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3 years ago

Is the force of gravity stronger on a piece of crumpled paper or a normal piece of flat paper?

Physics

1 answer:

GaryK [48]3 years ago

3 0

Yes, because a crumpled piece of paper has more center mass, allowing it to fall faster.

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Assume a beam of light hits the boundary separating medium 1, with index of refraction n1 and medium 2, with index of refraction

Reptile [31]

Answer:

option C

Explanation:

Given,

Refractive index of medium 1 = n₁

Refractive index of medium 2 = n₂

For total internal reflection to take place light should move from denser medium to the rarer medium.

Here Total internal reflection take place at the boundary of medium 1 and medium 2 so, the refractive index of medium 1 is more than medium 2

n₁ > n₂

The correct answer is option C

3 0

3 years ago

If you add together all of forces exerted on an object and get a non zero value this is called the _____ force on the object

butalik [34]

Your answer would be the; NET force on the object. Refer to Newton's Laws of Forces and Motion.

Hope that helps!!!!!!!!!!!!!!!! : )

8 0

3 years ago

Read 2 more answers

Power Rating of a Resistor. The power rating of a resistor is the maximum power the resistor can safely dissipate without too gr

IgorLugansk [536]

(a) 273.9 V

The power rating of the resistor is given by

$P=\frac{V^2}{R}$

where

P is the power rating

V is the potential difference across the resistor

R is the resistance

If the maximum power rating is $P=5.0 W$ , and the resistance of the resistor is $R=15 k\Omega = 15000 \Omega$ , then we can find the maximum potential difference across the resistor by re-arranging the previous equation for V:

$V=\sqrt{PR}=\sqrt{(5.0 W)(15000 \Omega)}=273.9 V$

(b) 1.6 W

In this case, we have:

$R=9.0 k\Omega = 9000 \Omega$ is the resistance of the resistor

$V=120 V$ is the potential difference across the resistor

So we can find the power rating by using the same formula of part (a):

$P=\frac{V^2}{R}=\frac{(120 V)^2}{9000 \Omega}=1.6 W$

(c) Maximum voltage: 14.1 V; Rate of heat: 2.00 W and 3.00 W

Here we have two resistors of

$R_1 = 100 \Omega\\R_2 = 150 \Omega$

and each resistor has a power rating of

P = 2.00 W

So the greatest potential difference allowed in the first resistor is

$V=\sqrt{PR_1}=\sqrt{(2.00 W)(100 \Omega)}=14.1 V$

While the greatest potential difference allowed in the second resistor is

$V=\sqrt{PR_2}=\sqrt{(2.00 W)(150 \Omega)}=17.3 V$

So the greatest potential difference allowed not to overheat either of the resistor is 14.1 V.

In this condition, the power dissipated on the first resistor is 2.00 W, while the power dissipated on the second resistor is

$P_2 = \frac{V^2}{R_2}=\frac{(14.1 V)^2}{150 \Omega}=1.33 W$

And this corresponds to the rate of heat generated in the first resistor (2.00 W) and in the second resistor (1.33 W).

4 0

3 years ago

As temperature increase what energy increases

garik1379 [7]

Kinetic energy would increase sir.

8 0

3 years ago

State the reason and tell whether true or false. The SPEED OF LIGHT IS THE SAME IN ALL MEDIA.

Sphinxa [80]

Answer:

The statement is true .

Explanation:

I hope this helps.

4 0

3 years ago