A chain of 5 link, each of mass 0.1 kg , is lifted vertically with a constant acceleration 1.2m/s^2 . The force of interaction between the top link and the one immediately below it isa) 1.10 N b) 0.88N c) 0.66N d) 0.44N

Speed of the aircraft,$v=720km/h=7205/18=200m/s$.Acceleration due to gravity,g=10m/s2Angle of banking =15oFor radius,$r$, of the loop, we have the relation:$Ncos\theta =mg$Nsinθ=Rmu2tanθ=u2/Rgr=u2/gtan$\theta$=2002/ (10tan15o)$=4000/0.26$$=14.92km$

A) For pulling a cart or for running, the horse pushes the earth with its feet and reaction of the earth makes it move in the forward direction. Since in empty space there is no reaction force, therefore the horse cannot run in empty space.

B) Due to inertia of motion.

C)While pulling a lawn mower, a force at an angle is applied on it, as shown in the given figure.

The vertical component of this applied force acts upward. This reduces the effective weight of the mower.

On the other hand, while...

A system of coplanar forces are in equilibrium when their resultant is zero. This implies that the vector sum of these forces along x and y axes should be zero.

total initial momentum=0

final momentum=final momentum of gun + final momentum of bullet

=m_gv_g + m_bv_b

=10*x+(50/1000)*500

=10x+25

BY PRINCIPLE OF CONSERVATON OF LINEAR MOMENTUM,

INITIAL MOMENTUM=FINAL MOMENTUM

0=10x+25

-25=10x

x=-2.5ms-1

the negative sign shows that the object is moving in opp. direction.

so, the gun recoils with the velocity of 2.5 ms-1.

The resultant momentum is 0 kgms-1 as the gun and bullet were at rest initially.

The torque T produced by applying a force F at a distance r from the axis of rotation is T = rF. It follows from this equation that the distance r should be increased in order to increase the torque. Since the force is applied to the handle put near the circumference of the stone, the distance r is the greatest. For this reason, the greater amount of torque can produce a larger angular acceleration. In such a hand driven grinding machine, grains are ground comfortably

When we use two fingers to open the cap of a bottle, we apply the torque due to a couple which is the product of one force and the perpendicular distance between the forces, i.e., a distance equal to the diameter of the bottle neck. When we use one finger to open it, we apply the torque which is the product of the force and the distance equal to its radius. So the former torque is twice the latter torque and hence it is easier to open the cap of the bottle with two fingers than that with one...

When a man carries a bucket full of water on his right hand without leaning towards his left, the location of the center of gravity of the system (man + bucket full of water) changes and the system becomes unstable. This produces a net torque about the center of gravity of the man and this torque tries to rotate the man to the right. In order to cancel this torque, the man has to lean to the left so that the center of gravity of the man and the bucket full of water as a whole lies within the...

A ball thrown straight up has zero velocity at its highest point instantly. The ball is not in equilibrium at this point because it is always acted upon by the force of gravity and has acceleration. If it had gained zero acceleration at its highest point, it would have been in equilibrium.

For a stable equilibrium of a body, its center of gravity should lie as low as possible. Loading heavy loads first in a truck and loading light loads on the heavy ones can lower the center of gravity of the truck and load system. This ensures the stability of the system and the loads can be transported safely. Therefore the heavier loads must be loaded first in any transporting vehicles. For the same reason, the passengers are not allowed to stand in a boat and the bottom of a ship is made...

Solution

Given:

Height of balloon when stone was dropped (s)= -50 m/s [since the stone was dropped downwards]

Initial velocity (u) = 5.0 m/s

Acceleration due to gravity (a) =10 m/s²

Using formula,

s= ut + 1/2 * at²

-50 = 5t + 1/2 (-10)t²

-50 = 5t - 5t²

-50 = 5(t - t²)

-50/5 = t - t²

-10 = t - t²

t²- t - 10 = 0

t²- (3.7 - 2.7) t - 10 = 0

t²- 3.7 t + 2.7 t - 10= 0

( t - 3.7 ) ( t + 2.7 ) = 0

Either,

t = 3.7

or,

t = - 2.7

Since time ( t ) can never be negative,

t = 3.7 sec

Height of the balloon when stone hits the...

1.(c) w1 + w2

2.(b) the ground on the horse

3.(d) the road

4.(c) mg cosθ

5.(b) mg/cosθ

6.(d) remain standing

7.(a) zero

8.(a) F1 must be equal to F2

9.(b) A will go higher than B

10.(c) remain at the top of the wedge

11.(b) t1 > t2

12.(b) cannot be an inertial frame because the earth is

revolving around the sun

And

(d) cannot be an inertial frame because the earth is

rotating about its axis.

Torque or turning effect due to a force is maximum when r is maximum. We prefer to use a wrench with long arm because when the length of the arm(r) is long, the force (F) required to produce a given turning effect ( x ) is smaller. Hence, a nut can be unscrewed easily.

Yes.

We can take example of a body thrown upwards.
when a body projected upwards reaches a highest point, its velocity is zero but the acceleration is equal to g-the acceleration due the gravity.

Yes, a body can be said to be at rest and in motion at the same time. No body remains absolute rest or motion i.e. they are comparitive body that depends on the reference frame/point. For example: a passenger travelling in a bus is at the state of rest for others passenger in a same bus but, the same passenger is at motion with respect to the pole, tress and houses outside.

When a woman in an elevator lets go of her briefcase but it does not fall, then the elevator is accelerating downward with greater acceleration than the acceleration due to gravity 'g' ( i.e. a>g). The reaction, R is negative and the briefcase stick in the celling of lift rather than falling.

Let F be the Force applied to pull a body of mass m as shown in fig (i) . F Cosθand F Sinθare the components of mass F, whereθis the angle between the force F and horizontal. The Normal reactionis given by R₁ = mg - FSinθ. Ifμ_kis the coefficientof kinetic friction, then frictionforce is given by,
f₁=μ_kR₁ =μ_k(mg -F Sinθ) .......................... 1

If the same force F is applied to push the body the normal reaction,
R₂ =(mg +F Sinθ)
The kinetic friction is given by,
f₂=μ_kR₂=μ_k(mg +F Sinθ)...