Answer:
B. 7.5 m/s^2
Explanation:
To find acceleration you need to subtract the final velocity by the starting velocity then divide that by the time
a= v-v/t
a= 60-0/8
a= 60/8
a=7.5 m/s^2
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.
Answer:
A : hot and moist, maritime tropical
B: cold and dry, maritime polar
C: hot and moist , maritime tropical
D: cold and dry, continental polar
E: hot and moist , maritime tropical
F: cold and dry , maritime polar
Explanation:
Cold air is denser than warm air. The more water vapor that is in the air, the less dense the air becomes. That is why cold, dry air is much heavier than warm, humid air.
Maritime polar (mP) air masses are cool, moist, and unstable. Some maritime polar air masses originate as continental polar air masses over Asia and move westward over the Pacific, collecting warmth and moisture from the ocean.
Maritime tropical (mT) air masses are warm, moist, and usually unstable.
Answer:
a) Pb= 200 PA
b).work done= -3600 joules
c).3600joules
D).the system works under isothermal condition so no heat was transferred
Explanation:
2.0 moles of a monatomic ideal gas expands isothermally from state a to state b, Pa = 600 Pa, Va = 3.0 m3, and Vb = 9.0 m3.
a). PbVb= PaVa
Pb= (PaVa)/VB
Pb= (600*3)/9
Pb= 1800/9
Pb= 200 PA
b). work done= n(Pb-Pa)(Vb-Va)
Work done= 2*(200-600)(9-3)
Work done= -600(6)
Work done=- 3600 Pam³
work done= -3600 joules
C). Change in internal energy I the work done on the system
= 3600joules
D).the system works under isothermal condition so no heat was transferred
