A police car is driving down the street with it's siren on. You are standing still on the sidewalk beside the street. If the fre
1 answer:
Answer:
A) 1568.60 Hz
B) 1437.15 Hz
Explanation:
This change is frequency happens due to doppler effect
The Doppler effect is the change in frequency of a wave in relation to an observer who is moving relative to the wave source
where
C = the propagation speed of waves in the medium;
Vr= is the speed of the receiver relative to the medium,(added to C, if the receiver is moving towards the source, subtracted if the receiver is moving away from the source;
Vs= the speed of the source relative to the medium, added to C, if the source is moving away from the receiver, subtracted if the source is moving towards the receiver.
A) Here the Source is moving towards the receiver(C-Vs)
and the receiver is standing still (Vr=0) therefore the observed frequency should get higher
B)Here the Source is moving away the receiver(C+Vs)
and the receiver is still not moving (Vr=0) therefore the observed frequency should be lesser
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Answer:
A. The range of A and B are equal.
Explanation:
Let u be the initial speed of both the projectile.
The gravitational force acting in the projectile is in the downward direction, sp the speed of the projectile in the horizontal direction remains constant and equals to the initial horizontal speed.
For projectile, a projectile having initial velocity u at an angle \theta with the horizontal direction,
The speed in the horizontal direction
and the speed in the vertical direction is upward.
For A:
The speed in the horizontal direction
and the speed in the vertical direction is upward.
For B:
The speed in the horizontal direction
and the speed in the vertical direction is upward.
Let and
are the time of flight for projectile A and B respectively.
As the range is the horizontal distance traveled by the projectile, so
The range for the projectile A
The range for the projectile B =
At the highest point, the vertical velocity is 0.
Bu using the equation of motion
Here, the final velocity v=0, the initial velocity , h= vertical distance up to the highest point, and
(as per sign convention).
So,
For projectile A: The maximum height attained.
For projectile B: The maximum height attained.
As , the height of A is attained by A is more than the heigHt attained by B.
Now, the times required to reach the highest point from the ground and again form the highest point to the ground are the same.
So, the total time of flight = 2 x (Time to reach the highest point)
In a similar way, by using the equation of motion v=u+at,
The time to reach the highest point
where g is the acceleration due to gravity.
So, the total time of flight
The total time of flight for A
The total time of flight for A
Now, from equations (i) and (ii),
The range for the projectile A =
The range for the projectile B =
Both the projectile have the same range.
Hence, option (A) is correct.
Answer:
a) t = 2.0 s, b) x_f = - 24.56 m, Δx = 16.56 m
Explanation:
This is an exercise in kinematics, the relationship of position and time is indicated
x = 5 t³ - 9t² -24 t - 8
a) ask when the velocity is zero
speed is defined by
v =
let's perform the derivative
v = 15 t² - 18t - 24
0 = 15 t² - 18t - 24
let's solve the quadratic equation
t =
t1 = -0.8 s
t2 = 2.0 s
the time has to be positive therefore the correct answer is t = 2.0 s
b) the position and distance traveled for a = 0
acceleration is defined by
a = dv / dt
a = 30 t - 18
a = 0
30 t = 18
t = 18/30
t = 0.6 s
we substitute this time in the expression of the position
x = 5 0.6³ - 9 0.6² - 24 0.6 - 8
x = 1.08 - 3.24 - 14.4 - 8
x = -24.56 m
we see that all the movement is in one dimension so the distance traveled is the change in position between t = 0 and t = 0.6 s
the position for t = 0
x₀ = -8 m
the position for t = 0.6 s
x_f = - 24.56 m
the distance
ΔX = x_f - x₀
Δx = | -24.56 -(-8) |
Δx = 16.56 m