All categories

3 years ago

How much will the kinetic energy of a school bus increase if its velocity is tripled?

Physics

1 answer:

son4ous [18]3 years ago

4 0

Answer:

Explanation:

Given a school bus.

Let say initially the school bus is traveling with speed "v"

Let assume mass of school bus is "m"

Then, the initial kinetic energy is

K.E_initial = ½mv²

Now, if the initial velocity is tripled,

Then, the new velocity is

v_new = 3v.

Note: the mass of the school does not change it is constant

Then, new kinetic energy is

K.E_new = ½m(v_new)²

v_new = 3v

Then,

K.E_new = ½m(3v)²

K.E_new = ½m × 9v²

K.E_new = 9 × ½mv²

Since K.E = ½mv²

Then,

K.E_new = 9 × K.E

So, the new kinetic energy will be 9 times the initial kinetic energy.

So, option D is correct

D. It will be nine times greater.

You might be interested in

What is a force which resists all motion

yaroslaw [1]

The answer is friction

4 0

3 years ago

Read 2 more answers

You push a bike pedal 10cm and the bike moves 100cm. Calculate the Speed ratio ( show all work ) ( use the formula and divide th

scZoUnD [109]

Answer:

when you are pushing the pedal you are causing the pedal to move done and then you will move 100cm

Explanation:

10 cm= 100 cm moved so when you move you will move because you are timeing the 10 by 100 to get the spped

3 0

2 years ago

Problem 4: A uniform flat disk of radius R and mass 2M is pivoted at point P A point mass of 1/2 M is attached to the edge of th

brilliants [131]

From the case we know that:

The moment of inertia Icm of the uniform flat disk witout the point mass is Icm = MR².
The moment of inerta with respect to point P on the disk without the point mass is Ip = 3MR².
The total moment of inertia (of the disk with the point mass with respect to point P) is I total = 5MR².

Please refer to the image below.

We know from the case, that:

m = 2M

r = R

m2 = 1/2M

distance between the center of mass to point P = p = R

Distance of the point mass to point P = d = 2R

We know that the moment of inertia for an uniform flat disk is 1/2mr². Then the moment of inertia for the uniform flat disk is:

Icm = 1/2mr²

Icm = 1/2(2M)(R²)

Icm = MR² ... (i)

Next, we will find the moment of inertia of the disk with respect to point P. We know that point P is positioned at the arc of the disk. Hence:

Ip = Icm + mp²

Ip = MR² + (2M)R²

Ip = 3MR² ... (ii)

Then, the total moment of inertia of the disk with the point mass is:

I total = Ip + I mass

I total = 3MR² + (1/2M)(2R)²

I total = 3MR² + 2MR²

I total = 5MR² ... (iii)

Learn more about Uniform Flat Disk here: brainly.com/question/14595971

#SPJ4

8 0

1 year ago

What happens to the water particles in a wave?

lys-0071 [83]

Answer: waves transport energy, not water. As a wave crest passes, the water particles move in circular paths. The movement of the floating inner tube is simulacra to the movement of the water particles. Water particles rise as a wave crest approaches.

Explanation:

8 0

3 years ago

Read 2 more answers

block of mass 5kgriding on a horizontal frictionlessxy-plane surface is subjected tothree applied forces:→F1= 12√2N[ 45◦]→F2= (8

dsp73

Answer:

(i) See attached image for the drawing

(ii) net force given in component form: (20, 20)N with magnitude: $\sqrt{800} \,\,\,N$

Explanation:

First try to write all forces in vector component form:

The force F1 acting at 45 degrees would have multiplication factors of $\frac{\sqrt{2} }{2}$ on both axes, to take care of the sine and cosine projections. Therefore, the:

x-component of F1 is $F1_x=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N$

y-component of F1 is $F1_y=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N$

As far as force F2, it is given already in x and y components, then:

x-component of F2 = 8 N

y-component of F2 = -6 N (negative meaning pointing down the y-axis)

Force F3 has only component (upwards) in the y-direction

x-component of F3 = 0 N

y-component of F3 =14 N

The additions of all these component by component, gives the resultant force (R) acting on the 5 kg mass:

x-component of R = 12 + 8 = 20 N

y-component of R = 12 + 14 - 6 = 20 N

Therefore, the acceleration that the mass receives due to this force is given in component form as:

x-component of acceleration: 20 N / 5 kg = $4\,\,\,m/s^2$

y-component of acceleration: 20 N / 5 kg = $4\,\,\,m/s^2$

Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:

x-component of the velocity is: $v_x=4\,*\,2=8\,m/s$

y-component of the velocity is: $v_y=4\,*\,2=8\,m/s$

7 0

3 years ago