Eye strain and self confidence issues.
Explanation:
a) d = ½.a.t²
200 = ½(4)t²
200 = 2t²
t² = 200/2
t² = 100
t =√100 = 10 s
b) Vt = a. t
= 4(10)
= 40 m/s
c) V av. = d/t = 200/10 = 20m/s
The distance at which the man slips is 0.3 m
Newton's Second Law, F = ma, is used to calculate the braking distance. By dividing the mass of the car by the gravitational acceleration, one may determine its weight. The weight of the car multiplied by the coefficient of friction equals the brake force.
Given-
mass of man= 70 kg
frictional coefficient μ=0.02
mass of body thrown= m2 = 3kg
let s be the stopping distance
we know that frictional force = F= μN
=μMg= 0.02 x 70 x 10
=14 N
∴acceleration, a= 14/70 = 0.2 m/s²
now on applying conservation of linear momentum
pi=pf pi=0 (initially at rest)
0=m1v1-m2v2 (v1= velocity of man) (v2=velocity of body= 8m/s
v1= m2v2 /m1= 0.3 m/s
we know,
v²- u² = -2as
0- (0.3) ²= -2 x 0.2 x 5
s= 0.09/0.4 ≈ 0.3 m
Learn more about distance here-
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It’s either snow or rain if it’s a riddle sort of.
Answer:
The time it can operate between chargins in minutes is
Explanation:
Given: 
, 
, 
a). The rotational kinetic energy
b). The power average 0.8kW un range time can be find
Solve to t'
