Answer:
80×5×10=4000J
so therefore, work done on the body is 4000J
1. False
2. Positive
I’m just joking I don’t know the answersss
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.
Learn more about electric field here: brainly.com/question/14372859
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Answer:
ac = 2.86 m / s²
Explanation:
Image can detail the system to determine the force in the FA to understand the system into the applicated force
m = 100 kg , L = 3 m
∑ F = 0 ⇒ Ay - 100 kg + P * cos (45) = 0
Ay = 768.86 N
∑ Mₐ = α * I ₐ
I ₐ = m * L² / 3 ⇒ I ₐ = 100 kg * 4² m / 3
Replacing
P * sin (45) * 3 = α * 100 kg * 4² m / 3
α = 1.193 rad / s²
ac = α *2 ⇒ ac = 1.193 rad / s² * 2
ac = 2.86 m / s²
Hello!
Because as you get closer to the surface of the earth, the more air that is on top of you. At the top of the atmosphere, there is less air, and everything is a vacuum, where you have no weight. When you get close to the earth, the weight of the air builds until it when you're at the very lowest point of the earths surface, all the air in the atmosphere above you is pressing down.
Thank You!
