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2 years ago

15

Describe a procedure that would increase the potential energy of two magnets if like poles are used. Explain why the energy of t

he system increases.

1 answer:

zalisa [80]2 years ago

5 0

Answer:

If you apply a force to separate 2 opposite poles, the potential energy of the system increases.

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Alfred wegener was the one who made contiental drift theory ?

navik [9.2K]

Yep. he discovered that coastline from south america and africa fit together like a puzzle, which later became a part of the continential drift theory

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2 years ago

A light-year measures the _______ that light travels in 1 year.

o-na [289]

<span>A light-year measures the distance that light travels in 1 year.

Answer : B ) Distance

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4 0

3 years ago

Read 2 more answers

You punt a ball straight up at 20 m/s. What is the balls hangtime

Natalija [7]

The hang time of the ball is 4.08 s

Explanation:

The ball is in free fall motion: this means that it is acted upon gravity only, so its acceleration is the acceleration of gravity,

$a=g=-9.8 m/s^2$

downward (the negative sign refers to the downward direction).

Since this is a uniformly accelerated motion, we can solve the problem by using the following suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the time

First we calculate the time it takes for the ball to reach the maximum height, where the velocity is zero:

v = 0

Substituting:

u = +20 m/s

$a=-9.8 m/s^2$

we find t

$t=\frac{v-u}{a}=\frac{0-20}{-9.8}=2.04 s$

The motion of the ball is symmetrical, so the total time of flight is just twice the time needed to reach the maximum height, therefore:

$T=2t=2(2.04)=4.08 s$

Learn more about free fall:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

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4 0

2 years ago

A charged particle A exerts a force of 2.45 μN to the right on charged particle B when the particles are 12.2 mm apart. Particle

Brilliant_brown [7]

Answer:

$F_2 = 1.10 \mu N$

Explanation:

As we know that the electrostatic force is a based upon inverse square law

so we have

$F = \frac{kq_1q_2}{r^2}$

now since it depends inverse on the square of the distance so we can say

$\frac{F_1}{F_2} = \frac{r_2^2}{r_1^2}$

now we know that

$r_2 = 18.2 mm$

$r_1 = 12.2 mm$

also we know that

$F_1 = 2.45 \mu N$

now from above equation we have

$F_2 = \frac{r_1^2}{r_2^2} F_1$

$F_2 = \frac{12.2^2}{18.2^2}(2.45\mu N)$

$F_2 = 1.10 \mu N$

5 0

3 years ago

Ratio of resistances of two bulbs is 2:3. If they are connected in series to a supply, then the ratio of voltages across them is

V125BC [204]

Answer:

Explanation:

Given that,

Two resistor has resistance in the ratio 2:3

Then,

R1 : R2 = 2:3

R1 / R2 =⅔

3 •R1 = 2• R2

Let R2 = R

Then,

R1 = ⅔R2 = 2/3 R

So, if the resistor are connected in series

Let know the current that will flow in the circuit

Series connection will have a equivalent resistance of

Req = R1 + R2

Req = R + ⅔ R = 5/3 R

Req = 5R / 3

Let a voltage V be connect across then, the current that flows can be calculated using ohms law

V = iR

I = V/Req

I = V / (5R /3)

I = 3V / 5R

This the current that flows in the two resistors since the same current flows in series connection

Now, using ohms law again to calculated voltage in each resistor

V= iR

For R1 = ⅔R

V1 =i•R1

V1 = 3V / 5R × 2R / 3

V1 = 3V × 2R / 5R × 3

V1 = 2V / 5

For R2 = R

V2 = i•R2

V2 = 3V / 5R × R

V2 = 3V × R / 5R

V2 = 3V / 5

Then,

Ratio of voltage 1 to voltage 2

V1 : V2 = V1 / V2 = 2V / 5 ÷ 3V / 5

V1 : V2 = 2V / 5 × 5 / 3V.

V1 : V2 =2 / 3

V1:V2 = 2:3

The ratio of their voltages is also 2:3

5 0

3 years ago