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3 years ago

At any given time in space, how much of the moon is lit by the sun? PLEASE ANSWER QUICKLY!!!

Physics

1 answer:

nikklg [1K]3 years ago

3 0

Answer: 50%

Explanation: The sun and the moon go around Earth and The sun is way bigger than the moon so 50% of the sun lights up the moon at night

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A snowball is launched horizontally from the top of a building at v = 16.9 m/s. If it lands d = 44 meters from the bottom, how h

Aloiza [94]

$v_x=16,9\frac{m}{s}\\
s=44m\\
g=10\frac{m}{s^2}\\\\
v_x=\frac{s}{t}\Rightarrow t=\frac{s}{v_x}\\\\
t=\frac{44}{16,9}\\\\
t=2,6s\\\\
h=\frac{gt^2}{2}\\\\
h=\frac{10*2,6^2}{2}\\\\
h=5*6,76\\\\
h=33,8m$

6 0

3 years ago

An ice skater of mass m = 60 kg coasts at a speed of v = 0.8 m/s past a pole. At the distance of closest approach, her center of

IrinaVladis [17]

Answer:

1.78 rad/s

1.70344 rad/s

Explanation:

v = Velocity = 0.8 m/s

m = Mass of person = 60 kg

$r_1$ = Distance between center of mass of person and pole = 0.45 m

$r_2$ = New distance between center of mass of person and pole = 0.46 m

I = Moment of inertia

Angular speed is given by

$\omega_1=\dfrac{v}{r_1}\\\Rightarrow \omega_1=\dfrac{0.8}{0.45}\\\Rightarrow \omega_1=1.78\ rad/s$

The angular speed is 1.78 rad/s

In this system the angular momentum is conserved

$L_1=L_2\\\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow mr_1^2\omega_1=mr_2^2\omega_2\\\Rightarrow \omega_2=\dfrac{r_1^2\omega_1}{r_2^2}\\\Rightarrow \omega_2=\dfrac{0.45^2\times 1.78}{0.46^2}\\\Rightarrow \omega_2=1.70344\ rad/s$

The new angular speed is 1.70344 rad/s

7 0

3 years ago

I NEED THIS ASAP

Stells [14]

Answer:

i believe its E.......

7 0

3 years ago

Read 2 more answers

A space vehicle is traveling at 2980 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The

Strike441 [17]

Answer:

3054.4 km/h

Explanation:

Using the conservation of momentum

momentum before separation = 5M × 2980 Km/h where M represent the mass of the module while 4 M represent the mass of the motor

initial momentum = 14900 M km/h

let v be the new speed of the motor so that the

new momentum = 4Mv and the new momentum of the module = M ( v + 94 km/h )

total momentum = 4Mv + Mv + 93 M = 5 Mv + 93M

initial momentum = final momentum

14900 M km/h = 5 Mv + 93M

14900 km/h = 5v + 93

14900 - 93 = 5v

v = 2961.4 km/h

the speed of the module = 2961.4 + 93 = 3054.4 km/h

8 0

3 years ago

A technician at a semiconductor facility is using an oscilloscope to measure the AC voltage across a resistor in a circuit. The

d1i1m1o1n [39]

Answer:

The value to be reported is 5.48V

Explanation:

The RMS (root mean square) is defined as the value of voltage that will produce the same heating effect, or power dissipation, in circuit, as this AC voltage.

The RMS voltage is also called effective voltage because it is just as effective as DC voltage in providing power to an element.

It is expressed as $V_{rms}$ = $\frac{V_{m} }{\sqrt{2} }$

where Vm is the maximum or peak value of the voltage

In calculating the RMS of the voltage , we simply divide the peak voltage by square root of 2 (√2)

$V_{rms}$ = $\frac{7.75}{\sqrt{2} }$

= $\frac{7.75}{1.414}$

= 5.48 V

6 0

3 years ago