Answer:
Time take to fill the standing wave to the entire length of the string is 1.3 sec.
Explanation:
Given :
The length of the one end 
, frequency of the wave 
= 2.3 Hz, wavelength of the wave λ = 1 m.
Standing wave is the example of the transverse wave, standing wave doesn't transfer energy in a medium.
We know,
∴ 
λ
Where 
speed of the standing wave.
also, ∴ 
where 
time take to fill entire length of the string.
Compare above both equation,
⇒ 
sec
Therefore, the time taken to fill entire length 0f the string is 1.3 sec.
Answer:
Explanation:
The car is rolling without slipping so Vcm= R×ω = 0.325×49 = 16
Answer:
a) When the sides of the joint are close together, the particles have more kinetic energy than they do when sides are farther apart.
Explanation:
To find the total number of miles traveled by a person, we add the distance that he has traveled: 3.0 + 5.00 + 4.000.
Now, to find the accurate number of significant figures when adding measurements, the basic rule for addition is to use the least number of decimal places when reporting the result.
Now, since 3.0 has the least number of decimal places, we report the sum with 1 decimal place and have 12.0 miles as the total distance traveled by the person to reach his destination.
Answer: 12.0 miles
-- We know that the y-component of acceleration is the derivative of the
y-component of velocity.
-- We know that the y-component of velocity is the derivative of the
y-component of position.
-- We're given the y-component of position as a function of time.
So, finding the velocity and acceleration is simply a matter of differentiating
the position function ... twice.
Now, the position function may look big and ugly in the picture. But with the
exception of 't' , everything else in the formula is constants, so we don't even
need any fancy processes of differentiation. The toughest part of this is going
to be trying to write it out, given the text-formatting capabilities of the wonderful
envelope-pushing website we're working on here.
From the picture . . . . . y (t) = (1/2) (a₀ - g) t² - (a₀ / 30t₀⁴ ) t⁶
First derivative . . . y' (t) = (a₀ - g) t - 6 (a₀ / 30t₀⁴ ) t⁵ = (a₀ - g) t - (a₀ / 5t₀⁴ ) t⁵
There's your velocity . . . /\ .
Second derivative . . . y'' (t) = (a₀ - g) - 5 (a₀ / 5t₀⁴ ) t⁴ = (a₀ - g) - (a₀ /t₀⁴ ) t⁴
and there's your acceleration . . . /\ .
That's the one you're supposed to graph.
a₀ is the acceleration due to the model rocket engine thrust
combined with the mass of the model rocket
'g' is the acceleration of gravity ... 9.8 m/s² or 32.2 ft/sec²
t₀ is how long the model rocket engine burns
Pick, or look up, some reasonable figures for a₀ and t₀
and you're in business.
The big name in model rocketry is Estes. Their website will give you
all the real numbers for thrust and burn-time of their engines, if you
want to follow it that far.
