The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
where
is the charge density
is the vacuum permittivity
We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
<span>Newton's law of gravitation is attractive, whereas Coulomb's law is attractive or repulsive. Both are proportional to the inverse square of distance.</span>
Explanation:
a) How much work is done by gravity?
- w = f x d
- w = 950 x 10 x 5.5 = 52250j
b) How much work is done by tension?
- v²=u²+2as
- 0.75²=0.25²+2a x5.5
- 0.56=0.06+2a x5.5
- 2a x5.5 = 0.56 - 0.06
- 2a x 5.5 =0.5
- 11a=0.5
- a = 0.5/11 = 0.05m/s²
w = f x d
w = 950 x 0.05 x 5.5 = 261.25j
Answer:
Light includes ALL of these answers: Radio/Microwaves. Visible light and X-rays/Gamma rays.
Answer:
Explanation:
Question 1
An arrow weighing 20g shortly after firing has a speed of 50m / s. Calculate the work done by the athlete. What is the potential energy of the elasticity of the tensed string?
mass m = 20g = 20/1000 = 0.02kg
speed v = 50m / s
P.E = K.E = ½mv²
P.E = ½ × 0.02 × 50²
P.E = 25 J
work done = P.E = 25J
Qestion 2
A 80 kg athlete stood on a trampoline with a coefficient of elasticity of k = 2 kN / m. As far as the edge of the trampoline lowers.
force of elasticity
F = -kx
x = F / k
in our case F will be the force of pressure or gravity
F = mg
g is gravitational acceleration, and according to Newton's second law, acceleration is force through mass - unit of force N, unit of mass kg. Acceleration either in m / s ^ 2 or N / kg
F = 80kg * 10N / kg = 800 N
x = 800N / -2000N = -0.4
The trampoline will lower, so from the level by 0.4 meters and hence this minus
