It's velocity is not constant as direction is changing.

We know, velocity is speed with direction, so if direction is changing, velocity can't be constant, doesn't matter that speed is constant.

Hope this helps!

8 0

3 years ago

A proton is initially at rest. After some time, a uniform electric field is turned on and the proton accelerates. The magnitude

marusya05 [52]

Answer:

a) 8.83*10⁵ m/s b) 2.80*10⁶ m/s

Explanation:

a) Assuming no other forces acting on the proton, the acceleration on it is produced by the electric field.

By definition, the force due to the electric field is as follows:

F = q*E = e*E (1)

where e is the elementary charge, the charge carried by only one proton, and is e = 1.6*10⁻¹⁹ C.

According to Newton's 2nd law, this force is at the same time, the product of the mass of the proton, times the acceleration a:

F = mp*a (2)

From (1) and(2), being left sides equal, right sides must be equal too:

$F = e*E = mp*a$

Solving for a:

$a = \frac{e*E}{mp} =\frac{1.6e-19C*1.36e5N/C}{1.67e-27kg} =1.3e13 m/s2$

⇒ a = 1.3*10¹³ m/s²

As we have the value of a (which is constant due to the field is uniform), the displacement x, and we know that the initial velocity is 0, in order to get the value of the speed, we can use the following kinematic equation:

$vf^{2} -vo^{2} = 2*a*x$

Replacing by v₀ = 0, a= 1.3*10¹³ m/s² and x = 0.03 m, we can find vf as follows:

$vf =\sqrt{2*(1.3e13 m/s2)*0.03m} = 8.83e5 m/s$

⇒ vf = 8.83*10⁵ m/s

b) We can just repeat the equation from above, replacing x=0.03 m by x=0.3 m, as follows:

$vf =\sqrt{2*(1.3e13 m/s2)*0.3m} = 2.80e6 m/s$

⇒ vf = 2.80*10⁶ m/s

4 0

3 years ago