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Artyom0805 [142]
2 years ago
13

A solid disc with a radius of 5.00 m and a mass of 20.0 kg is initially at rests and lies on the plane of the paper. A smaller s

olid disc with a radius of 2.50 m and a mass of 10.0 kg is spinning at 3500. rpm. The smaller disc is carefully pressed against the larger disc (flat side to flat side) so that they spin together without slipping. What is the angular velocity of the large disc
Physics
1 answer:
Maksim231197 [3]2 years ago
4 0

Answer:

Explanation:

This problem is based on conservation of angular momentum.

moment of inertia of larger  disc I₁ = 1/2 m r²  , m is mass and r is radius of disc . I

I₁ = .5 x 20 x 5²

= 250 kgm²

moment of inertia of smaller  disc I₂ = 1/2 m r²  , m is mass and r is radius of disc . I

I₂ = .5 x 10 x 2.5²

= 31.25 kgm²

3500 rmp = 3500 / 60 rps

n = 58.33 rps

angular velocity of smaller disc ω₂ = 2πn

= 2π x 58.33

= 366.3124 rad /s

applying conservation of angular momentum

I₂ω₂  = ( I₁ +I₂) ω  , ω is the common angular velocity

31.25 x 366.3124 = ( 250 +31.25) ω

ω = 40.7 rad / s .

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A company has been in business for 40 years. It has tens of thousands of customer addresses in three different computer systems
alexira [117]

The person in charge of information management is the database administrator.

<h3>What is a database?</h3>

The term database refers to the arranging of the data that concerns clients in relevant entry point such as computers and file cabinets. This is the way by which companies are bale to keep track of their customers.

The person that is most helpful in the management of the tens of thousands of customer addresses in three different computer systems and written on paper in filing cabinets is the database administrator.

Learn more about database:brainly.com/question/6447559

#SPJ1

7 0
1 year ago
1) On the way to the moon, the Apollo astro-
kramer
(1) You must find the point of equilibrium between the two forces,

<span>G * <span><span><span>MT</span><span>ms / </span></span><span>(R−x)^2 </span></span>= G * <span><span><span>ML</span><span>ms / </span></span><span>x^2
MT / (R-x)^2 = ML / x^2

So,

x = R * sqrt(ML * MT) - ML / (MT - ML)
R = is the distance between Earth and Moon.

</span></span></span>The result should be,
x = 3.83 * 10^7m
from the center of the Moon, and 

R - x = 3.46*10^8 m
from the center of the Earth.


(2) As the distance from the center of the Earth is the number we found before,
d = R - x = 3.46*10^8m
The acceleration at this point is
g = G * MT / d^2
g = 3.33*10^-3 m/s^2
6 0
2 years ago
Here on Earth you hang a mass from a vertical spring and start it oscillating with amplitude 1.9 cm. You observe that it takes 3
Vlada [557]

Answer:

T = 3.23 s

Explanation:

In the simple harmonic movement of a spring with a mass the angular velocity is given by

               w = √ K / m

With the initial data let's look for the ratio k / m

The angular velocity is related to the frequency and period

           w = 2π f = 2π / T

            2π / T = √ k / m

            k₀ / m₀ = (2π / T)²

            k₀ / m₀ = (2π / 3.0)²

            k₀ / m₀ = 4.3865

The period on the new planet is

          2π / T = √ k / m

           T = 2π √ m / k

In this case the amounts are

           m = 6 m₀

           k = 10 k₀

We replace

            T = 2π√6m₀ / 10k₀

            T = 2π √6/10 √m₀ / k₀

            T = 2π √ 0.6  √1 / 4.3865

            T = 3.23 s

5 0
3 years ago
A 60 kg sprinter has a momentum of +600 kg-m/s when he crosses the finish
MakcuM [25]

Answer:

10 ms⁻¹

Explanation:

The amount of momentum that an object has is dependent upon two factors

  • mass of the moving object  
  • speed of motion

In terms of an equation,

Momentum (P) = Mass(m)×velocity(v)

                     P = m×v

                 600 = 60 × v ⇒ v = 10 ms⁻¹

3 0
3 years ago
With detailed explaniation
belka [17]
  • Ø=37°
  • Initial velocity=u=20m/s
  • g=10m/s²

#A

\\ \rm\Rrightarrow H_{max}=\dfrac{u^2sin^2\theta}{2g}

\\ \rm\Rrightarrow H_{max}=\dfrac{20^2(sin37)^2}{2(10)}

\\ \rm\Rrightarrow H_{max}=\dfrac{400sin^237}{20}

\\ \rm\Rrightarrow H_{max}=20sin^237

\\ \rm\Rrightarrow H_{max}=7.2m

#B

\\ \rm\Rrightarrow R=\dfrac{u^2sin2\theta}{g}

\\ \rm\Rrightarrow R=\dfrac{20^2sin74}{10}

\\ \rm\Rrightarrow R=40sin74

\\ \rm\Rrightarrow R=38.5m

#C

\\ \rm\Rrightarrow T=\dfrac{2usin\theta}{g}

\\ \rm\Rrightarrow T=\dfrac{2(20)sin37}{10}

\\ \rm\Rrightarrow T=4sin37

\\ \rm\Rrightarrow T=2.4s

Now

\\ \rm\Rrightarrow v=u-gt

\\ \rm\Rrightarrow v=20-10(2.4)

\\ \rm\Rrightarrow v=20-24

\\ \rm\Rrightarrow v=-4m/s

4 0
2 years ago
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