When the outer rail of a curved railway track is raised over the inner, the horizontal component of the normal reaction of the rails, provides the necessary centripetal force for the train to enable it moving along the curved path.

When we go up a mountain, the opposing forces of friction F = μR = μmg cos θ
Where θ is angle of slope with horizontal. To avoid skidding, F should be large.
∴ cos θ shpuld be large and hence, θ must be small. 
Therefore, mountain roads are generally made winding upwards. The roads straight up would have large slope.

When a body is moving along a circular path, speed always remains constant and a centripetal force is acting on the body.

As the applied force increases, the static friction also increases and becomes equal to the applied force to make the object stationery. That is why static friction is called a self-adjusting force.

It is given that:
Mass of body, m=25 gm =0.025 kg.
It is also given that the object is moving with constant velocity, v=5 m / s
Now, we need to find out the force acting on the body .
Also, we know, Force, F= ma
Mass is given to us and we need to find the acceleration .
We know acceleration is change in velocity per unit time.
Since, velocity is constant here. Therefore, acceleration is zero.
Therefore, force acting is also zero.

Step 1: Given that:
The mass(m) of the car= 1500kg
Initial velocity of the car(u) = 90kmh⁻¹=90 × 518 ms⁻¹ =25ms⁻¹
Final velocity of the car(v) = 0
Step 2: Calculation of impulse:
We have,
Impulse= Change in momentum of the body = mv−mu
Thus, Impulse on the car will be;
= 1500kg × 0 − 1500kg × 25ms⁻¹
=−37500kgms⁻¹
=−3.75×10⁴kgms⁻¹
Thus,Impulse on the car will be 3.75 × 10⁴kgms⁻¹

When a child falls on a cement floor, her body comes to rest instantly.
BUt F x \(\triangle \) t = cahnge in momentum = constant.
As time of stopping  \(\triangle \) t  decreases, therefore F increases and hence child feel more pain. When she falls on a soft muddy ground in the garden, the time of stopping increases and hence F decreases and she feels lesser pain.

Given, mass, m = 72.2 kg
Gravity acceleration, g = 9.8 m/s² 
Scale reading = apparent weight = R = ?
(i)  While descending with constant velocity, a = 0
\(\therefore \)    R =  mg
R = 72.2 x 9.8
\(\Longrightarrow \)  R = 707.56 N
(ii)  While ascending with a = 3.2 m/s² 
R = m ( g + a )
R = 72.2 ( 9.8 + 3.2 ) = 938.6 N

No change in speed, but change in direction is possible. Forces acting on a body in circular motion is an example.

zero, 3.6 N

Here, \(m=10 \mathrm{~kg}, \theta=30^{\circ}, \mu=\frac{1}{\sqrt{3}}\)
As is clear from force required just to push the body up the inclined plane is
\( F=m g \sin \theta+f \)
\(=m g \sin \theta+\mu R \)
\(=m g \sin \theta+\mu m g \cos \theta \)
\(=m g(\sin \theta+\mu \cos \theta) \)
\(=10 \times 9.8\left(\sin 30^{\circ}+\frac{1}{\sqrt{3}} \cos 30^{\circ}\right) \)
\( F=98\left(0.5+\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{2}\right)=98 N \)

Given, r =10⁴ km = 10⁴ \(\times \) 1000 m = 10⁷ m,
m = 2500 kg
v = 4 kms^-1 = 4\(\times \)10³ ms^-1
Now,centripeta force is F = \(\frac { m{ v }^{ 2 } }{ r } \) 
F = \(\frac { 2500\times (4\times { 10 }^{ 3 }{ ) }^{ 2 } }{ { 10 }^{ 7 } } =\frac { 2500\times 16\times { 10 }^{ 6 } }{ { 10 }^{ 7 } }\)
\( \\ F=\quad \frac { 250\times 16\times { 10 }^{ 7 } }{ { 10 }^{ 7 } }\)
\(= 4000N\)