An automobile approaches a barrier at a speed of 20 m/s along a level road. The driver locks the brakes at a distance of 50 m fr
1 answer:
Answer:
μ = 0.408
Explanation:
given,
speed of the automobile (u)= 20 m/s
distance = 50 m
final velocity (v) = 0 m/s
kinetic friction = ?
we know that,
v² = u² + 2 a s
0 = 20² + 2 × a × 50
a = 4 m/s²
We know
F = ma = μN
ma = μ mg
a = μ g
μ = 0.408
hence, Kinetic friction require to stop the automobile before it hit barrier is 0.408
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Complete Question
A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?
Answer:
Explanation:
From the question we are told that
Mass of ball
Length of string
Wind force
Generally the equation for is mathematically given as
Max angle =
Generally the equation for max Height is mathematically given as
Generally the equation for Equilibrium Height is mathematically given as
Answer:
a). 53.78 m/s
b) 52.38 m/s
c) -75.58 m
Explanation:
See attachment for calculation
In the c part, The negative distance is telling us that the project went below the lunch point.
