All categories

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

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 .

You might be interested in

What type of energy does the horse have because of its motion?

Alekssandra [29.7K]

is the horse running/moving? if so then kinetic i believe

7 0

3 years ago

A 37 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar

Ne4ueva [31]

Answer:

The frictional force acting on the block is 14.8 N.

Explanation:

Given that,

Weight of block = 37 N

Coefficients of static = 0.8

Kinetic friction = 0.4

Tension = 24 N

We need to calculate the maximum friction force

Using formula of friction force

$f=\mu mg$

Put the value into the formula

$f=0.8\times37$

$f=29.6\ N$

So, the tension must exceeds 29.6 N for the block to move

We need to calculate the frictional force acting on the block

Using formula of frictional force

$f = \mu N$

Put the value in to the formula

$f=0.4\times37$

$f=14.8\ N$

Hence, The frictional force acting on the block is 14.8 N.

6 0

3 years ago

A Tennis ball falls from a height 40m above the ground the ball rebounds

worty [1.4K]

If the ball is dropped with no initial velocity, then its velocity v at time t before it hits the ground is

v = -g t

where g = 9.80 m/s² is the magnitude of acceleration due to gravity.

Its height y is

y = 40 m - 1/2 g t²

The ball is dropped from a 40 m height, so that it takes

0 = 40 m - 1/2 g t²

==> t = √(80/g) s ≈ 2.86 s

for it to reach the ground, after which time it attains a velocity of

v = -g (√(80/g) s)

==> v = -√(80g) m/s ≈ -28.0 m/s

During the next bounce, the ball's speed is halved, so its height is given by

y = (14 m/s) t - 1/2 g t²

Solve y = 0 for t to see how long it's airborne during this bounce:

0 = (14 m/s) t - 1/2 g t²

0 = t (14 m/s - 1/2 g t)

==> t = 28/g s ≈ 2.86 s

So the ball completes 2 bounces within approximately 5.72 s, which means that after 5 s the ball has a height of

y = (14 m/s) (5 s - 2.86 s) - 1/2 g (5 s - 2.86 s)²

==> (i) y ≈ 7.5 m

(ii) The ball will technically keep bouncing forever, since the speed of the ball is only getting halved each time it bounces. But y will converge to 0 as t gets arbitrarily larger. We can't realistically answer this question without being given some threshold for deciding when the ball is perfectly still.

During the first bounce, the ball starts with velocity 14 m/s, so the second bounce begins with 7 m/s, and the third with 3.5 m/s. The ball's height during this bounce is

y = (3.5 m/s) t - 1/2 g t²

Solve y = 0 for t :

0 = (3.5 m/s) t - 1/2 g t²

0 = t (3.5 m/s - 1/2 g t)

==> (iii) t = 7/g m/s ≈ 0.714 s

As we showed earlier, the ball is in the air for 2.86 s before hitting the ground for the first time, then in the air for another 2.86 s (total 5.72 s) before bouncing a second time. At the point, the ball starts with an initial velocity of 7 m/s, so its velocity at time t after 5.72 s (but before reaching the ground again) would be

v = 7 m/s - g t

At 6 s, the ball has velocity

(iv) v = 7 m/s - g (6 s - 5.72 s) ≈ 4.26 m/s

4 0

4 years ago

What is the kinetic energy of a 150 gram object moving at a velocity of 100 m/s?

Alex Ar [27]

Answer:

750 J

Explanation:

lets convert mass into kg first , 150 /1000 = 0.15 kg

kinetic energy = $\frac{mv^{2} }{2}$ = $\frac{0.15*100^{2} }{2} = 750 J$

3 0

3 years ago

Which formula correctly expresses the property density?

Olenka [21]

Answer:

Option A

D = m/v

Explanation:

Density is defined as mass per unit volume of an object. Therefore, D=m/v where m is the mass of the object and v is the volume

Therefore, option A is the right option

6 0

3 years ago