Answer:<em> tried my best to answer your ?s:</em>
<em>B</em>
<em>A</em>
<em>B</em>
<em>D</em>
<em>A</em>
<em>A</em>
<em>B?</em>
<em>Hope this helps!!</em>
Given:
Mass of the rail road car, m = 2 kg
velocity of the three cars coupled system, v' = 1.20 m/s
velocity of first car, = 3 m/s
Solution:
a) Momentum of a body of mass 'm' and velocity 'v' is given by:
p = mv
Now for the coupled system according to law of conservation of momentum, total momentum of a system before and after collision remain conserved:
(1)
where,
= velocity of the first car
= velocity of the 2 coupled cars after collision
Now, from eqn (1)
v' = 1.80 m/s
Therefore, the velocity of the combined car system after collision is 1.80 m/s
One of the efficient concepts that can help us find the number of turns of the cable is through the concept of induced voltage or electromotive force given by Faraday's law. The electromotive force or emf can be described as,
Where,
N = Number of loops
B = Magnetic Field
A = Cross-sectional Area
= Angular velocity
Re-arrange to find N,
Our values are given as,
Replacing at our equation we have:
Therefore the number of loops of wire should be wound on the square armature is 32 loops
Answer: Option (a) is the correct answer.
Explanation:
Resources which are created by human beings are known as man-made resources.
For example, glass, rayon, nylon etc are all man-made resources.
Whereas resources which are naturally created are known as natural resources.
For example, wind, air, water etc are all natural resources.
Thus, we can conclude that out of the given options time is not a natural resource.
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-
brainly.com/question/15172156
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