The magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.
<h3>Electric field on the master charge</h3>
E = kq/r²
where;
- q is magnitude of master charge
- r is distance of separation
- k is Coulomb's constant
E = (9 x 10⁹ x 0.63)/(0.75²)
E = 1.008 x 10¹⁰ N/C
<h3>Force on the test charge</h3>
F = Eq
where;
- E is electric field
- q is the test charge
F = (1.008 x 10¹⁰) x (0.5)
F = 5.04 x 10⁹ N
Thus, the magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.
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Answer:
Halfway between B and A on the return leg.
Explanation:
Your average SPEED for the entire trip will equal your constant speed as the time and distance increase at proportionate rates.
Your average VELOCITY will equal your constant speed while you travel from A to B because time and displacement are increasing at proportionate rates.
When you turn around at B to return, your Displacement is now decreasing while your travel time continues to increase, so your average velocity decreases.
Lets say the distance from A to B is 90 km and your constant speed is 30 km/hr.
your average speed is 30 km/hr because you took 6 hrs to travel 180 km
We want to find your position when your average velocity is 30/3 = 10 km/hr
it took 3 hrs to go 90 km from A to B. Let t be the time lapsed since turn around
your displacement is given by d = 90 - 30(t)
and your total time of travel is t + 3 hrs
v = d/t
10 = (90 - 30t) / (t + 3)
10(t + 3) = (90 - 30t)
10t + 30 = 90 - 30t
40t = 60
t = 1.5 hrs
This will occur when you are halfway between B and A
Answer:
21 psi
Explanation:
The weight of the car is:
W = mg
W = 1000 kg * 9.8 m/s²
W = 9800 N
Divided by 4 tires, each tire supports:
F = W/4
F = 9800 N / 4
F = 2450 N
Pressure is force divided by area, so:
P = F / A
P = (2450 N) / (0.13 m × 0.13 m)
P ≈ 145,000 Pa
101,325 Pa is the same as 14.7 psi, so:
P ≈ 145,000 Pa × (14.7 psi / 101,325 Pa)
P ≈ 21 psi
