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ohaa [14]
3 years ago
13

The flash on a high-tech camera contains a capacitor of 750 μ F. The battery in the camera supplies 330 V. (a) Determine the ene

rgy that is used to produce the flash. (b) Assuming that the flash lasts for 5 ms, find the power of the flash.
Physics
1 answer:
Arada [10]3 years ago
3 0
Two equations for this one: Q=CV  & PE=(1/2)QV. 
C=750 * 10^-6 F , V=330 V . 
so if you use Q=CV=(<span>750 * 10^-6)(330). Q=0.2475

Then solve for PE. PE=(1/2)QV= (1/2)(0.2475)(330)= 40.8 J
So PE=40.8 J</span>
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_____________ circular motion occurs when an object is traveling with constant speed in a circle.
emmasim [6.3K]
Your answer is ''Uniform''.

Hope this helps :)
7 0
2 years ago
the time required for one cycle, a complete motion that returns to its starting point, it called the_____ medium frequency perio
telo118 [61]

Answer:

The correct answer to the following question will be "Period".

Explanation:

The Period seems to be the time deemed necessary for such a perfect cycle of vibration to transfer a particular moment. Because as the amplitude of the wave raises, the wavelength falls.

It is denoted by "T" and its formula will be:

⇒  T  = \frac{1}{F}

Where, T = Period

            F = Frequency

The other given choices are not related to the given circumstances. So that the above would be the right answer.

4 0
3 years ago
In a hydroelectric power plant, 65 m3 /s of water flows from an elevation of 90-m to a turbine-generator, where electricity is g
Mariulka [41]

The electric output of the plant is 48.19 MW

First we need to calculate the water power, it is given by the formula

WP=ρQgh

Here, ρ=1000 kg/m3 is density of water,Q is the flow rate, g is the gravity, and h is the water head

Therefore, WP=1000*65*9.81*90=57388500 W=57.38 MW

Now the overall efficiency of the hydroelectric power plant is given as

η=\frac{electric power}{water power}

Plugging the values in the above equation

0.84=EP/57.38

EP=48.19 MW

Therefore, the electric output of the plant is 48.19 MW.

3 0
2 years ago
The 45-g arrow is launched so that it hits and embeds in a 1.40 kg block. The block hangs from strings. After the arrow joins th
worty [1.4K]

Question: How fast was the arrow moving before it joined the block?

Answer:

The arrow was moving at 15.9 m/s.

Explanation:

The law of conservation of energy says that the kinetic energy of the arrow must be converted into the potential energy of the block and arrow after it they join:

\dfrac{1}{2}m_av^2 = (m_b+m_a)\Delta Hg

where m_a is the mass of the arrow, m_b is the mass of the block, \Delta H of the change in height of the block after the collision, and v is the velocity of the arrow before it hit the block.

Solving for the velocity v, we get:

$v = \sqrt{\frac{2(m_b+m_a)\Delta Hg}{m_a} } $

and we put in the numerical values

m_a = 0.045kg,

m_b = 1.40kg,

\Delta H = 0.4m,

g= 9.8m/s^2

and simplify to get:

\boxed{ v= 15.9m/s}

The arrow was moving at 15.9 m/s

6 0
3 years ago
Green plants convert light energy from the sun into ______. * a. gravitational potential energy b. chemical potential energy c.
katrin [286]

Answer:

chemical potiential energy

6 0
3 years ago
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