All categories

3 years ago

Until a train is a safe distance from the station, it must travel at 5 m/s. Once the train is on open track, it can speec

Physics

1 answer:

nadezda [96]3 years ago

4 0

Answer:

The acceleration of the train is 5 m/s².

Explanation:

Given:

let the initial velocity of a train = 5 m/s and

final velocity of a train = 45 m/s

time taken = 8 s

To find:

acceleration: ?

Solution:

We define acceleration as change in velocity per unit time that is the difference between the final velocity and initial velocity divided by time.

$Acceleration = \frac{\textrm{final velocity} - \textrm{initial velocity}}{time} \\$

On substituting the above values we get the required acceleration

$Acceleration = \frac{45 - 5}{8}\\ =\frac{40}{8}\\ =5\ m/s^{2}$

Therefore,the acceleration of the train is 5 m/s².

You might be interested in

A 15.0cm string is how many times shorter than the 30.0cm string?

Alex Ar [27]

Easy. answers B. Because 15.00x2=30

3 0

2 years ago

Read 2 more answers

A circular loop of radius 11.7 cm is placed in a uniform magnetic field. (a) If the field is directed perpendicular to the plane

Mama L [17]

Answer:

Magnetic field, B = 0.199 T

Explanation:

It is given that,

Radius of circular loop, r = 11.7 cm = 0.117 m

Magnetic flux through the loop, $\phi=8.6\times 10^{-3}\ T/m^2$

The magnetic flux linked through the loop is :

$\phi=B.A$

$\phi=BA\ cos\theta$

Here, $\theta=0$

$B=\dfrac{\phi}{A}$

$B=\dfrac{\phi}{\pi r^2}$

$B=\dfrac{8.6\times 10^{-3}}{\pi (0.117 )^2}$

B = 0.199 T

So, the strength of the magnetic field is 0.199 T. Hence, this is the required solution.

4 0

3 years ago

A4.00-kg box sits on a ramp that is inclined at 30°above the horizontal.The coefficient of kinetic friction between the box and

enot [183]

From the calculation, we can see that the force that moves the box forward is 5.7 N.

<h3>What is frictional force?</h3>

The term frictional force refers to the force that opposes motion. Given that;

ma = ukmgcosθ + F

Let F be the forward force.

Hence;

F = ma - ukmgcosθ

F = 4 * 3 - (0.32 * 4.00 * 9.8 * sin 30)

F = 5.7 N

Learn more about frictional force:brainly.com/question/14662717

#SPJ1

3 0

1 year ago

. A helicopter is 8.50 m above the ground and climbing at 2.5 m/s. It drops a package from rest (relative to the helicopter). At

Rudik [331]

Answer:

2.5 m/s

Explanation:

The velocity of the package relative to the ground = the velocity of the package relative to the helicopter + the velocity of the helicopter relative to the ground

v = 0 m/s + 2.5 m/s

v = 2.5 m/s

At the moment it is released, the package is rising at 2.5 m/s.

4 0

2 years ago

What is the acceleration of a 11 kg box sliding across the floor under the action of a net force 35N

Fofino [41]

Answer:

<h3>The answer is 3.18 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

where

f is the force

m is the mass

We have

$a = \frac{35}{11} \\ = 3.181818...$

We have the final answer as

Hope this helps you

8 0

3 years ago