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tester [92]
3 years ago
14

The car A has a weight of 4000 lb and is traveling to the right at 3 ft/s. Meanwhile a 2000 lb car B is traveling at 6 ft/s to t

he left. If the cars crash head-on, and at a time instant during the crash the impact force on A is 900 lb to the left, what is the magnitude and direction of the impact force exerted on B at the instant?
(A) 900 lb to the left
(B) 450 lb to the left
(C) 450 lb to the right
(D) 1800 lb to the left
(E) 900 lb to the right
Physics
1 answer:
iVinArrow [24]3 years ago
6 0

To solve this problem we should apply Newton's third law for which it is defined that there must be an equal reaction in the opposite direction.

From this law, if Car A generates a force on car B, that car must have opposed a force exactly the same but in the opposite direction. The car A moved to the left and generates a force of 900lb so the magnitude of the force in the car B is also 900lb but to the right (opposite direction to the first car)

The correct option is 900lb to the right.

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The velocities of light in air and glass are 3.0 x 10^8ms and 2.0×10^8ms respectively. If the angle of refraction is 30°, the si
JulijaS [17]

Answer:

0.75

Explanation:

refractive \: index \:  =  \frac{3.0 \times  {10}^{2} }{2.0 \times  {10}^{2} }

= 1.5

refractive \: index =  \frac{ \sin(angle \: of \: incidence) }{ \sin(angle \: of \: refraction) }

1.5 =  \ \frac{ \sin(i) }{ \sin(30) }

1.5 × ½ = sin(i)

\sin(i)  =  0.75

5 0
3 years ago
Under the influence of its drive force, a snowmobile is moving at a constant velocity along a horizontal patch of snow. When the
balandron [24]

Answer:

a) Δx = 11.6 m

b) t = 3.9 s

Explanation:

a)

  • Since the snowmobile is moving at constant speed, and the drive force is 195 N, this means that thereis another force equal and opposite acting on it, according to Newton's 2nd Law, due to there is no acceleration present in the horizontal direction .
  • This force is just the force of kinetic friction, and is equal to -195 N (assuming the positive direction as the direction of the movement).
  • Once the drive force is shut off, the only force acting on the snowmobile remains the friction force.
  • According Newton's 2nd Law, this force is causing a negative acceleration (actually slowing down the snowmobile) that can be found as follows:

       a = \frac{F_{fr} }{m} = \frac{-195N}{128kg} = -1.5 m/s2 (1)

  • Assuming the friction force keeps constant, we can use the following kinematic equation in order to find the distance traveled under this acceleration before coming to an stop, as follows:

       v_{f} ^{2}  -v_{o} ^{2} = 2* a* \Delta x (2)

  • Taking into account that vf=0, replacing by the given (v₀) and a from (1), we can solve for Δx, as follows:

       \Delta x =- \frac{v_{o}^{2}}{2*a} =- \frac{(5.90m/s)^{2}}{2*(-1.5m/s2)} = 11.6 m (3)

b)

  • We can find the time needed to come to an stop, applying the definition of acceleration, as follows:

       v_{f} = v_{o} + a*\Delta t (4)

  • Since we have already said that the snowmobile comes to an stop, this means that vf = 0.
  • Replacing a and v₀ as we did in (3), we can solve for Δt as follows:

       \Delta t = \frac{-v_{o} }{a} = \frac{-5.9m/s}{-1.5m/s2} = 3.9 s   (5)

7 0
3 years ago
An inflatable raft (unoccupied) floats down a river at an approximately constant speed of 5.6 m/s. A child on a bridge, 71 m abo
saveliy_v [14]

Answer:

21.28 m

Explanation:

height, h = 71 m

velocity of raft, v = 5.6 m/s

let the time taken by the stone to reach to raft is t.

use second equation of motion for stone

h = ut + \frac{1}{2}at^{2}

u = 0 m/s, h = 71 m, g = 9.8 m/s^2

71 = 0 + 0.5 x 9.8 x t^2

t = 3.8 s

Horizontal distance traveled by the raft in time t

d = v x t = 5.6 x 3.8 = 21.28 m

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