Answer:

Explanation:
we know angular velocity in terms of moment of inertia and angular speed
ω .... (1)
moment of inertia of rod rotating about its center of length b
........ .(2)
using v = ωr
where w is angular velocity
and r is radius of rod which is equal to b
so we get 2v = ωb
ω = 2v/b ................. (3)
here velocity is two time because two opposite ends are moving opposite with a velocity v so net velocity will be 2v
put second and third equation in ist equation
×
so final answer will be 
Answer:
Wet surfaces areaA=+25.3ft^2
Explanation:
Using F= K×A× S^2
Where F= drag force
A= surface area
S= speed
Given : F=996N S=20mph A= 83ft^2
K = F/AS^2=996/(83×20^2)
K= 996/33200 = 0.03
1215= (0.03)× A × 18^2
1215=9.7A
A=1215/9.7=125.3ft^2
The net force is negative, and there is a change in motion.
Initial speed, u = 15 m/s
Final speed, v = 10 m/s
Distance traveled, s = 6.0 m
The acceleration, a, is determined from
u² + 2as = v²
(15 m/s)² + 2*(a m/s²)*(6.0 m) = (10 m/s)²
225 + 12a = 100
12a = -125
a = -10.4167 m/s²
The time, t, for the velocity to change from 15 m/s to 10 m/s is given by
(10 m/s) = (15 m/s) - (10.4167 m/s²)*(t s)
10 = 15 - 10.4167t
t = 0.48 s
The average speed is
(6.0 m)/(0.48 s) = 12.5 m/s
Answer: 12.5 m/s
Answer:
6.78 X 10³ N/C
Explanation:
Electric field near a charged infinite plate
= surface charge density / 2ε₀
Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.
Field due to charge density of +95.0 nC/m2
E₁ = 95 x 10⁻⁹ / 2 ε₀
Field due to charge density of -25.0 nC/m2
E₂ = 25 x 10⁻⁹ / 2ε₀
Total field
E = E₁ + E₂
= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ / 2ε₀
= 6.78 X 10³ N/C