Answer:
Force on front axle = 6392.85 N
Force on rear axle = 8616.45 N
Explanation:
As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels
Now we know that
now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle
so we can write torque balance about its center of mass
now from above equation
now we have
now the other force is given as
Answer: 17.83 AU
Explanation:
According to Kepler’s Third Law of Planetary motion <em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>
(1)
Talking in general, this law states a relation between the <u>orbital period</u> 
of a body (moon, planet, satellite, comet) orbiting a greater body in space with the <u>size</u> 
of its orbit.
However, if is measured in <u>years</u>, and is measured in <u>astronomical units</u> (equivalent to the distance between the Sun and the Earth: 
), equation (1) becomes:
(2)
This means that now both sides of the equation are equal.
Knowing 
and isolating from (2):
![a=\sqrt[3]{T^{2}}=T^{2/3}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7BT%5E%7B2%7D%7D%3DT%5E%7B2%2F3%7D)
(3)
(4)
Finally:
(5)
Answer:
Explanation:
This problem can be solved with the conservation of the momentum.
If the ball is fired upward, the momentum before and after the ball is fired must conserve. Hence, the speed of the ball is the same that the speed of the car just in the moment in wich the ball is fired.
Hence, the result depends of the acceleration of the car. If the change in the speed is higher than the speed of the ball, it is probably that the ball will be behind the car or it will come back to the car.
If the ball is fired forward, and if the change in the speed of the car is not enogh, the ball will be in front of the car.
HOPE THIS HELPS!!
