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3 years ago

The product of charge through, and potential across, an electrical device is:

2 answers:

5 0

The option is Work.

The product of charge and potential is equal to the energy. Adn, as we know work is related to energy as the capacity to do work.

Alos, because, Potential is given as, V = E/q
or E = Vq
Thus, t<span>he product of charge through, and potential across, an electrical device is:work
</span>

gregori [183]3 years ago

3 0

Answer:

The product of the charge and voltage is the work .

Explanation:

According to Ohm's law

The voltage is equal to the product of the current and resistance. Its unit is volt.

$V= I\times R$

Power :

Power is the defined as the electric energy per unit time. its unit is J/s or watt.

$P=\dfrac{E}{t}$

According to Ohm's law,

The power is equal to the square of voltage divided by the resistance.

$P=\dfrac{V^2}{R}$

The power equal to the product of resistance and square of current.

$P = R\times I^2$

The product of current and voltage is equal to the power.

$P=I\times V$

Current :

The current is equal to the voltage divided by the resistance.

The ampere is the unit of current.

Work : Work is related to the energy.

The Voltage is equal to the energy divided by the charge.

$V=\dfrac{E}{q}$

The product of charge and potential is called energy.

$E=q\times V$

Hence, The product of the charge and voltage is the work .

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Assuming a 8 kilogram bowling ball moving at 2 m/s bounces off a spring at the same speed that had before bouncing what is the a

Naya [18.7K]

a) 32 kg m/s

Assuming the spring is initially at rest, the total momentum of the system before the collision is given only by the momentum of the bowling ball:

$p_i = m u = (8 kg)(2 m/s)=16 kg m/s$

The ball bounces off at the same speed had before, but the new velocity has a negative sign (since the direction is opposite to the initial direction). So, the new momentum of the ball is:

$p_{fB}=m v_b =(8 kg)(-2 m/s)=-16 kg m/s$

The final momentum after the collision is the sum of the momenta of the ball and off the spring:

$p_f = p_{fB}+p_{fS}$

where $p_{fS}$ is the momentum of the spring. For the conservation of momentum,

$p_i = p_f\\p_i = p_{fB}+p_{fS}\\p_{fS}=p_i -p_{fB}=16 kg m/s -(-16 kg m/s)=32 kg m/s$

b) -32 kg m/s

The change in momentum of bowling ball is given by the difference between its final momentum and initial momentum:

$\Delta p=p_{fb}-p_i=-16 kg m/s - 16 kg m/s=-32 kg m/s$

c) 64 N

The change in momentum is equal to the product between the average force and the time of the interaction:

$\Delta p=F \Delta t$

Since we know $\Delta t=0.5 s$ , we can find the magnitude of the force:

$F=\frac{\Delta p}{\Delta t}=\frac{-32 kg m/s}{0.5 s}=-64 N$

The negative sign simply means that the direction of the force is opposite to the initial direction of the ball.

d) The force calculated in the previous step (64 N) is larger than the force of 32 N.

5 0

3 years ago

What types of sound waves diffract more than others

cricket20 [7]

Answer:

Infrasound

but the longer the wavelength of a soundwave, the more it diffracts

Explanation:

7 0

2 years ago

Two parallel slits are illuminated by light composed of two wavelengths. One wavelength is λA = 622nm. The other wavelength is λ

Sergio039 [100]

Answer:

$\lambda_{B}=414.67 nm$

Explanation:

In this question we have given

$\lambda_{A}=622nm$

we have to find

$\lambda_{B}=?$

We know that

optical path difference for bright fringe is given as $=n\lambda$

Here,

n is order of fringe

and optical path difference for dark fringe is given as $=(n+.5)\lambda$

since the light with wavelength $\lambda_{A}$ produces its third-order bright fringe at the same place where the light with wavelength $\lambda_{B}$ produces its fourth dark fringe

it means

optical path difference for 3rd order bright fringe= optical path difference for forth order dark fringe

Therefore,

$3\lambda_{A}=(4+.5)\lambda_{B}$ ...............(1)

Put value of $\lambda_{A}$ in equation (1)

$3 \times 622=(4+.5)\lambda_{B}$

$1866=4.5\lambda_{B}$

$\lambda_{B}=414.67 nm$

3 0

2 years ago

A ball is hit with a paddle, causing it to travel straight upward. It takes 2.90 s for the ball to reach its maximum height afte

zmey [24]

Answer:

A. 28.42 m/s

B. 41.21 m

Explanation:

From the question given above, the following data were obtained:

Time (t) to reach the maximum height = 2.90 s

Initial velocity (u) =?

Maximum height (h) =?

A. Determination of the initial velocity of the ball.

Time (t) to reach the maximum height = 2.90 s

Final velocity (v) = 0 m/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =?

v = u – gt (since the ball is going against gravity)

0 = u – (9.8 × 2.9)

0 = u – 28.42

Collect like terms

0 + 28.42 = u

u = 28.42 m/s

Thus, the initial velocity of the ball is 28.42 m/s

B. Determination of the maximum height reached by the ball.

Final velocity (v) = 0 m/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) = 28.42 m/s

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 28.42² – (2 × 9.8 × h)

0 = 807.6964 – 19.6h

Collect like terms

0 – 807.6964 = – 19.6h

– 807.6964 = – 19.6h

Divide both side by – 19.6

h = – 807.6964 / – 19.6

h = 41.21 m

Thus, the maximum height reached by the ball is 41.21 m

5 0

3 years ago

It is known that the population mean for the verbal section of the SAT is 500 with a standard deviation of 100. In 2006, a sampl

kvasek [131]

Answer

given,

SAT is 500 with a standard deviation of 100.

a sample of 400 students whose family income was between $70,000 and $80,000 had an average verbal SAT score of 511.

sample mean = $\dfrac{standard\ deviation}{\sqrt{n}}$

= $\dfrac{100}{\sqrt{400}}$

= 5

95% confidence level is achieved within +/- 1.960 standard deviations.

1.960 standard deviations x 5 is equal to +/- 9.8

confidence interval = 511 - 9.8 --- 511 + 9.8

= 501.2-----520.8

4 0

2 years ago