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Zinaida [17]
3 years ago
14

How is most of the electricity we use at home generated?

Physics
1 answer:
Sveta_85 [38]3 years ago
3 0
Nuclear power plants, wind farms, water farms, and geothermal heating
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What is the driving force for (a) heat transfer, (b) electric current, and (c) fluid flow?
lara31 [8.8K]

Answer:

The driving force for (a) heat transfer is temperature difference. (b) electric current is voltage difference. (c) fluid flow is pressure or hydraulic head difference.

Explanation: (a) The driving force for heat transfer is temperature difference. Heat transfer between two mediums is possible only if the two mediums are at different temperature, the higher the temperature, the higher the heat transfer.

(b) The driving force for electric current is voltage difference. Voltage difference is defined as the potential difference in charge between two points in electrical field. For electric current to occur,the voltage must be high.

(c) The driving force for fluid flow is pressure difference or hydraulic head difference. For fluid to move upward,it requires energy.

3 0
3 years ago
If an object is moving to the right and there is a net force acting on it to the left, what will happen to the object?
Ksju [112]
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7 0
3 years ago
The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
Sav [38]

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = \dfrac{2c}{3b},

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

7 0
3 years ago
A horizontal line on a displacement-time graph means the object is
r-ruslan [8.4K]

Answer:

traveling further away

Explanation:

4 0
2 years ago
Which of the following is not dimensionally correct?<br> E=mc2 B. vf=vi+at<br> S=vt2 D. S=1/2 at2
Elodia [21]

Answer:

s=vt2 just simplify all into metric units first

5 0
2 years ago
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