Answer:
0.75 grams
Explanation:
Today we have 6 grams
In 30 years, 3 grams remain
In 60 years 1.5 grams remain
in 90 years 0.75 grams remain
mathematically It would look like this
m = 6(0.5⁽ⁿ/³⁰⁾) where n is number of years
It transfers and changes into different types of energy, this is why the ground feels hot when something moves fast over it.
Answer:
12 m
Explanation:
The object is in uniformly accelerated motion, so the distance covered can be found using the following suvat equation:
where
s is the distance
u is the initial velocity
t is the time
a is the acceleration
For this problem,
and
u = 0, since we are considering the first second of motion
So, substituting t = 1 s, we find
Answer:
B) 12 m
Explanation:
Gravitational potential energy is:
PE = mgh
Given PE = 5997.6 J, and m = 51 kg:
5997.6 J = (51 kg) (9.8 m/s²) h
h = 12 m
Angry sound level = 70 db
Soothing sound level = 50 db
Frequency, f = 500 Hz
Assuming speed of sound = 345 m/s
Density (assumed) = 1.21 kg/m^3
Reference sound intensity, Io = 1*10^-12 w/m^2
Part (a): Initial sound intensity (angry sound)
10log (I/Io) = Sound level
Therefore,
For Ia = 70 db
Ia/(1*10^-12) = 10^(70/10)
Ia = 10^(70/10)*10^-12 = 1*10^-5 W/m^2
Part (b): Final sound intensity (soothing sound)
Is = 50 db
Therefore,
Is = 10^(50/10)*10^-12 = 18*10^-7 W/m^2
Part (c): Initial sound wave amplitude
Now,
I (W/m^2) = 0.5*A^2*density*velocity*4*π^2*frequency^2
Making A the subject;
A = Sqrt [I/(0.5*density*velocity*4π^2*frequency^2)]
Substituting;
A_initial = Sqrt [(1*10^-5)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-8 m = 69.7 nm
Part (d): Final sound wave amplitude
A_final = Sqrt [(1*10^-7)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-9 m = 6.97 nm
