Work done = force x distance = 40 x 2 = 80 Joules.
Answer
given,
Speed of car A = 95 Km/h
= 95 x 0.278 = 26.41 m/s
Speed of Car B = 121 Km/h
= 121 x 0.278 = 33.64 m/s
Distance between Car A and B at t=0 = 41 Km
a) Distance travel by car B
d = 26.41 t + 41000
speed of the car A = 33.64 m/s
distance = s x t
26.41 t + 41000 = 33.64 x t
7.23 t = 41000
t = 5670.82 s
time taken by Car B to cross Car A is equal to t = 5670.82 s
distance traveled by car A
D = s x t = 26.41 x 5670.82 = 149766.25 m = 149.76 Km
b) distance travel by the car B in 30 s after overtaking car A
D' = s x t = 33.64 x 30 = 1009.2 m = 1 Km
Answer:
570 N
Explanation:
Draw a free body diagram on the rider. There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.
The rider is moving at constant speed, so acceleration is 0.
Sum of the forces in the x direction:
∑F = ma
F cos 30° - T cos 15° = 0
F = T cos 15° / cos 30°
Sum of the forces in the y direction:
∑F = ma
F sin 30° - W - T sin 15° = 0
W = F sin 30° - T sin 15°
Substituting:
W = (T cos 15° / cos 30°) sin 30° - T sin 15°
W = T cos 15° tan 30° - T sin 15°
W = T (cos 15° tan 30° - sin 15°)
Given T = 1900 N:
W = 1900 (cos 15° tan 30° - sin 15°)
W = 570 N
The rider weighs 570 N (which is about the same as 130 lb).
Answer:
magnitude of force on charge 2Q =
Direction of force on charge = 61 ⁰
Explanation:
The magnitude on the force on the charge can be evaluated by finding the net force acting on the charge 2Q i.e x-component of the net force and the y-component of the net force
║F║ = = after considering the forces coming from Q, 3Q and 4Q AND APPLYING COULOMBS LAW
magnitude of force acting on 2Q =
The direction of the force on charge 2Q is calculated as
tan ∅ = = 1.8284
therefore ∅ = 1.8284
= 61⁰