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matrenka [14]
3 years ago
11

How much of the Moon is always illuminated one time? Explain your answer.

Physics
1 answer:
Natalija [7]3 years ago
6 0

Answer:

50% of it .

Explanation:

50% of it is illuminated by the Sun.

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Two resistors A and B are arranged in series in one branch of a parallel arrangement. The other branch contains a single resisto
Ganezh [65]

Answer:

Explanation:

A and B are in series , Total resistance = Ra + Rb

This resistance is in parallel with single resistor C

Equivalent resistance Re = Rc x ( Ra + Rb ) / [Rc + ( Ra + Rb )]

Now this combination is in series in single resistance D .

Total resistance = Rd + Re

= Rd + { Rc x ( Ra + Rb ) / [Rc + ( Ra + Rb )] }

5 0
2 years ago
This formula equation is unbalanced.
Grace [21]
The answer is:  [C]:  "4" .
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Note:  To balance this equation, the coefficient, "4", should be placed in front of the PCl₃ ;  and the coefficient, "6", should be placed in front of the Cl₂ .
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The balanced equation is:
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             P₄ (s) + 6 Cl₂ (g)  <span>→ 4 </span>PCl₃ (l)   .
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6 0
3 years ago
What is the value of acceleration due to gravity at heght h=4re​
Norma-Jean [14]

Answer: Formula for Acceleration Due to Gravity

These two laws lead to the most useful form of the formula for calculating acceleration due to gravity: g = G*M/R^2, where g is the acceleration due to gravity, G is the universal gravitational constant, M is mass, and R is distance.please mark as brainliest

Explanation:

5 0
3 years ago
The number of magnetic field force lines passing through the given surface determines: 1. magnetic flux 2. magnetic induction 3.
Assoli18 [71]
A magnetic flux would be the correct answer

8 0
3 years ago
A box rests on top of a flat bed truck. The box has a mass of m = 16.0 kg. The coefficient of static friction between the box an
3241004551 [841]

Answer:

1) 1.31 m/s2

2) 20.92 N

3) 8.53 m/s2

4) 1.76 m/s2

5) -8.53 m/s2

Explanation:

1) As the box does not slide, the acceleration of the box (relative to ground) is the same as acceleration of the truck, which goes from 0 to 17m/s in 13 s

a = \frac{\Delta v}{\Delta t} = \frac{17 - 0}{13} = 1.31 m/s2

2)According to Newton 2nd law, the static frictional force that acting on the box (so it goes along with the truck), is the product of its mass and acceleration

F_s = am = 1.31*16 = 20.92 N

3) Let g = 9.81 m/s2. The maximum static friction that can hold the box is the product of its static coefficient and the normal force.

F_{\mu_s} = \mu_sN = mg\mu_s = 16*9.81*0.87 = 136.6N

So the maximum acceleration on the block is

a_{max} = F_{\mu_s} / m = 136.6 / 16 = 8.53 m/s^2

4)As the box slides, it is now subjected to kinetic friction, which is

F_{\mu_s} = mg\mu_k = 16*9.81*0.69 = 108.3 N

So if the acceleration of the truck it at the point where the box starts to slide, the force that acting on it must be at 136.6 N too. So the horizontal net force would be 136.6 - 108.3 = 28.25N. And the acceleration is

28.25 / 16 = 1.76 m/s2

5) Same as number 3), the maximum deceleration the truck can have without the box sliding is -8.53 m/s2

3 0
3 years ago
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