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rewona [7]
2 years ago
12

A car of mass M traveling with velocity v strikes a car of mass M that is at rest. The two cars’ bodies mesh in the collision. T

he loss of the kinetic energy the moving car undergo in the collision is
a) a quarter of the initial kinetic energy.
b) half of the initial kinetic energy.
c) all the initial kinetic energy.
d) zero.
Physics
1 answer:
krok68 [10]2 years ago
4 0

Answer:

the correct answer is B

Explanation:

Let's propose the solution of the problem, for this we form a system formed by the two cars, so that the forces during the collision are internal, the momentum is conserved

instantly starts. Before the crash

         p₀ = M v +0

final instant. After the crash

        m_f = (M + M) v_f

the moment is preserved

        p₀ = p_f

        M v = 2 M v_f

        v_f = v / 2

let's look for kinetic energy

before the crash

       K₀ = ½ M v²

after the crash

       K_f = ½ 2M (v_f)²

       K_f = ½ 2M (v/2)²

       K_f = (½ M v²) ½

       K_f = K₀ / 2

therefore the correct answer is B

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Suppose you monitor a large number (many thousands) of stars over a period of 3 years, searching for planets through the transit
nasty-shy [4]

Answer:

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Explanation:

The transit method is a very important and effective tool for discovering new exoplanets (the planets orbiting other stars out of the solar system). In this method the stars are observed for a long duration. When the exoplanet will cross in front of theses stars as seen from Earth, the brightness of the star will dip. To observe this dip following conditions must be met:

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2. The brightness of the star must be observed precisely as the period of transit can be less than a second as seen from Earth. Also the dip in brightness depends on the size of the planet. If the planet is not that big the intensity dip will be very less.

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4 0
3 years ago
Carbon dioxide in a piston-‐‐cylinder is expanded in a polytropic manner. The initialtemperature and pressure are 400 K and 550
Arisa [49]

Answer:

 q_poly = 14.55 KJ/kg

Explanation:

Given:

Initial State:

P_i = 550 KPa

T_i = 400 K

Final State:

T_f = 350 K

Constants:

R = 0.189 KJ/kgK

k = 1.289 = c_p / c_v

n = 1.2   (poly-tropic index)

Find:

Determine the heat transfer per kg in the process.

Solution:

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                            q_poly = w_poly*(k - n)/(k-1)

- Evaluate w_poly:

                            w_poly = R*(T_f - T_i)/(1-n)

                            w_poly = 0.189*(350 - 400)/(1-1.2)

                            w_poly = 47.25 KJ/kg

-Hence,

                           q_poly = 47.25*(1.289 - 1.2)/(1.289-1)

                           q_poly = 14.55 KJ/kg

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vladimir2022 [97]
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6 0
3 years ago
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