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Tresset [83]
3 years ago
6

An ambulance is traveling north at 55.9 m/s, approaching a car that is also traveling north at 28.4 m/s. The ambulance driver he

ars his siren at a frequency of 983 Hz. The velocity of sound is 343 m/s. At what frequency does the driver of the car hear the ambulance's siren?
Physics
1 answer:
wlad13 [49]3 years ago
6 0

Answer:

915 Hz

Explanation:

The observed frequency from a sound source is given as

f₀ = f [(v + v₀)/(v+vₛ)]

where

f₀ = observed frequency of the sound by the observer = ?

f = actual frequency of the sound wave = 983 Hz

v = actual velocity of the sound waves = 343 m/s

vₛ = velocity of the source of the sound waves = 55.9 m/s

v₀ = velocity of the observer = 28.4 m/s

f₀ = 983 [(343+28.4)/(343+55.9)]

f₀ = 915.2 Hz = 915 Hz

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Answer:

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Calculate the recoil speed of a 1.4 kg rifle shooting 0.006 kg bullets with muzzle speed of 800 m/a.3.43 m/s
Angelina_Jolie [31]

Answer:

a) 3.43 m/s

Explanation:

Due to the law of conservation of momentum, the total momentum of the bullet - rifle system must be conserved.

The total momentum before the bullet is shot is zero, because they are both at rest, so:

p_i = 0

Instead the total momentum of the system after the shot is:

p_f = mv+MV

where:

m = 0.006 kg is the mass of the bullet

M = 1.4 kg is the mass of the rifle

v = 800 m/s is the velocity of the bullet

V is the recoil velocity of the rifle

The total momentum is conserved, therefore we can write:

p_i = p_f

Which means:

0=mv+MV

Solving for V, we can find the recoil velocity of the rifle:

V=-\frac{mv}{M}=-\frac{(0.006)(800)}{1.4}=-3.43 m/s

where the negative sign indicates that the velocity is opposite to direction of the bullet: so the recoil speed is

a) 3.43 m/s

5 0
3 years ago
a flowerpot falls from a windowsill 25m above the sidewalk how fast is the flowerpot moving when it strikes the ground
Mariana [72]

s = 25m

v = ?

a = 9.81m/s²

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3 years ago
The highest speed of a cheetah is 100 km/hr and the highest speed of a gazelle is 80 km/hr. If at t = 0 both animals are running
Tju [1.3M]

Answer:

t=18s

Explanation:

The final position of an object moving at constant speed is given by the formula x=x_0+vt, where x_0 is its initial position, v its speed and t the time elapsed.

For the cheetah we have x_c=x_{0c}+v_ct, and for the gazelle x_g=x_{0g}+v_gt. We want to know at which t their positions are equal, that is, x_c=x_g, which means,

x_{0c}+v_ct=x_{0g}+v_gt

Where we can do:

v_ct-v_gt=x_{0g}-x_{0c}

(v_c-v_g)t=x_{0g}-x_{0c}

t=\frac{x_{0g}-x_{0c}}{v_c-v_g}

We then substitute the values we have (the initial position of the cheetah is 0m), writing the meters in km so distance units cancel out correctly:

t=\frac{0.1km-0km}{100km/hr-80km/hr}=0.005hr=18s

On the last step we just multiply by 3600 because is the number of seconds in an hour.

3 0
3 years ago
An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.200 rev/s . The magnitude
blagie [28]

Answer:

(A). the angular velocity of fan is 0.380 rev/s.

(B). The number of revolution is 0.059 rad.

(C). The tangential speed of the blade is 0.907 m/s.

(D). The tangential acceleration of the blade is 2.10 m/s²

Explanation:

Given that,

Initial angular velocity = 0.200 rev/s

Angular acceleration = 0.883 rev/s²

Diameter = 0.760 m

Time = 0.204 s

(A). We need to calculate the angular velocity of fan

Using equation of angular motion

\omega_{f}=\omega_{i}+\alpha t

Put the value into the formula

\omega_{f}=0.200+0.883\times0.204

\omega_{f}=0.380\ rev/s

(B). We need to calculate the number of revolution

Using formula of angular displacement

\theta=\omega_{i}t+\dfrac{1}{2}\alpha t^2

Put the value into the formula

\theta=0.200\times0.204+\dfrac{1}{2}\times0.883\times(0.204)^2

\theta=0.059\ rad

(C). We need to calculate the tangential speed of the blade

Using formula of speed

v=\dfrac{d}{2}\omega_{f}

Put the value into the formula

v=\dfrac{0.760}{2}\times0.380\times2\pi

v=0.907\ m/s

(D). We need to calculate the tangential acceleration of the blade

Using formula of tangential acceleration

a_{t}=\alpha r

Put the value into the formula

a_{t}=0.883\times0.38\times2\pi

a_{t}=2.10\ m/s^2

Hence, (A). the angular velocity of fan is 0.380 rev/s.

(B). The number of revolution is 0.059 rad.

(C). The tangential speed of the blade is 0.907 m/s.

(D). The tangential acceleration of the blade is 2.10 m/s²

3 0
3 years ago
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