The force of ___ always turns mechanical energy into thermal energy

You are operating a powerboat at night. Your red sidelight must be visible to boats approaching from which direction(s)

Dmitriy789 [7]

The answer is the red sidelight on a powerboat should be visible from the front and from the left (port side).

What are Sidelights?

There is various combinations of lights that must be used on a boat when it is dark, and these are:
Sidelights: These lights are called combination lights and are red and green. The red sidelight must be visible from the port side and the green light indicates the right side (the starboard).
Stern light: The stern light is seen at the back end of the vessel.
Masthead Light: The masthead light is a white light that shines forwards and on all sides of the vessel. All powered vessels must use this light.
All-Round white light: This light is the major light that is used to join the masthead light and the stern light. This single light is visible to all vessels from all directions.
Thus, as a rule for a boat rider, he should show the vision of red light and it should be visible from the front and from the left (port side).

To learn more about Sidelights visit:

brainly.com/question/28205057

#SPJ4

4 0

1 year ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

KE = Kinetic Energy and PE = Potential Energy

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

3 years ago

None of these make sense someone help

vfiekz [6]

I would say B because the rest of them are false.

Because Saturn is made of gas

Saturn is smaller than Jupiter

And it takes 29 years for Saturn to orbit the sun

Hope This Helps! Have A Nice Day!!

6 0

2 years ago

Read 2 more answers

1. Which statement about friction is true?(1 point) A. Static friction and kinetic friction in a system always act in the same d

ArbitrLikvidat [17]

Answer:

true 1Ay 2D

Explanation:

1) In this exercise you are asked to investigate which statements are true

A) True. The friction force opposes the movement caused by the external force,

B) False. Mantuano in the opposite direction force

C) False. The static and scientific friction force act in the same direction, since the second appears when the movement does not start and the static friction decreases.

D) Fale the static and kinetic friction forces act in the same direction

2) How to overcome friction on a ramp

A) False. If the texture of the surface becomes rough, the friction force increases

B) False. Pressing the brick against the surface increases the normal and as the friction is proportional to the normal, it also increases

C) False. By lowering the table the weight component in the friction direction decreases

D) True. When lifting the board, the weight component increases and therefore can become greater than the friction

Wx-fr = ma

W sin tea - my mg cos tea = m a

As it increases, the sine increases and the cosine decreases.

3 0

2 years ago

When is an object moving in uniform circular motion?

ad-work [718]

When its tangential speed is constant
Although the speed of an object that has a uniform circular motion is constant, its velocity is not constant. Not only that, but it is actually changing constantly.

6 0

3 years ago

Read 2 more answers