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zalisa [80]
3 years ago
10

How to solve this step by step.

Physics
1 answer:
brilliants [131]3 years ago
7 0

Answer:

a = (v² − v₀²) / (2 (x − x₀))

Explanation:

v² = v₀² + 2a(x − x₀)

Subtract v₀² from both sides.

v² − v₀² = 2a(x − x₀)

Divide both sides by 2 (x − x₀).

(v² − v₀²) / (2 (x − x₀)) = a

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Which division of time on the geologic time scale is the shortest?
Scrat [10]

Explanation:

In geology, there are 5 divisions of time on the time scale. They are:

  1. eons
  2. eras
  3. periods
  4. epochs
  5. ages

Eons are the largest time period, while ages are the shortest time period. The rest of the above listed are in between the two.

8 0
3 years ago
A rock dropped on the moon will increase its speed from 0m/s to 8.15 m/s in 5 seconds what is the acceleration of the rock
sergejj [24]

speed  Δv     =  v₂ =  8.15ms⁻¹

                     v₁ = 0ms⁻¹

time  =t = 5s

acceleration  =  a = speed / time taken

                          a =    Δv / Δt           ( as Δv = v₂ - v₁)

                          a  =   8.15ms⁻¹  /  5 s

                            a  = 1.6ms⁻²

                               

3 0
4 years ago
A/an ___ is a machine which tells us about the strength and speed of sisemec waves.
svp [43]
The answer is Seismograph
3 0
4 years ago
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When work is done on an object , the results is an increase in its what
Ksju [112]

Answer:Im guessing Mechanical Energy

Explanation:

We are learning that work and energy work hand in hand so im completely guessing this

5 0
3 years ago
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What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.
valkas [14]
<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.

This Law is originally expressed as follows:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=6.39(10)^{23}kg is the mass of Mars

a=3.4(10)^{6}m  is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2)

T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}    (3)

T=\sqrt{11581157.44 s^{2}}    (4)

Finally:

T=3403.1099s=56.718min    This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0
3 years ago
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