Ask question

Ask question

All categories

2 years ago

9

Kara Less was applying her makeup when she drove into South's busy parking lot last Friday morning. Unaware that Lisa Ford was s

topped in her lane 30 feet ahead, Kara rear-ended Lisa's rented Taurus. Kara's 1300-kg car was moving at 11 m/s and stopped in 0.14 seconds.a. Determine the momentum change of Kara's car.b. Determine the impulse experienced by Kara's car.c. Determine the magnitude of the force experienced by Kara's car.

2 answers:

SOVA2 [1]2 years ago

6 0

Answer:

Explanation:

Given that,

Mass of kara car m =1300kg

Velocity at which kara car was moving Vi =11m/s

The car stopped after t= 0.14sec

Therefore the final velocity is Vf = 0m/s

a. What is the change in momentum?

Change is momentum can be determine using

∆p = ∆MV

∆p = M∆V

∆p = m(Vf-Vi)

∆p = 1300(0-11)

∆p = 1300×-11

∆p = —14,300 kgm/s

The change in momentum of kara's car is —14,300kgm/s.

b. Impulse felt be kara car?

Impulse can be determine by using Newton second law of motion

Ft = ∆p

Impulse is Ft

I = ∆p = -14,300kgm/s.

Then, the impulse felt by Kara's car is -14,300kgm/s

c. Magnitude of force experienced by Kara's car?

From the impulse formula,

Ft = ∆p

Therefore,

F = ∆p/t

F = -14,300/0.14

F = -102,142.9N.

F ≈ —102,143N

The force experienced by kara's is -102,143N.

We can also determined the deceleration of Kara's car

From Newton second law

F=ma

Then, a = F/m

a = -102,143/1300

a = -78.57m/s²

The negative sign show deceleration

We can also calculate the distance the car moved before coming to halt.

Using equation of motion

X =ut+½at²

X = 11×0.14 + ½(-78.57) × 0.14²

X = 1.54 —0.77

X = 0.77m

NikAS [45]2 years ago

3 0

Answer

given,

Mass of Kara's car = 1300 Kg

moving with speed = 11 m/s

time taken to stop = 0.14 s

final velocity = 0 m/s

distance between Lisa ford and Kara's car = 30 m

a) change in momentum of Kara's car

Δ P = m Δ v

$\Delta P = m (v_f-v_i)$

$\Delta P = 1300 (0 - 11)$

Δ P = - 1.43 x 10⁴ kg.m/s

b) impulse is equal to change in momentum of the car

I = - 1.43 x 10⁴ kg.m/s

c) magnitude of force experienced by Kara

I = F x t

I is impulse acting on the car

t is time

- 1.43 x 10⁴= F x 0.14

F = -1.021 x 10⁵ N

negative sign represents the direction of force

You might be interested in

Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin.

Sergio039 [100]

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge $q_{1}= 3.8\times10^{-6}\ C$

Second charge $q_{2}=3.2\times10^{-6}\ C$

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

$E_{1}=E_{2}$

$\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}$

$\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}$

$\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}$

$x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}$

Put the value into the formula

$x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}$

$x=1.69\ m$

Hence, The distance is 1.69 m.

5 0

2 years ago

Is the magnetic field inside a toroid uniform?

victus00 [196]

If I am not wrong i thinks it is in the toroid uniforms

4 0

2 years ago

a school bus has stopped to allow children to get off the bus which graph shows the motion of the bus?

Usimov [2.4K]

Answer: what is it in

Explanation:

4 0

2 years ago

A 50-gram sample of water is initially at a temperature of 22 °C. The sample is heated until the temperature is 32 °C The specif

forsale [732]

Answer:

500cal

Explanation:

Given parameters:

Mass of water = 50g

Initial temperature = 22°C

Final temperature = 32°C

Specific heat of water = 1cal/g

Unknown:

Amount of heat absorbed by the water in calories = ?

Solution:

To solve this problem, we use the expression below:

H = m c Ф

H is the amount of heat absorbed

m is the mass

c is the specific heat capacity

Ф is the temperature change

H = 50 x 1 x (32 - 22) = 500cal

5 0

2 years ago

What is the static friction force??

Paul [167]

55 Kg has a weight of 55x9.8= 539 N
That is equal to the Normal force.
The static friction = 0.19 x 539 = 102.4 N

6 0

2 years ago