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3 years ago

11

What is the starting and final energy for a battery?

1 answer:

EleoNora [17]3 years ago

8 0

Electrical energy is the starting and final energy of a battery

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The force it would take to accelerate a 900-kg car at a rate of 3 m/s2 is

Vlad1618 [11]

F = ma

F = (3)(900)

F = 2700 N

4 0

3 years ago

Read 2 more answers

Tcecarenko [31]

The answer of this question is B.

5 0

3 years ago

Which of the following accurately describes the behavior of water when subject to temperature change? A. The volume of water wil

nata0808 [166]

Choices 'C' and 'D' are both correct.

(Except in 'C', changing the temperature from 1°C to 3°C is not usually
described as 'cooling', and it's not the water's 'mass' that changes. But
water does contract in volume during that change.)

8 0

3 years ago

We decided to make an iced latte by adding ice to a 200 mL hot latte at 45 °C. The ice starts out at 0 C. How much ice do we nee

tankabanditka [31]

Answer:

m = 77.75 g

Explanation:

Here we know that at equilibrium the temperature of the system will be 10 degree C

so heat given by hot latte = heat absorbed by the ice

now we have

heat given by latte = $m s\Delta T$

$Q_1 = (200)(4.186)(45 - 10)$

$Q_1 = 29302 J$

now heat absorbed by ice is given as

$Q_2 = mL + ms\Delta T$

$Q_2 = m(335 + 4.186(10 - 0))$

$Q_2 = m(376.86)$

now by heat balance we have

$Q_1 = Q_2$

$29302 = m(376.86)$

$m = 77.75 g$

6 0

3 years ago

a sphere of diameter 6•0cm is moulded into a thin uniform wire of diameter 0•2mm.calculate the length of the wire in metres

Scilla [17]

The length of the wire is 36 m.

<u>Explanation:</u>

Given, Diameter of sphere = 6 cm

We know that, radius can be found by taking the half in the diameter value. So,

$\text { sphere radius, } R=\frac{D}{2}=\frac{6}{2}=3 \mathrm{cm}=3 \times 10^{-2} \mathrm{m}$

Similarly,

$\text { wire radius, } r=\frac{0.2}{2}=0.1 \mathrm{mm}=1 \times 10^{-3} \mathrm{m}$

We know the below formulas,

$\text {volume of sphere}=\frac{4}{3} \times \pi \times R^{3}$

$\text {volume of wire}=\pi \times r^{2} \times l$

When equating both the equations, we can find length of wire as below, where $\pi=\frac{22}{7}$

$\frac{4}{3} \times \pi \times R^{3}=\pi \times r^{2} \times l$

$\frac{4}{3} \times \frac{22}{7} \times\left(3 \times 10^{-2}\right)^{3}=\frac{22}{7} \times\left(1 \times 10^{-3}\right)^{2} \times l$

The $\pi$ value gets cancelled as common on both sides, we get

$\frac{4}{3} \times 27 \times 10^{-6}=10^{-6} \times l$

The $10^{-6}$ value gets cancelled as common on both sides, we get

$l=4 \times 9=36 m$

7 0

3 years ago