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Ann [662]
2 years ago
9

in an automobile crash, a vehicle that was stopped at a red light is rear-ended by another vehicle. The vehicles have the same m

ass. If the tire marks show that the two vehicles moved after the collision at 4 m/s, what was the speed of the vehicle before the collision
Physics
1 answer:
baherus [9]2 years ago
6 0

The initial speed of the vehicle before the collision is 8 m/s.

  • Let the mass of the vehicle = m
  • Let the initial speed of the vehicle stopped = u
  • The initial speed of the vehicle parked at the red light = 0

<h3>Principle of conservation of linear momentum</h3>
  • The initial speed of the vehicle before the collision is calculated by applying principle of conservation of linear momentum as follows;

m_1 u_1 + m_2 u_2 = v(m_1 + m_2)\\\\mu + m(0) = 4(m+ m)\\\\mu = 4(2m)\\\\mu = 8m\\\\u = 8 \ m/s

Thus, the initial speed of the vehicle before the collision is 8 m/s.

Learn more about inelastic collision here: brainly.com/question/7694106

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Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 1.6×104 m/s when at a distance
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Answer:

v₂ = 7.6 x 10⁴ m/s

Explanation:

given,

speed of comet(v₁) = 1.6 x 10⁴ m/s

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\dfrac{1}{2}mv_1^2-\dfrac{GMm}{r_1} = \dfrac{1}{2}mv_2^2-\dfrac{GMm}{r_2}

\dfrac{1}{2}v_1^2-\dfrac{GM}{r_1} = \dfrac{1}{2}v_2^2-\dfrac{GM}{r_2}

v_2^2= v_1^2 + \dfrac{2GM}{r_2}-\dfrac{2GM}{r_1}

v_2= \sqrt{v_1^2 +2GM(\dfrac{1}{r_2}-\dfrac{1}{r_1})}

v_2= \sqrt{(1.6\times 10^4)^2 +2\times 6.67 \times 10^{-11}\times 1.99 \times 10^{30}(\dfrac{1}{4.8\times 10^{10}}-\dfrac{1}{2.7\times 10^{11}})}

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3 0
3 years ago
If a roller coaster train has a potential energy of 1,500 J and a kinetic energy of 500 J as it starts to travel downhill, its t
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4 0
3 years ago
Read 2 more answers
The filament of a certain lamp has a resistance that increases linearly with temperature. When a constant voltage is switched on
alekssr [168]

Answer:

The change in temperature is \Delta T  = 1795 K

Explanation:

From the question we  are told that

   The temperature coefficient is  \alpha  =  4 * 10^{-3 }\  k^{-1 }

The resistance of the filament is mathematically represented as

           R  =  R_o [1 + \alpha  \Delta T]

Where R_o is the initial resistance

Making the change in temperature the subject of the formula

     \Delta T = \frac{1}{\alpha } [\frac{R}{R_o} - 1 ]

Now from ohm law

           I = \frac{V}{R}

This implies that current varies inversely with current so

           \frac{R}{R_o} = \frac{I_o}{I}

Substituting this we have

       \Delta T  = \frac{1}{\alpha } [\frac{I_o}{I} - 1 ]

From the question we are told that

    I  = \frac{I_o}{8}

Substituting this we have

   \Delta T  = \frac{1}{\alpha } [\frac{I_o}{\frac{I_o}{8} } - 1 ]

=>     \Delta T  = \frac{1}{3.9 * 10^{-3}} (8 -1 )

        \Delta T  = 1795 K

6 0
3 years ago
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