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
Hello,
To solve we need to know the formula for speed
The formula is D/T=S (Distance of time=speed)
Now all we have to do is plug in the numbers.
20/40= 1/2 or 0.5
SO the speed is 0.5 m/s
Have a great day!
Answer:
No work is performed or required in moving the positive charge from point A to point B.
Explanation:
Lets take
Q= Positive charge which move from point A to point B along
Voltage difference,ΔV =V₁ - V₂
The work done
W = Q . ΔV
Given that charge is moved from point A to point B along an equipotential surface.It means that voltage difference is zero.
ΔV = 0
So
W = Q . ΔV
W = Q x 0
W= 0 J
So work is zero.
-- We're going to be talking about the satellite's speed.
"Velocity" would include its direction at any instant, and
in a circular orbit, that's constantly changing.
-- The mass of the satellite makes no difference.
Since the planet's radius is 3.95 x 10⁵m and the satellite is
orbiting 4.2 x 10⁶m above the surface, the radius of the
orbital path itself is
(3.95 x 10⁵m) + (4.2 x 10⁶m)
= (3.95 x 10⁵m) + (42 x 10⁵m)
= 45.95 x 10⁵ m
The circumference of the orbit is (2 π R) = 91.9 π x 10⁵ m.
The bird completes a revolution every 2.0 hours,
so its speed in orbit is
(91.9 π x 10⁵ m) / 2 hr
= 45.95 π x 10⁵ m/hr x (1 hr / 3,600 sec)
= 0.04 x 10⁵ m/sec
= 4 x 10³ m/sec
(4 kilometers per second)
It’s either C or D. let me know but I would do C!
