F(of spring)=230x=ma=3.5(5)=17.5=230x; x=0.07m.
Answer:
The rate at which bus 1 is going is 55 mph
The rate at which bus 1 is going is 35 mph
Explanation:
As per the question:
Suppose, the distance traveled by Bus 1 be 'd' at the rate R after a time, t = 3h
Thus
Suppose, the distance traveled by Bus 1 be 'd'' at the rate, R'20 mph slower than the rate of Bus 1 after the same time.
R' = R - 20
The distance is given as the product of rate and time:
d = Rt (1)
Now, the total distance given is 270 miles:
d + d' = 270
Now, using eqn (1):
Rt + R't = 270
3(R + R - 20) = 270
6R = 270 + 60
R = 55 mph
R' = R - 20 = 55 - 20 = 35 mph
A series circuit is a closed circuit in whicj the current flows through one path.
<span>a 205 kg mass object will weight 1143.43 lbs on Jupiter.
Looking up the surface gravity of Jupiter, you can find that it's 2.53 times that of earth. So the 205 kg object will weigh
205 * 2.53 = 518.65 kg on Jupiter.
Now we need to convert from kg to pounds. This is done by multiplying by 2.20462
518.65 kg * 2.20462 lb/kg = 1143.43 lb</span>
Answer:
The centripetal acceleration of the runner as he runs the curved portion of the track is 0.91 m/s²
Explanation:
Given;
distance traveled in the given time = 200 m
time to cover the distance, t = 29.6 s
speed of the runner, v = d / t
v = 200 / 29.6
v = 6.757 m/s
The centripetal acceleration of the runner is given by;
where;
r is the radius of the circular arc, given as 50 m
Substitute the givens;
Therefore, the centripetal acceleration of the runner as he runs the curved portion of the track is 0.91 m/s².
