All categories

3 years ago

A car going 30 m/s to 70 m/s in 10 seconds calculate the acceleration of the car

Physics

2 answers:

Zina [86]3 years ago

7 0

70m/s-30m/s is eqal to 40m/s. Divide 40 by 10 and you get 4. This would be the accleration of the car.

Anastasy [175]3 years ago

6 0

4 should be the answer for the acceleration for the car. I hope this helps!

You might be interested in

A truck going 15 kmôŹ°€h has a head-on collision with a small car going 30 kmôŹ°€h. Which statement best describes the situation

zlopas [31]

1. e) None of the above is necessarily true.

2.d) Without knowing the mass of the boat and the sack, we cannot tell.

5 0

3 years ago

Imagine you have to design a safe, useful device that uses an electric discharge to make your life better. How would you describ

Norma-Jean [14]

Answer:

i would describe it as some thing that keeps ur belongings safe and can also make my life better and not so sad

Explanation:

4 0

3 years ago

Read 2 more answers

A 47.2 kg girl is standing on a 177 kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat,

Softa [21]

Answer:

v_g,i = 1.208 m/s

Explanation:

We are given;

Mass of girl; m_g = 47.2 kg

Mass of plank; m_p = 177 kg

Let the velocity of girl to ice be v_g,i

Let the velocity of plank to ice be v_p,i

Since the velocity of the girl is 1.53 m/s relative to the plank, then;

v_g,i + v_p,i = 1.53

From conservation of momentum;

m_g × v_g,i = m_p × v_p,i

Thus;

47.2(v_g,i) = 177(v_p,i)

Dividing both sides by 47.2 gives;

v_g,i = 3.75(v_p,i)

v_pi = (v_g,i)/3.75

Thus, from v_g,i + v_p,i = 1.53, we have;

v_g,i + ((v_g,i)/3.75) = 1.53

v_g,i(1 + 1/3.75) = 1.53

1.267v_g,i = 1.53

v_g,i = 1.53/1.267

v_g,i = 1.208 m/s

5 0

3 years ago

Unpolarized light with an average intensity of 845 W/m2 enters a polarizer with a vertical transmission axis. The transmitted li

RideAnS [48]

The concept to develop this problem is the Law of Malus. Which describes what happens with the light intensity once it passes through a polarized material.

Mathematically this can be expressed as

$I = I_0 cos^2\theta$

Where

I = New intensity after pass through the Polarizer

$I_0$ = Original intensity

$\theta$ = Indicates the angle between the axis of the analyzer and the polarization axis of the incident light.

When the light passes perpendicularly through the first polarizer, the light intensity is reduced by half which will cause the intensity to be $225W / m ^ 2$ at the output of the new polarizer, mathematically: