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umka2103 [35]
3 years ago
12

How are the sun, the moon, and Earth related during a solar eclipse?

Physics
1 answer:
german3 years ago
4 0

B. The moon is located between the Sun and Earth

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A billiard ball is moving in the x-direction at 30.0 cm/s and strikes another billiard ball moving in the y-direction at 40.0 cm
____ [38]

Answer:

53.13 °

Explanation:

In order to do this, we just need to apply the following:

tanα = Dy/Dx

Where:

Vy: speed of the ball in the y axis.

Vx: speed of the ball in the x axis.

At this point we do not need the speed of the first ball after the collision because in that moment is already heading in the direction that we are looking for. Therefore, we just need to use the innitial data to calculate the direction which the first ball will go.

According to this, then:

tanα = (40/30)

tanα = 1.3333

α = tan⁻¹(1.3333)

<h2>α = 53.13°</h2>

This means that the final direction of the first ball is 53.13° and in the x axis because the starting momentum of this ball in the x axis has not dissapeared.

Hope this helps

6 0
2 years ago
The forces acting on this object are _______ and the net force is ________ Balanced, 0 N Balanced, 10 N Unbalanced, 10 N Unbalan
Amanda [17]
1st Blank Answer: Balanced   
2nd Blank Answer: 0n 

I took the test and got it right
I would appreciate a Brainliest...
5 0
3 years ago
Read 2 more answers
What are some similarities between electricity and gravity?
kow [346]
Similarity : inverse square law for strength of force compared with distance. 
5 0
3 years ago
A pyramid was built with approximately 2.3 million stone blocks, each weighing 2.4 tons (1 ton = 2,000 lbm). Find the mass of th
pogonyaev

Total weight in tons

\\ \sf\longmapsto 2.3\times 10^6\times 2.4

\\ \sf\longmapsto 5.52\times 10^6

\\ \sf\longmapsto 5520000ton

  • 1ton=1000kg

\\ \sf\longmapsto 5520000\times 1000

\\ \sf\longmapsto 5520000000kg

\\ \sf\longmapsto 5.52\times 10^9Kg

4 0
3 years ago
Read 2 more answers
A girl is sledding down a slope that is inclined at 30º with respect to the horizontal. The wind is aiding the motion by providi
OleMash [197]

Answer:

The sled required 9.96 s to travel down the slope.

Explanation:

Please, see the figure for a description of the problem. In red are the x and y-components of the gravity force (Fg). Since the y-component of Fg (Fgy) is of equal magnitude as Fn but in the opposite direction, both forces get canceled.

Then, the forces that cause the acceleration of the sled are the force of the wind (Fw), the friction force (Ff) and the x-component of the gravity force (Fgx).

The sum of all these forces make the sled move. Finding the resulting force will allow us to find the acceleration of the sled and, with it, we can find the time the sled travel.

The magnitude of the friction force is calculated as follows:

Ff = μ · Fn

where :

μ = coefficient of kinetic friction

Fn =  normal force

The normal force has the same magnitude as the y-component of the gravity force:

Fgy = Fg · cos 30º = m · g · cos 30º

Where

m = mass

g = acceleration due to gravity

Then:

Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º

Fgy = 744 N

Then, the magnitude of Fn is also 744 N and the friction force will be:

Ff = μ · Fn = 0.151 · 744 N = 112 N

The x-component of Fg, Fgx, is calculated as follows:

Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N

The resulting force, Fr, will be the sum of all these forces:

Fw + Fgx - Ff = Fr

(Notice that forces are vectors and the direction of the friction force is opposite to the other forces, then, it has to be of opposite sign).

Fr = 161 N + 430 N - 112 N = 479 N

With this resulting force, we can calculate the acceleration of the sled:

F = m·a

where:

F = force

m = mass of the object

a = acceleration

Then:

F/m = a

a = 479N/87.7 kg = 5.46 m/s²

The equation for the position of an accelerated object moving in a straight line is as follows:

x = x0 + v0 · t + 1/2 · a · t²

where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the sled starts from rest and the origin of the reference system is located where the sled starts sliding, x0 and v0 = 0.

x = 1/2· a ·t²

Let´s find the time at which the position of the sled is 271 m:

271 m = 1/2 · 5.46 m/s² · t²

2 · 271 m / 5.46 m/s² = t²

<u>t = 9.96 s </u>

The sled required almost 10 s to travel down the slope.

8 0
3 years ago
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