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2 years ago

7

When are magnets attracted and when are they repelled?

2 answers:

ohaa [14]2 years ago

6 0

<em>when </em><em>South </em><em>and </em><em>North </em><em>poles </em><em>are </em><em>placed </em><em>together </em><em>they </em><em>attract </em><em>as </em><em>they </em><em>are </em><em>opposite</em><em> </em><em>poles. </em><em>In</em><em> </em><em>magnets </em><em>opposite </em><em>po</em><em>l</em><em>es </em><em>are </em><em>attracted </em><em>and </em><em>like </em><em>poles </em><em>are </em><em>repelled. </em>

yarga [219]2 years ago

6 0

Magnets are attracted when easy of the different sides, most commonly known as “north” and “south”, are facing eachother. They repel when north and north, or south and south are facing eachother.

hope this helps! :)

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A motorcycle starts from rest and has a constant acceleration. In a time interval t, it undergoes a displacement x and attains a

iren [92.7K]

Answer:

√(6ax)

Explanation:

Hi!

The question states that during a time t the motorcyle underwent a displacement x at constant acceleration a starting from rest, mathematically we can express it as:

x=(1/2)at^2

Then the we need to find the time t' for which the displacement is 3x

3x=(1/2)a(t')^2

Solving for t':

t'=√(6x/a)

Now, the velocity of the motorcycle as a function of time is:

v(t)=a*t

Evaluating at t=t'

v(t')=a*√(6x/a)=√(6*x*a)

Which is the final velocity

Have a nice day!

3 0

2 years ago

Suppose an object in freefall is dropped from a building it’s starting velocity is 0m/s. Ignoring the facts of air resistance wh

Alja [10]

v = v₀ + at

v = final speed, v₀ = initial speed, a = acceleration, t = elapsed time

Given values:

v₀ = 0m/s (starts from rest), a = 9.81m/s², t = 3s

Plug in and solve for v:

v = 0 + 9.81(3)

v = 29.4m/s

3 0

2 years ago

If anyone here can help me with physics please comment. its about coulombs law

Minchanka [31]

I may be able to help you with This subject

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2 years ago

An object with a mass M = 250 g is at rest on a plane that makes an angle θ = 30 o above the horizontal. The coefficient of kine

liubo4ka [24]

Answer:

v = 79.2 m/s

Solution:

As per the question:

Mass of the object, m = 250 g = 0.250 kg

Angle, $\theta = 30^{\circ}$

Coefficient of kinetic friction, $\mu_{k} = 0.100$

Mass attached to the string, m = 0.200 kg

Distance, d = 30 cm = 0.03 m

Now,

The tension in the string is given by:

$Mgsin\theta + \mu_{k}Mgcos\theta + Ma = T$ (1)

Also

T = m(g + a)

Thus eqn (1) can be written as:

$Mgsin\theta + \mu_{k}Mgcos\theta + Ma = m(g - a)$

$Mgsin\theta + \mu_{k}Mgcos\theta + Ma = mg + ma$

$mg - Mgsin\theta - \mu_{k}Mgcos\theta = (M - m)a$

$a = \frac{0.2\times 9.8 - 0.250\times 9.8\times sin30^{\circ} - 0.1\times 0.250\times 9.8\times cos30^{\circ}}{0.250 - 0.200}$

$a = 10.45\ m/s^{2}$

Now, the speed is given by the third eqn of motion with initial velocity being zero:

$v^{2} = u^{2} + 2ad$

where

u = initial velocity = 0

Thus

$v = \sqrt{2ad}$

$v = \sqrt{2\times 10.45\times 0.03} = 0.792\ m/s$

3 0

2 years ago

DOUBLE POINTS!!!!

Nataly_w [17]

Explanation:

C.

Object A will require more force to be set in motion but will travel faster than object B.

2. true

7 0

2 years ago