Answer:
73.5 m/s
Explanation:
The position of the first ball is:
y = y₀ + v₀ t + ½ at²
y = h + (0)(18) + ½ (-9.8)(18)²
y = h − 1587.6
The position of the second ball is:
y = y₀ + v₀ t + ½ at²
y = h + (-v) (18−6) + ½ (-9.8)(18−6)²
y = h − 12v − 705.6
Setting the positions equal:
h − 1587.6 = h − 12v − 705.6
-1587.6 = -12v − 705.6
1587.6 = 12v + 705.6
882 = 12v
v = 73.5
The second ball is thrown downwards with a speed of 73.5 m/s
Electric Energy = Current x Voltage x time = 207 x 9.4 = 1,945.8 J
Answer:
The centripetal acceleration of the runner is
.
Explanation:
Given that,
A runner completes the 200 m dash in 24.0 s and runs at constant speed throughout the race. We need to find the centripetal acceleration as he runs the curved portion of the track. We know that the centripetal acceleration is given by :

v is the velocity of runner

Centripetal acceleration,

So, the centripetal acceleration of the runner is
. Hence, this is the required solution.
Answer:
317.22
Explanation:
Given
Circular platform rotates ccw 93.1kg, radius 1.93 m, 0.945 rad/s
You 69.7kg, cw 1.01m/s, at r
Poodle 20.2 kg, cw 1.01/2 m/s, at r/2
Mutt 17.7 kg, 3r/4
You
Relative
ω = v/r
= 1.01/1.93
= 0.522
Actual
ω = 0.945 - 0.522
= 0.42
I = mr^2
= 69.7*1.93^2
= 259.6
L = Iω
= 259.6*0.42
= 109.4
Poodle
Relative
ω = (1.01/2)/(1.93/2)
= 0.5233
Actual
ω = 0.945- 0.5233
= 0.4217
I = m(r/2)^2
= 20.2*(1.93/2)^2
= 18.81
L = Iω
= 18.81*0.4217
= 7.93
Mutt
Actual
ω = 0.945
I = m(3r/4)^2
= 17.7(3*1.93/4)^2
= 37.08
L = Iω
= 37.08*0.945
= 35.04
Disk
I = mr^2/2
= 93.1(1.93)^2/2
= 173.39
L = Iω
= 173.39*0.945
= 163.85
Total
L = 109.4+ 7.93+ 36.04+ 163.85
= 317.22 kg m^2/s
Answer:
The magnitud of the velocity is

and the direccion:
degrees from the horizontal.
Explanation:
Fist we define our variables:

The letters i and j represent the direction of the movement, i in this case is the horizontal direction, and j is perpendicular to i.
velocities with sub-index 1 are the speeds before the crash, and with sub-index 2 are the velocities after the crash.
Using conservation of momentum:

Clearing for the velocity of the stone after the crash:

Substituting known values:

The magnitud of the velocity is :

and the direction:

this is -28.3 degrees from the +i direction or the horizontal direcction.
Note: i and j can also be seen as x and y axis.