Answer:
7.344 s
Explanation:
A = 0.15 x 0.3 m^2 = 0.045 m^2
N = 240
e = - 2.5 v
B1 = 0.1 T
B2 = 1.8 T
ΔB = B2 - B1 = 1.8 - 0.1 = 1.7 T
Δt = ?
e = - dФ/dt
e = - N x A x ΔB/Δt
- 2.5 = - 240 x 0.045 x 1.7 / Δt
2.5 = 18.36 / Δt
Δt = 7.344 s
The total effective resistance of several resistors in SERIES is just the sum of all the individual resistances.
So the effective resistance of ten 120-ohm resistors in series is
(120 + 120 + 120 + 120 + 120 + 120 + 120 + 120 + 120 + 120) .
A much easier way to write it is . . . (10) x (120) .
Total effective resistance = <em>1,200 ohms</em> .
Answer:
Explanation:
let A represent me and B represent my friend
A speed=30 m/hr toward west
B speed =40 m/hr toward south
after 1/2 hr
total distance cover by A =1/2 * 30 miles = 15 miles
total distance by B = 1/2*40 miles = 20 miles
now from figure
Differentiate above equation
Answer:
(a) N₁= 1100 N
(b) N₂ = 716.66 N
(c) μs =0.65
Explanation:
We apply Newton's first law:
∑Fx = 0
∑Fy =0
∑F : algebraic sum of the forces in Newton (N)
∑MA=0
∑MA : algebraic sum of moments at point A.(N*m)
Look at the free body diagram in the attached graph
(a) Find the magnitude of the upward normal force applied by the ground on the ladder (N₁)
∑Fy = 0 positive (+) is upward
N₁-500-600=0
N₁= 500+600 = 1100 N
(b) Find the magnitude of the horizontal normal force (N₂) applied by the wall on the ladder.
H= 3m
∑MA=0 :positive moment (+) counterclockwise
Ff(3)-N₁(4)+500(2.1)+600(2) = 0 ,Ff: friction force
(3)Ff = 1100(4)-500(2.1)-600(2)
∑Fx = 0 ,positive (+) to the right
Ff-N₂ = 0
Ff=N₂
N₂ = 716.66 N
(c) Determine the minimum possible coefficient of static friction (μs) for the ladder-ground interaction, for the ladder to remain at rest.
Ff = μs *N₁
μs = Ff /N₁
μs =716.66 / 1100
μs =0.65