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ludmilkaskok [199]
3 years ago
6

A power plant supplies 1,100 megawatts of power to the electric grid. how many joules of energy does it supply each second?

Physics
1 answer:
slavikrds [6]3 years ago
5 0
The answer is it will supply 1.1 x 10⁹ J of energy each second.
we can calculate this by using the following equation;
P = W/t 
<span>W = P x t 
</span><span>and by work energy relation;
E = W = P x t
</span>1 watt = 1j/s
1megawatt = 1000000 = 10⁶ j/s
<span>E = 1100 x 106 x 1 </span>
E = 1.1 x 10⁹ J
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Initially sliding with a speed of 1.9 m/s, a 1.8 kg block collides with a spring and compresses it 0.35 m before coming to rest.
Alika [10]
Let k =  the force constant of the spring (N/m).

The strain energy (SE) stored in the spring when it is compressed by a distance x=0.35 m is
SE = (1/2)*k*x²
     = 0.5*(k N/m)*(0.35 m)²
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The KE (kinetic energy) of the sliding block is
KE = (1/2)*mass*velocity²
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3 0
3 years ago
One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories
ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = 9.11\times 10^{-31}\ kg

B = Magnetic field = 0.044 T

q = Charge of electron = 1.6\times 10^{-19}\ C

The centripetal force and the magnetic forces are conserved

m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}

Velocity of first electron

v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s

Velocity of second electron

v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s

Total kinetic energy is given by

K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J

Converting to eV

1\ J=\frac{1}{1.6\times 10^{-19}}\ eV

1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV

The energy of incident electron is 114.92749 keV

5 0
3 years ago
The first periodic table of elements Mendeleev created contained how many elements
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Mendeleev created a periodic table with 63 initial elements.
He left gaps for unknown elements.
To this date, there are 118 elements in the periodic table.
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