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2 years ago

ANSWER FOR BRAINLIEST

Physics

1 answer:

Nikitich [7]2 years ago

5 0

A (b) would be 42 (c) from x to Y mark me as brainlist thanks

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Calculate the period of a satellite orbiting the Moon, 98 kmkm above the Moon's surface. Ignore effects of the Earth. The radius

beks73 [17]

Answer:

3.6*10^18s

Explanation:

To find the period of the satellite

We need to apply kephler's third law

Which is

MP² = (4π²/G) d³

d=semi-major axis which is the distance from center of moon = 98km+1740km = 1838km

where M= mass of the moon = 7.3x10^22kg

P=period

G=newtonian gravatational constant= 6.67x10^-11

To find the Period solve for P

P = √[(4π²/G M)xd³]

P=√(4 π²/6.67x10^-22*7.3x10^22kg) x (1.838x10^6m)³]

= 3.6*10^18s

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3 years ago

How do I find magnitude of acceleration?

Leno4ka [110]

Divide the change in speed by the time for the change.

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2 years ago

A backyard swimming pool holds 165 cubic yards (yd3) of water. what is the weight of the water in pounds?

asambeis [7]

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8 0

2 years ago

A car travels 20 km South, then turns and travels 30 km East? What is the total displacement?

Ne4ueva [31]

Answer:

Explanation:36.05 km

Given

First car travels $r_1=20\ km$ South

then turns and travels $r_2=30\ km$ east

Suppose south as negative y axis and east as positive x axis

So, $r_1=-20\hat{j}$

$r_2=30\hat{i}$

Displacement is the shortest between initial and final point

Dispalcement $=r=r_1+r_2$

Displacement $=-20\hat{j}+30\hat{i}$

Displacement $=30\hat{i}-20\hat{j}$

Magnitude $=\sqrt{30^2+(-20)^2}$

Magnitude $=36.05\ km$

4 0

2 years ago

A capacitor is charged until it holds 5.0 j of energy. it is then connected across a 10-kω resistor. in 13.6 ms , the resistor d

prohojiy [21]

Answer:

The capacite is C=5.32 uF using the equations of voltage and energy in capacitance

Explanation:

The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J

Using:

$V_{t}= V_{o}e^{\frac{-t}{R*C} }$

Voltage in this case is the energy dissipated so

$E_{t}= E_{o}e^{\frac{-t}{R*C} }$

$\frac{\sqrt{E_t} }{\sqrt{E_o} } = e^{\frac{-t}{R*C} }$

$\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }$

Using the equation to find capacitance

$ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }$

$C= 5.32x10^{-6}$ F

C= 5.32 uF because u is the symbol for micro that is equal to $10^{-6}$

8 0

3 years ago