Ask question

Ask question

All categories

3 years ago

5

a car accelerates uniformly from rest to a speed of 51.9mi/h in 9.37s. find the constant acceleration of the car. answer in unit

s of m/s^2

1 answer:

Darina [25.2K]3 years ago

3 0

Here is my step-by-step-work. Let me know if you have any questions! :)

You might be interested in

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 2.0 rev/s in 10.0 s. At th

Arte-miy333 [17]

Answer:

22 revolutions

Explanation:

2 rev/s = 2*(2π rad/rev) = 12.57 rad/s

The angular acceleration when it starting

$\alpha_a = \frac{\Delta \omega}{\Delta t} = \frac{12.57}{10} = 1.257 rad/s^2$

The angular acceleration when it stopping:

$\alpha_o = \frac{\Delta \omega}{\Delta t} = \frac{-12.57}{12} = -1.05 rad/s^2$

The angular distance it covers when starting from rest:

$\omega^2 - 0^2 = 2\alpha_a\theta_a$

$\theta_a = \frac{\omega^2}{2\alpha_a} = \frac{12.57^2}{2*1.257} = 62.8 rad$

The angular distance it covers when coming to complete stop:

$0 - \omega^2 = 2\alpha_o\theta_o$

$\theta_o = \frac{-\omega^2}{2\alpha_o} = \frac{-12.57^2}{2*(-1.05)} = 75.4 rad$

So the total angular distance it covers within 22 s is 62.8 + 75.4 = 138.23 rad or 138.23 / (2π) = 22 revolutions

6 0

3 years ago

A researcher measures the thickness of a layer of benzene (n = 1.50) floating on water by shining monochromatic light onto the f

NNADVOKAT [17]

Answer:

Explanation:

This problem relates to interference of light in thin films .

The condition of bright fringe in thin films which is sandwitched by two layers of medium having lesser refractive index is as follows.

2nt = (2n+1) λ / 2 , n is refractive index of thin layer , t is its thickness , λ is wavelength of light .

2 x 1.5 t = λ / 2 , if n = 0 for minimum thickness.

2 x 1.5 t = 600 / 2 nm

t = 100 nm .

5 0

3 years ago

How much Tim in minutes will it take a car driving at 90km/hr to travel 27 kilometers

astraxan [27]

Answer:

18min

Explanation:

v=d/t

t=d/v= 27/90 =0.3hrs =18min

6 0

3 years ago

When you push a 1.90-kg book resting on a tabletop, you have to exert a force of 2.10 N to start the book sliding. Once it is sl

zimovet [89]

Static frictional force = ƒs = (Cs) • (Fɴ)
2.26 = (Cs) • m • g
2.26 = (Cs) • (1.85) • (9.8)
Cs = 0.125
kinetic frictional force = ƒκ = (Cκ) • (Fɴ)
1.49 = (Cκ) • m • g
1.49 = (Cκ) • (1.85) • (9.8)
Cκ = 0.0822

3 0

3 years ago

12. An organ pipe that is 1.75 m long and open at both ends produces sound of

podryga [215]

Answer:

354 m/s

Explanation:

For the second overtune (Third harmonic) of an open pipe,

λ = 2L/3................................ Equation 1

Where L = Length of the open pipe, λ = Wave length.

Given: L = 1.75 m.

Substitute into equation 1

λ = 2(1.75)/3

λ = 1.17 m.

From the question,

V = λf.......................... Equation 2

V = speed of sound in the room, f = frequency

Given: f = 303 Hz.

Substitute into equation 2

V = 1.17(303)

V = 353.5

V ≈ 354 m/s

Hence the right answer is 354 m/s

8 0

3 years ago