The question is incomplete. The mass of the object is 10 gram and travelling at a speed of 2 m/s.
Solution:
It is given that mass of object before explosion is,m = 10 g
Speed of object before explosion, v = 2 m/s
Let
be the masses of the three fragments.
Let
be the velocities of the three fragments.
Therefore, according to the law of conservation of momentum,


So the x- component of the velocity of the m2 fragment after the explosion is,

∴ 
Answer:
0.572
Explanation:
First examine the force of friction at the slipping point where Ff = µsFN = µsmg.
the mass of the car is unknown,
The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.
First the tangential direction
∑Ft =Fft =mat
And then in the centerward direction ∑Fc =Ffc =mac =mv²t/r
Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2
So going backwards and plugging in Ffc =m2atπr/ 2r =πmat
Ff = √(F2ft +F2fc)= matp √(1+π²)
µs = Ff /mg = at /g √(1+π²)=
1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572
ANSWER: NATURE
EXPLAINTION:
Mass of 1 staple = 6.8 g/210 staples
mass of 1 staple = 0.032380952 g
Hope that helps!!
The correct answer to the question is : 9375 N.
CALCULATION:
As per the question, the mass of the car m = 1500 Kg.
The diametre of the circular track D = 200 m.
Hence, the radius of the circular path R = 
= 
= 100 m.
The velocity of the truck v = 25 m/s.
When a body moves in a circular path, the body needs a centripetal force which helps the body stick to the orbit. It acts along the radius and towards the centre.
Hence, the force acting on the car is centripetal force.
The magnitude of the centripetal force is calculated as -
Force F = 
= 
= 9375 N. [ANS}
The centripetal force is provided to the car in two ways. It is the friction which provides the necessary centripetal force. Sometimes friction is not sufficient. At that time, the road is banked to some extent which provides the necessary centripetal force.