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3 years ago

A battery derives electric energy from _____energy?

Physics

2 answers:

snow_lady [41]3 years ago

5 0

The answer is d. chemical energy

sammy [17]3 years ago

4 0

A). nuclear
No. There were batteries long long before we learned
how to use nuclear energy. Also, there is no danger of
exposure to radioactivity when you're working with a battery.

b). mechanical
No. A battery has no moving parts.

c). gravitational
No. No matter how high you take a battery in an airplane, or
how far you lower it into a mine-shaft, its characteristics don't
change. In fact, batteries even work on things that are in orbit.

d). chemical
Bingo.

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Sedaia [141]

Answer:

V = 20.5 m/s

Explanation:

Given,

The mass of the cart, m = 6 Kg

The initial speed of the cart, u = 4 m/s

The acceleration of the cart, a = 0.5 m/s²

The time interval of the cart, t = 30 s

The final velocity of the cart is given by the first equation of motion

v = u + at

= 4 + (0.5 x 30)

= 19 m/s

Hence the final velocity of cart at 30 seconds is, v = 19 m/s

The speed of the cart at the end of 3 seconds

V = 19 + (0.5 x 3)

= 20.5 m/s

Hence, the final velocity of the cart at the end of this 3.0 second interval is, V = 20.5 m/s

6 0

3 years ago

PLEASE HELP ME,, I WOULD BE SO HAPPY

Juliette [100K]

Answer:

Energy is force times distance. For your problem, no matter how long you push, the wall still goes nowhere, so there is no obvious energy transfer. so in conclusion, you actually didn't do anything :(

Explanation:

6 0

3 years ago

Read 2 more answers

g Suppose Howard is pulling a bucket of bricks up along the side of a building with a rope. The bricks have a mass of 20 kg and

cupoosta [38]

Answer:

= 236N

Explanation:

tension T = mg + ma

Given that,

m = 20kg

g = 9.8 m/s²

a = 2.0 m/s²

T = m(g + a)

T = 20( 9.8 + 2.0)

= 20(11.8)

= 236N

4 0

4 years ago

A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

-BARSIC- [3]

Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be:

$\displaystyle q(t) = q_0 \, e^{-t / \tau}$ .

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{8}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{8}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{8}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{8}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 8 - \ln 9$ .

$t = - \tau \, \left(\ln 8 - \ln 9\right) = \tau(\ln 9 - \ln 8)$ .

$\displaystyle q(t) = \left(1 - \frac{1}{9}\right) \, q_0 = \frac{7}{9}\, q_0$ .

Apply the equation $\displaystyle q(t) = q_0 \, e^{-t / \tau}$ :

$\displaystyle \frac{7}{9}\, q_0 = q_0 \, e^{-t/\tau}$ .

The goal is to solve for $t$ in terms of $\tau$ . Rearrange the equation:

$\displaystyle e^{-t/\tau} = \frac{7}{9}$ .

Take the natural logarithm of both sides:

$\displaystyle \ln\, e^{-t/\tau} = \ln \frac{7}{9}$ .

$\displaystyle -\frac{t}{\tau} = \ln 7 - \ln 9$ .

$t = - \tau \, \left(\ln 7 - \ln 9\right) = \tau(\ln 9 - \ln 7)$ .

7 0

3 years ago

A resistor R1 is wired to a battery, then resistor R2 is added in series. Are (a) the potential difference across R1 and (b) the

tia_tia [17]

Answer:

Explanation:

a ) Earlier emf of cell applied on R₁ but now emf will be distributed among R₁ and R₂

Potential difference on R₁ will become less .

b ) Current is inversely proportional to resistance of the circuit. As resistance increases , current will be less . So current through R₁ will become less.

c )

When resistance is added in series , they are added up to obtain equivalent resistance . So equivalent resistance R₁₂ will be more than R₁ OR R₂.

6 0

3 years ago