(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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<h3>Answer</h3>
m/s^2 (meter per sec square)
Explanation:
acc = change in velocity/time
= distance/time
----------------
time
= m/s
------
s
=m/s^2
To solve this problem we will apply the concepts related to the kinematic equations of linear motion. For this purpose we will define the speed as the distance traveled in a given period of time. Here the distance is equivalent to the orbit traveled around the earth, that is, a circle. Approaching the height of the aircraft with the radius of the earth, we will have the following data,



The circumference of the earth would be

Velocity is defined as,


Here
, then

Therefore will take
s or 506 hours, 19 minutes, 17 seconds
The orbiting velocity of the satellite is 4.2km/s.
To find the answer, we need to know about the orbital velocity of a satellite.
<h3>What's the expression of orbital velocity of a satellite?</h3>
- Mathematically, orbital velocity= √(GM/r)
- r = radius of the orbital, M = mass of earth
<h3>What's the orbital velocity of a satellite orbiting earth with a radius 3.57 times the earth radius?</h3>
- M= 5.98×10²⁴ kg, r= 3.57× 6.37×10³ km = 22.7×10⁶m
- Orbital velocity= √(6.67×10^(-11)×5.98×10²⁴/22.7×10⁶)
=4.2km/s
Thus, we can conclude that the orbiting velocity of the satellite is 4.2km/s.
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Answer:
there are go fella hope u understood