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blondinia [14]
3 years ago
8

Disk A, with a mass of 2.0 kg and a radius of 60 cm , rotates clockwise about a frictionless vertical axle at 20 rev/s . Disk B,

also 2.0 kg but with a radius of 40 cm , rotates counterclockwise about that same axle, but at a greater height than disk A, at 20 rev/s . Disk B slides down the axle until it lands on top of disk A, after which they rotate together.After the collision, what is their common angular speed (in rev/s) and in which direction do they rotate?
Physics
1 answer:
Vladimir79 [104]3 years ago
6 0

Answer:

Explanation:

  • let m1 = 2 kg
  • r1 = 60 cm = 0.6 m
  • w1 = -20 rev/s (clockwise)

I1 = 0.5 x m1 x r1²

= 0.5 x 2 x 0.6²

= 0.36 kg.m²

m2 = 2 kg

r2 = 40 cm = 0.4 m

w2 = 20 rev/s (counter clockwise)

I2 = 0.5 x m2 x r2²

= 0.5 x 2 x 0.4²

= 0.16 kg.m²

let wf be the final angular velocity of the two disks.

Applying conservation of angular momentum ;

(I1 + I2) x wf = I1 x w1 + I2 x w2

wf = (I1 x w1 + I2 x w2) / (I1 + I2)

= (0.36 x (-20) + 0.16 x 20 )/(0.36 + 0.16)

= -7.69rev/s

The negative sign indicates the disks rotate clockwise direction.

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A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
galina1969 [7]

Answer:

44.64 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.2\times 1180+80.6^2}\\\Rightarrow v=128.01\ m/s

v=u+at\\\Rightarrow 128.01=80.6+4.2t\\\Rightarrow t=\frac{128.01-80.6}{4.2}=11.29\ s

<u>Time taken to reach 1180 m is 11.29 seconds</u>

v=u+at\\\Rightarrow 0=128.01-9.8t\\\Rightarrow t=\frac{128.01}{9.8}=13.06\ s

<u>Time the rocket will keep going up after the engines shut off is 13.06 seconds.</u>

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128.01^2}{2\times -9.8}\\\Rightarrow s=836.05\ m

The distance the rocket will keep going up after the engines shut off is 836.05 m

Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m

The rocket will fall from this height

s=ut+\frac{1}{2}at^2\\\Rightarrow 2016.05=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2016.05\times 2}{9.8}}\\\Rightarrow t=20.29\ s

<u>Time taken by the rocket to fall from maximum height is 20.29 seconds</u>

Time the rocket will stay in the air is 11.29+13.06+20.29 = 44.64 seconds

5 0
3 years ago
Mengapa indra peraba tidak dapat digunakan untuk membandingkan suhu dengan tepat?
Debora [2.8K]
Answer it ur self if u have internet
3 0
3 years ago
A skateboarder starts from rest and maintains a constant acceleration of 0.50 m/s² for 8.4 s. What is the rider's displacement d
gregori [183]

Answer:

4.2m/s

Explanation:

0.50x8.4=4.2m/s

3 0
2 years ago
A vase of flowers is stationary on a table. which of the following statements best explains why the vase is stationary?
goldenfox [79]
D. The table pushes up on the vase with the same amount of force as gravity pulling it down.
7 0
3 years ago
A tank contains 350 liters of fluid in which 10 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu
aleksandr82 [10.1K]

Answer:

A(t) = -340e^{-t/70} + 350

Explanation:

Since fluid is pumping in and out at the same rate (5L/min), the total fluid volume in the tank stays constant at 350L. Only the amount of salt and its concentration changed overtime.

Let A(t) be the amount of salt (g) at time t and C(t) (g/L) be the concentration at time t

A(0) = 10 g

Brine with concentration of 1g/L is pouring in at the rate of 5L/min so the salt income rate is 5 g/min

The well-mixed solution is pouring out at the rate of 5L/min at concentration C(t) so the salt outcome rate is 5C g/min

But the concentration is total amount of salt over 350L constant volume

C = A / 350

Therefore our rate of change for salt A' is

A' = 5 - 5A/350 = 5 - A/70

This is a first-order linear ordinary differential equation and it has the form of y' = a + by. The solution of this is

y = ce^{bt} + \frac{a}{b}

So A = ce^{\frac{-t}{70}} + \frac{5}{1/70} = ce^{-t/70} + 350

with A(0) = 10

c + 350 = 10

c = 10 - 350 = -340

A(t) = -340e^{-t/70} + 350

4 0
3 years ago
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