Solution:
The relation between the potential difference and the electric field between the plates of the parallel plate capacitor is given by :
Differentiating on both the sides with respect to time, we get
Therefore, the rate of the electric field changes between the plates of the parallel plate capacitor is given by :
V/m-s
Answer:
The distance of the goggle from the edge is 5.30 m
Explanation:
Given:
The depth of pool (d) = 3.2 m
let 'i' be the angle of incidence
thus,
i = 
i = 67.75°
Now, Using snell's law, we have,
n₁ × sin(i) = n₂ × 2 × sin(r)
where,
r is the angle of refraction
n₁ is the refractive index of medium 1 = 1 for air
n₂ is the refractive index of medium 1 = 1.33 for water
now,
1 × sin 67.75° = 1.33 × sin(r)
or
r = 44.09°
Now,
the distance of googles = 2.2 + d×tan(r) = 2.2 + (3.2 × tan(44.09°) = 5.30 m
Hence, <u>the distance of the goggle from the edge is 5.30 m</u>
Hello
This is a problem of accelerated motion, where the acceleration involved is the gravitational acceleration:
, and where the negative sign means it points downwards, against the direction of the motion.
Therefore, we can use the following formula to solve the problem:
where
is the initial vertical velocity of the athlete,
is the vertical velocity of the athlete at the maximum height (and
at maximum height of an accelerated motion) and S is the distance covered between the initial and final moment (i.e., it is the maximum height). Re-arranging the equation, we get
Answer
given,
mass of the crate = 53 Kg
force applied by the worker = 180 N
Angle made with the horizontal = 35°
crate moves = 2.9 m
a) The work done equals the force in the direction of the displacement, times the displacement
W = F_x ×d
W = 180 cos 35° × 2.9
W = 427.6 J
b) A force that is perpendicular to the direction of the displacement does not do any work so work done by gravitational force is zero
c) Similarly for the normal force the work done will be zero.
d) Total work done by the crate is equal to 427.6 J
