Complete Question
The question image is in the first uploaded image
Answer:
Explanation:
From the question we are told that
Distance b/w Q mid point and P is given as x
Generally the equation for magnitude of the electric field at the point P is given as
where
Therefore
Therefore equation for magnitude of the electric field at the point P is
Walking at a speed of 2.1 m/s, in the first 2 s John would have walked
(2.1 m/s) (2 s) = 4.2 m
Take this point in time to be the starting point. Then John's distance from the starting line at time <em>t</em> after the first 2 s is
<em>J(t)</em> = 4.2 m + (2.1 m/s) <em>t</em>
while Ryan's position is
<em>R(t)</em> = 100 m - (1.8 m/s) <em>t</em>
where Ryan's velocity is negative because he is moving in the opposite direction.
(b) Solve for the time when they meet. This happens when <em>J(t)</em> = <em>R(t)</em> :
4.2 m + (2.1 m/s) <em>t</em> = 100 m - (1.8 m/s) <em>t</em>
(2.1 m/s) <em>t</em> + (1.8 m/s) <em>t</em> = 100 m - 4.2 m
(3.9 m/s) <em>t</em> = 95.8 m
<em>t</em> = (95.8 m) / (3.9 m/s) ≈ 24.6 s
(a) Evaluate either <em>J(t)</em> or <em>R(t)</em> at the time from part (b).
<em>J</em> (24.6 s) = 4.2 m + (2.1 m/s) (24.6 s) ≈ 55.8 m
(1).....opaque -object that not allow light to pass through it
we can't look through opaque objects
ex..wall
translucent... objects that all some light to pass through it
we can partially see through it
ex...coloured glass window
transparent... objects that wholly allow light to pass through it
we can clearly see through it
ex...glass of specs
(2)..the bouncing back of light from a smooth polished surface is called reflection of light
(3)/.....there must be a solid ,opaque to block light
there must be a source of light
there must be a surface to have shadow on it..
(6)...reflected images r formed by light reflection from a smooth polished surface while shadows are formed by blockage of light by an solid,opaque object...
Answer:
a) w = 31.4 rad / s, b) a = 118.4 m / s²
Explanation:
a) let's reduce to the SI system
w = 5 rev / s (2pi rad / 1 rev)
w = 31.4 rad / s
b) the expression for the centripetal acceleration is
a = v² / r
linear and angular variables are related
v = w r
we substitute
a = w² r
a = 31.4² 0.120
a = 118.4 m / s²
Answer:
0.6m/s²
Explanation:
Given parameters:
Initial velocity = 3m/s
Final velocity = 4.5m/s
Time taken = 2.4s
Unknown:
Average acceleration = ?
Solution:
To solve this problem, we use the expression:
Acceleration =
v is the final velocity
u is the initial velocity
t is the time taken
Acceleration = = 0.6m/s²