mv² / R = 0.5Fg + Fg

mv² /R = 1.5Fg

mv² = 1.5Fg * R

F = mg

mv² = 1.5* mg * R

v² = 1.5gR

v = sqrt(1.5gR)

V = sqrt(1.5 * 9.8 * 3.2 * 10^3)

V = sqrt(47.04^3)

V = 6858.5712 m/s

6 0

2 years ago