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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.

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3 years ago

What is meant by the statement,the linear expansivity of copper is 0.000017k

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Answer:

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3 years ago

How high would you have to lift a 1000kg car to give it a potential energy of:

Elza [17]

Given parameters:

Mass of the car = 1000kg

Unknown:

Height = ?

To find the heights for the different amount potential energy given, we need to understand what potential energy is.

Potential energy is the energy at rest due to the position of a body.

It is mathematically expressed as:

P.E = mgh

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g is the acceleration due to gravity = 9.8m/s²

h is the height of the car

Now the unknown is h, height and we make it the subject of the expression to make for easy calculation.

h = $\frac{P.E }{mg}$

For 2.0 x 10³ J;

h = $\frac{2000}{1000 x 9.8}$ = 0.204m

For 2.0 x 10⁵ J;

h = $\frac{200000}{9.8 x 1000}$ = 20.4m

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3 years ago

In an electric vehicle, each wheel is powered by its own motor. The vehicle weight is 4,000 lbs. By regenerative braking, its sp

Slav-nsk [51]

Answer:

the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

Explanation:

Given that;

weight of vehicle = 4000 lbs

we know that 1 kg = 2.20462

m = 4000 / 2.20462 = 1814.37 kg

Initial velocity $V_{i}$ = 60 mph = 26.8224 m/s

Final velocity $V_{f}$ = 30 mph = 13.4112 m/s

now we determine change in kinetic energy

Δk = $\frac{1}{2}$ m( $V_{i}$ ² - $V_{f}$ ² )

we substitute

Δk = $\frac{1}{2}$ ×1814.37( (26.8224)² - (13.4112)² )

Δk = $\frac{1}{2}$ × 1814.37 × 539.5808

Δk = 489500 Joules

we know that; 1 kilowatt hour = 3.6 × 10⁶ Joule

Δk = 489500 / 3.6 × 10⁶

Δk = 0.13597 ≈ 0.136 kWh

Therefore, the theoretical maximum energy in kWh that can be recovered during this interval is 0.136 kWh

4 0

3 years ago