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vladimir1956 [14]
3 years ago
5

A disk rotates freely on a vertical axis with an angular velocity of 50 rpm . An identical disk rotates above it in the same dir

ection about the same axis, but without touching the lower disk, at 20 rpm . The upper disk then drops onto the lower disk. After a short time, because of friction, they rotate together.The final angular velocity of the disks is1. 49 rpm2. 21 rpm3. 77 rpm4. 35 rpm5. 42 rpm
Physics
1 answer:
True [87]3 years ago
5 0

Answer:

Final angular velocity is 35rpm

Explanation:

Angular velocity is given by the equation:

I1w1i + I2w2i = I1w1f -I2w2f

But the two disks are identical, so Ii =I2

wf can be calculated using

wf = w1i - w2i/2

Given: w1i =50rpm w2i= 30rpm

wf= (50 + 20) / 2

wf= 70/2 = 35rpm

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Kepler’s first law states that:
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Answer:

Kepler's first law means that planets move around the Sun in elliptical orbits. An ellipse is a shape that resembles a flattened circle. ... It is zero for a perfect circle.

6 0
2 years ago
On a 100km track , a train travels the first 30km with a speed of 30km/h . How fast the train travel the next 70 km if the avera
nirvana33 [79]

Solution :-

Given :

Distance 1 = 30 km

Distance 2 = 70 km

We know that speed = distance/time

and, Average speed = total distance/total time taken

When the train acquired a speed of 30 km/hr, the time taken = 30/30 = 1 hour

Average speed = 9distance 1 + distance 2)/(time 1 + time 2)

AS time 2 or t2 is time taken for the second part of the journey of 70 km

⇒ 40 = 100/(1 + t2)

⇒ 40 + 40t2 = 100

⇒ 40t2 = 100 - 40

⇒ 40t2 = 60

⇒ t2 = 60/40

⇒ t2 = 1.5

So, t2 or time taken to travel the second part of the journey is 1.5 hours.

Speed of the second part of the journey = distance 2/time 2

⇒ 70/1.5

⇒ 46.666 km/hr or 46.7 km/hr.

Hence the answer is = 46.666 km/hr or 46.7 km/hr.

Hope it helped u if yes mark me BRAINLIEST!

Tysm!

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3 0
3 years ago
One uniform ladder of mass 30 kg and 10 m long rests against a frictionless vertical wall and makes an angle of 60o with the flo
yuradex [85]

Answer:

   μ = 0.37

Explanation:

For this exercise we must use the translational and rotational equilibrium equations.

We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive

let's write the rotational equilibrium

           W₁  x/2 + W₂ x₂ - fr y = 0

where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances

             cos 60 = x / L

where L is the length of the ladder

              x = L cos 60

            sin 60 = y / L

           y = L sin60

the horizontal distance of man is

            cos 60 = x2 / 7.0

            x2 = 7 cos 60

we substitute

         m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0

         fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60

let's calculate

         fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)

         fr = (735 + 2450) / 8.66

         fr = 367.78 N

the friction force has the expression

         fr = μ N

write the translational equilibrium equation

         N - W₁ -W₂ = 0

         N = m₁ g + W₂

         N = 30 9.8 + 700

         N = 994 N

we clear the friction force from the eucacion

        μ = fr / N

        μ = 367.78 / 994

        μ = 0.37

3 0
3 years ago
A body moving with an acceleration 2 m/s?then what is the change in velocity in 4sec.​
enot [183]

Answer:

As Per Provided Information

Moving body has 2m/s² acceleration

Time taken by body is 4 second

We are asked to find the 'change in velocity' ( ∆V) by the body.

<u>Formula Used here</u>

\boxed{\bf{\Delta \: V \:  =  acceleration \:  \times time \:}}

<u>Substituting </u><u>the </u><u>given </u><u>value</u>

<u>\sf\longrightarrow\Delta\:V \:  = 2 \times 4 \\  \\  \\ \sf\longrightarrow\Delta\:V \:  =8m {s}^{ - 1}</u>

<u>Therefore</u><u>,</u>

  • <u>Change </u><u>in </u><u>velocity </u><u>is </u><u>8</u><u> </u><u>m/</u><u>s</u>
7 0
2 years ago
Find the length of a pendulum that oscillates with a frequency of 0.16 hz. the acceleration due to gravity is 9.81 m/s 2 . answe
Vsevolod [243]
The period of the pendulum is the reciprocal of the frequency:
T= \frac{1}{f}= \frac{1}{0.16 Hz}=6.25 s

The period of the pendulum is given by
T=2 \pi  \frac{L}{g}
where L is the length of the pendulum, and g the acceleration of gravity. By re-arranging the formula and using the value of T we found before, we can  calculate the length of the pendulum L:
L=g  \frac{T^2}{(2 \pi)^2}=(9.81 m/s^2)  \frac{(6.25 s)^2}{(2 \pi)^2}=9.71 m
7 0
3 years ago
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