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dalvyx [7]
3 years ago
13

As a bat flies toward a wall at a speed of 6.0 m/s, the bat emits an ultrasonic sound wave with frequency 30.0 kHz. What frequen

cy does the bat hear in the reflected wave?
Physics
1 answer:
Leto [7]3 years ago
7 0

Answer:

f_b = 29.98 Hz

Explanation:

speed of bat = 6 m/s

sound wave frequency emitted by bat = 30.0 kHz

as we know,

speed of sound (c)= 343 m/s

f_w = f(\dfrac{c+v_r}{c+v_s})

f_w = 30(\dfrac{343+0}{343+6})

f_w = 29.48 Hz

now frequency received by bat is equal to  

f_b = f(\dfrac{c+v_r}{c+v_s})

f_b = 29.48(\dfrac{343+6}{343+0})

f_b = 29.98 Hz

hence the frequency hear by bat will be 29.98 Hz

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  • Height=h=102m
  • Mass=m=120g
  • Acceleration due to gravity=g=10m/s^2

\\ \sf\longmapsto P.E=mgh

  • P E means potential energy

\\ \sf\longmapsto P.E=102(120)(10)

\\ \sf\longmapsto P.E=122400J

7 0
3 years ago
If a 4 engine jet accelerates down a runway at 8.7 m/s^2, Suppose now that all 4 engins are operational on the jet from the prev
soldi70 [24.7K]

Answer:

5.22m/s^2

Explanation:

One of the first propulsion characteristics given in the example is that all engines are equal.

In this way if we have 4 engines running at the same time, it means that its capacity is 100%.

Under this premise, if 100% is found, the Jet is capable of reaching a speed of 8.7m / s ^ 2.

However, the question is, what would happen if 2.4 "Engines" now work.

To do this then we make a simple equivalence,

If 4 engines is the equivalent of 100%, when would it be 2.4 engines?

X = \frac{2.4 * 100\%} { 4 }= 60\%

In this way it would mean that the body could be driven to 60% of its total.

So

Speed_{Decreased} = 8.7 * 60\% = 5.22 \frac{m}{s^2}

3 0
3 years ago
A circular loop of wire 78 mm in radius carries a current of 114 A. Find the (a) magnetic field strength and (b) energy density
Finger [1]

Explanation:

It is given that,

Radius of loop, r = 78 mm = 0.078 m

Current, I = 114 A

(a) Magnetic field strength of the circular loop is given by :

B=\dfrac{\mu_oI}{2R}

B=\dfrac{4\pi \times 10^{-7}\times 114}{2\times 0.078}

B = 0.000918 T

or

B=9.18\times 10^{-4}\ T

(b) Energy density at the center of the loop is given by :

U=\dfrac{B^2}{2\mu_o}

U=\dfrac{(0.000918)^2}{2\times 4\pi \times 10^{-7}}

U=0.335\ J/m^3

Hence, this is the required solution.

7 0
3 years ago
Consider a spherical planet of uniform density rho. The distance from the planet's center to its surface (i.e., the planet's rad
alisha [4.7K]

Answer:

a) g(r) = 4\pi \cdot G \cdot \rho\cdot r, b) g = 4\pi \cdot G \cdot \rho\cdot r_{P}

Explanation:

a) The acceleration due to gravity inside the planet is:

dg = G\cdot \frac{\rho \cdot dV}{r^{2}}

dg = G\cdot \frac{\rho \cdot dV}{r^{2}}

dg = G\cdot \frac{4\pi\cdot \rho \cdot r^{2}\,dr}{r^{2}}

dg = 4\pi\cdot G\cdot \rho \,dr

g(r) = 4\pi \cdot G \cdot \rho\cdot r

b) The acceleration at the surface of the planet is:

g = 4\pi \cdot G \cdot \rho\cdot r_{P}

5 0
3 years ago
until a train is a safe distance from the station, it must travel at 5 m/s. Once the train is on open track , it can speed up to
kiruha [24]

Answer:

5

Explanation:

6 0
3 years ago
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