Answer:Gravity
Explanation:Gravity is the force that pulls everything down instead of up because if we didn’t have gravity we would be floating upwards
Answer:
0.15 s
Explanation:
From the question given above, the following data were obtained:
Speed of sound (v) = 330 m/s
Distance (x) = 25 m
Time (t) =?
The time taken for the echo of the sound to the bat can be obtained as follow:
v = 2x / t
330 = 2 × 25 / t
330 = 50 / t
Cross multiply
330 × t = 50
Divide both side by 330
t = 50 / 330
t = 0.15 s
Thus, it will take 0.15 s for the echo of the sound to the bat
Answer:
vo=5.87m/s
Explanation:
Hello! In this problem we have a uniformly varied rectilinear movement.
Taking into account the data:
α =69.2
vf = 10m / s
h=2.7m
g=9.8m/s2
We know we want to know the speed on the y axis.
We calculate vfy
vfy = 10m / s * (sen69.2) = 9.35m / s
We can use the following equation.
We clear the vo (initial speed)
vo=5.87m/s
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: 10m/s
Explanation: Since a single vector with length of 1cm expresses 5m/s, the vector which has a length doubled from the original vector should have the speed which is also doubled.
