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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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<h2>Answer:</h2>

16°C

<h2>Explanation:</h2>

The equation relating temperature with resistance is given as follows;

R_{T} = R_{0} [\alpha(T^{} - T_{0}) + 1]         --------------(i)

<em>Where;</em>

T_{0} = Temperature at some reference point

R_{0} = Resistance at the reference point

T = Temperature at some other point

R_{T} = Resistance at the other point

\alpha = constant called "temperature coefficient of the resistor material"

<em>From the question; </em>

Let's take the reference point temperature (T_{0}) as 4.00°C;

Therefore;

R_{0} = Resistance at the reference point = 217.4Ω

R_{T} = Resistance at the other point = 216.0Ω

\alpha = Temperature coefficient of the resistor material (carbon) = -0.0005/°C

<em>Now substitute these values into equation (i) as follows;</em>

216.0 = 217.4 [(-0.0005)(T^{} - 4) + 1]

216.0 = 217.4 [-0.0005T^{} + 0.002 + 1]

216.0 = 217.4 [-0.0005T^{} + 1.002]

<em>Divide through by 217.4 as follows;</em>

0.994 = 1.002 - 0.0005T

<em>Solve for T;</em>

0.0005T = 1.002 - 0.994

0.0005T = 0.008

T = 0.008 / 0.0005

T = 16°C

Therefore, the temperature on a spring day at that resistance is 16°C

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The final velocity after the collision is 8.2 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, if we consider the system to be isolated (=no external unbalanced forces), the total momentum of the raindrop+mosquito must be conserved before and after the collision.

If the collision is perfectly inelastic, moreover, the raindrop and the mosquito stick together and travel at the same velocity v after the collision.

Mathematically:

p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v  

where:  

m_1 is the mass of the first mosquito

u_1 = 0 is the initial velocity of the mosquito

m_2 = 50 m_1 is the mass of the raindrop

u_2 = 8.4 m/s is the initial velocity of the raindrop

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Re-arranging the equation and substituting, we find:  

m_1 u_1 + 50 m_1 u_2 = (m_1 + 50 m_1) v\\50 m_1 u_2 = 51 m_1 v\\50 u_2 = 51 v\\v=\frac{50}{51}u_2 = \frac{50}{51}(8.4)=8.2 m/s

Learn more about momentum here:

brainly.com/question/7973509  

brainly.com/question/6573742  

brainly.com/question/2370982  

brainly.com/question/9484203  

#LearnwithBrainly

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The blue car is moving with a velocity of -14 m/s using the red car as a reference frame.
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