Regions in the milky way where density waves have caused gas clouds to crash into each other are called clumps.Clumps are molecular clouds (interstellar clouds) with higher density,where lots of dust and gs cores resides. These clouds are the beginning of stars.
Answer:
Vf = 69.56 cm/s
Explanation:
In order to find the final speed of the ramp, we will use the equations of motion. First we use second equation of motion to find out the acceleration of marble:
s = Vi t + (1/2)at²
where,
s = distance traveled = 160 cm
Vi = Initial Speed = 0 cm/s (since, marble starts from rest)
t = time interval = 4.6 s
a = acceleration = ?
Therefore,
160 cm = (0 cm/s)(4.6 s) + (1/2)(a)(4.6 s)²
a = (320 cm)/(4.6 s)²
a = 15.12 cm/s²
Now, we use first equation of motion:
Vf = Vi + at
Vf = 0 cm/s + (15.12 cm/s²)(4.6 s)
<u>Vf = 69.56 cm/s</u>
Given Information:
Pendulum 1 mass = m₁ = 0.2 kg
Pendulum 2 mass = m₂ = 0.6 kg
Pendulum 1 length = L₁ = 5 m
Pendulum 2 length = L₂ = 1 m
Required Information:
Affect of mass on the frequency of the pendulum = ?
Answer:
The mass of the ball will not affect the frequency of the pendulum.
Explanation:
The relation between period and frequency of pendulum is given by
f = 1/T
The period of pendulum is given by
T = 2π√(L/g)
Where g is the acceleration due to gravity and L is the length of the string
As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.
Bonus:
Pendulum 1:
T₁ = 2π√(L₁/g)
T₁ = 2π√(5/9.8)
T₁ = 4.49 s
f₁ = 1/T₁
f₁ = 1/4.49
f₁ = 0.22 Hz
Pendulum 2:
T₂ = 2π√(L₂/g)
T₂ = 2π√(1/9.8)
T₂ = 2.0 s
f₂ = 1/T₂
f₂ = 1/2.0
f₂ = 0.5 Hz
So we can conclude that the higher length of the string increases the period of the pendulum and decreases the frequency of the pendulum.
When she starts out, he is (40x2.5)= 100 miles ahead of her.
She gains (65-40)= 25 miles on him every hour.
It takes her (100/25)= 4 hours to catch up to him.
