According to Coulomb’s Law, the force will quadruple.
However…
Let’s assume, for the sake of argument, that both proton and electron behave entirely like particles.
The trajectory of the electron will stretch, and its distance from the nucleus will oscilate until is estabilizes again at the original value.
Answer:
Explanation:
Let v be the terminal velocity of the bar .
emf induced in the bar of length L
= B L v where B is the value of magnetic field.
current i in the circuit containing resistance R
i = induced emf / R
BLv / R
Magnetic force in upward direction in the bar
F = BiL
= BL x BLv / R
B²L²v / R
For attainment of uniform velocity
magnetic force = weight
B²L²v / R = mg
so current
i = BLv / R
Answer:
FALSE
Explanation:
When object float in water then it means that weight of the object is counterbalanced by the buoyancy force.
Here as we know that buoyancy force is the weight of water displaced by the object.
So here when object is submerged into the water then the submerged part of the volume of object will displace exactly the same volume of water and that volume of water will exert upwards buoyancy force on the object.
So here this statement is not true as it says that volume of water displaced is volume above the surface of water while it must be like volume of water displaced must be equal to volume of submerged part of the object.
<span>Answer:
First we need to find the acceleration.
torque on cylinder Ď„ = T * r where T is the string tension;
T = m(g - a) where a is the acceleration of the cylinder. Then
Ď„ = m(g - a)r
But also τ = Iα. For a solid cylinder, I = ½mr²,
and if the string doesn't slip, then α = a / r, so
τ = ½mr² * a/r = ½mra.
Since Ď„ = Ď„, we have
m(g - a)r = ½mra → m, r cancel, leaving
g - a = ½a
g = 3a/2
a = 2g/3 where g, of course, is gravitational acceleration.
We know that v(t) = a*t, so for our cylinder
v(t) = 2gt / 3 â—„ linear velocity
and ω = v(t) / r = 2gt / 3r ◄ angular velocity</span>
Answer:
The value is 
Explanation:
Generally from the calorimetry principle we have that
So here heat gained water is mathematically represented as i.e
substituting 0.150 kg for 
, 4200 J/kg.°C for 
, 100°C for 
and 75°C for T
We have
The Heat loss by the ice is mathematically represented as
Here 
is the energy to move the ice to its melting point which is evaluated as
Here 
is the mass of ice
is the specific heat of ice with value 
temperature of ice at melting point with value 0°C
is the temperature of ice with value -12°C
is the energy to move the ice from its its melting point to liquid which is evaluated as
Here L is the Latent heat of melting of ice with value 
is the energy to move the ice from liquid to the equilibrium temperature which is evaluated as
So
![Q_i = m_i [ c_i * ( T_o -T_i) + L + c_w * ( T -T_o) ]](https://tex.z-dn.net/?f=Q_i%20%20%3D%20m_i%20%5B%20c_i%20%2A%20%28%20T_o%20-T_i%29%20%2B%20L%20%20%2B%20c_w%20%2A%20%28%20T%20-T_o%29%20%5D%20)
=> ![Q_i = m_i [ 2.05 * 10^3 * ( 0 -(-12)) + 334 * 10^3 + 4200 * ( 75 - 0) ]](https://tex.z-dn.net/?f=Q_i%20%20%3D%20m_i%20%5B%202.05%20%2A%2010%5E3%20%2A%20%28%200%20-%28-12%29%29%20%2B%20334%20%2A%2010%5E3%20%20%2B%20%204200%20%2A%20%28%2075%20-%200%29%20%5D%20)
From
We have that
=> 
=>
