Answer:
distance and length are the same quantity
Answer:
1341.03 V/m
Explanation:
The power output per unit area is the intensity and also the is the magnitude of the Poynting vector.
= cε₀
⇒
= cε₀
Where;
P is the power output
A is the area of the beam
c is speed of light
ε₀ is permittivity of free space 8.85 × 10⁻¹² F/m
is the average (rms) value of electric field
Making electricfield
the subject of the equation
= P / Acε₀
= √(P / Acε₀)
But area A = πr²
= √(P / πr²cε₀)
Given:
Output power, P = 15 mW = 0. 015 W
Diameter, d = 2 mm = 0.002 m
⇒ Radius,
Solving for average (rms) value of electric field;
= 1341.03 V/m
Answer:
<em>The force that would be applied on the rope just to start moving the wagon is 122 N</em>
Explanation:
Frictional force opposes motion between two surfaces in contact. It is the force that must be applied before a body starts to move. Static friction opposes the motion of two bodies that are in contact but are not moving. The magnitude of static friction to overcome for the body to move can be calculated using equation 1.
F = μ x mg .............................. 1
where F is the frictional force;
μ is the coefficient of friction ( μs, in this case, static friction);
m is mass of the object and;
g is the acceleration due to gravity( a constant equal to 9.81 m/
)
from the equation we are provide with;
μs = 0.25
m = 50 kg
g = 9.81 m/
F =?
Using equation 1
F = 0.25 x 50 kg x 9.81 m/
F = 122.63 N
<em>Therefore a force of 122 N must be applied to the rope just to start the wagon.</em>
Answer:
Explanation:
Given
Initial speed is u=V
Maximum height of Pebble is H
Deriving maximum height of Pebble and considering motion in vertical direction

where v=final velocity
u=initial velocity
a=acceleration
s=Displacement
Final velocity will be zero at maximum height


i.e. maximum height is dependent on square of initial velocity
for twice the height

on comparing
Answer:
Explanation:
Force of friction acting on the body = μ mg cosθ
= .4 x 70 x 9.8 x cos30
= 237.63 N
component of weight = mgsinθ
= 70 x 9.8 x sin30
= 343 N
Net upward force = 600 - mgsinθ - μ mg cosθ
= 600 - 343 - 237.63
= 105.37 N
acceleration in upward direction = 105.37 / 70
= 1.5 m /s²
s = ut + 1/2 a t²
= 0 + .5 x 1.5 x 3²
= 6.75 m .