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Maksim231197 [3]
3 years ago
6

A whistle you use to call your hunting dog has a frequency of 21 kHz, but your dog is ignoring it. You suspect the whistle may n

ot be working, but you can’t hear sounds above 20 kHz. To test it, you ask a friend to blow the whistle, then you hop on your bicycle. In which direction should you ride (toward or away from your friend) and at what minimum speed to know if the whistle is working?
Physics
1 answer:
uranmaximum [27]3 years ago
3 0

The Doppler effect is the right concept to solve this problem. The Doppler effect is understood as the change in apparent frequency of a wave produced by the relative movement of the source with respect to its observer. Mathematically it can be described as,

f = (1-\frac{v_0}{v})f_0

Here,

f_0 = Frequency of the sound from the Whistle

f = Frequency of sound heard

v = Speed of the sound in the Air

Replacing we have that

1- \frac{v_0}{343} = \frac{20kHz}{21kHz}

\frac{v_0}{343} = 1-\frac{20}{21}

\frac{v_0}{343} = \frac{1}{21}

v_0 = \frac{1}{21}(343)

v_0 = 16.33m/s

Therefore the minimum speed to know if the whistle is working is 16.33m/s

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Can lamp that works on a 2.5 v work on a 1.12 v ?​
12345 [234]

Answer:

Explanation:

Thinking about the logics it can but it may be dim because 1.12 is lower than 2,5v so this will mean u lamp may not work or may work very dimely due to the low voltage it is receiving.

5 0
2 years ago
Alan and Monica tested how fertilizer X affected the growth of a plant. Alan and Monica put 100 grams of fertilizer X in pot 1.
BARSIC [14]

Answer:

D.

a control group

Explanation:

In a scientific experiment such as the one above, there is an experimental group and a control group. The experimental group is the group that receives the treatment while the control group does not receive any treatment. The control group helps the researcher to observe if the treatment had any significant effect.

In this case, it will help Alan and Monica to determine if fertilizer X actually had an effect on the plant. Therefore, the pot with o grams of fertilizer in it is the control group.

7 0
3 years ago
A bead slides without friction around a loopthe-loop. The bead is released from a height 21.9 m from the bottom of the loop-the-
wariber [46]

Answer:

Part a)

v = 12.45 m/s

Part B)

F_n = 0.05 N

Explanation:

Part A)

As we know that the point A lies on the top of the loop

so we will have by energy conservation

mgH = \frac{1}{2}mv^2 + mg(2R)

so the speed at point A is given as

mg(H - 2R) = \frac{1}{2}mv^2

v = \sqrt{2g(H - 2R)}

v = \sqrt{2(9.81)(21.9 - 2\times 7)}

v = 12.45 m/s

Part B)

Now the force equation at point A is given as

F_n + mg = \frac{mv^2}{R}

F_n = \frac{mv^2}{R} - mg[/tex]

F_n = 0.004(\frac{12.45^2}{7} - 9.81)

F_n = 0.05 N

6 0
3 years ago
How much force is needed to accelerate a 1-kilogram toy car at a rate of 2 meters per second per second?
kozerog [31]
F = ma,    where m = mass in kg, a = acceleration in m/s², F = Force in Newton

F = 1 * 2

F = 2 N

Force needed is 2 Newtons.
5 0
3 years ago
Read 2 more answers
A softball is thrown from the origin of an X-Y coordinate system with an initial speed of 18 m/s at an angle of 35 degrees above
Goshia [24]
Vo = 18 m/s
angle 35 degrees

1) Components of the initial velocity
 
Vox = Vo*cos(35) = 18*cos(35) m/s = 14.74 m/s
Voy = Vo* sin(35) = 18*sin(35) m/s = 10.32 m/s

2) Equations of postion:

x = Vox*t
y = Voy*t - gt^2 / 2

3) Calculations

A) t = 0.5 s, t = 1.0 st = 1.5 s, t = 2.0 s

x = 14.74 * t

t = 0.5 s => x = 14.74 m/s * 0.5s = 7.37 m

t = 1.0 s => x = 14.74 m/s * 1.0s = 14.74 m

t = 1.5s => x = 22.11 m

t = 2s => x = 29.48 m

B)

y = Voy*t - gt^2 / 2

Voy = 10.32 m/s
g = 10 m/s (approximation)

y = 10.32*t - 5t^2

t = 0.5 s=> y = 3.91m

t = 1 s => y = 5.32m

t = 1.5 s => y = 4.23m

t = 2 s => y = 0.64 m



7 0
3 years ago
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