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disa [49]
2 years ago
6

What frequency is received by a stationary mouse just before being dispatched by a hawk flying at it at 24.7 m/s and emitting a

screech of frequency 3400 Hz? Assume room temperature air.
Physics
1 answer:
Montano1993 [528]2 years ago
6 0

Answer:

   f' = 3665.51 Hz

Explanation:

given,

speed of the hawk = 24.7 m/s

frequency of screech emitted by the hawk = 3400 Hz

speed of sound = 331 m/s

By Doppler's effect

f' = (\dfrac{v}{v-v_s}})f

f' is the frequency received by the mouse

v is the speed of the sound

v_s is the speed of the hawk

now,

f' = \dfrac{341}{341-24.7}}\times 3400

   f' = 1.078 x 3400

   f' = 3665.51 Hz

The frequency received by the stationary mouse is equal to 3665.51 Hz

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kow [346]

The brightness of the lamp is proportional to the current flowing through the lamp: the larger the current, the brighter the lamp.

The current flowing through the lamp is given by Ohm's law:

I=\frac{V}{R}

where

V is the potential difference across the lamp, which is equal to the emf of the battery, and R is the resistance of the lamp.

The problem says that the battery is replaced with one with lower emf. Looking at the formula, this means that V decreases: if we want to keep the same brightness, we need to keep I constant, therefore we need to decrease R, the resistance of the lamp.

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2 years ago
9. Calculate the distance (in km) that Charlie runs if he maintains the average
Karo-lina-s [1.5K]

<u>Correct Question:</u>

Calculate the distance (in km) charlie runs if he maintains an average speed of 8 km/hr for 1 hour

<u>Answer:</u>

The total distance covered by Charlie is 8 km in 1 hour.

<u>Explanation:</u>

The average velocity as given in the question is,

v = 8 km/hr

Total time taken,

$t=1 hour

As we know the formula to evaluate the total distance d when the average velocity and time is given;

v=\frac{d}{t}

d=v \times t

d=8 \times 1

d=8 k m

Hence, the total distance covered by Charlie in 1 hour will be 8 km.

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What is the <br>newton's first law of motion​
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A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.
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To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

F = \frac{kq_1q_2}{d^2}

Here

k = Coulomb's Constant

q_{1,2} = Charge of each object

d = Distance

Our values are given as,

q_1 = 1 \mu C

q_2 = 6 \mu C

d = 1 m

k =  9*10^9 Nm^2/C^2

a) The electric force on charge q_2 is

F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}

F_{12} = 54 mN

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b) As the force exerted on q_2 will be equal to that act on q_1,

F_{21} = F_{12}

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Force is positive i.e. repulsive

c) If q_2 = -6 \mu C, a negative sign will be introduced into the expression above i.e.

F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}

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3 years ago
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