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3 years ago

How many cities could 8 TeV power for a week?

Physics

2 answers:

otez555 [7]3 years ago

7 0

Your answer is unable to be answered, only because I researched the question and nothing came up.

HACTEHA [7]3 years ago

6 0

science hasnt figured it out yet

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Green sea turtles eat sea grass that grows on the sea floor. Seagrass, a foundation species, provides an important breeding grou

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The answer is D. <span> There would be a decrease in the population of marine organisms. </span>

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3 years ago

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A mother pats a child every time he throws a chocolate wrapper in the dustbin. This is an example of (A) Observational Learning

TEA [102]

D. I think, not sure

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4 years ago

2. A rock is shot straight up into the air with a slingshot that had been stretched 0.30 m. Assume

Luba_88 [7]

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3 years ago

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Kelli weighs 425 N, and she is sitting on a playground swing that hangs 0.36 m above the ground. Her mom pulls the swing back an

kodGreya [7K]

Answer:

V = 3.54 m/s

Explanation:

Using the conservation of energy:

$E_i = E_f$

so:

$wh = \frac{1}{2}mV^2$

where w is te weigh of kelly, h the distance that kelly decends, m is the mass of kelly and V the velocity in the lowest position.

So, the mass of kelly is:

m = 425N/9.8 = 43.36 Kg

and h is:

h = 1m-0.36m =0.64m

then, replacing values, we get:

$(425N)(0.64m) = \frac{1}{2}(43.36kg)v^2$

Solving for v:

V = 3.54 m/s

7 0

3 years ago

A bug is 12 cm from the center of a turntable that is rotating with a frequency of 45 rev/min . What minimum coefficient frictio

Agata [3.3K]

Answer:

The minimum coefficient of friction is 0.27.

Explanation:

To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.

Centripetal force is written as

$F_c = m\frac{v^2}{r}$

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

$v = r\omega$

with ω denoting the angular velocity, which we are given. With that, the above becomes:

$F_c = m\frac{v^2}{r}=m\omega^2 r$

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

$F_r=\mu mg \geq m\omega^2 r = F_c\implies\\\mu \geq \frac{\omega^2 r}{g}$

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

$\frac{1 rev}{1 min}\cdot\frac{\frac{2\pi rad}{rev}}{\frac{60s}{1 min}}=\frac{2\pi}{60}\frac{rad}{s}$

and so 45 rev/min = 4.71 rad/s.

$\mu \geq \frac{\omega^2 r}{g}=\frac{4.71^2\frac{1}{s^2}\cdot 0.12m}{9.8\frac{m}{s^2}}=0.27$

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.

3 0

3 years ago