Find the work w1 done on the block by the force of magnitude f1 = 75.0 n as the block moves from xi = -1.00 cm to xf = 3.00 cm .
1 answer:
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Answer:
m/s
Explanation:
Parameters given:
distance of the proton form the insulating sheet = 0.360mm
speed of the proton, = 990m/s
Surface charge density, σ = 2.34 x C/
We need to calculate the speed at time, t = 7.0 * s.
We know that the proton is moving parallel to the sheet, hence, we can say it is moving in the x direction, with a speed on the axis.
The electric force acting on the proton moves in the y direction, so this means it is moving with velocity in the y axis.
Hence, the resultant velocity of the proton is given by:
= 990m/s from the question. We need to find
and then the resultant velocity v.
Electric field is given in terms of surface charge density, σ as:
E = σ/ε0
where ε0 = permittivity of free space
=>
E = 132 N/C
Electric Force, F is given in terms of Electric field:
F = eE
where e = electronic charge
=> F = ma = eE
∴ a = eE/m
where
a = acceleration of the proton
m = mass of proton
a = 1.3 * m/
Therefore, at time, t = 7.0 * , we can use one of the equations of linear motion to find the velocity in the y axis:
= 0 + (1.3 *
* 7.0 *
)
= 910 m/s
∴
v = 1344.69 m/s = m/s