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3 years ago

I need help ASAP!

Physics

1 answer:

san4es73 [151]3 years ago

7 0

Answer:

Explanation:

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telo118 [61]

i believe the answer is C.

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4 years ago

Explain the different types of energy in a working wind turbine

tatiyna

Answer:

A wind turbine converts the kinetic energy of the wind into mechanical power.

The wind moves the turbine and the turbine produce energy.

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3 years ago

A drag racer starts from rest and accelerates at 7.4 m/s2. How far will he travel in 2.0 seconds?

Leto [7]

Using the kinematic equation below we can determine the distance traveled if t=2, a=7.4m/s^2. First we must determine the final velocity:

$v_{final}=v_{initial}+\frac{1}{2}at\\\\v_{final}=0+(7.4m/s^2)(2s)=34.8m/s$

Now we will determine the distance traveled:

$v_{final}^2=v_{initial}^2+2a \Delta x\\\\\Delta x = \frac{v_{final}^2}{2a} =\frac{(34.8)^2}{(2)(7.4)}=81.83 m$

Therefore, the drag racer traveled 81.83 meters in 2 seconds.

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3 years ago

The hottest recorded temperature in the history of the United States is 134 °F, which is 57 °C. What would the speed of sound at

guapka [62]

Answer:

365 m/s

Explanation:

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2 years ago

The terminal velocity is not dependent on which one of the following properties? the drag coefficient 1 the force of gravity 2 c

ahrayia [7]

<h2>Answer: the falling time</h2>

Explanation:

When a body or object falls, basically two forces act on it:

1. The force of air friction, also called "drag force" $D$ :

$D={C}_{d}\frac{\rho V^{2} }{2}A$ (1)

Where:

$C_ {d}$ is the drag coefficient

$\rho$ is the density of the fluid (air for example)

$V$ is the velocity

$A$ is the transversal area of the object

So, this force is proportional to the transversal area of the falling element and to the square of the velocity.

2. Its weight due to the gravity force $W$ :

$W=m.g$

(2)

Where:

$m$ is the mass of the object

$g$ is the acceleration due gravity

So, at the moment when the drag force equals the gravity force, the object will have its terminal velocity:

$D=W$ (3)

${C}_{d}\frac{\rho V^{2} }{2}A=m.g$ (4)

$V=\sqrt{\frac{2m.g}{\rho A{C}_{d}}}$ (5) This is the terminal velocity

As we can see, there is no "falling time" in this equation.

Therefore, the terminal velocity is not dependent on the falling time.

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3 years ago