(a) The required magnitude of the electric field when the point charge is an electron is 5.57 x 10⁻¹¹ N/C.
(b) The required magnitude of the electric field when the point charge is an proton is 1.02 x 10⁻⁷ N/C.
<h3>
Magnitude of electric field </h3>
The magnitude of electric field is given by the following equation.
F = qE
But F = mg
mg = qE
E = mg/q
where;
- E is the electric field
- m is mass of the particle
- g is acceleration due to gravity
- q is charge of the particle
<h3>For an electron</h3>
E = (9.11 x 10⁻³¹ x 9.8)/(1.602 x 10⁻¹⁹)
E = 5.57 x 10⁻¹¹ N/C
<h3>For proton</h3>
E = (1.67 x 10⁻²⁷ x 9.8)/(1.602 x 10⁻¹⁹)
E = 1.02 x 10⁻⁷ N/C
Thus, the required vertical electric field is greater when the charge is proton.
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Answer:
Explanation:
in series, resistors just add up
27+33+51+1200+240= 1551 ohms
Speed=frequency x wavelength
if the medium is the same density and chemical make up all the way through then the speed is fixed because frequency is always unaffected. If density increases then speed decreases which means wavelength will decrease too.
Answer:
Part a)
Part B)
Explanation:
Part A)
At the top of the hump the force on the rider is
1) Normal force
2) weight
so here we know that
Part B)
At the top of the loop we will have
in order to remain in contact the normal force must be just greater than zero
so we will have
Given data
mass (m) =8 Kg ,
speed (v) =10 m/s ,
time (t) = 5 s
determine magnitude of force (Fnet) = ?
we knoe that
Fnet = m.a
= 8 × (v/t) since a = v/t
= 8 × (10/5)
= 16 N
<em>Magnitude of net force is 16 N</em>
