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4 years ago

How much is the moon in the first quarter actually lit

Physics

2 answers:

grandymaker [24]4 years ago

8 0

Half of the moon is illuminated.

slamgirl [31]4 years ago

7 0

There's always 50% (half) of the moon lit by the sun. The thing that changes is how much of the lit part we can see.

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A 8.8 cm diameter circular loop of wire is in a 1.04 T magnetic field. The loop is removed from the field in 0.30 s . Assume tha

denis23 [38]

Answer:

0.021 V

Explanation:

The average induced emf (E) can be calculated usgin the Faraday's Law:

$E = - \frac{N*\Delta \phi}{\Delta t}$

Where:

N = is the number of turns = 1

ΔΦ = ΔB*A

Δt = is the time = 0.3 s

A = is the loop of wire area = πr² = πd²/4

ΔB: is the magnetic field = (0 - 1.04) T

Hence the average induced emf is:

$E = - \frac{\Delta B*A}{\Delta t} = - \frac{(0- 1.04 T) \pi (0.088 m)^{2}}{4*0.3 s} = 0.021 V$

Therefore, the average induced emf is 0.021 V.

I hope it helps you!

8 0

4 years ago

All circuits include

valina [46]

Answer:

a battery, wires, and a switch.

Explanation:

All circuits include?

4 0

2 years ago

if the vapor's volume were to be incorrectly recorded as 125ml, how will this error affect the calculated molar mass of the unkn

diamong [38]

So base on your question that as if the vapors volume were to incorrectly recorded as 125ml, the effect of the error to calculate the molar mass is the same as the error in measuring the volume of the vapor. I hope you are satisfied with my answer and feel free to ask for more

7 0

3 years ago

An ideal heat engine absorbs 97.2 kJ of heat and exhausts 83.8 kJ of heat in each cycle. What is the efficiency of the engine?

r-ruslan [8.4K]

Answer:

13.7%

Explanation:

Given that,

Heat absorbed by the engine = 97.2 kJ

Heat exhausted by the engine in each cycle = 83.8 kJ

We need to find the efficiency of the engine. It is calculated by the formula.

$\eta=1-\dfrac{Q_e}{Q_a}\\\\=1-\dfrac{83.8}{97.2}\\\\=0.137\\\\=13.7\%$

so, the efficiency of heat engine is 13.7%.

5 0

3 years ago

If one of two interacting charges is doubled, the force between the charges will _____________.

malfutka [58]

If one of two interacting charges is doubled, the force between the charges will double.

Explanation:

The force between two charges is given by Coulomb's law

$F=\frac{k q1 q2}{r^{2}}$

K=constant= 9 x 10⁹ N m²/C²

q1= charge on first particle

q2= charge on second particle

r= distance between the two charges

Now if the first charge is doubled,

we get $F'=\frac{k (2q1) q2}{r^{2}}$

F'= 2 F

Thus the force gets doubled.

4 0

3 years ago