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vladimir2022 [97]
3 years ago
12

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:

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Why is random sampling and random assignment in experimental research important?
Hunter-Best [27]

So the result is not biased or affected in some way

6 0
3 years ago
What is the equivalent resistance of a circuit that contains two 50.00
Whitepunk [10]

Answer:

A

Explanation:

Resistors in series add. There is only one path the current can take. That's why Christmas Tree lights sometimes give a lot of trouble. If a bulb burns out, it could be any one of them and time is needed to find the burned out bulb.

That being the case R = R1 + R2

R1 = 50 ohms

R2 = 50 ohms

R = 50 + 50

R = 100 ohms

Answer A

4 0
3 years ago
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El tiempo que habria que esperar para que el dia fuera 1 hora mas largo que es hoy
myrzilka [38]

Answer: el tiempo que habria que esperar para que el dia fuera 1 hora mas largo que es hoy

Explanation:

8 0
3 years ago
The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ
prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

          \nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})

                           = (30)^{2} + 2(-0.25(412 - 0))

                           = 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

          y = 200 e^{\frac{x}{1000}}

      \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}

Now, second derivative of the train is calculated as follows.

         \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}      

       \frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}    

Radius of curvature of the train is as follows.  

   \rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}

               = \frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}

              = 3808.96 m

Now, we will calculate the normal component of the train as follows.

            a_{n} = \frac{\nu^{2}_{B}}{\rho}

                        = \frac{(26.34)^{2}}{3808.96}

                        = 0.1822 m/s^{2}

The magnitude of acceleration of train is calculated as follows.

            a = \sqrt{(a_{t})^{2} + (a_{n})^{2}}

               = \sqrt{(-0.25)^{2} + (0.1822)^{2}}

              = 0.309 m/s^{2}

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is 0.309 m/s^{2}.

6 0
3 years ago
9. To completely describe the motion of an object, you need
pav-90 [236]
I think it’s D) all of the above
4 0
3 years ago
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