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salantis [7]
2 years ago
14

The security alarm on a parked car goes off and produces a frequency of 769 Hz. The speed of sound is 343 m/s. As you drive towa

rd this parked car, pass it, and drive away, you observe the frequency to change by 69.5 Hz. At what speed are you driving
Physics
1 answer:
uysha [10]2 years ago
6 0

Answer:

Explanation:

ASSUMING your speed is constant

f₀ = f(v + vo)/(v + vs)

   Δf = f approach - f depart

69.5 = (769(343 + vo)/(343 + 0)) - (769(343 - vo)/(343 + 0))

69.5 = 769(2vo/343)

  vo = 15.5 m/s

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Yakvenalex [24]

Answer:

So the minimum force is

32.2Newton

Explanation:

To solve for the minimum force, let us assume it to be F (N)

So

F=mgsinA

But

=>>>> coefficient of static friction x (F + mgcosA

=>3 x 9.8 x sin35 = 0.3 x (F + 3 x 9.8 x cos35)

So making F subject of formula

F + 24.0 = 56.2

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2 years ago
Explain why the top of the loop cannot be the same height as (or higher than) the top of the first hill. Assume the roller coast
Ivahew [28]

Answer:

By conservation of energy, it can climb up to a height equal to that it went down before. However, due to the friction in the machines, the total mechanical energy of the roller coaster will decrease. As a result, the first "hill" of many roller coasters are the highest, but the followings will have decreasing heights.

Explanation:

7 0
2 years ago
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The combination of an applied force and a friction force produces a constant total torque of 35.5 N · m on a wheel rotating abou
Cerrena [4.2K]

Answer:

a) I = 19.799\,kg\cdot m^{2}, b) T = -3.405\,N\cdot m, c) n_{T} \approx 54.842\,rev

Explanation:

a) The net torque is:

T = I\cdot \alpha

Let assume a constant angular acceleration, which is:

\alpha = \frac{\omega-\omega_{o}}{t}

\alpha = \frac{10.4\,\frac{rad}{s} - 0\,\frac{rad}{s} }{5.80\,s}

\alpha = 1.793\,\frac{rad}{s^{2}}

The moment of inertia of the wheel is:

I = \frac{T}{\alpha}

I = \frac{35.5\,N\cdot m}{1.793\,\frac{rad}{s^{2}} }

I = 19.799\,kg\cdot m^{2}

b) The deceleration of the wheel is due to the friction force. The deceleration is:

\alpha = \frac{\omega-\omega_{o}}{t}

\alpha = \frac{0\,\frac{rad}{s} - 10.4\,\frac{rad}{s}}{60.4\,s}

\alpha = - 0.172\,\frac{rad}{s^{2}}

The magnitude of the torque due to friction:

T = (19.799\,kg\cdot m^{2})\cdot (-0.172\,\frac{rad}{s^{2}} )

T = -3.405\,N\cdot m

c) The total angular displacement is:

\theta_{T} = \theta_{1} + \theta_{2}

\theta_{T} = \frac{(10.4\,\frac{rad}{s} )^{2}-(0\,\frac{rad}{s} )^{2}}{2\cdot (1.793\,\frac{rad}{s^{2}} )} + \frac{(0\,\frac{rad}{s} )^{2}-(10.4\,\frac{rad}{s} )^{2}}{2\cdot (-0.172\,\frac{rad}{s^{2}} )}

\theta_{T} = 344.580\,rad

The total number of revolutions of the wheel is:

n_{T} = \frac{\theta_{T}}{2\pi}

n_{T} = \frac{344.580\,rad}{2\pi}

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5 0
3 years ago
A hammer taps on the end of a 3.4-m-long metal bar at room temperature. A microphone at the other end of the bar picks up two pu
OLEGan [10]

Answer:

S = 2266.67 m/s

Explanation:

Given,

length of the metal = 3.4 m

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speed of sound in air= 343 m/s

speed of sound in this metal = ?

time taken

t = \dfrac{distance}{speed}

t = \dfrac{3.4}{343}

t = 9.9 ms

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t = 9.9 - 8.4 = 1.5 ms

time for which sound is in metal is equal to 1.5 ms

speed of sound in metal

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S = \dfrac{3.4}{1.5 \times 10^{-3}}

S = 2266.67 m/s

Speed of sound in metal is equal to S = 2266.67 m/s

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