Using conservation of energy and momentum we get m1*v1=(m1+m2)*v2 so rearranging for v2 and plugging the given values in we get:
(200000kg*1.00m/s)/(21000kg)=.952m/s
Answer:
When the ball goes to first base it will be 4.23 m high.
Explanation:
Horizontal velocity = 30 cos17.3 = 28.64 m/s
Horizontal displacement = 40.5 m
Time
Time to reach the goal posts 40.5 m away = 1.41 seconds
Vertical velocity = 30 sin17.3 = 8.92 m/s
Time to reach the goal posts 40.5 m away = 1.41 seconds
Acceleration = -9.81m/s²
Substituting in s = ut + 0.5at²
s = 8.92 x 1.41 - 0.5 x 9.81 x 1.41²= 2.83 m
Height of throw = 1.4 m
Height traveled by ball = 2.83 m
Total height = 2.83 + 1.4 = 4.23 m
When the ball goes to first base it will be 4.23 m high.
Answer:
93 km/h
Explanation:
Given that a bus took 8 hours to travel 639 km. For the first 5 hours, it travelled at an average speed of 72 km/h
Let the first 5 hours journey distance = F
From the formula of speed,
Speed = distance/time
Substitute speed and time
72 = F/5
F = 72 × 5 = 360 km
The remaining distance will be:
639 - 360 = 279km
The remaining time will be:
8 - 5 = 3 hours
Speed = 279/3
Speed = 93 km/h
Therefore, the average speed for the remaining time of the journey is equal to 93 km/h
Answer:
The last one, nicotine from tobacco
Answer:
a.Distance = 150 m
b. Displacement = 50 m
Time lapsed = 5 seconds
Explanation:
a. Distance is the change in the position of an object.
The distance covered by the car = 100 + 50
= 150 m
b. Since displacement is a vector quantity,
Displacement of the car = 100 - 50
= 50 m due east
c. Time elapsed is the time taken for the motion of the car starting from when its starts to when it stops.
Thus, the time elapsed = 4 + 1
= 5 seconds