All categories

3 years ago

A school bus moves at speed of 35 mi/hr for 20 miles. How long will it take the bus to get to school?

Physics

2 answers:

Irina18 [472]3 years ago

7 0

The time required for the bus to get to school (in hours) is

<em>(the distance from here to school) divided by (35 mi/hr) </em> .

In order to find the actual number, we would need to know the distance from here to school.

mafiozo [28]3 years ago

4 0

Answer:

Time, t = 0.57 hours

Explanation:

It is given that,

Speed of the school bus, v = 35 mi/hr

Distance covered by the bus, d = 20 miles

We need to find the time taken by the bus to get to school. Time taken by the bus is given by :

$t=\dfrac{d}{v}$

$t=\dfrac{20\ mi}{35\ mi/hr}$

t = 0.57 hours

So, the time taken by the bus to reach school is 0.57 hours. Hence, this is the required solution.

You might be interested in

How are calculation for velocity and speed different

BARSIC [14]

Only velocity uses direction of travel in its calculations.

5 0

3 years ago

Match the term to the correct description.

Sedaia [141]

1. Proton
2. negative ion
3. electric charge
4. electron
5. repel
6. attract
7. positive ion

7 0

3 years ago

Read 2 more answers

Give an example of a reputable website

Leni [432]

Well, almost any website can be, but if you are meaning for money, then google. if you are meaning reliable, then look for something with .org, its history and stuff

6 0

3 years ago

Read 2 more answers

In a large centrifuge used for training pilots and astronauts, a small chamber is fixed at the end of a rigid arm that rotates i

RSB [31]

a) The length of the arm of the centrifuge is 10.9 m

b) The angular acceleration is $2.7 rad/s^2$

Explanation:

In a uniform circular motion, the centripetal acceleration is given by

$a_c=\omega^2 r$

where:

$\omega$ is the angular speed of the circular motion

r is the radius of the circle

For the centrifuge in this problem, we have:

$\omega=1.7 rad/s$ is the angular speed

The centripetal acceleration is 3.2 times the acceleration due to gravity ( $g=9.8 m/s^2$ ), so:

$a_c=3.2 g = 3.2(9.8)=31.4 m/s^2$

Therefore, we can re-arrange the previous equation to find r, the radius of the circle (which corresponds to the length of the arm of the centrifuge):

$r=\frac{a_c}{\omega^2}=\frac{31.4}{1.7^2}=10.9 m$

In the second part of the exercise, the centrifuge speeds up from an initial angular speed of 0 to a final angular speed of 1.7 rad/s. The total acceleration experienced at the final moment is

$a=4.4 g$