Which equation contains only scalar quantities

An aluminium object has a mass of 27kg and density of 2700 the object is atached to a spring and immerged in a tank of water det

laiz [17]

Answer:

0.01m^3

Explanation:

m=27

density=2700

density=mass/volume

2700=27/volume

volume=27/2700=0.01

8 0

1 year ago

The form of energy that can move from place to place across the universe is . On Earth, the main source of this energy is .

serg [7]

The form of energy that can move from place to place across the universe is light energy. On earth, the main source of this energy is Sun. Most of the light energy comes from the sun because it is the primary source of all the energies. The food, fossil fuels, movement of winds, etc all exists due to Sun. Without sun, there won't be any light energy on the earth. In all the processes which occur on earth has a direct or indirect involvement of light energy which comes from sun.

5 0

2 years ago

Read 2 more answers

A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev

tatuchka [14]

Answer:

$W_f = 2.319 rad/s$

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

$L_i = L_f$

so:

$I_mW_m = I_sW_f$

where $I_m$ is the moment of inertia of the merry-go-round, $W_m$ is the initial angular velocity of the merry-go-round, $I_s$ is the moment of inertia of the merry-go-round and the child together and $W_f$ is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = $\frac{1}{2}M_mR^2$

I = $\frac{1}{2}(115 kg)(2.5m)^2$

I = 359.375 kg*m^2

Where $M_m$ is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2 $\pi$ rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

$I_s = \frac{1}{2}M_mR^2+mR^2$

$I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2$

$I_s = 506.25kg*m^2$

Finally we replace all the data:

$(359.375)(3.2672) = (506.25)W_f$

Solving for $W_f$ :

$W_f = 2.319 rad/s$

7 0

2 years ago

When does a star reach equilibrium?

LenaWriter [7]

A protostar reaches equilibrium when gravity is balanced by the outward force of nuclear fusion.

6 0

2 years ago

A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above t

elena55 [62]

Answer:

The first part can be solved via conservation of energy.

$mgh = mg2R + K\\K = mg(h-2R)$

For the second part,

the free body diagram of the car should be as follows:

- weight in the downwards direction

- normal force of the track to the car in the downwards direction

The total force should be equal to the centripetal force by Newton's Second Law.

$F = ma = \frac{mv^2}{R}\\mg + N = \frac{mv^2}{R}$

where $N = 0$ because we are looking for the case where the car loses contact.

$mg = \frac{mv^2}{R}\\v^2 = gR\\v = \sqrt{gR}$

Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.

$mgh = mg2R + \frac{1}{2}mv^2\\mgh = mg2R + \frac{1}{2}m(gR)\\gh = g2R + \frac{1}{2}gR\\h = 2R + \frac{R}{2}\\h = \frac{5R}{2}$

Explanation:

The point that might confuse you in this question is the direction of the normal force at the top of the loop.

We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.

7 0

2 years ago