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Ymorist [56]
2 years ago
11

Two swift canaries fly toward each other, each moving at 15.0 m/s relative to the ground, each warbling a note of frequency 1750

Hz. (a) What frequency note does each bird hear from the other one? (b) What wavelength will each canary measure for the note from the other one?
Physics
1 answer:
lesya692 [45]2 years ago
3 0

Answer:

Part a)

f = 1911.5 Hz

Part b)

\lambda = 0.186 m

Explanation:

Here the source and observer both are moving towards each other

so we know that the apparent frequency is given as

f' = f_0 (\frac{v + v_o}{v - v_s})

here we know that

f_0 = 1750 Hz

v_o = 15 m/s

v_s = 15 m/s

now we will have

f' = (1750)(\frac{340 + 15}{340 - 15})

f' = 1911.5 Hz

Part b)

Apparent wavelength is given by the formula

\lambda = \frac{v_{relative}}{f_{app}}

here we will have

\lambda = \frac{340 + 15}{1911.5}

\lambda = 0.186 m

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Answer:

A) vertically upward

Explanation:

Since the tyre is rotating with uniform angular speed and moving with constant linear speed

So as soon as a small stone is stuck into the groove of the tyre the speed of the stone is same as that of the tyre

so now we can say that stone will start revolving with the tyre of the car at constant angular speed and moving with uniform speed also

so here just after that the tangential acceleration of the stone must be zero while radial acceleration must be towards the center of the tyre given as

a_c = \omega^2 R

so we will have direction of net acceleration is towards its center so correct answer will be

A) vertically upward

7 0
2 years ago
Ebo throws a ball into the air its velocity at the start is 18m/s at an angle of 37° to the ground. What is the range of the bal
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PLS help ASAP I DONT have time to answer this, it also detects if it’s right or wrong.
6 0
2 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m.how much work
aksik [14]
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F=k_e  \frac{q_e q_e}{d^2}
So, in order for the additional electron to cross this point, it is required an infinite amount of work, which is impossible.
5 0
2 years ago
If you stand next to a wall on a frictionless skateboard and push the wall with a force of 40n, how hard does the wall push on y
pentagon [3]
The wall pushes back with the same force you exert so 40N.

to show your acceleration is .5m/s^2 use newtons second law F=ma

so plugging in numbers gives 40N=80kg*a knowing that a newton is equal 1kgm/s^2 we could write 40kg*m/s^2= 80kg*a so solving for a gives

40kg*m/s^2/80kg = a we see the kg's cancel and we're left with
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4 0
2 years ago
Some gliders are launched from the ground by means of a winch, which rapidly reels in a towing cable attached to the glider. Wha
aliina [53]

Answer:

P=627.47W

Explanation:

To solve this problem we have to take into account, that the work done by the winch is

W=Fh

the force, at least must equal the gravitational force

F=Mg=(156kg)(9.8\frac{m}{s^2})=1258.8N

with force the tension in the cable makes the winch go up.

The work done is

W=(1258.8N)(58.0m)=73010.4J

To calculate the power we need to know what is the time t. But first we have to compute the acceleration

The acceleration will be

v_f^2=v_0+2ah\\a=\frac{v_f^2}{2h}=\frac{(24.9\frac{m}{s})}{2(58.0m)}=0.214\frac{m}{s^2}

and the time t

v_f=v_0+at\\t=\frac{v_f}{a}=116.35s

The power will be

P=\frac{W}{t}=\frac{73010.4J}{116.35s}=627.47W

HOPE THIS HELPS!!

6 0
2 years ago
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