Answer:
Option D. 4.4 m/s²
Explanation:
The following data were obtained from the question:
Velocity (v) = 21 m/s
Radius (r) = 100 m
Centripetal acceleration (a) =.?
The centripetal acceleration of the car can be obtained as follow:
Centripetal acceleration (a) = Velocity square (v²) / radius (r)
a = v²/r
a = 21²/100
a = 441/100
a = 4.41 ≈ 4.4 m/s²
Therefore, the centripetal acceleration of the car is 4.4 m/s².
Answer:
Work done, 
Explanation:
Displacement,

Force, 
Work done by the opponent do on the football player is given by :

So, the work done by the opponent do on the football player is
.
Answer:
At focus
Explanation:
A concave mirror is converging in nature. In a mirror, concave in nature, the rays which are parallel to the principal axis are supposed to be coming from very large distances or we assume the source to be placed at infinity for such rays which are parallel to the principal axis.
These rays, parallel to the principal axis, coming from infinity, converges at the focus of the mirror concave in nature after reflecting from the concave mirror
Answer:
a) the one with a lower orbit b) the one with a higher orbit
Explanation:
Let's consider orbital mechanics. To get an object in orbit, we need it to fall to earth parallel to the earth's surface. To understand it easily imagine a projectile thrown horizontally further and further away, at one point, the projectile hits the cannon from behind. Considering there is no wind resistance, that would be a projecile in orbit.
In other words, the circular orbits of some objects around a massive body are due to the equality between centrifugal acceleration and gravity acceleration.
.
so the velocity is

where "G" is the gravitational constant, "M" the mass of the massive body and "r" the distance between the object and the center of gravity of mass M. As you can note, if "r" increase, "v" decrease.
The orbital period of any object in orbit is

where "a" is length of semi-major axis (a = r in circular orbits). So if "r" increase, "T" increase.
Answer:
The explosive force experienced by the shell inside the barrel is 23437500 newtons.
Explanation:
Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:
(1)
Where:
- Explosive force, measured in newtons.
- Barrel length, measured in meters.
- Mass of the shell, measured in kilograms.
,
- Initial and final speeds of the shell, measured in meters per second.
If we know that
,
,
and
, then the explosive force experienced by the shell inside the barrel is:

![F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B%281250%5C%2Ckg%29%5Ccdot%20%5Cleft%5B%5Cleft%28750%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%29%5E%7B2%7D-%5Cleft%280%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%29%5E%7B2%7D%5Cright%5D%7D%7B2%5Ccdot%20%2815%5C%2Cm%29%7D)

The explosive force experienced by the shell inside the barrel is 23437500 newtons.