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Alexeev081 [22]
3 years ago
12

To drive a typical car at 40 mph on a level road for one hour requires about 3.2 × 107 J of energy. Suppose we tried to store th

is much energy in a spinning, solid, uniform, cylindrical flywheel. A large flywheel cannot be spun too fast or it will fracture. If we used a flywheel of diameter 1.2 m and mass 400 kg, what angular speed would be required to store 3.2 × 107 J?
Physics
1 answer:
tatiyna3 years ago
6 0

Answer:

9000RPM

Explanation:

"Angular velocity" is directly related to kinetic energy, that is, the Kinetic energy equation would allow an approximation to the resolution investigated in the problem.

The equation for KE is given by:

KE = \frac{1}{2} lw ^ 2

Now, starting from there towards the <em>Angular equation of kinetic energy</em>, the moment of inertia (i) is used instead of mass (m), and angular velocity (w) instead of linear velocity (V)

That's how we get

KE_{Angular} = \frac{1}{2} Iw^2

calculating the inertia for a solid cylindrical disk, of

m = 400kg

r = 1.2 / 2 = 0.6m

I_{disk} = \frac{1}{2} mr^2 = (0.5) (400) (0.6)^2 = 72 kgm^2

We understand that the total kinetic energy is 3.2 * 10 ^ 7J, like this:

3.2*10^7 = \frac{1}{2} Iw^2 = (0.5) (72) w^2 = 36w^2w^2 = 3.2*10^7 / 36 = 0.0888*10^7 = 88.8*10^4

w = 9.43*10^2 = 943 rad / s

Thus,

943 rad / s ≈ 9000 rpm

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A 1.5m wire carries a 7 A current when a potential difference of 68 V is applied. What is the resistance of the wire?
PSYCHO15rus [73]

Working...

length of wire L = 1.5 m

current I = 7 A

potential difference V = 68 Volt

According to Ohm's Law

V = IR

R = V/I

R = 68/7

R = 9.7 Ω

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3 years ago
How did amphibians adapt to their changing environment ?
kobusy [5.1K]

Answer:

I think it is B

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A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio
Angelina_Jolie [31]

a) -0.259 rad/s/y

b) 1732.8 years

c) 0.0069698 s

Explanation:

a)

The angular acceleration of a rotating object is equal to the rate of change of angular velocity of the object.

Mathematically, it is given by

\alpha=\frac{\Delta \omega}{\Delta t}

where

\Delta \omega is the change in angular velocity

\Delta t is the time elapsed

The angular velocity can be written as

\omega=\frac{2\pi}{T}

where T is the period of rotation of the object.

Therefore, the change in angular velocity can be written as

\Delta \omega = \frac{2\pi}{T'}-\frac{2\pi}{T}=2\pi (\frac{1}{T'}-\frac{1}{T})

In this problem:

T = 0.0140 s is the initial period of the pulsar

The period increases at a rate of 8.09 x 10-6 s/y, so after 1 year, the new period is

T'=T+8.09\cdot 10^{-6} =0.01400809 s

Therefore, the change in angular velocity after 1 year is

\Delta \omega =2\pi (\frac{1}{0.01400809}-\frac{1}{0.0140})=-0.259 rad/s

So, the angular acceleration of the pulsar is

\alpha = \frac{-0.259 rad/s}{1 y}=-0.259 rad/s/y

b)

To solve this part, we can use the following equation of motion:

\omega'=\omega + \alpha t

where

\omega' is the final angular velocity

\omega is the initial angular velocity

\alpha is the angular acceleration

t is the time

For the pulsar in this problem:

\omega=\frac{2\pi}{T}=\frac{2\pi}{0.0140}=448.8 rad/s is the initial angular velocity

\omega'=0, since we want to find the time t after which the pulsar stops rotating

\alpha = -0.259 rad/s/y is the angular acceleration

Therefore solving for t, we find the time after which the pulsar stops rotating:

t'=-\frac{\omega}{\alpha}=-\frac{448.8}{-0.259}=1732.8 y

c)

As we said in the previous part of the problem, the rate of change of the period of the pulsar is

\frac{\Delta T}{\Delta t}=8.09\cdot 10^{-6} s/y

which means that the period of the pulsar increases by

\Delta T=8.09\cdot 10^{-6} s

For every year:

\Delta t=1 y

From part A), we also know that the current period of the pulsar is

T = 0.0140 s

The current period is related to the initial period of the supernova by

T=T_0+\frac{\Delta T}{\Delta t}\Delta t

where T_0 is the original period and

\Delta t=869 y

is the time that has passed; solving for T0,

T_0=T-\frac{\Delta T}{\Delta t}\Delta t=0.0140 - (8.09\cdot 10^{-6})(869)=0.0069698 s

6 0
3 years ago
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Answer:

F<em>lift=</em>5.00×10³ N

F<em>horizontal</em>=0.50×10³ N

F<em>g</em>=4.80×10³ N

Explanation:

A free-body diagram of the glider.

pretend this is a number line

^

|  F<em>lift</em>

|  

|

|---->F<em>horizontal</em>

|

| F<em>g</em>

|

∨

hope this helps! :)

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The interaction of light that distorted the image of where Bill's goggles appear to be is Refraction.

<h3>What happens to the speed of light when it hits water in a swimming pool?</h3>

The light slows down when it passes from the less dense air into the water, this slow down of the ray of light which  causes the ray of light to change direction and then the changes the speed of the light that causes refraction.

The swimming pool appears shallow than it appear to be  because the rays of light coming from the bottom of the swimming pool that is refracted by the water making the height of the swimming pool to be above the actual swimming pool bed and there by misleading the observer(Bill).

Get more information about Refraction of light herebrainly.com/question/15838784

#SPJ1

4 0
1 year ago
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