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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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Spacetime interval: What is the interval between two events if in some given inertial reference frame the events are separated b
shepuryov [24]

Answer:

  • a. \Delta s ^2 = 8.0888 \ 10^{17} m^2
  • b. \Delta s ^2 = 3.0234 \ 10^{16} m^2
  • c. \Delta s ^2 = 3.0234 \ 10^{20} m^2

Explanation:

The spacetime interval \Delta s^2 is given by

\Delta s ^2 = \Delta (c t) ^ 2 - \Delta \vec{x}^2

please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:

\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2.

Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.

<h3>a.</h3>

\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2

\Delta \vec{x}^2 = 5,625 m^2

\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2

\Delta (c t) ^ 2 = (899,377,374 \ m)^2

\Delta (c t) ^ 2 = 8.0888 \ 10^{17} m^2

so

\Delta s ^2 = 8.0888 \ 10^{17} m^2 - 5,625 m^2

\Delta s ^2 = 8.0888 \ 10^{17} m^2

<h3>b.</h3>

\Delta \vec{x}^2 = (5 \ 10 \ m)^2

\Delta \vec{x}^2 = 2,500 m^2

\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2

\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2

\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2

so

\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2

\Delta s ^2 = 3.0234 \ 10^{16} m^2

<h3>c.</h3>

\Delta \vec{x}^2 = (5 \ 10 \ m)^2

\Delta \vec{x}^2 = 2,500 m^2

\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2

\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2

\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2

so

\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2

\Delta s ^2 = 3.0234 \ 10^{20} m^2

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Which of the following is not a part of dalton s atomic theory?
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<span>c. atoms are always in motion..............</span>
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A baseball pitcher throws a ball at 90.0 mi/h in the horizontal direction. How far does the ball fall vertically by the time it
Lisa [10]

Answer:

Vertical distance=  3.3803ft

Explanation:

First with the speed of the ball and the distance traveled horizontally we can determine the flight time to reach the plate:

Velocity= (90 mi/h) × (1 mile/5280ft) = 475200ft/h

Distance= Velocity × time⇒ time= 60.5ft / (475200ft/h) = 0.00012731h

time=  0.00012731h × (3600s/h)= 0.458316s

With this time we can determine the distance traveled vertically taking into account that its initial vertical velocity is zero and its acceleration is that of gravity, 9.81m/s²:

Vertical distance= (1/2) × 9.81 (m/s²) × (0.458316s)²=1.0303m

Vertical distance= 1.0303m × (1ft/0.3048m) = 3.3803ft

This is the vertical distance traveled by the ball from the time it is thrown by the pitcher until it reaches the plate, regardless of air resistance.

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A roller coaster travels around a vertical 8-m radius loop. Determine the speed at the top of the loop if the normal force exert
strojnjashka [21]

Answer:

v=10m/sec

Explanation:

From the question we are told that

Radius of vertical r= 8m

Force exerted by passengers is 1/4 of weight

Generally the net force acting on top of the roller coaster is give to be

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F_N =forceof the normal

Fg= force due to gravity

Generally the net force is given to be FC(force towards center)

F_C =F_N + Fg

F_N =F_C -Fg

F_N=F_C-F_g

F_N=\frac{mv^2}{R} -mg

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\frac{5mg}{4} =\frac{mv^2}{8}

v^2 =\frac{40*10}{4}

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Answer:

Heterogeneous

Explanation:

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