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Burka [1]
3 years ago
14

Two asteroids are 100,000 m apart. One has a mass of 3.5 x 106 kg. If the

Physics
2 answers:
max2010maxim [7]3 years ago
7 0

Answer:

mass of the other asteroid = 4.49*10^9kg\\

Explanation:

We use the definition for the force between two celestial objects under the action of the gravity they produce using newton's general gravitational constant: G=6.674*10^{-11} \frac{N*m^2}{kg^2}

The force between the two asteroids will then be given by:

F_G=G*\frac{M_1*M_2}{d^2}

where G is Newton's gravitational constant, the asterioid's masses are M1 and M2 respectively, and d is the distance between them.

We replace the known values in he equation above, and solve for the missing mass:

F_G=G*\frac{M_1*M_2}{d^2}\\1.05*10^{-4}N=6.674*10^{-11} \frac{N*m^2}{kg^2} \frac{3.5*10^6kg*M_2}{(10^5m)^2} \\1.05*10^{-4}=2.3359*10^{-14} * M_2\\M_2=\frac{1.05*10^{-4}}{2.3359*10^{-14}} =4.49*10^9kg

Since the units for the given quantities are all in the SI system, our resultant units for the unknown mass of the asteroid will be in kg.

hram777 [196]3 years ago
5 0

Answer:

4.5*10^9

Explanation:

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Explanation:

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10.764 T + 731.85 T = 62207.25 + 236.808

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A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o
gulaghasi [49]

A. 0.77 A

Using the relationship:

P=\frac{V^2}{R}

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega

The two light bulbs are connected in series, so their equivalent resistance is

R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega

The two light bulbs are connected to a voltage of

V  = 240 V

So we can find the current through the two bulbs by using Ohm's law:

I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A

B. 142.3 W

The power dissipated in the first bulb is given by:

P_1=I^2 R_1

where

I = 0.77 A is the current

R_1 = 240 \Omega is the resistance of the bulb

Substituting numbers, we get

P_1 = (0.77 A)^2 (240 \Omega)=142.3 W

C. 42.7 W

The power dissipated in the second bulb is given by:

P_2=I^2 R_2

where

I = 0.77 A is the current

R_2 = 72 \Omega is the resistance of the bulb

Substituting numbers, we get

P_2 = (0.77 A)^2 (72 \Omega)=42.7 W

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

P=\frac{E}{t}

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

7 0
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