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Genrish500 [490]
3 years ago
9

In which layer of the Earths atmosphere does outer space begin?

Physics
1 answer:
sergij07 [2.7K]3 years ago
5 0
The answer would be c
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A 120 kg tackler moving at 3.0 m/s meets head-on (and tackles) a 91 kg halfback moving at 7.5 m/s. What will be their mutual vel
Vesna [10]

Explanation:

It is given that,

Mass of the tackler, m₁ = 120 kg

Velocity of tackler, u₁ = 3 m/s

Mass, m₂ = 91 kg

Velocity, u₂ = -7.5 m/s

We need to find the mutual velocity immediately the collision. It is the case of inelastic collision such that,

v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}

v=\dfrac{120\ kg\times 3\ m/s+91\ kg\times (-7.5\ m/s)}{120\ kg+91\ kg}

v = -1.5 m/s

Hence, their mutual velocity after the collision is 1.5 m/s and it is moving in the same direction as the halfback was moving initially. Hence, this is the required solution.

3 0
3 years ago
One scientist is using an electromagnetic wave to map the dust between stars. The electromagnetic wave has shorter wavelength th
m_a_m_a [10]
Infrared light

it is also found under the name IR lights. although it’s technically invisible, it can still be seen with machinery up to at least 1050 nm in experiments.
3 0
3 years ago
slader How much energy is required to move a 1040 kg object from the Earth's surface to an altitude four times the Earth's radiu
andrew-mc [135]

Answer:

ΔU = 5.21 × 10^(10) J

Explanation:

We are given;

Mass of object; m = 1040 kg

To solve this, we will use the formula for potential energy which is;

U = -GMm/r

But we are told we want to move the object from the Earth's surface to an altitude four times the Earth's radius.

Thus;

ΔU = -GMm((1/r_f) - (1/r_i))

Where;

M is mass of earth = 5.98 × 10^(24) kg

r_f is final radius

r_i is initial radius

G is gravitational constant = 6.67 × 10^(-11) N.m²/kg²

Since, it's moving to altitude four times the Earth's radius, it means that;

r_i = R_e

r_f = R_e + 4R_e = 5R_e

Where R_e is radius of earth = 6371 × 10³ m

Thus;

ΔU = -6.67 × 10^(-11) × 5.98 × 10^(24)

× 1040((1/(5 × 6371 × 10³)) - (1/(6371 × 10³))

ΔU = 5.21 × 10^(10) J

7 0
3 years ago
A student conducts an experiment to determine how the temperature of water affects the time for sugar to dissolve. In each trial
Sveta_85 [38]
B- the student changed too many variables
7 0
3 years ago
Moving a charged object against an electrical force causes it to
Alinara [238K]

Decelerate

<u>Explanation:</u>

If a charged object is moving against an electric force, the electric force would cause the charged object to decelerate. Rate of deceleration would depend on the amount of the charge the object posses and amount of the opposing electric force.

This could be understood by visualising a hypothetical situation where a charged object is moving against an electric force. Since the object is charged, it would exert a force in its direction of motion which would be opposed by the electric force, thus causing it to decelerate  

5 0
3 years ago
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