Answer:
The car must be moving away from the person.
Explanation:
From Doppler's Effect, we know that when a sound source moves towards a stationary observer, the apparent frequency of that sound increases. While the apparent frequency decreases if the source moves away from the stationary observer.
The audible range of frequencies for a human ear is 20 Hz to 20000 Hz. Therefore, in order for the sound of a loud speaker to be audible for the person, the frequency must decrease below 20000 Hz.
<u>Due to this reason, the car must be moving away from the person.</u>
Answer:
Explanation:
a ) starting from rest , so u = o and initial kinetic energy = 0 .
Let mass of the skier = m
Kinetic energy gained = potential energy lost
= mgh = mg l sinθ
= m x 9.8 x 70 x sin 30
= 343 m
Total kinetic energy at the base = 343 m + 0 = 343 m .
b )
In this case initial kinetic energy = 1/2 m v²
= .5 x m x 2.5²
= 3.125 m
Total kinetic energy at the base
= 3.125 m + 343 m
= 346.125 m
c ) It is not surprising as energy gained due to gravitational force by the earth is enormous . So component of energy gained due to gravitational force far exceeds the initial kinetic energy . Still in a competitive event , the fractional initial kinetic energy may be the deciding factor .
The centripetal force exerted on the automobile while rounding the curve is ![1.31\times10^4 N](https://tex.z-dn.net/?f=1.31%5Ctimes10%5E4%20N)
<u>Explanation:</u>
given that
![Mass\ of\ the\ automobile\ m =1.46\times 10^3 kg\\radius\ of\ the\ curve\ r =25 m\\velocity\ of\ the\ automobile\ v=15m/s\\](https://tex.z-dn.net/?f=Mass%5C%20of%5C%20the%5C%20automobile%5C%20m%20%20%3D1.46%5Ctimes%2010%5E3%20kg%5C%5Cradius%5C%20of%5C%20the%5C%20curve%5C%20r%20%3D25%20m%5C%5Cvelocity%5C%20of%5C%20the%5C%20automobile%5C%20v%3D15m%2Fs%5C%5C)
Objects moving around a circular track will experience centripetal force towards the center of the circular track.
![centripetal\ force=mv^2/r\\=1.46\times10^3\times15^2/25\\=1.46\times 10^3\times 225/15\\=1.31\times 10^4 N](https://tex.z-dn.net/?f=centripetal%5C%20force%3Dmv%5E2%2Fr%5C%5C%3D1.46%5Ctimes10%5E3%5Ctimes15%5E2%2F25%5C%5C%3D1.46%5Ctimes%2010%5E3%5Ctimes%20225%2F15%5C%5C%3D1.31%5Ctimes%2010%5E4%20N)
Answer:
The current on the water layer = 1.64×10^-3A
Explanation:
Let's assume that the radius given for the string originates from the centre of the string. The equation for determining the current in the water layer is given by:
I = V × pi[(Rwater + Rstring)^2 - (Rstring)^2/ ( Resitivity × L)
I =[ 166×10^6 ×3.142[(0.519×10^-4) + (2.15×10^-3])^2 - ( 2.15×10^-3)^2] / ( 183 × 831)
I =[ 521572000(4.848×10^6)- 4.623×10^-6]/ 154566
I = 252.83 -(4.623×10^-6)/ 154566
I = 252.83/154566
I = 1.64× 10^-3A