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kkurt [141]
2 years ago
8

an input force of 50 Newtons is applied through a distance of 10 meters to machine with mechanical advantage of 3. If the work o

utput for the machine is 450 newton x meters and this work is applied through a distnace of 3 meters, what is the output force of the machine
Physics
1 answer:
gladu [14]2 years ago
5 0
The output of the machine is

                                      (output work) =  (output force) x (distance)

                                        450 N-m      =  (output force) x (3 meters)

Divide each side
by  3 meters:                Output force = (450 N-m) / (3 m)

                                                           =    150 newtons .

With all the information given about the output work, we don't need
to know anything about the input work, or even the fact that we're
dealing with a machine.

It's comforting, though, to look back and notice that the output work
(450 N-m) is not more than the input work (500 N-m).  So everything
is nice and hunky-dory.
___________________________________

Well, my goodness !
I didn't even need to go through all of that.

Given:

-- The input force to the machine is 50 newtons.

-- The mechanical advantage of the machine is 3 .

That right there tells us that

-- The output force of the machine is 150 newtons.

We don't need any of the other given information.
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ollegr [7]
To make the motor turn faster we can:
(a) increase the current
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8 0
2 years ago
A single slit, which is 0.050 mm wide, is illuminated by light of 550 nm wavelength. What is the angular separation between the
likoan [24]

Answer:

The separation between the first two minima on either side is 0.63 degrees.

Explanation:

A diffraction experiment consists on passing monochromatic light trough a small single slit, at some distance a light diffraction pattern is projected on a screen. The diffraction pattern consists on intercalated dark and bright fringes that are symmetric respect the center of the screen, the angular positions of the dark fringes θn can be find using the equation:

a\sin \theta_n=n\lambda

with a the width of the slit, n the number of the minimum and λ the wavelength of the incident light. We should find the position of the n=1 and n=2 minima above the central maximum because symmetry the angular positions of n=-1 and n=-2 that are the angular position of the minima below the central maximum, then:

for the first minimum

a\sin \theta_1=(1)\lambda

solving for θ1:

\theta_1=\arcsin (\frac{\lambda}{a})=\arcsin (\frac{550\times10^{-9}}{0.05\times10^{-3}})

\theta_1=0.63 degrees

for the second minimum:

a\sin \theta_2=(2)\lambda

\theta_2=\arcsin (\frac{2\lambda}{a})=\arcsin (\frac{2*550\times10^{-9}}{0.05\times10^{-3}})

\theta_2=1.26 degrees

So, the angular separation between them is the rest:

\Delta \theta =1.26-0.63

\Delta \theta=0.63

4 0
2 years ago
A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta
BartSMP [9]

Answer:

35%

Explanation:

Assuming no external torques present during the collision between the record and the turntable, total angular momentum must be conserved.

For a rotating body with some angular velocity and moment of inertia, the angular momentum can be expressed as follows;

L = I* ω

So, as initial angular momentum and final angular momentum must be the same, we have:

Li = Lf

⇒ I₁ * ω₁ = I₂ * ω₂ (1)

where I₁ is the rotational inertia of the turntable, and I₂, is the combined rotational inertia of the turntable and the record:

I₂ = I₁ + 0.54 I₁ = 1.54 I₁

We can solve (1) for the new common angular speed, as follows:

ω₂ = ω₁ / 1.54 (2)

The initial rotational kinetic energy is given by definition for the following equation:

Kroti = 1/2 * I₁ * ω₁² (3)

The final rotational kinetic  energy takes into account the new rotational inertia and the common final angular speed:

Krotf = 1/2* I₂ * ω₂² = 1/2* 1.54 I₁* (ω₁/1.54)² (4)

Dividing both sides in (3) and (4), we get:

Krotf/Kroti = 1/1.54 = 0.65

This means that the final rotational kinetic energy, has reduced to 0.65 of the initial value, or that has lost 35% of the initial kinetic energy.

8 0
2 years ago
In a circus act, an acrobat rebounds upward from the surface of a trampoline at the exact moment that another acrobat, perched 9
slega [8]

Answer:

1.6 secs

Explanation:

In a circus act, an acrobat upwards from the surface of a trampoline

At that same moment another acrobat perched 9.0m above him

A ball is released from rest

While still in motion the acrobat catches the ball

He left the ball with a trampoline of 5.6m/s

Since the ball is falling downwards from a distance then acceleration will be negative

a= -9.8

s= d

s= 1/2at^2

= 1/2 × (-9.8)t^2

= 0.5× (-9.8)t^2

d = -4.9t^2

Therefore the time the acrobat stays in the air before catching the ball can be calculated as follows

9 - 4.9t^2= 5.6t + 1/2(-9.8)t^2

9 - 4.9t^2= 5.6t + (-4.9)t^2

9 - 4.9t^2= 5.6t - 4.9t^2

9= 5.6t

t= 9/5.6

t= 1.6 secs

6 0
2 years ago
A car came to a stop from a speed of 35 m/s in a time of 8.1 seconds. What was the acceleration of the car?
uranmaximum [27]
Simply subtract the two velocities and divide by 8.1,

\frac{0 - 35}{8.1} = - 4.32

~~

I hope that helps you out!!

Any more questions, please feel free to ask me and I will gladly help you out!!

~Zoey
5 0
2 years ago
Read 2 more answers
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