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2 years ago

How many joules is 3 lightning bolts?

Physics

2 answers:

pogonyaev2 years ago

6 0

Answer:

one billion to ten billion joules.

Explanation:

To keep a 100-watt light bulb going for one second, one hundred joules of energy will be used.

xxTIMURxx [149]2 years ago

4 0

Answer:

In 3 lightning bolts, there is approximately 1 billion to ten billion joules.

Explanation:

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In a Millikan experiment, a droplet of mass 4.7 x 10^-15 kg floats in an electric field of 3.20 x 104 N/C.

andrezito [222]

Answer:

a. The force of gravity on the droplet is approximately 4.606 × 10⁻¹⁴ Newtons

b. The electric force that balances the force of gravity is approximately -4.606 × 10⁻¹⁴ Newtons

c. The excess charge is approximately -1.439375 × 10⁻¹⁸ C

d. There are approximately 9 excess electrons on the droplet

Explanation:

The parameters of the Millikan experiment are;

The mass of the droplet, m = 4.7 × 10⁻¹⁵ kg

The electric field in which the droplet floats, E = 3.20 × 10⁴ N/C

a. The force of gravity on the droplet, F = The weight of the droplet, W = m × g

Where;

g = The acceleration due to gravity ≈ 9.8 m/s²

W = 4.7 × 10⁻¹⁵ kg × 9.8 m/s² = 4.606 × 10⁻¹⁴ Newtons

∴ The force of gravity on the droplet = W = 4.606 × 10⁻¹⁴ Newtons

b. The electric force that balances the force of gravity, $F_v$ = -W = -4.606 × 10⁻¹⁴ Newtons

c. The excess charge is given as follows;

$F_v$ = q·E

∴ The electric force that balances the force of gravity, $F_v$ = q·E = -4.606 × 10⁻¹⁴ N

q·E = -4.606 × 10⁻¹⁴ N

q = -4.606 × 10⁻¹⁴ N/E

∴ q = -4.606 × 10⁻¹⁴ N/(3.20 × 10⁴ N/C) ≈ -1.439375 × 10⁻¹⁸ C

The excess charge, q ≈ -1.439375 × 10⁻¹⁸ C

d. The charge of one electron, e = 1.602176634 × 10⁻¹⁹C

The number of excess electrons in the droplet, n, is given as follows;

n = 1.439375 × 10⁻¹⁸ C/(1.602176634 × 10⁻¹⁹C) = 8.98387212405 electrons

∴ n ≈ 9 electrons.

7 0

2 years ago

In 8,450 seconds, the number of radioactive nuclei decreases to 1/16 of the number present initially. What is the half-life (in

almond37 [142]

Answer:

<h2>2113 seconds</h2>

Explanation:

The general decay equation is given as $N = N_0e^{-\lambda t} \\\\$ , then;

$\dfrac{N}{N_0} = e^{-\lambda t} \\$ where;

$N/N_0$ is the fraction of the radioactive substance present = 1/16

$\lambda$ is the decay constant

t is the time taken for decay to occur = 8,450s

Before we can find the half life of the material, we need to get the decay constant first.

Substituting the given values into the formula above, we will have;

$\frac{1}{16} = e^{-\lambda(8450)} \\\\Taking\ ln\ of \both \ sides\\\\ln(\frac{1}{16} ) = ln(e^{-\lambda(8450)}) \\\\\\ln (\frac{1}{16} ) = -8450 \lambda\\\\\lambda = \frac{-2.7726}{-8450}\\ \\\lambda = 0.000328$

Half life f the material is expressed as $t_{1/2} = \frac{0.693}{\lambda}$

$t_{1/2} = \frac{0.693}{0.000328}$

$t_{1/2} = 2,112.8 secs$

Hence, the half life of the material is approximately 2113 seconds

7 0

2 years ago

You pour 250 g of tea into a Styrofoam cup, initially at 80∘C and stir in a little sugar using a 100-g aluminum 20∘C spoon and l

jok3333 [9.3K]

To solve this problem it is necessary to apply the concepts related to the conservation of energy and heat transferred in a body.

By definition we know that the heat lost must be equal to the heat gained, ie

$Q_g = Q_l$

Where,

Q = Heat exchange

The heat exchange is defined as

$Q = c_p m \Delta T$

Where,

$c_p =$ Specific heat

m = mass

$\Delta T=$ Change in Temperature

Therefore replacing we have that

$Q_g = Q_l$

$c_{p-tea} m \Delta T = c_{p-al} m \Delta T$

Replacing with our values we have that

$0.25*4180*(80-T) = 0.1*900*(T-20)$

$11.61*(80-T) = T-20$

$T= \frac{948.8}{11.61}$

$T = 75.24\°C$

Therefore the highest possible temperature of the spoon when you finally take it out of the cup is 75.24°C

3 0

2 years ago

A speed-time graph is shown below:

VladimirAG [237]

Answer: $0.5 m/s^{2}$

Explanation:

Average acceleration $a_{ave}$ is the variation of velocity $\Delta V$ over a specified period of time $\Delta t$ :

$a_{ave}=\frac{\Delta V}{\Delta t}}$

Where:

$\Delta V=V_{f}-V_{o}$ being $V_{o}=0 cm/s$ the initial velocity and $V_{f}=4 cm/s$ the final velocity (according to the information given from the described graph)

$\Delta t=8 s$

Then:

$a_{ave}=\frac{4 cm/s -0 cm/s}{8 s}}$

$a_{ave}=0.5 m/s^{2}$

5 0

2 years ago

e force acting between two charged particles A and B is 5.2 × 10-5 newtons. Charges A and B are 2.4 × 10-2 meters apart. If the

anastassius [24]

The force acting between the particles is

$F=k \frac{Q_{1}Q_{2}}{r^2}$
Then
$Q_{2}= \frac{5.2 \times 10^-^5 \times 0.024^2}{ 9.0 \times 10^9 7.2 \times 10^-^8} =4.622 \times 10^-^1^1C$

7 0

2 years ago

Read 2 more answers