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3 years ago

Hydroelectric dams use ------- to produce electricity. gravitational potential energy of falling water

Physics

1 answer:

Eduardwww [97]3 years ago

7 0

Answer:

Gravitational potential energy

Explanation:

Hydroelectric dams are the power plants which generates electricity by using the energy of falling water from a great height.

A the water is stored in big reservoirs and at a great height, it contain lot of potential energy due to the height. As it falls downwards, the potential energy is converted into kinetic energy of the water. this kinetic energy of the falling water is used to run the turbine, and then the electric energy is generated.

So, in hydroelectric power stations, the potential energy of water is converted into the electric energy.

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If Vector A is (3, 0) and Vector B is (-3, 3), what is the resultant?

Archy [21]

Answer:resultant vector R = (0, 3)

Explanation: vector A = (3, 0)

vector B =(-3, 3)

Vectors are added such that those in same directions are added together. The resultant vector R is the given by R = (3-3, 0+3)

= (0, 3)

6 0

3 years ago

How do i find the number of leptons in an atom??

Serjik [45]

Answer:

In particle physics, a lepton is an elementary particle of half-integer spin (spin 1⁄2) that does not undergo strong interactions.[1] Two main classes of leptons exist: charged leptons (also known as the electron-like leptons or muons), and neutral leptons (better known as neutrinos). Charged leptons can combine with other particles to form various composite particles such as atoms and positronium, while neutrinos rarely interact with anything, and are consequently rarely observed. The best known of all leptons is the electron.

8 0

3 years ago

A 100-kg tackler moving at a speed of 2.6 m/s meets head-on (and holds on to) an 92-kg halfback moving at a speed of 5.0 m/s. Pa

DIA [1.3K]

Given that,

Mass of trackler, m₁ = 100 kg

Speed of trackler, u₁ = 2.6 m/s

Mass of halfback, m₂ = 92 kg

Speed of halfback, u₂ = -5 m/s (direction is opposite)

To find,

Mutual speed immediately after the collision.

Solution,

The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :

$m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{100\times 2.6+92\times (-5)}{(100+92)}\\\\V=-1.04\ m/s$

So, the mutual speed immediately after the collision is 1.04 m/s but in opposite direction.

3 0

4 years ago

The pressure exerted by a gas is 2.0 atm while it has a volume of 350 mL. What would be the volume of this sample of gas at stan

34kurt

Answer:

700 mL or 0.0007 m³

Explanation:

P₁ = Initial pressure = 2 atm

V₁ = Initial volume = 350 mL

P₂ = Final pressure = 1 atm

V₂ = Final volume

Here the temperature remains constant. So, Boyle's law can be applied here.

P₁V₁ = P₂V₂

$\frac{P_1V_1}{P_2}=V_2\\\Rightarrow V_2=\frac{2\times 350}{1}\\\Rightarrow V_2=700\ mL$

So, volume of this sample of gas at standard atmospheric pressure would be 700 mL or 0.0007 m³

7 0

3 years ago

To what potential should you charge a 3.0 μf capacitor to store 1.0 j of energy?

Nimfa-mama [501]

The energy stored in a capacitor is given by:
$U= \frac{1}{2}CV^2$
where
U is the energy
C is the capacitance
V is the potential difference

The capacitor in this problem has capacitance
$C=3.0 \mu F = 3.0 \cdot 10^{-6} F$
So if we re-arrange the previous equation, we can calculate the potential V that should be applied to the capacitor to store U=1.0 J of energy on it:
$V= \sqrt{ \frac{2U}{C} }= \sqrt{ \frac{2 \cdot 1.0 J}{3.0 \cdot 10^{-6}F} }=816 V$

8 0

3 years ago