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4 years ago

what is the acceleration of a car that slows down from 60m/s to 40 m/s during a period of 10 seconds?

Physics

1 answer:

algol [13]4 years ago

5 0

The acceleration of the car is stated as -2 m/s², while the deceleration of the car is termed as 2 m/s².

Answer:

Explanation:

Acceleration exerted by any object in order of decrease in the speed is termed as deceleration. In other words , negative acceleration or acceleration acting in opposite direction to stop the motion of any object is termed as deceleration. The magnitude of acceleration and deceleration is same only the direction is completely opposite to each other. So the negative sign will be obtained for acceleration value of any object whose velocity slows down from initial velocity.

Here , the initial velocity was 60 m/s and the final velocity is slowed down to 40 m/s in time 10 s.

So acceleration = (Final velocity - Initial velocty)/Time

Acceleration = (40-60)/10 = -20/10 = -2 m/s².

Thus, the acceleration of the car is stated as -2 m/s², while the deceleration of the car is termed as 2 m/s².

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A system of releases 125kJ of heat while 104kJ of work is done in the system. Calcilate the change om imternal energy (in kJ)

artcher [175]

Answer:

DU = 21 KJ

Explanation:

Given the following data;

Quantity of heat = 125 KJ

Work = 104 KJ

To find the change in internal energy;

Mathematically, the change in internal energy of a system is given by the formula;

DU = Q - W

Where;

DU is the change in internal energy.

Q is the quantity of energy.

W is the work done.

Substituting into the formula, we have;

DU = 125 - 104

DU = 21 KJ

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3 years ago

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Evgen [1.6K]

Answer:

factory:

-mechanical energy

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-gravitational energy

Explanation:

8 0

3 years ago

What is formed from at least two types of chemically combined atoms?

SashulF [63]

A compound. For example, hydrogen and oxygen atoms form water.

7 0

3 years ago

A hamster eats a carrot before using its hamster wheel. The hamster wheel is connected to a generator which powers a light bulb.

Andrei [34K]

Answer: Chemical → Mechanical → Electrical → Radiant

Explanation:

First, the Hamster eats the carrot, then the hamster is getting chemical energy.

Now the hamster starts using his wheel, then he "transforms" the chemical energy into mechanical energy.

Now the mechanical energy is connected to a generator, this means that the mechanical energy (the rotation of the wheel) is being converted into electrical energy.

And we know that there is a light bulb powered by this electrical energy, then we have electrical energy being transformed into radiant energy.

Then the correct option is:

Chemical → Mechanical → Electrical → Radiant

6 0

3 years ago

Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by

Ulleksa [173]

A) $v=\sqrt{\frac{2qV}{m}}$

B) $r=\frac{mv}{qB}$

C) $T=\frac{2\pi m}{qB}$

D) $\omega=\frac{qB}{m}$

E) $r=\frac{\sqrt{2mK}}{qB}$

Explanation:

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

$\Delta U = qV$

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

$\Delta K=\frac{1}{2}mv^2$

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

$qV=\frac{1}{2}mv^2$

And solving for v, we find the speed v at which the particle enters the cyclotron:

$v=\sqrt{\frac{2qV}{m}}$

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

$F=qvB$

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

$F=m\frac{v^2}{r}$

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

$qvB=m\frac{v^2}{r}$

And solving for r, we find the radius of the orbit:

$r=\frac{mv}{qB}$ (1)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference ( $2\pi r$ ) and the velocity of the particle (v):

$T=\frac{2\pi r}{v}$ (2)

From eq.(1), we can rewrite the velocity of the particle as

$v=\frac{qBr}{m}$

Substituting into(2), we can rewrite the period of revolution of the particle as:

$T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}$

And we see that this period is indepedent on the velocity.

The angular frequency of a particle in circular motion is related to the period by the formula

$\omega=\frac{2\pi}{T}$ (3)

where T is the period.

The period has been found in part C:

$T=\frac{2\pi m}{qB}$

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

$\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}$

And we see that also the angular frequency does not depend on the velocity.

For this part, we use again the relationship found in part B:

$v=\frac{qBr}{m}$

which can be rewritten as

$r=\frac{mv}{qB}$ (4)

The kinetic energy of the particle is written as

$K=\frac{1}{2}mv^2$

So, from this we can find another expression for the velocity:

$v=\sqrt{\frac{2K}{m}}$

And substitutin into (4), we find:

$r=\frac{\sqrt{2mK}}{qB}$

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

6 0

4 years ago