Explanation:
It is given that,
Force applied to object, 
Position, 
(b) The cross product of force and position vector is used to find the net torque about the z axis. It is given by :
or
The torque is acting in -z axis.
(a) The magnitude of torque is given by :
Hence, this is the required solution.
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
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Answer:
Oppositely charged objects attract each other
When we say "<span>The moon's surface gravity is one-sixth that of the earth.",
we mean that the acceleration of gravity on the Moon's surface is 1/6 of
the acceleration of gravity on the Earth's surface.
The acceleration of gravity is (9.8 m/s</span>²) on the Earth's surface, so
<span>it would be (9.8/6 m/s</span>²) on the Moon's surface.
<span>
The weight of any object, right now, is
(object's mass) </span>· (acceleration of gravity where the object is located now) .
<span>
If the object's mass is 24 kg and the object is on the Moon right now,
then its weight is
(24 kg) </span>· (9.8/6 m/s²)
= (24 · 9.8 / 6) kg-m/s²
= 39.2 Newtons
