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3 years ago

How would amperage and voltage affect the power of the fence

1 answer:

8 0

So it's easier to break amps and volts down as if they were water through a pipe. Voltage is the pressure at which the water is forced through so you can survive high voltage. While Amperes would be the amount or volume of water that flows out over a period of time it doesn't take much amps to kill you as goes the saying "amps give the cramps and volts give the jolts"

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What is the work done when a 400.n force is use to lift a 400.n object 3.5 meters strait up

alexandr1967 [171]

Answer:

i think it's 1400

Explanation:

w = Force x displacement

w = 400 x 3.5

3 0

3 years ago

9) If a wave has a speed of 362 m/s and a period of 4.17 s, what is its wavelength?

kakasveta [241]

Answer:

Option B (1.51 m)

Explanation:

U 2 can help me by marking as brainliest........

5 0

2 years ago

Read 2 more answers

A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet

dimaraw [331]

Answer:

$v = 2.8898 \frac{m}{s}$

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

$E_{ep} = \frac{1}{2} k (\Delta x)^2$

where $\Delta x$ is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

$\vec{F} = - k \Delta \vec{x}$

as we need 9.12 N to compress 4.60 cm, this means:

$k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}$

$k = 198.26 \ \frac{ N}{m}$

So, the elastic energy of the compressed spring is:

$E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2$

$E_{ep} = 0.209759 \ Joules$

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

$\Delta E_{gp} = m g \Delta h$

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

$\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m$

$\Delta E_{gp} = 0.00224 \ Joules$

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

$E_k = \frac{1}{2} m v^2$

so, for the pellet will be

$E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

<h3>All together</h3>

By conservation of energy, we know:

$E_{ep} = \Delta E_{gp} + E_k$

$0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.209759 \ Joules - 0.00224 \ Joules$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.207519 \ Joules$

$v = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }$

$v = 2.8898 \frac{m}{s}$

7 0

3 years ago

Who speaks the line "Lord, what fools these mortals be"?

ivanzaharov [21]

The answer is D.Puck.

♡♡Hope I helped!!! :)♡♡

3 0

3 years ago

Which statement correctly defines power?

Anna007 [38]

Answer:

$Power=V\, I$ which corresponds to the second option shown: "voltage times amperage"

Explanation:

The electric power is the work done to move a charge Q across a given difference of potential V per unit of time.

Since such electrical work is the product of the potential difference V times the charge that moves through that potential, and this work is to be calculated by the unit of time, we need to divide the product by time (t) which leads to the following final simple equation:

$Power=\frac{V\,Q}{t} =V\,\frac{Q}{t} = V\, I$

Notice that we replaced the quotient representing charge per unit of time (Q/t) by the actual current running through the circuit.

This corresponds to the second option shown in the question: "Voltage times amperage".

6 0

3 years ago