Explanation :
It is given that, the driver accelerates from a stop sign, cruises for 20 s at a constant speed of 60 km/h, and then brakes to come to a stop 40 s after leaving the stop sign.
We know that acceleration is defined as the rate of change of velocity.
Where
dv is the change in velocity, dv = 0 - 60 m/s = -60 m/s
dt is the change in time, dt = 40 s - 30 s = 10 s
So, 
From the graph it is clear that, from 30 s to 40 s the car is decelerating. So, at every second within this time the value of acceleration will be same i.e. 
.
So mathematical harmonics are based around a divergent set of fractions. Sigma(1/n)
with the 1st harmonic being... well 1, or 1 full wavelength.The second harmonic is exactly 1/2 the wavelength of the 1st with the third being 1/3 the wavelength. As Wavelengths go down, frequencies go up in a perfect ratio.
Second Harmonic has double the Frequency of the 1st or base note. Third Harmonic is triple and so on.
So the Harmonic set of 375 is.
1. 375
2. 375×2=750
3. 375×3= 1125
.
.
.
etc (: I hope this helps.
Answer:
henry moseley was an english physicist, whose contribution to the science of physics was the justification from physical laws of the previous chemical concept of the atomic number. one of his developments were of moseley's law in x-ray spectra.
Explanation:
Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ = 
/ vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m
