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4 years ago

A massless beam supports two weights as shown.

Physics

1 answer:

Vsevolod [243]4 years ago

3 0

<h2>The weight W = 471 N</h2><h2>Explanation:</h2>

Taking moments about B

N x L'' = W x L'

Here the moment is = force x perpendicular distance between the axis of rotation and the point of applied force .

Here L' = L/3 and L'' = 9

Thus from figure

471 x 9 = W x $\frac{L}{3}$

But L'' = $\frac{1}{2}$ ( L - $\frac{L}{3}$ ) = $\frac{L}{3}$ = 9

Thus W = 471 N

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a 20 kg sig is pulled by a horizontal force such that the single rope holding the sign make an angle of 21 degree with the verti

True [87]

Answer:

T= 210.15 N

F= 75.31 N

Explanation:

Let the tension in string be T newton.

According to the question

⇒T×cos21°= mg

⇒T= mg/cos21°

⇒T=20×9.81/cos21

⇒T= 210.15 N

now, the magnitude of horizontal force

F= Tsin21°

⇒F= 210.15×sin21°

=75.31 N

6 0

3 years ago

Help plz im lost

siniylev [52]

Gravitational potential energy=mass*gravitational acceleration*height
Kinetic energy = 0.5*mass*velocity²
So with the given data
K.E 0.5*1*x²=12.5 v²=12.5÷(0.5*1)
v=√12.5÷(0.5*1) v=5
GPE mass*gravitational acceleration*height
1*9.81*h=98
h=98÷(9.81*1)
h= 9.98 m

3 0

3 years ago

I am so confused by this! Someone, please help me! This isn't a test or homework question, just me being really confused!

Molodets [167]

Relativistic mean its so fast its a fraction of the sped of light.

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4 years ago

A mass on the end of a spring undergoes simple harmonic motion. At the instant when the mass is at its equilibrium position, wha

NeX [460]

Answer:

V = ω A sin ω t can be used to describe SHM

When sin ω t = 1 (the maximum value possible)

V = ω A at equilibrium sin ω t = 1

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2 years ago

A girl moves quickly to the center of a spinning merry-go-round, traveling along the radius of the merry-go-round. Which of the

DochEvi [55]

Answer:

The angular speed of the system increases.

The moment of inertia of the system decreases.

Explanation:

As we know that the girl is going towards the center of the circle so here the moment of inertia of the girl is given as

$I = mr^2$

here we know that

r = position of the girl from the center of the disc

now we know that the girl is moving towards the center so its distance will continuously decreasing

So the moment of inertia of the girl will decrease

Now we know that that with respect to the center of the disc there is no torque on the disc + girl system

So here we can use angular momentum conservation

So we have

$I\omega = constant$

since moment of inertia is decreasing for the system

so angular speed will increase

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4 years ago