<span>Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.</span>
On driving your motorcycle in a circle of radius 75 m on wet pavement, the fastest you can go before you lose traction, assuming the coefficient of static friction is 0.20 is 147m/s
Friction helps to maintain the slipping of the vehicle on the road hence lays a very important role.
Maximum velocity of a road with friction is given by the formula,
v = μRg
where, v is the maximum velocity
μ is the coefficient of static friction
R is the radius of the circle road
g is the acceleration due to gravity
Given,
μ = 0.20
R = 75m
g = 9.8m/s²
On substituting the given values in the above formula,
v = 0.20× 75 ×9.8
v = 147m/s
So, the Maximum velocity of the wet road is 147m/s.
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Answer:
Explanation:
We know that Impulse = force x time
impulse = change in momentum
change in momentum = force x time
Force F = .285 t -.46t²
Since force is variable
change in momentum = ∫ F dt where F is force
= ∫ .285ti - .46t²j dt
= .285 t² / 2i - .46 t³ / 3 j
When t = 1.9
change in momentum = .285 x 1.9² /2 i - .46 x 1.9³ / 3 j
= .514i - 1.05 j
final momentum
= - 3.1 i + 3.9j +.514i - 1.05j
= - 2.586 i + 2.85j
x component = - 2.586
y component = 2.85
Answer:
Re=160ohm
Explanation:
Step#1
Rt=R1+R2 ( because both are in series)
Rt=(100+220 ) ohm
Rt=320 ohm
Step#2
Rt and R3 are parallel so,
Re= (Rt× R3) ÷ (Rt+R3)
Re= (320×320)÷( 320+320)
Re = 102,400÷ 640
Re=160ohm
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