Answer:
36,67 degrees Celsius
Explanation:
The simplest way to approach this problem, given the information provided, is to simply start with the speed difference.
Goal: 353 m/s
Start: 343 m/s (at 20 degrees Celsius).
Difference: 10 m/s
Variation rate: 0.60 m/s/d (d = degree)
So, 16,67 degrees more than the starting point.
The temperature will then be 36.67 degrees Celsius, when the sound travels at the speed of 353 m/s.
Answer:
The grating spacing of the beetle is 
Explanation:
The concept to solve this problem is relate to interference effect given in the Young's Slits. Here was demonstrated that the length of the side labelled \lambda is known as the path difference. The equation is given by,
Where,
= wavelenght of light
N = a positive integer: 1,2,3...
= Angle from the center of the wall to the dark spot
d= width of the slit
Replacing our values we have that for n=1,
Therefore the grating spacing of the beetle is
ur answer is A or also known as
When you push a child on a swing, you are doing work on the child because you are pushing against the force of gravity
hope this helps :)
Answer:
A) a = 73.304 rad/s²
B) Δθ = 3665.2 rad
Explanation:
A) From Newton's first equation of motion, we can say that;
a = (ω - ω_o)/t. We are given that the centrifuge spins at a maximum rate of 7000rpm.
Let's convert to rad/s = 7000 × 2π/60 = 733.04 rad/s
Thus change in angular velocity = (ω - ω_o) = 733.04 - 0 = 733.04 rad/s
We are given; t = 10 s
Thus;
a = 733.04/10
a = 73.304 rad/s²
B) From Newton's third equation of motion, we can say that;
ω² = ω_o² + 2aΔθ
Where Δθ is angular displacement
Making Δθ the subject;
Δθ = (ω² - ω_o²)/2a
At this point, ω = 0 rad/s while ω_o = 733.04 rad/s
Thus;
Δθ = (0² - 733.04²)/(2 × 73.304)
Δθ = -537347.6416/146.608
Δθ = - 3665.2 rad
We will take the absolute value.
Thus, Δθ = 3665.2 rad
