Incomplete question as the car's speed is missing.I have assumed car's speed as 6.0m/s.The complete question is here
An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v 6.0m/s
Answer:

Explanation:
Set up force equation
∑F=ma
∑F=W+FB
The minus sign for downward direction
Given that,
Acceleration, a = 9.71 m/s²
Force, F = 5050 N
mass, m = ?
Since, we know that
F=ma
m= F/a
m= 5050/9.71
m= 520.08 kg
The mass of hi car is 520.08 kilograms.
Responder:
Velocidad = 41.5m / s
Espacio recorrida = 352.5 metros
Explicación:
Dado lo siguiente:
Velocidad inicial (u) = 19.8 km / h
Aceleración (a) = 2.4m / s ^ 2
Tiempo de viaje (t) = 15 s
A.) velocidad después de 15 s
Velocidad inicial = (19.8 × 1000) m / 3600s Velocidad inicial = 19800m / 3600 = 5.5m / s
Usando la ecuación: v = u + at, donde v es la velocidad
v = 5.5 + 2.4 (15)
v = 5.5 + 36
v = 41.5m / s
Espacio recorrida:
v ^ 2 = u ^ 2 + 2aS; donde S es la distancia recorrida
41.5 ^ 2 = 5.5 ^ 2 + 2 × (2.4) × S
1722.25 = 30.25 + 4.8S
1722.25 - 30.25 = 4.8S
1692 = 4.8S S = 1692 / 4.8 S = 352.5m
Answer:
based on my opinion
The smart guys at NASA designed the trajectory such that as they passed Jupiter, they gained some speed by being dragged along by Jupiter
This is called a gravity assist . Voyager 2 picked up about 18 km / s of velocity from that Jupiter gravity assist ...
<em>Hope</em><em> this</em><em> helps</em>