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ratelena [41]
3 years ago
14

Why does moon appear to have alot of phases

Physics
1 answer:
Bingel [31]3 years ago
4 0
It depends on the angle of the earth, and our point of view. A full moon would occur if we were to be right in front of it.
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What is the magnitude of the force needed to keep a 60 newton rubber block moving across level,dry asphalt in a straight line at
pantera1 [17]

let Coefficients of Friction of Rubber on asphalt (dry) =0.7

F= Coefficients of Friction * normal force = 0.7 * 60 =42 N

so the net force of the rubber is zero, meaning it will travel at a constant speed.

When the rubber is travel at 2m/s, 42N is required to keep moving at constant speed

6 0
3 years ago
Read 2 more answers
If 1 meter = 3.28 feet, what is the height of the Washington Monument in meters?
jasenka [17]
Since you didn't provide how tall the Monument was, I took the liberty to find it and it is 555 feet tall. So to convert to meters we must divide 555 by 3.28 or multiply it by 0.3048 (this is the method I used).
555 x 0.3048 = 169.164 meters
5 0
3 years ago
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A student pushes a 2.85 kg cart causing it to accelerate at a rate of 4.9 m/s squared .What amount of force must the student hav
uysha [10]
In order to find the force (F), you would have to use the formula for it:
F=ma
where m is mass and a is acceleration.
In the problem, the mass is 2.85kg and the acceleration is 4.9m/s^2.
Therefore,
F=2.85kg(4.9m/s^2)
F=13.965kg(m/s^2)
Since N=kg(m/s^2)
F=13.965N
And because the problem requires that we use only 2 significant figures,
F=13N
Therefore, the student must exert 13N of force.

7 0
3 years ago
1. When in the past have you pushed your personal limits? Give at least one
Yuri [45]

Answer:

Umm that's a personal question. All u have to do is say when have u pushed your personal limits....... Ummm one for me is when i had to try out for a select soccer and that is past my comfort zone.  

Explanation:

4 0
3 years ago
How much work is required to stop an electron (m = 9.11 \times 10^ - 31 kg) which is moving with a speed of 2.10
Karo-lina-s [1.5K]

Answer:

-2.00876\times 10^{-18}\ J

Explanation:

v = Speed of electron = 2.1\times 10^6\ m/s (generally the order of magnitude is 6)

m = Mass of electron = 9.11\times 10^{-31}\ kg

Work done would be done by

W=K_i-K_f\\\Rightarrow W=0-\dfrac{1}{2}mv^2\\\Rightarrow W=0-\dfrac{1}{2}\times 9.11\times 10^{-31}\times (2.1\times 10^6)^2\\\Rightarrow W=-2.00876\times 10^{-18}\ J

The work required to stop the electron is -2.00876\times 10^{-18}\ J

6 0
3 years ago
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