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3 years ago

Explain how a battery in a circuit is similar to a water pump.

1 answer:

5 0

<span>A battery creates a potential difference, or voltage, which is really just storing electrical potential energy just as a pump would by moving water up against gravity, thereby storing gravitational potential energy. To complete the analogy, as current flows through the circuit, this potential energy is released and can be used by the components in the circuit. This is the same as water falling back down and turning some wheel or turbine and converting its gravitational potential energy into mechanical energy.</span>

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What is the current if a charge of 25 coulombs passes in 5 seconds?

Jobisdone [24]

Answer: 5 amps

Explanation: I = Q/t

8 0

3 years ago

What is the magnitude of the change in potential energy of the block-spring system when it travels from its lowest vertical posi

maksim [4K]

Answer:

ΔU = 2 mg h

Explanation:

In a spring mass system the potential energy is U = m g h

where h is measured from the equilibrium point of the spring

the potential energy at the highest point is

U₁ = m g h

the potential energy at the lowest point is

U₂ = m g (-h)

instead in this energy it is

ΔU = 2 mg h

In this two points the kinetic energy is zero, but there is elastic potential energy that has the same value in the two points, so its change is zero

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3 years ago

True or false? 1:The pH scale measures the concentration of hydroxide ions.

lukranit [14]

That would be false hope this helps

6 0

3 years ago

Read 2 more answers

A mover brings a box up the stairs in 10 seconds. If he applied a force of 20 N over a distance 10 m on the box, calculate the p

Softa [21]

Answer:

20 Watts

Explanation:

Work = force × distance

W = (20 N) (10 m)

W = 200 Joules

Power = work / time

P = 200 J / 10 s

P = 20 Watts

8 0

3 years ago

Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless

drek231 [11]

Answer:

$v_f = 15 \frac{m}{s}$

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum $\vec{L}$ is

$\vec{L} = \vec{r} \times \vec{p}$

where $\vec{r}$ is the position and $\vec{p}$ the linear momentum.

We also know that the torque is

$\vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d}{dt} ( \vec{r} \times \vec{p} )$

$\vec{\tau} = \frac{d}{dt} \vec{r} \times \vec{p} + \vec{r} \times \frac{d}{dt} \vec{p}$

$\vec{\tau} = \vec{v} \times \vec{p} + \vec{r} \times \vec{F}$

but, as the linear momentum is $\vec{p} = m \vec{v}$ this means that is parallel to the velocity, and the first term must equal zero

$\vec{v} \times \vec{p}=0$

$\vec{\tau} = \vec{r} \times \vec{F}$

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

$\vec{\tau}_{rod} = 0$