The answer to this question is:
C-"That moving clocks run slower"
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km x h = km/h
First trial: 6 x 1 = 6km/h
Second trial: 9 x 2 = 18km/h
6 + 18 = <u>24km/h</u> (Total)
Or
6 + 9 = 15 km
2 + 1 = 3h
15 + 3 = 18
15 x 2 = 30
3 x 2 = 6
30 - 6 = <u>24km/h</u>
Answer:
R = 0.0503 m
Explanation:
This is a projectile launching exercise, to find the range we can use the equation
R = v₀² sin 2θ / g
How we know the maximum height
² =
² - 2 g y
= 0
= √ 2 g y
= √ 2 9.8 / 15
= 1.14 m / s
Let's use trigonometry to find the speed
sin θ =
/ vo
vo =
/ sin θ
vo = 1.14 / sin 60
vo = 1.32 m / s
We calculate the range with the first equation
R = 1.32² sin(2 60) / 30
R = 0.0503 m
Given
m1(mass of red bumper): 225 Kg
m2 (mass of blue bumper): 180 Kg
m3(mass of green bumper):150 Kg
v1 (velocity of red bumper): 3.0 m/s
v2 (final velocity of the combined bumpers): ?
The law of conservation of momentum states that when two bodies collide with each other, the momentum of the two bodies before the collision is equal to the momentum after the collision. This can be mathemetaically represented as below:
Pa= Pb
Where Pa is the momentum before collision and Pb is the momentum after collision.
Now applying this law for the above problem we get
Momentum before collision= momentum after collision.
Momentum before collision = (m1+m2) x v1 =(225+180)x 3 = 1215 Kgm/s
Momentum after collision = (m1+m2+m3) x v2 =(225+180+150)x v2
=555v2
Now we know that Momentum before collision= momentum after collision.
Hence we get
1215 = 555 v2
v2 = 2.188 m/s
Hence the velocity of the combined bumper cars is 2.188 m/s