Which waves give off heat?

What does fitness mean to you?

Lana71 [14]

Answer:

fitness is a physical exercise to mainten the body healthy and fit

5 0

3 years ago

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You dribble a basketball while walking on a basketball court. List and describe at least 3

Airida [17]

Answer:

1. You push on the ball and the ball pushes on your hand .

2. The ball hits the ground and the ground pushes back on the ball .

3. You walk on the ground with your feet and the ground pushes back on you.

Explanation:

4 0

4 years ago

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A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?What is the total energy stores in the

Rama09 [41]

1) $1.11\cdot 10^{-7} J$

The capacitance of a parallel-plate capacitor is given by:

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0$ is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

$r=\frac{2.0 cm}{2}=1.0 cm=0.01 m$

so the area is

$A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2$

While the separation between the plates is

$d=0.50 mm=5\cdot 10^{-4} m$

So the capacitance is

$C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F$

And now we can find the energy stored,which is given by:

$U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J$

2) 0.71 J/m^3

The magnitude of the electric field is given by

$E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m$

and the energy density of the electric field is given by

$u=\frac{1}{2}\epsilon_0 E^2$

and using

$E=4\cdot 10^5 V/m$ , we find

$u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3$

7 0

4 years ago

How do you find displacement when all you have is time with no velocity, and then in order to find velocity you have to have dis

saw5 [17]

If you really have nothing else but time, then you can't. There must be some other shred of information. Search around. Look under rocks.

3 0

4 years ago

The electric output of a power plant is 716 MW. Cooling water is the main way heat from the powerplant is rejected, and it flows

Stels [109]

Answer:

(a) 83475 MW

(b) 85.8 %

Explanation:

Output power = 716 MW = 716 x 10^6 W

Amount of water flows, V = 1.35 x 10^8 L = 1.35 x 10^8 x 10^-3 m^3

mass of water, m = Volume x density = 1.35 x 10^8 x 10^-3 x 1000

= 1.35 x 10^8 kg

Time, t = 1 hr = 3600 second

T1 = 25.4° C, T2 = 30.7° C

Specific heat of water, c = 4200 J/kg°C

(a) Total energy, Q = m x c x ΔT

Q = 1.35 x 10^8 x 4200 x (30.7 - 25.4) = 3 x 10^12 J

Power = Energy / time

Power input = $P = \frac{3 \times 10^{12}}{3600}=8.35 \times 10^{8}W$

Power input = 83475 MW

(b) The efficiency of the plant is defined as the ratio of output power to the input power.

$\eta =\frac{Power output}{Power input}$

$\eta =\frac{716}{83475}=0.858$

Thus, the efficiency is 85.8 %.

7 0

3 years ago