Answer:
Here the source is moving away from the observer so frequency will be smaller than the actual frequency and since the speed is increasing so the frequency is decreasing with time so correct answer is
D) lower than the original pitch and decreasing as he falls.
Explanation:
As we know by the Doppler's effect of sound we have
so we will have

so here when source moves away from the observer with a some speed then the frequency of the sound observed by the observer is smaller than the actual frequency
Here we know that the speed of the source is increasing with time as the source is falling under gravity
So we can say that the pitch of the sound will decrease with time
There is one mistake in the question.The Correct question is here
A cat falls from a tree (with zero initial velocity) at time t = 0. How far does the cat fall between t = 1/2 and t = 1 s? Use Galileo's formula v(t) = −9.8t m/s.
Answer:
y(1s) - y(1/2s) = - 3.675 m
The cat falls 3.675 m between time 1/2 s and 1 s.
Explanation:
Given data
time=1/2 sec to 1 sec
v(t)=-9.8t m/s
To find
Distance
Solution
As the acceleration as first derivative of velocity with respect to time
So
acceleration(-g)= dv/dt
Solve it
dv = a dt
dv = -g dt
v - v₀ = -gt
v= dy/dt
dy = v dt
dy = ( v₀ - gt ) dt
y(1s) - y(1/2s) = ( v₀ ) ( 1 - 1/2 ) - ( g/2 )[ ( t1)² -( t1/2s )² ]
y(1s) - y(1/2s) = ( - 9.8/2 ) [ ( 1 )² - ( 1/2 )² ]
y1s - y1/2s = ( - 4.9 m/s² ) ( 3/4 s² )
y(1s) - y(1/2s) = - 3.675 m
The cat falls 3.675 m between time 1/2 s and 1 s.
Answer: The center of gravity is 1.1338 m away from the left side of the barbell
Explanation:
Length of the barbell = 1.90 m
The distance center of gravity from left = x
Mass on the left side = 25 kg
The distance center of gravity from right = 1.90 - x
Mass on the right side = 37 kg
At the balance point: 


The center of gravity is 1.1338 m away from the left side of the barbell
Explanation:
it's B =) hhggusucvgaugcavsjssnd
we know that center of mass is given as
r = (m₁
+ m₂
)/(m₁ + m₂)
taking derivative both side relative to "t"
dr/dt = (m₁ d
/dt + m₂ d
/dt)/(m₁ + m₂)
v = (m₁
+ m₂
)/(m₁ + m₂)
taking derivative again relative to "t" both side
dv/dt = (m₁ d
/dt + m₂ d
/dt)/(m₁ + m₂)
a= (m₁
+ m₂
)/(m₁ + m₂)