All categories

3 years ago

To calculate how much rest energy an object contains, you multiply

Physics

1 answer:

Rina8888 [55]3 years ago

5 0

This is what e=mc² means.
So rest energy is RH side of equation that is its mass times speed of light squared.

You might be interested in

A ball was dropped from a height of 10 feet. Each time it hits the ground, it bounces 4/5 of its previous height. Find the total

Shtirlitz [24]

Answer:

d = 90 ft

Explanation:

As we know that after each bounce it reaches to 4/5 times of initial height

so we can say

$h_2 = \frac{4}{5}h$

so the distance covered is given as

$d = h + 2(\frac{4}{5}h) + 2(\frac{4}{5})^2h + 2(\frac{4}{5})^3h........$

here we know that

h = 10 feet

$d = h + 2(\frac{4}{5}h)(1 + \frac{4}{5} + (\frac{4}{5})^2 + ...........)$

$d = 10 + 2(\frac{4}{5}(10))(\frac{1}{1 - \frac{4}{5}})$

$d = 90 ft$

8 0

3 years ago

The resistance and the magnitude of the current depend on the path that the current takes. The drawing shows three situations in

maria [59]

Answer: (a). Resistance = 0.4286ohms and Current (I) = 7A

(b). Resistance (R) = 0.027 ohms and Current (I) = 111.1A

(c). Resistance (R) = 0.1071 ohms and Current (I) = 28A

Explanation:

From the question, given that;

ρ = 1.5*10-2ῼ.m

Lo = 7cm = 0.07m

V = 3V

From the formula R = ρL/A, where A is the area of cross section, L is the length of material and ρ is the resistivity.

(A)

L = 4Lo and A = 2Lo*Lo

R = ρL/A

R = ρ4Lo/(2Lo*Lo)

R = 2ρ /Lo = 2*1.5*10-2/0.07

R = 0.4286 ῼ

From this the current becomes;

I = V/R = 3/ 0.4286 = 6.99 = 7A

(B)

L = Lo and A = 4Lo * 2Lo

R = ρL/A

R = ρLo/ (4Lo*2Lo) after eliminating Lo from both sides we get,

R = ρ/8Lo = 1.5*10-2 / 8*0.07

R = 0.027

Current (I) = V/R = 3/0.027 = 111.1A

(C)

L = 2Lo and A = Lo * 4Lo

R = ρL/A

R = ρ2Lo/ (Lo*4Lo) eliminating Lo from both sides we get,

R = ρ/2Lo = 1.5*10-2 / 2*0.07 = 0.1071

The current becomes;

I = V/R = 3/0.1071 = 28A

4 0

3 years ago

Read 2 more answers

How will a current change if the resistance of a circuit remains constant while the voltage across the circuit decreases to half

beks73 [17]

Answer:

1. The current will drop to half of its original value.

Explanation:

The problem can be solved by using Ohm's law:

$V=RI$

where

V is the voltage across the circuit

R is the resistance of the circuit

I is the current

We can rewrite it as

$I=\frac{V}{R}$

In this problem, we have:

- the resistance of the circuit remains the same: R' = R

- the voltage is decreased to half of its original value: $V'=\frac{V}{2}$

So, the new current will be

$I'=\frac{V'}{R'}=\frac{V/2}{R}=\frac{1}{2}\frac{V}{R}=\frac{I}{2}$

so, the current will drop to half of its original value.

4 0

3 years ago

Suppose you observe that at night, the air just above the ground feels cooler than the air above it. then in the middle of a sun

yanalaym [24]

The answer is convection

6 0

3 years ago

A current of 9 A flows through an electric device with a resistance of 43 Ω. What must be the applied voltage in this particular

blondinia [14]

The answer is D. Hope this helps you!

6 0

3 years ago