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GalinKa [24]
3 years ago
5

1. During which phase of the moon may a solar eclipse occur? (Points : 1)

Physics
2 answers:
kotykmax [81]3 years ago
7 0
1. New moon
This is because the moon comes between the Earth and sun and this is only possible during its new moon phase.

2. Full moon
This is because the Earth comes between the Moon and the sun and the effect is only visible when there is a full moon.

3. Corona
The corona is the outer layer and is the only one visible when there is an eclipse.
shtirl [24]3 years ago
5 0

1).  A solar eclipse, if it happens, can only happen at the time of New Moon.

2).  A lunar eclipse, if it happens, can only happen at the time of Full Moon.

3).  The sun's corona is only visible during a total solar eclipse.

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When u write on a piece of glass sheet with a piece of chalk , the writing is not clear explain .
svetlana [45]
Becuse your weighting with chalk that has pigment
6 0
3 years ago
A boy finds an abandoned mine shaft in the woods, and wants to know how deep the hole is. He drops in a stone, and counts 4 seco
oee [108]
S=(0x4)+(0.5x4.81x4x4)
S=0.78.48

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(My brain hurts now) :P Good Luck!
4 0
3 years ago
7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N
artcher [175]

Answer:

20.4 m/s^{2}

Explanation:

To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.

First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.

Applied force = 1055 N (490 + 565)

Friction force= Unknown

To find the friction force, use the kinetic friction formula, Friction = μkN

μk is the coefficient, which the problem includes- it is 0.613.

N is the normal force, which we have to find.

*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.

Weight = mg = (40)(9.8) = 392 N

Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

Friction = 240.296 N

Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

Horizontal Net Force: 814.704 N

Now that we know the net force, plug in the numbers for the formula

∑F= ma.

814.704 = (40.0)(a)

*Divide on both sides)

a = 20.3676 m/s^2

Round it to 3 significant figures, to get:

20.4 m/s^{2}

7 0
3 years ago
Students can take the aspire test in ninth and grade
Simora [160]

<em>The answer is </em>Ninth <em>and </em>Tenth <em>grade so the answer would be</em> B

<em>I hope this helps you </em>


3 0
2 years ago
Read 2 more answers
4. O/N 15/P11/Q11
anygoal [31]

Answer:

B) 4500 Pa

Explanation:

As pressure is force per unit area,

P = F/A

It stands to reason that the smallest pressure for a given force is when it is shared by the largest area.

The possible areas are

0.30(0.40) = 0.12 m²

0.30(0.50) = 0.15 m²

0.40(0.50) = 0.20 m²

The pressure when the face with the largest area (0.20 m²) is down is

P = 900 / 0.20 = 4500 N/m² or 4500 Pa

the other possible pressures would be

900/0.15 = 6000 Pa

900/0.12 = 7500 Pa

which are both larger than our solution.

5 0
3 years ago
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