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2 years ago

At which moon position would a person on Earth see the entire half of the moon(full-moon)?

Physics

2 answers:

iragen [17]2 years ago

8 0

Wouldn't that be the full moon position...its position 5...idk know what its asking for exactly..what are your choices?

jeka942 years ago

3 0

Sun-earth-moon in a straight line. Earth in the 'middle'.

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In addition to possibly releasing harmful chemicals in the environment, mining is considered

ddd [48]

In addition to possibly releasing harmful chemicals in the environment, mining is considered B. The most dangerous job in the United States.

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2 years ago

Batman chased the joker on his batbike for 20 minutes, traveling at the speed of 30 kilometers per minute. How far did Batman go

AURORKA [14]

Answer:

123 kilometer

Explanation:

none

8 0

2 years ago

150c passes through cell in 30 seconds. cell has a potential difference of 12v. what is current in the circuit

zzz [600]

The current in the circuit is 5 A

Explanation:

The intensity of current is given by the equation:

$I=\frac{q}{t}$

where

I is the current

q is the amount of charge passing through a given point of the circuit in a time interval of t

For the cell in this problem, we have

q = 150 C is the charge

t = 30 s is the time interval

Substituting into the equation, we f ind

$I=\frac{150}{30}=5 A$

Learn more about current:

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6 0

2 years ago

What action force is used when sitting down

Genrish500 [490]

When sitting down there is gravity , when sat down the chair is pushing back at a equal but opposite force

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3 years ago

A 65 kg person jumps from a window to a fire net 18 m below, which stretches the net 1.1 m. Assume the net behaves as a simple s

posledela

Answer:

0.03167 m

1.52 m

Explanation:

x = Compression of net

h = Height of jump

g = Acceleration due to gravity = 9.81 m/s²

The potential energy and the kinetic energy of the system is conserved

$P_i=P_f+K_s\\\Rightarrow mgh_i=-mgx+\frac{1}{2}kx^2\\\Rightarrow k=2mg\frac{h_i+x}{x^2}\\\Rightarrow k=2\times 65\times 9.81\frac{18+1.1}{1.1^2}\\\Rightarrow k=20130.76\ N/m$

The spring constant of the net is 20130.76 N

From Hooke's Law

$F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{65\times 9.81}{20130.76}\\\Rightarrow x=0.03167\ m$

The net would strech 0.03167 m

If h = 35 m

From energy conservation

$65\times 9.81\times (35+x)=\frac{1}{2}20130.76x^2\\\Rightarrow 10065.38x^2=637.65(35+x)\\\Rightarrow 35+x=15.785x^2\\\Rightarrow 15.785x^2-x-35=0\\\Rightarrow x^2-\frac{200x}{3157}-\frac{1000}{451}=0$

Solving the above equation we get

$x=\frac{-\left(-\frac{200}{3157}\right)+\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}, \frac{-\left(-\frac{200}{3157}\right)-\sqrt{\left(-\frac{200}{3157}\right)^2-4\cdot \:1\left(-\frac{1000}{451}\right)}}{2\cdot \:1}\\\Rightarrow x=1.52, -1.45$

The compression of the net is 1.52 m

4 0

3 years ago