The resistance of the lamp is apparently 50V/2A = 25 ohms.
When the circuit is fed with more than 50V, we want to add
another resistor in series with the 25-ohm lamp so that the
current through the combination will be 2A.
In order for 200V to cause 2A of current, the total resistance
must be 200V/2A = 100 ohms.
The lamp provides 25 ohms, so we want to add another 75 ohms
in series with the lamp. Then the total resistance of the circuit is
(75 + 25) = 100 ohms, and the current is 200V/100 ohms = 2 Amps.
The power delivered by the 200V mains is (200V) x (2A) = 400 watts.
The lamp dissipates ( I² · R ) = (2² · 25 ohms) = 100 watts.
The extra resistor dissipates ( I² · R) = (2² · 75 ohms) = 300 watts.
Together, they add up to the 400 watts delivered by the mains.
CAUTION:
300 watts is an awful lot of power for a resistor to dissipate !
Those little striped jobbies can't do it.
It has to be a special 'power resistor'.
300 watts is even an unusually big power resistor.
If this story actually happened, it would be cheaper, easier,
and safer to get three more of the same kind of lamp, and
connect THOSE in series for 100 ohms. Then at least the
power would all be going to provide some light, and not just
wasted to heat the room with a big moose resistor that's too
hot to touch.
Answer:
a)The approximate radius of the nucleus of this atom is 4.656 fermi.
b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527
Explanation:
= Constant for all nuclei
r = Radius of the nucleus
A = Number of nucleons
a) Given atomic number of an element = 25
Atomic mass or nucleon number = 52
The approximate radius of the nucleus of this atom is 4.656 fermi.
b) 
k=
= Coulombs constant
= charges kept at distance 'a' from each other
F = electrostatic force between charges
Force of repulsion between two protons on opposite sides of the diameter
The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527
Answer:
The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.
Explanation:
The Impulse Theorem states that the impulse experimented by the hockey park is equal to the vectorial change in its linear momentum, that is:
(1)
Where:
- Impulse, in kilogram-meters per second.
- Mass, in kilograms.
- Initial velocity of the hockey park, in meters per second.
- Final velocity of the hockey park, in meters per second.
If we know that 
, ![\vec{v}_{1} = -10\,\hat{i}\,\left[\frac{m}{s}\right]](https://tex.z-dn.net/?f=%5Cvec%7Bv%7D_%7B1%7D%20%3D%20-10%5C%2C%5Chat%7Bi%7D%5C%2C%5Cleft%5B%5Cfrac%7Bm%7D%7Bs%7D%5Cright%5D)
and ![\vec {v_{2}} = 25\,\hat{i}\,\left[\frac{m}{s} \right]](https://tex.z-dn.net/?f=%5Cvec%20%7Bv_%7B2%7D%7D%20%3D%2025%5C%2C%5Chat%7Bi%7D%5C%2C%5Cleft%5B%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%5D)
, then the impulse applied by the stick to the park is approximately:
![I = (0.2\,kg)\cdot \left(35\,\hat{i}\right)\,\left[\frac{m}{s} \right]](https://tex.z-dn.net/?f=I%20%3D%20%280.2%5C%2Ckg%29%5Ccdot%20%5Cleft%2835%5C%2C%5Chat%7Bi%7D%5Cright%29%5C%2C%5Cleft%5B%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%5D)
![I = 7\,\hat{i}\,\left[\frac{kg\cdot m}{s} \right]](https://tex.z-dn.net/?f=I%20%3D%207%5C%2C%5Chat%7Bi%7D%5C%2C%5Cleft%5B%5Cfrac%7Bkg%5Ccdot%20m%7D%7Bs%7D%20%5Cright%5D)
The impulse applied by the stick to the hockey park is approximately 7 kilogram-meters per second.
The answer is: 120V
Power is the rate at which energy is supplied/transformed in time:
we can write:
V ddp in Volts represents Energy/Charge i.e. energy carried by each coulomb;
I current in Amperes represents Charge/time or coulombs passing each seconds.
combining them we have:
Power = energy/time = V • 1
or
1200 = V ⋅ 10
V = 1200/10 = 120V
Answer:
Magnitude of force on wheel B is 4 N
Explanation:
Given that
For wheel A
m= 1 kg
d= 1 m,r= 0.5 m
F=1 N
We know that
T= F x r
T=1 x 0.5 N.m
T= 0.5 N.m
T= I α
Where I is the moment of inertia and α is the angular acceleration
T= I α
0.5= 0.25 α
For Wheel B
m= 1 kg
d= 2 m,r=1 m
Given that angular acceleration is same for both the wheel
T= I α
T= 1 x 2
T= 2 N.m
Lets force on wheel is F then
T = F x r
2 = F x 1
So F= 2 N
Magnitude of force on wheel B is 2 N
