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Alex Ar [27]
3 years ago
5

Assuming a roughly spherical shape and a density of 4000 kg/m3, estimate the diameter of an asteroid having the average mass of

1017 kg.
Physics
1 answer:
vredina [299]3 years ago
8 0

Explanation:

It is given that,

Density of asteroid, \rho=4000\ kg/m^3

Mass of asteroid, m=10^{17}\ kg

We need to find the diameter of the asteroid. The formula of density is given by:

\rho=\dfrac{m}{V}

V is the volume of spherical shaped asteroid, V=\dfrac{4}{3}\pi r^3

r^3=\dfrac{3M}{4\pi \rho}

r^3=\dfrac{3\times 10^{17}\ kg}{4\pi \times 4000\ kg/m^3}

r^3=\sqrt{5.96\times 10^{12}}

r = 2441311.12 m

Diameter = 2 × radius

d = 4882622.24 m

or

d=4.88\times 10^6\ m

Hence, this is the required solution.

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What is the resultant of the vectors shown?
ad-work [718]

Answer:

Option B

Explanation:

Option A is the wrong answer because the horizontal vector is in the opposite direction.

Option C is the wrong answer as the horizontal vector is in the opposite direction and all the vectors are connected head to tail [of the arrows] [Triangle law of vector addition]

Option D is the wrong answer as the horizontal vector is in the opposite direction.

5 0
2 years ago
I need help with this too. (im not good at science or math)
mr_godi [17]

Answer:

i belive 1 m/s

Explanation:

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please tell me if right!

6 0
3 years ago
Read 2 more answers
A 26.4 g silver ring (cp = 234 J/kg·°C) is heated to a temperature of 66.2°C and then placed in a calorimeter containing 4.94 ✕
Slav-nsk [51]

Answer:

The final temperature of the mixture = 64.834 °C.

Explanation:

Heat lost by the silver ring = heat gained by the water + heat transferred to the surrounding.

c₁m₁(t₁-t₃) = c₂m₂(t₃-t₂) + Q..............Equation 1

Where c₁ = specific heat capacity of the silver copper, m₁ = mass of the silver copper, t₁ = initial temperature of the silver copper, t₃ = final temperature of the mixture. c₂ = specific heat capacity of water, t₂ = initial temperature of water, m₂ = mass of water, Q = energy transferred to the surrounding.

making t₃ the subject of the equation,

t₃ = [c₁m₁t₁+c₂m₂t₂-Q]/(c₁m₁+c₂m₂)........................ Equation 2

Given: c₁ = 234 J/kg.°C, m₁ = 26.4 g, t₁ = 66.2 °C, c₂ = 4200 J/K.°C, m₂ = 4.92×10⁻² kg, t₂ = 24.0 °C, Q = 0.136 J.

Substituting into equation 2

t₃ = [(234×26.4×66.2)+(4200×0.0492×24)-0.136]/[(234×26.4)+(4200×0.0492)]

t₃ = (408957.12+4959.36-0.136)/(6177.6+206.64)

t₃ = (413916.48-0.136)/6384.24

t₃ = 413916.34/6384.24

Thus the final temperature of the mixture = 64.834 °C.

6 0
3 years ago
I need some help on this
lord [1]

Answer: A balloon provides a simple example of how a rocket engine works. The air trapped inside the balloon pushes out the open end, causing the balloon to move forward. The force of the air escaping is the "action"; the movement of the balloon forward is the "reaction" predicted by Newton's Third Law of Motion.

Explanation: ur welcome

8 0
3 years ago
The physics of wind instruments is based on the concept of standing waves. When the player blows into the mouthpiece, the column
Fynjy0 [20]

Answer:

See explaination.

Explanation:

L = 80 cm

= 0.8 m

V = 343 m/s ( sound speed in air )

V1 = n V / 2 L

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= 1 X 343 / 2 X 0.8

V1 = 214.375 Hz

5 0
4 years ago
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