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Misha Larkins [42]
1 year ago
11

15) A 328-kg car moving at 19.1 m/s in the +x direction hits from behind a second car moving at 13 m/s in the same direction. If

the second car has a mass of 790 kg and a speed of 15.1 m/s right after the collision, what is the velocity of the first car after this sudden collision
Physics
1 answer:
kolbaska11 [484]1 year ago
5 0

Answer:

Explanation:

Given that,

Mass of first car

M1= 328kg

The car is moving in positive direction of x axis with velocity

U1 = 19.1m/s

Velocity of second car

U2 = 13m/s, in the same direction as the first car..

Mass of second car

M2 = 790kg

Velocity of second car after collision

V2 = 15.1 m/s

Velocity of first car after collision

V1 =?

This is an elastic collision,

And using the conservation of momentum principle

Momentum before collision is equal to momentum after collision

P(before) = P(after)

M1•U1 + M2•U2 = M1•V1 + M2•V2

328 × 19.1 + 790 × 13 = 328 × V1 + 790 × 15.1

16534.8 = 328•V1 + 11929

328•V1 = 16534.8—11929

328•V1 = 4605.8

V1 = 4605.8/328

V1 = 14.04 m/s

The velocity of the first car after collision is 14.04 m/s

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Inna Hurry is traveling at 6.8 m/s, when she realizes she is late for an appointment. She accelerates at 4.5 m/s^2 for 3.2 s. Wh
Alborosie

Answer:

1) v = 21.2 m/s

2) S = 63.33 m

3) s = 61.257 m

4) Deceleration, a = -4.32 m/s²

Explanation:

1) Given,

The initial velocity of Inna, u = 6.8 m/s

The acceleration of Inna, a = 4.5 m/s²

The time of travel, t = 3.2 s

Using the first equation of motion, the final velocity is

                v = u + at

                   = 6.8 + 4.5 x 3.2

                   = 21.2 m/s

The final velocity of Inna is, v = 21.2 m/s

2) Given,

The initial velocity of Lisa, u = 12 m/s

The final velocity of Lisa, v = 26 m/s

The acceleration of Lisa, a = 4.2 m/s²

Using the III equations of motion, the displacement is

                          v² = u² +2aS

                         S = (v² - u²) / 2a

                            = (26² -12²) / 2 x 4.2

                            = 63.33 m

The distance Lisa traveled, S = 63.33 m

3) Given,

The initial velocity of Ed, u = 38.2 m/s

The deceleration of Ed, d = - 8.6 m/s²

The time of travel, t = 2.1 s

Using the II equations of motion, the displacement is

                        s = ut + 1/2 at²

                           =38.2 x 2.1 + 0.5 x(-8.6) x 2.1²

                           = 61.257 m

Therefore, the distance traveled by Ed, s = 61.257 m

4) Given,

The initial velocity of the car, u = 24.2 m/s

The final velocity of the car, v = 11.9 m/s

The time taken by the car is, t = 2.85 s

Using the first equations of motion,

                         v = u + at

∴                        a = (v - u) / t

                            = (11.9 - 24.2) / 2.85

                            = -4.32 m/s²

Hence, the deceleration of the car, a = = -4.32 m/s²

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tensa zangetsu [6.8K]

Answer:

Smaller refractive power

Explanation:

The refractive power of an eye is the extent to which it can converge or diverge the light rays.

Near point is the the closest point for an eye such that when an object is placed at that point the image it forms is sharp and clearly visible to the eye.

A the person ages, the ciliary muscles of the eyes weakens and as a result the lens contracts and the formation of the image takes place behind the retina instead of forming at the retina.

Thus the near point also increases and the refractive power becomes smaller.

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In a car crash, large accelerations of the head can lead to severe injuries or even death. A driver can probably survive an acce
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Answer:

14.7 m/s

Explanation:

a = acceleration experienced by driver's head = 50 g = 50 x 9.8 m/s² = 490 m/s²

v₀ = initial speed of the driver = 0 m/s

v = final speed of the driver after 30 ms

t = time interval for which the acceleration is experienced = 30 ms = 0.030 s

Using the equation

v = v₀ + a t

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v = 0 + (490) (0.030)

v = 14.7 m/s

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You obtain a spectrum of an object in space. The spectrum consists of a number of sharp, bright emission lines. Is this object a
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Answer:

In hot gases , the atoms keeps colliding with each other and sometimes the energy liberated during collision takes the electron to a higher level,thus, .The object is a cloud of hot gas and finally the electron returns back emitting photon

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