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3 years ago

How long could you survive without electricity? What parts of your life would be affected by loss of electricity?

Physics

1 answer:

Law Incorporation [45]3 years ago

6 0

Answer:

Depends on the person, if you are hooked to your phone which most people are these days, you might go a little crazy. But think cavemen or anybody long ago, they survived without electricity. People in Africa bathe in the lake or any water nearby, they also walk for miles just to get drinking water. So yes, if you really tried you could go without electricity for possibly your whole life. It sucks if you have homework though because you don't have Brainly :)

Explanation:

You might be interested in

A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When

11111nata11111 [884]

Answer:

The value is $R_f = \frac{4}{5} R$

Explanation:

From the question we are told that

The initial velocity of the proton is $v_o$

At a distance R from the nucleus the velocity is $v_1 = \frac{1}{2} v_o$

The velocity considered is $v_2 = \frac{1}{4} v_o$

Generally considering from initial position to a position of distance R from the nucleus

Generally from the law of energy conservation we have that

$\Delta K = \Delta P$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K = K__{R}} - K_i$

=> $\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta P = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P = k * \frac{q_1 * q_2 }{R} - 0$

$\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0$

=> $\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}$

=> $- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )$

Generally considering from initial position to a position of distance $R_f$ from the nucleus

Here $R_f$ represented the distance of the proton from the nucleus where the velocity is $\frac{1}{4} v_o$

Generally from the law of energy conservation we have that

$\Delta K_f = \Delta P_f$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K_f = K_f - K_i$

=> $\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance $R_f$ from the nucleus , this is mathematically represented as

$\Delta P_f = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P_f = k * \frac{q_1 * q_2 }{R_f } - 0$

$\frac{1}{2} * m * \frac{1}{8} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R_f }$

=> $\frac{1}{2} * m *v_o^2 [-\frac{15}{16} ] = k * \frac{q_1 * q_2 }{R_f }$

=> $- \frac{15}{32} * m *v_o^2 = k * \frac{q_1 * q_2 }{R_f } ---(2)$

Divide equation 2 by equation 1

$\frac{- \frac{15}{32} * m *v_o^2 }{- \frac{3}{8} * m *v_0^2 } } = \frac{k * \frac{q_1 * q_2 }{R_f } }{k * \frac{q_1 * q_2 }{R } }}$

=> $-\frac{15}{32 } * -\frac{8}{3} = \frac{R}{R_f}$

=> $\frac{5}{4} = \frac{R}{R_f}$

=> $R_f = \frac{4}{5} R$

7 0

3 years ago

100 °C is a greater temperature than which of the following?

ohaa [14]

When somebody hands you a Celsius°, it's easy to find the equivalent Fahrenheit°.

Fahrenheit° = (1.8 · Celsius°) + 32° .

So 100°C works out to 212°F.

It's also easy to find the equivalent Kelvin. Just add 273.15 to the Celsius.

So now you can see that 100°C is equal to A and D,
and it's less than B .

The only one it's greater than is C .

6 0

2 years ago

Which is the mode of the following data?<br>120, 125, 130, 125, 135

Aleks [24]

The mode in this case would be 125 because it occurs the most in the sequence of numbers.

8 0

3 years ago

Read 2 more answers

A 75.0 kg diver falls from rest into a swimming pool from a height of 5.10 m. It takes 1.34 s for the diver to stop after enteri

vichka [17]

Answer:

559.5 N

Explanation:

Applying,

v² = u²+2gs............. Equation 1

Where v = final velocity,

From the question,

Given: s = 5.10 m, u = 0 m/s ( from rest)

Constant: 9.8 m/s²

Therefore,

v² = 0²+2×9.8×5.1

v² = 99.96

v = √(99.96)

v = 9.99 m/s

As the diver eneters the water,

u = 9.99 m/s, v = 0 m/s

Given: t = 1.34 s

Apply

a = (v-u)/t

a = 9.99/1.34

a = -7.46 m/s²

F = ma.............. Equation 2

Where F = force, m = mass

Given: m = 75 kg, a = -7.46 m/s²,

F = 75(-7.46)

F = -559.5 N

Hence the average force exerted on the diver is 559.5 N

7 0

2 years ago

CAN YOU SUBSCRIBE? PLEASE...

nadezda [96]

Answer:

just wait I'll subscribe yours

<h2>hope it helps you have a good day keep smiling be happy stay safe</h2>

4 0

3 years ago