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olga2289 [7]
2 years ago
5

A meteor is falling towards the earth. The mass and the radius of the earth is 6×10^24 kg and 6.4×10^3 km respectively. What is

the height of the meteor from the earth surface at which the acceleration due to gravity is 4m/s^2
Physics
1 answer:
joja [24]2 years ago
8 0

Answer:

3600km

Explanation:

gravitational constant is the gravitational attraction between any two things

G = gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²

g = acceleration due to gravity = 4 m/s²

M = mass of the earth = 6 × 10²⁴

r = radius of Earth + height of meteor

Since F = mg and F = GMm/r², mg = GMm/r²

g = GM/r²

r² = GM/g

r = √(GM/g)

r = √((6.67 × 10⁻¹¹ Nm²/kg²) × (6 × 10²⁴ kg))/(4 m/s²)) =

1 × 10⁷ m = 1 × 10⁴ km

To calculate the height of the meteor above the earth,

subtract the radius of the earth from the calculated radius.

height of meteor = 1 × 10⁴ km - 6 × 10³ km = 4 × 10³ km

quora Meave Gilchrist

snapsolve

quora

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Assume that the Deschutes River has straight and parallel banks and that the current is 0.75 m/s. Drifting down the river, you f
DedPeter [7]

Answer:

    d = 142.5 m

Explanation:

This is a vector exercise. Let's calculate how much the boat travels in the 40s

     d₀ = v_{b} t

    d₀ = 0.75 40

    d₀ = 30 m

Let's write the kinematic equations

Boat

     x = d₀  +  v_{b} t

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At the meeting point the coordinate is the same for both

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    t ( v_{h} -  v_{b}) = d₀  

    t = d₀  / ( v_{b}-  v_{h})

The two go in the same direction therefore the speeds have the same sign

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The distance traveled by man is

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3 0
4 years ago
A car is driving 25.5 m/s when it hits
IceJOKER [234]

Answer:

it takes the car 4.362 seconds to cover the distance of 88.4 m.

Explanation:

The distance the car covers is given by the function

x(t) = vt-\dfrac{1}{2}at^2,

where v = 25.5m/s , and a = 2.40m/s, putting  these in we get:

x(t) = 25.5t-\dfrac{1}{2}(2.4)t^2\\\\x(t) = 25.5t-1.2t^2

Now, when the car has moved to 88.4m, x(t) = 88.4 , or

x(t) = 25.5t-1.2t^2 = 88.4

which is a quadratic equation with solutions

t = 4.362s,\\t = 16.887s

We take the first solution t = 4.362s, <em>since at that time the car is still moving right and decelerating</em>. The second solution t =16.887 describes the situation where the car has stopped decelerating and is now moving leftwards because the decelerating is leftwards, <em>which is utterly wrong because we know that cars do not start moving backwards after the brakes have stopped them! </em>

Thus, it takes the car 4.362 seconds to cover the distance of 88.4 m.

6 0
4 years ago
In Anchorage, collisions of a vehicle with a moose are so common that they are referred to with the abbreviationMVC. Suppose a 1
lara31 [8.8K]

Answer:

Part a)

f = \frac{8}{9}

Part b)

f = \frac{120}{169}

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

Explanation:

Part a)

Let say the collision between Moose and the car is elastic collision

So here we can use momentum conservation

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

1000 v_o = 1000 v_{1f} + 500 v_{2f}

also by elastic collision condition we know that

v_{2f} - v_{1f} = v_o

now we have

2v_o = 2v_{1f} + v_o + v_{1f}

now we have

v_{1f} = \frac{v_o}{3}

Now loss in kinetic energy of the car is given as

\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)

\Delta K = \frac{1}{2}m(v_o^2 - \frac{v_o^2}{9})

so fractional loss in energy is given as

f = \frac{\Delta K}{K}

f = \frac{\frac{4}{9}mv_o^2}{\frac{1}{2}mv_o^2}

f = \frac{8}{9}

Part b)

Let say the collision between Camel and the car is elastic collision

So here we can use momentum conservation

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

1000 v_o = 1000 v_{1f} + 300 v_{2f}

also by elastic collision condition we know that

v_{2f} - v_{1f} = v_o

now we have

10v_o = 10v_{1f} + 3(v_o + v_{1f})

now we have

v_{1f} = \frac{7v_o}{13}

Now loss in kinetic energy of the car is given as

\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)

\Delta K = \frac{1}{2}m(v_o^2 - \frac{49v_o^2}{169})

so fractional loss in energy is given as

f = \frac{\Delta K}{K}

f = \frac{\frac{60}{169}mv_o^2}{\frac{1}{2}mv_o^2}

f = \frac{120}{169}

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

8 0
3 years ago
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