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sveta [45]
3 years ago
10

Which describes a train’s acceleration if it changes speed from 25 m/s to 10 m/s in 240 s? positive acceleration no acceleration

negative acceleration
Physics
1 answer:
OLEGan [10]3 years ago
3 0

To understand acceleration, you need to know speed and velocity.

Why?

Acceleration is the change in velocity, and the velocity equation is:

velocity = distance / time     Units: (m/s)

If a change in position means an object has a velocity, then that means that a change in velocity has an acceleration (I recommend you look at the graphs, it will help a lot to understand the relationship between each).

Now that we have some background, lets use it to answer the problem!

The change in velocity is from 25 m/s to 10 m/s, and we have a time. Lets plug it into the (average) acceleration equation so we can visualize it better!

The symbol delta means "change" : Δ

a = Δv / Δt

expanded to:

a = vf - vi / tf - ti  

(vf = velocity final, vi = velocity initial, tf = time final, ti = time initial)

So if the train's speed went from 25 m/s to 10 m/s, then that means that 10 m/s is our final velocity and 25 m/s is the initial:

a = 10 - 25 / 240

a = -15 / 240

a = -0.0625 m/s^2

Since we ended up with a negative number, it means that the train has a negative acceleration!!!!!

The fast way to do this:

25 m/s is bigger than 10 m/s, so that means that to go from 25 m/s to 10 m/s you have to slow down--otherwise known as putting on your brakes. So if the change in speed is negative then the acceleration will be negative.

In other words, if your final speed is less than your initial speed then your acceleration would be negative.

Sorry for the long explanation! I hope you never struggle with this again though!

Hope it helped,

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A satellite is in orbit 36000km above the surface of the earth. Its angular velocity is 7.27*10^-5 rad/s. What is the velocity o
bagirrra123 [75]

Answer:

v = 3.08 km/s

Explanation:

Given that,

The angular velocity of the satellite = \omega=7.27\times 10^{-5} rad/s

A satellite is in orbit 36000km above the surface of the earth.

The radius of the earth is 6400 km

Let v is the velocity of the satellite. It can be calculated as :

v=r\omega\\\\v=(36000\times 10^3+6400\times 10^3)\times 7.27\times 10^{-5}\\\\v=3082.48\ m/s\\\\v=3.08\ km/s

So, the velocity of the satellite is 3.08 km/s.

6 0
3 years ago
7.
jeka94

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5 0
3 years ago
A 5,000 kg satellite is orbiting the Earth in a circular path. The height of the satellite above the surface of the Earth is 800
Arada [10]

Explanation:

The given data is as follows.

            m = 5000 kg,            h = 800 km = 0.8 \times 10^{6} m

    R_{e} = 6.37 \times 10^{6} m,    r = R + h = 7.17 \times 10^{6} m

   M_{e} = 5.98 \times 10^{24} kg,   G = 6.67 \times 10^{-11} Nm^{2}/kg^{2}

As we know that,

              \frac{mv^{2}}{r} = \frac{GmM_{e}}{r^{2}}

                      v = \sqrt{\frac{GM_{e}}{r^{2}}}

And, it is known that formula to calculate angular velocity is as follows.

               \omega = \frac{v}{r} = \sqrt{\frac{GM_{e}}{r^{3}}}

                            v = \sqrt{\frac{GM_{e}}{r^{3}}}

                               = \sqrt{\frac{6.67 \times 10^{-11} Nm^{2}/kg^{2} \times 5.98 \times 10^{-24} kg^{-2}}{(7.17 \times 10^{6} m)^{3}}}

                                = 1.0402 \times 10^{-3} rad/s

Thus, we can conclude that speed of the satellite is 1.0402 \times 10^{-3} rad/s.

6 0
3 years ago
The initial kinetic energy imparted to a 0.25 kg bullet is 1066 J. The acceleration of gravity is 9.81 m/s 2 . Neglecting air re
lubasha [3.4K]

Answer:

The range of the bullet is 0.435 kilometers.

Explanation:

According to the problem, maximum height is equal to the range of the bullet. That is:

\Delta x = \Delta y

Where:

\Delta x - Range of the bullet, measured in meters.

\Delta y - Maximum height of the bullet, measured in meters.

By the Principle of Energy Conservation, gravitational potential energy reaches its maximum at the expense of the initial kinetic energy. That is to say:

K_{1} = U_{2}

Where:

K_{1} - Kinetic energy at point 1, measured in joules.

U_{1} - Gravitational potential energy at point 2, measured in joules, and:

U_{2} = m\cdot g \cdot \Delta y

Where:

m - Mass of the bullet, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

The maximum height is now cleared:

K_{1} = m\cdot g \cdot \Delta y

\Delta y = \frac{K_{1}}{m\cdot g}

If K_{1} = 1066\,J, m = 0.25\,kg and g = 9.81\,\frac{m}{s^{2}}, the maximum height is now computed:

\Delta y = \frac{1066\,J}{(0.25\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

\Delta y = 434.791\,m

\Delta y = 0.435\,km

Lastly, the range of the bullet is 0.435 kilometers.

3 0
3 years ago
Pls help me quickly ......​
marishachu [46]

Answer:

the last one: weight force

3 0
3 years ago
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