Answer:
-(0.330m/s² ) kˆ
Explanation:
given data:
Mass of particle 'm'= 1.81 x kg
Velocity 'v'= (3.00 x m/s)j
Charge of particle 'q'= 1.22 x C
Uniform magnetic field 'B' = (1.63iˆ + 0.980jˆ )T
In order to calculate particle's acceleration, we'll use Newton's second law of motion i.e F=ma
Also,the force a magnetic field exerts on a charge q moving with velocity v is called the magnetic Lorentz force. It is given by:
F = qv × B
F= ma = qV x B
a= --->eq(1)
Lets determine the value of (v x B) first
v x B= (3.00 x m/s)j x (1.63iˆ + 0.980jˆ )
v x B= 4.89 x
Plugging all the required values in eq(1)
a= [1.22 x x (4.89 x kˆ)] / 1.81 x
a= -(0.330m/s² ) kˆ
-ve sign is representing the opposite direction
Answer:
82
Explanation:
weight on earth-weight on air
Answer:
0.2631 N/C
Explanation:
Given that:
The radius of the wire r = 0.22 mm = 0.22 × 10⁻³ m
The radius of the thick wire r' = 0.55 mm = 0.55 × 10⁻³ m
The numbers of electrons passing through B, N = 6.0 × 10¹⁸ electrons
Electron mobility μ = 6.0 x 10-4 (m/s)/(N/C)
= 0.0006
The number of electron flow per second is calculated as follows:
The magnitude of the electric field is:
E =
E =
E =
E = 0.2631 N/C
Answer:
Explanation
There are two factors that can cause the fusion rate in the sun's core to increase.
1) Rise in the temperature of core:
If the temperature of the sun's core increases then it will increases the nuclear fusion reaction. The nuclear fusion reactions has such a strong dependency on temperature that even a smallest rise in temperature will results in the higher rate of reaction. That is why these reactions happen in the hottest core of the stars.
2) Reduction in the radius of the core:
Density plays a huge role in the nuclear fusion reactions. If the radius of the sun's core decrease then there will be an increase in the density of the core. Thus the gravitational pressure will also increases. In order to resist this increase in pressure the fusion reactions will speed up and their rate becomes higher.