Answer:
Explanation:
Given
Average Electric Field Strength
Considering charge distribution to be uniform and earth to be symmetrical
Using Gauss law
Also
here Electric Field is acting inward therefore
Answer:
5 N
Explanation:
From the question,
The magnitude of the force that would be required to just loosen the nut when the force is applied perpendicularly at the end of the handle is
Fy = Fsinθ................. Equation 1
Where Fy = force acting perpendicular at the end of the handle, F = Force applied to the handle, θ = angle of inclination of the force to the end of the handle.
Given: F = 10 N, θ = 30°
Substitute these values into equation 1
Fy = 10(sn30°)
Fy = 10(0.5)
Fy = 5 N.
Answer: Thus, the force is directed 72.5° above the horizontal.
Explanation:
The magnitude of the force, F = 15 N
let this force be detected at angle θ from the horizontal, then horizontal component of force is: F cos θ
and vertical component is F sin θ
Horizontal component is given,
F cos θ = 4.5 N
⇒15 N cos θ = 4.5 N
⇒ cos θ = 0.3
⇒ θ = cos⁻¹ 0.3 = 72.5°
Volume = mass / density = 45.6/10.5 = .... L
Answer:
speed for last lap is 247.89 km/h
Explanation:
given data
velocity v1 = 203 km/h
velocity v2 = 199 km/h
no of lap n = 10
to find out
average speed for last lap
solution
we consider here distance d for 1 lap
so in first 9 lap time taken is
t1 = distance / velocity v2
t1 = 9d / 199 ...............1
and
for 10 lap time is
t2 = 10d / 203 .............2
so from 1 and 2 equation time for last lap
last lap time t3 = t2 - t1
t3 = 10d / 203 - 9d / 199
t3 = 0.004034 d
so speed for last lap is
speed = distance / time
speed = d / 0.004034 d
speed = 247.89 km/h
so speed for last lap is 247.89 km/h