Answer:
The force on the proton will be twice in comparison with the force experienced by the electron.
Explanation:
The magnetic force acting on the moving charge particle is perpendicular to the velocity as well as the magnetic field. The factors upon which the magnitude of the force is depending and proportional are as follows:
1 Magnitude of the charge particle.
2 Magnitude of the velocity of moving charge.
3 Magnitude of magnetic field.
4 Sine of angle between the velocity and magnetic field.
As far as direction is concerned, it can be find out by right hand rule.
Lets consider,
Speed of electron = Ve
Speed of proton vp= 2ve
Magnitude of the force F = qvBsin∅
Force acting on electron Fe = qe.ve.Bsin∅
Force acting on proton Fp = qp.vp.Bsin∅
Fp = -2 qe.ve.Bsin∅
Fp = -2Fe
Negative sign shows the direction of Force on proton is opposite to the direction of force on electron.
The efficiency of the scissor is 200%.
<u>Explanation:</u>
Efficiency is defined as the ratio of output of any instrument or device or machine to the input supplied to it. So the greater the output the greater will be the efficiency of the device.
As here the work done by us on the system is said to be 10 J so this will be equal to the input work done on the system. And the work done by the system i.e., the scissor is 20 J, so this will be the output work.
So, the efficiency is the ratio of output to input as shown below.
Efficiency = 
= 200
So, the efficiency of the scissor is 200%.
Answer:
No, because the tornado goes faster than a car at 100 m per hour
Explanation:
<h2>
Answer:</h2>
(a) 6.95 x 10⁻⁸ C
(b) 6.25N/C
<h2>
Explanation:</h2>
The electric field (E) on a point charge, Q, is given by;
E = k x Q / r² ---------------(i)
Where;
k = constant = 8.99 x 10⁹ N m²/C²
r = distance of the charge from a reference point.
Given from the question;
E = 10000N/C
r = 0.250m
Substitute these values into equation(i) as follows;
10000 = 8.99 x 10⁹ x Q / (0.25)²
10000 = 8.99 x 10⁹ x Q / (0.0625)
10000 = 143.84 x 10⁹ x Q
Solve for Q;
Q = 10000/(143.84 x 10⁹)
Q = 0.00695 x 10⁻⁵C
Q = 6.95 x 10⁻⁸ C
The magnitude of the charge is 6.95 x 10⁻⁸ C
(b) To get how large the field (E) will be at r = 10.0m, substitute these values including Q = 6.95 x 10⁻⁸ C into equation (i) as follows;
E = k x Q / r²
E = 8.99 x 10⁹ x 6.95 x 10⁻⁸ / 10²
E = 8.99 x 10⁹ x 6.95 x 10⁻⁸ / 100
E = 6.25N/C
Therefore, at 10.0m, the electric field will be just 6.25N/C
