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Elan Coil [88]
3 years ago
11

Your friend says, “chemical changes are caused by an input in energy. In physical changes, there is no transfer of energy” is yo

ur friend correct? Why or why not?
Physics
1 answer:
nalin [4]3 years ago
3 0

Answer:

Ok, let's suppose the simplest of the physical changes:

We have an object that is not moving (so it is not accelerated)

and there is change, now the object moves.

Because there was a change, means that there was an acceleration, and by the second Newton's law.

Force equals mass times acceleration:

F = m*a

There must be a force.

So suppose that you pushed the object, then some energy that you had, you transferred it to the object, that now is moving and now has kinetic energy.

Now, is kinda true that in a closed system the total energy is always constant, but it depends on what is our system.

So if we think in our system as you and the object, then in the whole system the energy does not change because the energy that you lost is now on the object, but again, there was a transfer of energy.

So no, your friend is not correct.

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If a bus is running with a speed of 72 km/hr,calculate the distance travelled by it in 5 second.Which of the following is right
ratelena [41]

Answer:

100m

Explanation:

conversation km to meter

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7 0
2 years ago
A magnet in the form of a cylindrical rod has a length of 7.30 cm and a diameter of 1.5 cm. It has a uniform magnetization of 5.
yulyashka [42]

Answer:

Magnetic dipole moment is 0.0683 J/T.

Explanation:

It is given that,

Length of the rod, l = 7.3 cm = 0.073 m

Diameter of the cylinder, d = 1.5 cm = 0.015 m

Magnetization, M=5.3\times 10^3\ A/m

The dipole moment per unit volume is called the magnetization of a magnet. Mathematically, it is given by :

M=\dfrac{\mu}{V}

\mu=M\times \pi r^2\times l

Where

r is the radius of rod, r = 0.0075 m

\mu=5.3\times 10^3\ A/m\times \pi (0.0075)^2\times 0.073\ m

\mu=0.0683\ J/T

So, its magnetic dipole moment is 0.0683 J/T. Hence, this is the required solution.

6 0
3 years ago
Read 2 more answers
In a carrom game, a striker weighs three times the mass of the other pieces, the carrom men and the queen, which each have a mas
Mila [183]

Answer:

- The final velocity of the queen is (3/2) of the initial velocity of the striker. That is, (3V/2)

- The final velocity of the striker is (1/2) of the initial velocity of the striker. That is, (V/2)

Hence, the relative velocity of the queen with respect to the striker after collision

= (3V/2) - (V/2)

= V m/s.

Explanation:

This is a conservation of Momentum problem.

Momentum before collision = Momentum after collision.

The mass of the striker = M

Initial Velocity of the striker = V (+x-axis)

Let the final velocity of the striker be u

Mass of the queen = (M/3)

Initial velocity of the queen = 0 (since the queen was initially at rest)

Final velocity of the queen be v

Collision is elastic, So, momentum and kinetic energy are conserved.

Momentum before collision = (M)(V) + 0 = (MV) kgm/s

Momentum after collision = (M)(u) + (M/3)(v) = Mu + (Mv/3)

Momentum before collision = Momentum after collision.

MV = Mu + (Mv/3)

V = u + (v/3)

u = V - (v/3) (eqn 1)

Kinetic energy balance

Kinetic energy before collision = (1/2)(M)(V²) = (MV²/2)

Kinetic energy after collision = (1/2)(M)(u²) + (1/2)(M/3)(v²) = (Mu²/2) + (Mv²/6)

Kinetic energy before collision = Kinetic energy after collision

(MV²/2) = (Mu²/2) + (Mv²/6)

V² = u² + (v²/3) (eqn 2)

Recall eqn 1, u = V - (v/3); eqn 2 becomes

V² = [V - (v/3)]² + (v²/3)

V² = V² - (2Vv/3) + (v²/9) + (v²/3)

(4v²/9) = (2Vv/3)

v² = (2Vv/3) × (9/4)

v² = (3Vv/2)

v = (3V/2)

Hence, the final velocity of the queen is (3/2) of the initial velocity of the striker and is in the same direction.

The final velocity of the striker after collision

= u = V - (v/3) = V - (V/2) = (V/2)

The relative velocity of the queen withrespect to the striker after collision

= (velocity of queen after collision) - (velocity of striker after collision)

= v - u

= (3V/2) - (V/2) = V m/s.

Hope this Helps!!!!

3 0
3 years ago
Read 2 more answers
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kow [346]

Answer:

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Explanation:

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8 0
3 years ago
What are the three ways that a person can manipulate light?
BARSIC [14]

The three ways a person can manipulate light would be the following:, filter, and the time the photograph is taken

 

<span>1.    </span>Angle - <span>The </span>camera angle<span> <span>marks the specific location at which the movie </span></span>camera<span> <span>or video </span></span>camera<span> is placed to take a shot.</span>

<span>2.    </span>Filter - Camera<span> <span>lens </span></span>filters<span> <span>still have many uses in digital photography, and should be an important part of any photographer's </span></span>camera<span> bag.</span>

<span>3.    </span>Time the photograph is taken - The golden hour, sometimes called the "magic hour", is roughly the first hour of light after sunrise, and the last hour of light before sunset, although the exact duration varies between seasons. During these times the sun is low in the sky, producing a soft, diffused light which is much more flattering than the harsh midday sun that so many of us are used to shooting in.

 

I am hoping that these answers have satisfied your queries and it will be able to help you in your endeavors, and if you would like, feel free to ask another question.

6 0
3 years ago
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