Answer:
The correct answer is option 'd': The frequency decreases and the intensity of the sound decreases.
Explanation:
1) <u>Effect on Frequency </u>
According to Doppler's effect of sound we have
for a source of sound moving away from the observer the relation between the observed and the original frequency is given by
where
c = speed of sound in air
is the velocity of observer of sound
is the velocity of source of sound
is the original frequency of sound
As we see the ratio is less than 1 thus the frequency of sound that the observer receives is less than that of source.
2) <u>Effect on Intensity:</u>
At a distance 'r' from source emitting a wave of Power 'P' is given by
As we see on increasing 'r' intensity of sound decreases.
Answer: Normal force, N = 141.64 Newton
Explanation:
All the forces acting on the system and described in free body diagram are:
1) gravitational pull in downward direction
2) Normal force in upward direction
3) External force of 40 N acting at an angle of 37° with the horizontal can be resolved in two rectangular components:
i) F Cos 37° along the horizontal plane in forward direction and
ii) F Sin 37° along the vertical plane in downward direction
Applying the Newton's second law, net forces in the vertical plane are:
Net force, f = N - (mg + F Sin 37°)
As there is no acceleration in the vertical plane hence, net force f = 0.
So,
N - (mg + F Sin 37°) = 0
Adding (mg + F Sin 37°) both the sides in above equation, we get
N = mg + F Sin 37°
N = 12 
9.8 + 40 0.601 because (Sin 37° = 0.601)
N = 117.6 + 24.04
N = 141.64 Newton
Answer:
Magnification, m = 3
Explanation:
It is given that,
Focal length of the lens, f = 15 cm
Object distance, u = -10 cm
Lens formula :
v is image distance
Magnification,
So, the magnification of the lens is 3.
Answer:
Explanation:
angular momentum of the putty about the point of rotation
= mvR where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .
= .045 x 4.23 x 2/3 x .95 cos46
= .0837 units
moment of inertia of rod = ml² / 3 , m is mass of rod and l is length
= 2.95 x .95² / 3
I₁ = .8874 units
moment of inertia of rod + putty
I₁ + mr²
m is mass of putty and r is distance where it sticks
I₂ = .8874 + .045 x (2 x .95 / 3)²
I₂ = .905
Applying conservation of angular momentum
angular momentum of putty = final angular momentum of rod+ putty
.0837 = .905 ω
ω is final angular velocity of rod + putty
ω = .092 rad /s .
Answer:
L' = 1.231L
Explanation:
The transmission coefficient, in a tunneling process in which an electron is involved, can be approximated to the following expression:
L: width of the barrier
C: constant that includes particle energy and barrier height
You have that the transmission coefficient for a specific value of L is T = 0.050. Furthermore, you have that for a new value of the width of the barrier, let's say, L', the value of the transmission coefficient is T'=0.025.
To find the new value of the L' you can write down both situation for T and T', as in the following:
Next, by properties of logarithms, you can apply Ln to both equations (1) and (2):
Next, you divide the equation (3) into (4), and finally, you solve for L':
hence, when the trnasmission coeeficient has changes to a values of 0.025, the new width of the barrier L' is 1.231 L
