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vichka [17]
2 years ago
6

A school bus moves down a road, dropping off students after school. The bus slows down from a speed of 15 meters per second to a

full stop over a distance of 55 meters in 11 seconds. What is the average speed, in meters per second, of the school bus while the bus is slowing down? * 3.7 5.0 26 40
Physics
1 answer:
My name is Ann [436]2 years ago
5 0

Given :

A school bus moves down a road, dropping off students after school.

The bus slows down from a speed of 15 meters per second to a full stop over a distance of 55 meters in 11 seconds.

To Find :

The average speed, in meters per second, of the school bus while the bus is slowing down.

Solution :

Initial velocity, u = 15 m/s.

Distance travelled, d = 55 m.

Time taken, t = 11 s.

Final velocity, v = 0 m/s.

We know, average velocity is given by :

v_{avg}=\dfrac{distance}{time}\\\\v_{avg}= \dfrac{55}{11}\ m/s\\\\v_{avg}=5 \ m/s

Therefore, average velocity is 5 m/s.

Hence, this is the required solution.

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What is the net force of an object with a mass of 90.0 and accelerating at 3.0 m/s
Ksivusya [100]

The net force is 270 N

Explanation:

We can solve this problem by using Newton's second law, which states that the net force on an object is equal to the product between its mass and its acceleration:

F=ma

where

F is the force

m is the mass

a is the acceleration

In this problem, we have

m = 90.0 kg

a=3.0 m/s^2

Substituting, we find the net force on the object:

F=(90.0)(3.0)=270 N

Learn more about Newton's second law:

brainly.com/question/3820012

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3 0
3 years ago
A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s
Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g=4.67\times 10^{-3} kg

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1,m_1=1177g=1.177kg

Mass of block 2,m_2=1626 g=1.626 kg

Velocity of block 1,v_1=0.681m/s

(a)

Let velocity of the second block  after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

mv=m_1v_1+(m+m_2)v_2

4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2

1.66719=0.801537+1.63067v_2

1.66719-0.801537=1.63067v_2

0.865653=1.63067v_2

v_2=\frac{0.865653}{1.63067}

v_2=0.531m/s

Hence, the  velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

K_i=\frac{1}{2}mv^2

k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)

k_i=297.59 J

Final kinetic energy after collision

K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2

K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2

K_f=0.5028 J

Now, he ratio of the total kinetic energy after the collision to that before the collision

=\frac{k_f}{k_i}=\frac{0.5028}{297.59}

=0.00169

5 0
3 years ago
A ball is projected upward at time t=0.0s, from a point on a roof 90m above the ground. The ball rises, then falls and strikes t
chubhunter [2.5K]
As v becomes zero at the highest point, i prefer considering different travelling directions so it will become less complicated.
dont forget to add the total time up .

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5 What is the kinetic energy of a 28 kg car moving at a velocity of 7 m/s O 1963 98 686 J O 350 nocter diagram, at which point w
Shtirlitz [24]

Answer:

98 joules

Explanation:

1/2 of 28 = 14

14 multiply by 7= 98 joules

4 0
2 years ago
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