Answer:
Explanation:
It is given that,
Speed of one quoll around a curve, v = 3.2 m/s (maximum speed)
Radius of the curve, r = 1.4 m
On the curve, the centripetal force is balanced by the frictional force such that the coefficient of frictional is given by :



So, the coefficient of static friction between the quoll's feet and the ground in this trial is 0.74. Hence, this is the required solution.
I am assuming this is a true or false question, to which the answer would be True.
k = spring constant of the spring = 85 N/m
m = mass of the box sliding towards the spring = 3.5 kg
v = speed of box just before colliding with the spring = ?
x = compression the spring = 6.5 cm = 6.5 cm (1 m /100 cm) = 0.065 m
the kinetic energy of box just before colliding with the spring converts into the spring energy of the spring when it is fully compressed.
Using conservation of energy
Kinetic energy of spring before collision = spring energy of spring after compression
(0.5) m v² = (0.5) k x²
m v² = k x²
inserting the values
(3.5 kg) v² = (85 N/m) (0.065 m)²
v = 0.32 m/s
Answer:
Scalar product is between ║A║ ║ B║ and -║A║ ║ B║
Explanation:
Dot product between vec A and vec B is
A.B = ║A║ ║ B║ cos θ
Here, both ║A║ and ║B║ are positive and value of cos θ depends upon θ and lies between 1 and -1
So, Scalar product is between ║A║ ║ B║ and -║A║ ║ B║
Answer with Explanation:
We are given that


Charge on proton,q=
a.We have to find the electric potential of the proton at the position of the electron.
We know that the electric potential

Where 


B.Potential energy of electron,U=
Where
Charge on electron
=Charge on proton
Using the formula

