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3 years ago

14

What angle is formed by the sun, the earth, and the moon during an eclipse?.

1 answer:

Andrew [12]3 years ago

4 0

Answer:

The Sun-Earth-Moon system happens to exhibit a striking geometric coincidence, which we examine in the first problem. PROBLEM 1. To an observer on Earth, the Sun and the Moon subtend almost the same angle in the sky. The average angle is 0.52 degrees for the Moon and 0.53 degrees for the Sun.

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John bounces a ball off the ground, and it leaves the ground with a velocity of 10 m/s. How long will it take the ball to reach

Maksim231197 [3]

Explanation:

This is true because at maximum height, the velocity is 0

8 0

4 years ago

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Three cars with identical engines and tires start from rest, and accelerate at their maximum rate. Car X is the most massive, an

aksik [14]

Answer:C) car X

Explanation:

Given

All the cars have identical Engine thus Force Produced by car X will be equal to Y and Z

and $Force=mass\times acceleration$

Since Car X is most massive so acceleration associated with it will be minimum

acceleration of car X is minimum thus it will travel farthest

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3 years ago

Which is found farthest from the center of an atom?

xxTIMURxx [149]

Electron

Hope this helps

5 0

3 years ago

A phone cord is 4.89 m long. The cord has a mass of 0.212 kg. A transverse wave pulse is produced by plucking one end of the tau

slega [8]

Answer:

Tension in Cord=174 N

Explanation:

Given Data

L (Phone Cord Length)=4.89 m

m (Cord Mass)=0.212 Kg

T (Time for four trips)=0.617 s

Tension=?

Solution

V=λ×f

$V=\frac{8*4.89}{0.617}\\ V=63.4m/s$

$Sigma=\frac{mass}{length}\\ Sigma=\frac{0.212}{4.89}\\ Sigma=0.0433 \frac{kg}{m}$

$Wave Speed=\sqrt{\frac{Tension}{Sigma} }\\ \\V=\sqrt{\frac{T}{Sigma} }\\ V^{2}=\frac{T}{Sigma}\\ T=V^{2}*Sigma\\ T=(63.4)^{2}*(0.0433)\\ T=174 N$

5 0

3 years ago

4. A meter has a resistance of 100 Ω and gives a full scale deflection when it carries a current of 25 μA. (a) What resistor, Rx

frez [133]

Answer:

A=50mΩ

B≅50mΩ

Explanation:

A) To answer this question we have to use the Current Divider Rule. that rule says:

$Ix=.\frac{Req}{Rx} *Itotal$ (1)

Itotal represents the new maximun current, 50mA, Ix is the current going through the 100 ohms resistor, and Req. is the equivalent resitor.

We now have a set of two resistor in parallel, so:

$Req.=\frac{1}{\frac{1}{R1}+\frac{1}{R2} }$ (2)

where R1 is the resitor we have to calculate, and R2 is the 100 ohms resistor (25 uA).

substituting and rearranging (2)

$Req.=\frac{ 100*R1}{R1+100}$ (3)

Now substituting (3) in (1).

$25*10^{-6} =\frac{\frac{ 100*R1}{R1+100}}{100} *50*10^{-3}$

solving this, The value of R1 is: 50mΩ

This value of R1 will guaranty that the ammeter full reflection willl be at 50mA.

Given that R2 (100ohm) it too much bigger than 50mΩ, the equivalent resistor will tend to 50mΩ

If you substitude this values on (2) Req. will be 49.97 mΩ.

7 0

3 years ago