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2 years ago

13

A.

1 answer:

UNO [17]2 years ago

7 0

Calculate the minimum speed record at the point B in order for the real question to reach the top

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A motorcycle running on gasoline wastes a large amount of energy mainly as A) heat energy and sound energy. B) light energy and

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A motorcycle mainly wastes energy as heat <u>energy</u> and <u>sound</u> energy. In the engine, chemical energy is transformed into mechanical energy. However, the engine is inefficient and much of the chemical energy is lost as heat energy. Also, some of the energy is transformed to sound energy. This explains why the motorcycle is noisy and has an exhaust pipe.

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2 years ago

Du een Shem.<br>when we shake mango free, the mangoes fall<br>down, why?

Vanyuwa [196]

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2 years ago

Everything is perished.........(is it,isn't it, aren't they, are they

andrey2020 [161]

Answer:

everything is perished isn't it?

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2 years ago

Read 2 more answers

Please help me find this answer

djyliett [7]

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5 0

1 year ago

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Two carts, one twice as heavy as the other, are at rest on a horizontal track. A person pushes each cart for 8 s. Ignoring frict

GalinKa [24]

Answer:

The correct option is: B that is 1/2 K

Explanation:

Given:

Two carts of different masses, same force were applied for same duration of time.

Mass of the lighter cart = $m$

Mass of the heavier cart = $2m$

We have to find the relationship between their kinetic energy:

Let the KE of cart having mass m be "K".

and KE of cart having mass m be "K1".

As it is given regarding Force and time so we have to bring in picture the concept of momentum Δp and find a relation with KE.

Numerical analysis.

⇒ $KE = \frac{mv^2}{2}$

⇒ $KE = \frac{mv^2}{2}\times \frac{m}{m}$

⇒ $KE = \frac{m^2v^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(mv)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2}\times \frac{1}{m}$

⇒ $KE = \frac{(\triangle p)^2}{2m}=\frac{(F\times t)^2}{2m}$

Now,

Kinetic energies and their ratios in terms of momentum or impulse.

KE (K) of mass m.

⇒ $K=\frac{(F\times t)^2}{2m}$ ...equation (i)

KE (K1) of mass 2m.

⇒ $K_1=\frac{(F\times t)^2}{2\times 2m}$

⇒ $K_1=\frac{(F\times t)^2}{4m}$ ...equation (ii)

Lets divide K1 and K to find the relationship between the two carts's KE.

⇒ $\frac{K_1}{K} =\frac{(F\times t)^2}{4m} \times \frac{2m}{(F\times t)^2}$

⇒ $\frac{K_1}{K} =\frac{2m}{4m}$

⇒ $\frac{K_1}{K} =\frac{2}{4}$

⇒ $\frac{K_1}{K} =\frac{1}{2}$

⇒ $K_1=\frac{K}{2}$

⇒ $K_1=\frac{1}{2}K$

The kinetic energy of the heavy cart after the push compared to the kinetic energy of the light cart is 1/2 K.

7 0

2 years ago