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Leno4ka [110]
3 years ago
12

A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 4.60 m/s

in 0.888 s. What are the magnitudes of the linear impulse and the average total force experienced by a 74.0 kg passenger in the car during the time the car accelerates
Physics
1 answer:
serg [7]3 years ago
3 0

Answer:

(a).The value of force applied on the car during acceleration F = 383.32 N

(b).The value of impulse on the car  I = 340.4 N-sec

Explanation:

Acceleration of the car  a = \frac{4.6}{0.888} = 5.18 \frac{m}{s^{2} }

Mass of the passengers = 74 kg

Total force experienced by the car = mass  × acceleration

⇒ F = m × a

Put the values of mass & acceleration in the above formula we get

Force F = 74 × 5.18

⇒ F = 383.32 N

This is the value of force applied on the car during acceleration.

linear Impulse is given by I = F × Δt

Where F = force applied on the car

Δt  = time interval upon which the force is applied on the car

⇒ I = 383.32 × 0.888

⇒ I = 340.4 N-sec

This is the value of impulse on the car.

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You observe two cars traveling in the same direction on a long, straight section of Highway 5. The red car is moving at a consta
Romashka-Z-Leto [24]

Answer:

a) 3.66 s

b) 124.4 m

c) 3.12s

Explanation:

Given that

Speed of the Red Car, v₁ = 34 m/s

Speed of the Blue Car, v₂ = 28 m/s

Distance between the two cars, d = 22 m

The difference between the speed of the cars is: 34 - 28 = 6 m/s

From this, we can deduce that the red car will be catching up to the blue car at a speed of 6 m/s.

1)

If we divide the distance by the difference in speed. This becomes

d/v = 22/6 = 3.66 s. Which means, It takes 3.66 seconds for the red car to meet up with the blue car.

2

From the previous part, we were able to confirm that it took 3.66 seconds for the red car to meet up the blue car. Also, the speed with which it were traveling was travelling at, was constant, so we only need to multiply it by the time from (1) to get the distance.

v = d * t = 34 * 3.66 = 124.4

Therefore the red car travels at 124.4 m before catching up to the blue car.

3

If the red car starts to accelerate the moment we see it, the time it will take, to get to the blue car will be less than what we had gotten. We can find this using one of the equations of motion.

S = ut + ½gt², where

S = 22

u = 6

t = ?

g = 2/3

22 = 6t + 1/3t²

By using the quadratic formula, we find out the two answers listed below

t1 = 3.12 s

t2 = - 21.12 s

We all know that negative time is not possible, so the answer is t1. At 3.12 seconds

8 0
3 years ago
Ask a member of the family to help you.Do the following activities and identify the skill/skills being excited . use a separate
Archy [21]

Answer:

Ok. Thanks.

I'll try it out.

3 0
2 years ago
A car starts from rest with an acceleration of 8 m/s/s for a time of 4 s. What was the car's final velocity?
Aleks [24]
32m/s
8m/s x 4s= 32m/s
6 0
2 years ago
The experimental apparatus shown in the figure above contains a pendulum consisting of a 0.66 kg ball attached to a string of le
lara31 [8.8K]

The problem is solved and the questions are answered below.

Explanation:

a. To calculate the speed of the 0.66 kg ball just before the collision

V₀ + K₀ = V₁ + K₁

= mgh₀ = 1/2 mv₁²

where, h= r - r cosθ

V = \sqrt{2gh}

 V = 2.42 m/s

b. Calculate the speed of the 0.22 kg ball immediately after the collision

y = y₀ + Vy₀t - 1/2 gt²

0 = 1.2 - 1/2 gt²

t = 0.495 s

x = x₀ + Vx₀t

1.4 = 0 + vx₀ (0.495)

Vx₀ = 2.83 m/s

C. To Calculate the speed of the 0.66 kg ball immediately after the collision

m₁ v₁ = m₁ v₃ + m₂ v₄

(0.66)(2.42) = (0.66) v₃ + (0.22)(2.83)

V₃ = 1.48 m/s

D. To Indicate the direction of motion of the 0.66 kg ball immediately after the collision is to the right.

E. To Calculate the height to which the 0.66 kg ball rises after the collision

V₀ + k₀ = V₁ + k₁

1/2 mv₀² = mgh₁

h₁ = v₀²/2 g

  = 0.112 m

F. Based on your data, No the collision is not elastic.

Δk = 1/2 m₁v₃² =1/2 m₂v₄² - 1/2 m₁v₁²

     = 1/2 (0.66)(1.48)² + 1/2 (0.22)(2.83)² - 1/2 (0.66)(2.42)²

    = - 0.329 J

Hence, kinetic energy is not conserved.

8 0
3 years ago
Two point sources, vibrating in phase, produce an interferencepattern in a ripple tank. If the frequency is increased by 20%, th
Vera_Pavlovna [14]

Answer:

option A

Explanation:

given,

frequency is increased by 20%

we know,

\dfrac{x_n}{L}d = (n-\dfrac{1}{2})\times \lambda...........(1)

where

x_n is the perpendicular distance between the point the interference pattern is obtained,

L is the distance between the center of the two point sources

and λ is  the wavelength of light.

If the frequency is increased by 20%, then the number of nodal lines is increased by 20%.

From equation (1),we observe that the frequency is directly proportional to the number of nodal lines.

Hence, the correct answer is option A

8 0
3 years ago
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