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2 years ago

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

Physics

2 answers:

max2010maxim [7]2 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]2 years ago

5 0

Answer:

4.5*10^9

Explanation:

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A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re

Lady_Fox [76]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

$P_1$ = Pressure at reservoir = 10 atm

$T_1$ = Temperature at reservoir = 300 K

$P_2$ = Pressure at exit = 1 atm

$T_2$ = Temperature at exit

$R_s$ = Mass-specific gas constant = 287 J/kgK

$\gamma$ = Specific heat ratio = 1.4 for air

For isentropic flow

$\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K$

The temperature of the flow at the exit is 155.38424 K

From the ideal equation density is given by

$\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3$

The density of the flow at the exit is 2.2721 kg/m³

4 0

2 years ago

You are sitting in your car at rest at a traffic light with a bicyclist at rest next to you in the adjoining bicycle lane. As so

grigory [225]

Answer:

Explanation:

Time duration during which acceleration exists in bicycle =

23 / 12 = 1.91 s

Time duration during which acceleration exists in car

= 47 / 8 = 5.875 s

Distance covered by bicycle during acceleration ( t = 1.91 s )

= 1/2 x 12 x (1.91)²

= 21.88 mi

Distance covered by car during this time ( t = 1.91 s )

= 1/2 x 8 x (1.91)²

7.64 mi ,

velocity of car after 1.91 s

= 8 x 1.91 = 15.28 mi/h

Let after time 1.91 , time taken by them to meet each other be t

Total distance covered by cycle = total distance covered by car

21.88 + 23 t = 7.64 + 15.28t + 4 t²

21.88 = 7.64 - 7.72t +4 t²

4 t² -7.72 t -14.24 = 0

t = 2.83 s

Total time taken

= 2.83 + 1.91

= 4.74 s

So after 4.74 s they will meet each other.

b ) Maximum distance occurs when velocity of both of them becomes equal .

Velocity after 1.91 s of bicycle

12 x 1.91 = 23 mi/h

Velocity after 1.91 s of car

8 x 1.91 = 15.28 mi/h . Let after time t , the velocity of car becomes 23

15.28 + 8t = 23

t = .965 s

So after time .965 s , car has velocity equal to that of bicycle.

The bicycle will travel a distance of

= 21.88 + .965 x 23 = 44.075 mi

car will travel a distance of

7.64 + 15.28 x .965 + .5 x 8 x .965²

= 7.64 + 14.75 + 3.72

= 26.11 mi

Distance between car and bicycle

= 44.075 - 26.11 = 17.965 mi

= 17.965 x 1760

= 31618.4 ft.

5 0

2 years ago

A steel sphere sits on top of an aluminum ring. The steel sphere (a= 1.1 x 10^-5/degrees celsius) has a diameter of 4.000 cm at

mote1985 [20]

Answer:

Explanation:

To solve this question, we will need to develop an expression that relates the diameter 'd', at temperature T equals the original diameter d₀ (at 0 degrees) plus the change in diameter from the temperature increase ( ΔT = T):

d = d₀ + d₀αT

for the sphere, we were given

D₀ = 4.000 cm

α = 1.1 x 10⁻⁵/degrees celsius

we have D = 4 + (4x(1.1 x 10⁻⁵)T = 4 + (4.4x10⁻⁵)T EQN 1

Similarly for the Aluminium ring we have

we were given

d₀ = 3.994 cm

α = 2.4 x 10⁻⁵/degrees celsius

we have d = 3.994 + (3.994x(2.4 x 10⁻⁵)T = 3.994 + (9.58x10⁻⁵)T EQN 2

Since @ the temperature T at which the sphere fall through the ring, d=D

Eqn 1 = Eqn 2

4 + (4.4x10⁻⁵)T =3.994 + (9.58x10⁻⁵)T, collect like terms

0.006=5.18x10⁻⁵T

T=115.7K

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2 years ago

A cork floats on the surface of an incompressible liquid in acontainer exposed to atmospheric pressure. The container is thensea

e-lub [12.9K]

Answer:

The cork:

C) floats at the same height

Explanation:

The law of floatation states that an object will float when it displaces its own weight of fluid in which it floats e.g. the weight of the object is equal to the displaced liquid. With this definition it is clear that the cork will float at the same height because the evacuated air has little or no effect on the cork floating on the surface of the incompressible liquid.

5 0

2 years ago

suppose larry stands with more weight on one scale than the other if one scale reads 400 N. what does the other read?

Evgen [1.6K]

The other scale will either have a higher or lower number because 400 could be the bigger number or the smaller number but we don't know that so you cant exactly answer it.<span />

8 0

2 years ago

Read 2 more answers