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marusya05 [52]
3 years ago
13

Dams hold collected water back from flowing at their normal rate. What type of energy is in the collected water above the dam? c

hemical energy electrical energy potential energy kinetic energy
Physics
2 answers:
Andreyy893 years ago
8 0
It would be mechanical potential energy.
forsale [732]3 years ago
3 0
Potential energy because the moment the dam breaks, a lot of water will burst out. Obviously this isn't chemical energy because there is not chemical reaction taking place. It's not electrical energy because there is no electricity at the dam. This leaves us with potential and kinetic energy. A key word in here is holds back, aka stored water. Storing almost always leads to potential energy.
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An electric drill rated at 400 W is connected to a 240V power line. How much current does it draw?
disa [49]

Answer:

1.67 A

Explanation:

Given that,

→ Power (P) = 400 W

→ Potential difference (V) = 240 V

→ Current (I) = ?

The amount of current drawn will be,

→ P = V × I

→ I = P/V

→ I = 400/240

→ I = 1.66666666667

→ [ I = 1.67 A ]

Hence, the current drawn 1.67 A.

8 0
2 years ago
Read 2 more answers
How would you respond to a person who argued that the Big Bang theory is incorrect because it is only a theory and theories have
Umnica [9.8K]

Well, you would reply that that's not what theories are. Theories explain the how and the why, laws explain the what. So, the Big Bang theory isn't "just a theory". It's a theory, it explains the how. (Also, if someone tells you it's anti-God or whatever, tell them the thoery was created by a Catholic scientist. True fact.) I hoped this helped!!! (You don't have the include who created the theory if this is for homework.)

5 0
3 years ago
Read 2 more answers
Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.
shusha [124]

Answer:

F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

                                  r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2  }

- Then compute the angle that Force makes with the y axis:

                                 cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

                                 F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

                                F_net = F_1,2 + kq / y^2

- Hence,

                                F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

- Now for the limit y >>a:

                              F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2}  )

- Insert limit i.e a/y = 0

                              F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2})  \\\\F_n = 3*k*q/y^2

Hence the Electric Field is off a point charge of magnitude 3q.

8 0
3 years ago
How do you calculate resistance with coulombs, time and J's?
Mars2501 [29]
Hey user


The energy E in joules (J) is equal to the voltage V in volts (V), times the electrical charge Q in coulombs (C):

E(J) = V(V) ×<span> Q</span>(C)

So

joule = volt × coulomb

or

J = V × C

Example

What is the energy in joules that is consumed in an electrical circuit with voltage supply of 15V and charge flow of 4 coulombs?

E = 15V × 4C = 60J

4 0
3 years ago
A certain lightning bolt moves 10.0 C of charge. How many charges have been moved if the fundamental unit of charge is 1.6 x 10-
Andrej [43]

To solve this problem, use the ratio given by the total number of electrons or protons that exist as a function of the total charge, and inversely proportional to the value of the fundamental charge. The number of fundamental unit  |q_e| that constitutes a charge of 40.0C can be calculated as

N = \frac{|Q|}{|q_e|}

Here,

q_e = Value of charge and it is the fundamental charge

Q = Total Charge

N = Total number of electron or protons

The number of fundamental units is calculated as follows

N = \frac{10.0C}{1.6*10^{-19}C}

N = 6.25*10^{19}

Therefore the number of fundamental charge units moved by lightning bolt is 6.25*10^{19}

8 0
3 years ago
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