Which example is a naturally occurring magnet

What is the test for whether or not a moving object is in equilibrium?

Sphinxa [80]

It is in equilibrium if its velocity is not changing.

3 0

3 years ago

A current i travels counterclockwise through a closed copper wire loop which has the shape of a rectangle with sides 'a' and 'b'

Maru [420]

<span>A current i travels counterclockwise through a closed copper wire loop which has the shape of a rectangle with sides</span>

4 0

3 years ago

Is winking a genetic and why?

PolarNik [594]

Yes,and because not everyone can wink and often that someone can only wink with one eye only

4 0

4 years ago

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A mass m at the end of a spring vibrates with a frequency

Wittaler [7]

Answer:

m = 0.59 kg.

Explanation:

First, we need to find the relation between the frequency and mass on a spring.

The Hooke's law states that

$F = -kx$

And Newton's Second Law also states that

$F = ma = m\frac{d^2x}{dt^2}$

Combining two equations yields

$a = -\frac{k}{m}x$

The term that determines the proportionality between acceleration and position is defined as angular frequency, ω.

$\omega = \sqrt{\frac{k}{m}}$

And given that ω = 2πf

the relation between frequency and mass becomes

$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$ .

Let's apply this to the variables in the question.

$0.88 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\\0.60 = \frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}\\\frac{0.88}{0.60} = \frac{\frac{1}{2\pi}\sqrt{\frac{k}{m}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}}\\1.4667 = \frac{\sqrt{m+0.68}}{\sqrt{m}}\\2.15m = m + 0.68\\1.15m = 0.68\\m = 0.59~kg$

6 0

3 years ago

A 14.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 55.0°-angle with the horizontal. (a

Nikitich [7]

Answer:

Check attachment for the free body diagram

Explanation:

Given that,

Ladder length L = 14m

Weight of ladder W= 490N

The weight will act at the midpoint

i.e at 14/2 = 7m, L1 = 7m

The ladder makes an angle of 55° with the horizontal. θ=55°

Weight of firefighter Wf =810N

The firefighter is at 3.9m from the horizontal ground, L2 =3.9

The wall exerts a force on the ladder, let It be Nw

The ground exerts a force on the ladder, let it be Ng

The let Ff be the frictional force that opposes motion.

a. We want to find the horizontal and vertical force the ground exerted on the ladder i.e Ng and Ff.

Using Newton second law

ΣFy= m•ay

ay=0, since the body is not accelerating

ΣFy = 0

Ng — Wf —W = 0

Ng = Wf + W

Ng = 490 + 810

Ng = 1300 N.

Also,

ΣFx= m•ax

ax=0, since the body is not accelerating

ΣFx = 0

Ff — Nw = 0

Ff = Nw

Now,

Let take moment about point A(ground), but note before we take moment, the forces must be perpendicular to the ladder.

applying condition of equilibrium of moment

Clockwise moment = anti-clockwise

WfCosθ•L2 + WCosθ•L1 = NwSinθ•L

(810Cos55)•3.9 + (490Cos55)•7 = (NwSin55)•14

1811.93 + 1967.37 = 11.47Nw

3779.3 = 11.47Nw

Then, Nw = 3779.3/11.47

Nw= 329.49N

Since Nw = Ff

Then Ff = 329.49N

So the required reaction exerted by the ground on the ladder are

Ng = 1300 N

Ff = 329.49 N

b. Now the firefighter is a distance of 9.4m from the horizontal and the ladder is about to slip

So we need to calculate the coefficient of static friction μs

Check attachment for new diagram,

So calculating for Nw again, since the firefighter have new position

Now the firefighter is at 9.4m from the ground. Therefore, L2 = 9.4m

applying condition of equilibrium of moment

Clockwise moment = anti-clockwise

WfCosθ•L2 + WCosθ•L1 = NwSinθ•L

(810Cos55)•9.4 + (490Cos55)•7 = (NwSin55)•14

4367.21 + 1967.37 = 11.47Nw

6334.58 = 11.47Nw

Then, Nw = 6334.58/11.47

Nw= 552.274N

Since Nw = Ff

Then Ff = 552.274N

Then, using frictional law

Ff = μs•Ng

The Ng doesn't change

Then, 552.274 = 1300μs

μs = 552.274/1300

μs = 0.42

c. Now, we want to know the maximum distance of the firefighter if the coefficient of static friction is reduce by half

Then, μs = 0.42/2

μs = 0.21

The assume the firefighter is at L2 from the horizontal.

Then, the frictional force is

Ff = μsNg

Ff = 0.21 × 1300

Ff = 273N

Then, Nw = Ff = 273N

Taking moment about point A

Clockwise moment = anti-clockwise

WfCosθ•L2 + WCosθ•L1 = NwSinθ•L

(810Cos55)•L2 + (490Cos55)•7 = (273Sin55)•14

464.6•L2 + 1967.37 = 3130.8

464.6•L2 = 3130.8—1967.37

464.6L2 = 1163.43

L2 = 1163.43/464.6

L2 = 2.5m

The maximum distance before the ladder begin to slip is 2.5m.

3 0

3 years ago

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