Light as a wave: Light can be described (modeled) as an electromagnetic wave. ... This changing magnetic field then creates a changing electric field and BOOM - you have light. Unlike many other waves (sound, water waves, waves in a football stadium), light does not need a medium to “wave” in.
Explanation:
A bridge supported by vertical cables which then leads to more support from larger cables.
Answer:
Fnet = 0
Explanation:
- Since the block slides across the floor at constant speed, this means that it's not accelerated.
- According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
- This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:
- In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:
⇒ 169 N + Fn = Fg = 216 N (3)
- This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
- Fn = 216 N - 169 N = 47 N (4)
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
We need to see what forces act on the box:
In the x direction:
Fh-Ff-Gsinα=ma, where Fh is the horizontal force that is pulling the box up the incline, Ff is the force of friction, Gsinα is the horizontal component of the gravitational force, m is mass of the box and a is the acceleration of the box.
In the y direction:
N-Gcosα = 0, where N is the force of the ramp and Gcosα is the vertical component of the gravitational force.
From N-Gcosα=0 we get:
N=Gcosα, we will need this for the force of friction.
Now to solve for Fh:
Fh=ma + Ff + Gsinα,
Ff=μN=μGcosα, this is the friction force where μ is the coefficient of friction. We put that into the equation for Fh.
G=mg, where m is the mass of the box and g=9.81 m/s²
Fh=ma + μmgcosα+mgsinα
Now we plug in the numbers and get:
Fh=6*3.6 + 0.3*6*9.81*0.8 + 6*9.81*0.6 = 21.6 + 14.1 + 35.3 = 71 N
The horizontal force for pulling the body up the ramp needs to be Fh=71 N.
