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3 years ago

Which heavenly bodies are involved in causing the earths tides?

Physics

1 answer:

Elis [28]3 years ago

4 0

All of them except any planet means "Earth, Moon and the Sun"

so, option C is your answer.

Hope this helps!

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Question 13 of 20

Dominik [7]

Answer:

B. It has a central nucleus composed of 29 protons and 35 neutrons,

surrounded by an electron cloud containing 29 electrons.

Explanation:

Protons and neutrons are the only subatomic particles with mass, and they are located in the nucleus. If this atom has an atomic number of 29 and is copper, it must have 29 protons (protons define which element is being observed). This means all remaining mass is from neutrons. 64-29 = 35.

Electrons have no mass and orbit the nucleus in the electron cloud. Since this copper atom is neutral (we are not told it has a charge), there must be an equal number of protons and electrons.

6 0

2 years ago

Read 2 more answers

A railroad car moving at a speed of 3.46 m/s overtakes, collides and couples with two coupled railroad cars moving in the same d

I am Lyosha [343]

Answer:

$2.09\ \text{m/s}$

$22298.4\ \text{J}$

Explanation:

m = Mass of each the cars = $1.6\times 10^4\ \text{kg}$

$u_1$ = Initial velocity of first car = 3.46 m/s

$u_2$ = Initial velocity of the other two cars = 1.4 m/s

v = Velocity of combined mass

As the momentum is conserved in the system we have

$mu_1+2mu_2=3mv\\\Rightarrow v=\dfrac{u_1+2u_2}{3}\\\Rightarrow v=\dfrac{3.46+2\times 1.4}{3}\\\Rightarrow v=2.09\ \text{m/s}$

Speed of the three coupled cars after the collision is $2.09\ \text{m/s}$ .

As energy in the system is conserved we have

$K=\dfrac{1}{2}mu_1^2+\dfrac{1}{2}2mu_2^2-\dfrac{1}{2}3mv^2\\\Rightarrow K=\dfrac{1}{2}\times 1.6\times 10^4\times 3.46^2+\dfrac{1}{2}\times 2\times 1.6\times 10^4\times 1.4^2-\dfrac{1}{2}\times 3\times 1.6\times 10^4\times 2.09^2\\\Rightarrow K=22298.4\ \text{J}$

The kinetic energy lost during the collision is $22298.4\ \text{J}$ .

6 0

2 years ago

. A dog walks 14 m to the east and then 20 m back to the west. What is the distance? What is the displacement?

Crank

Answer:

Hi...

14-20=-6...displacement

14+20=34...distance

3 0

2 years ago

My Exams are coming.so, please tell me some ways to score good marks?

Citrus2011 [14]

Instead of asking this question go and study

5 0

1 year ago

Read 2 more answers

To counter the effects of centrifugal force and reduce vehicle traction it is important to to counter the effects of centrifugal

tatiyna

Answer: Add an incline or grade to the road track.

Explanation:
Refer to the figure shown below.

When a vehicle travels on a level road in a circular path of radius r, a centrifugal force, F, tends to make the vehicle skid away from the center of the circular path.
The magnitude of the force is
F = mv²/r
where
m = mass of the vehicle
v = linear (tangential) velocity to the circular path.

The force that resists the skidding of the vehicle is provided by tractional frictional force at the tires, of magnitude
μN = μW = μmg
where
μ = dynamic coefficient of friction.

At high speeds, the frictional force will not overcome the centrifugal force, and the vehicle will skid.

When an incline of θ degrees is added to the road track, the frictional force is augmented by the component of the weight of the vehicle along the incline.
Therefore the force that opposes the centrifugal force becomes
μN + Wsinθ = W(sinθ + μ cosθ).

5 0

2 years ago