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3 years ago

15

Automobile traveling at 65 mph constant on the road described below. Find rate at which radar must rotate when theta = 15 deg. A

ns: theta_dot = 0.219 rad/s.

1 answer:

Finger [1]3 years ago

3 0

Answer:

The rate at which radar must rotate is 0.335 rad/s.

Explanation:

Given that,

Velocity = 65 m/h = 29.0576 m/s

Angle = 15°

Suppose, the radius given by

$r=(100\cos2\theta)\ m$

We need to calculate the rate at which radar must rotate

Using formula of linear velocity

$v=r\omega$

$\omega=\dfrac{v}{r}$

Where, v = velocity

r = radius

Put the value into the formula

$\omega=\dfrac{29.0576}{100\cos30}$

$\omega=0.335\ rad/s$

Hence, The rate at which radar must rotate is 0.335 rad/s.

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3 years ago

An object is at rest in front of a compressed spring. It travels over a surface that exerts a kinetic frictional force on it and

PolarNik [594]

Answer:

the object will travel 0.66 meters before to stop.

Explanation:

Using the energy conservation theorem:

$E_i+K_i+W_f=K_f+U_f$

The work done by the friction force is given by:

$W_f=F_f*d\\W_f=\µ*m*g*d\\W_f=0.35*4*9.81*d\\W_f=13.7d[J]$

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3 years ago

A physicist drives through a stop light. When he is pulled over, he tells the police officer that the Doppler shift made the red

Alik [6]

Answer:

Speed of physicist car is 0.036c or 1.08 x 10⁷ m/s .

Explanation:

Doppler Effect is defined as the change in frequency or wavelength of the wave as the source or/and observer moving away or towards each other.

In this case, the Doppler effect equation in terms of wavelength is :

$\lambda_{s} = \lambda_{o}\sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } }$ ......(1)

Here $\lambda_{s}$ is source wavelength, $\lambda_{o}$ is observed wavelength, v is speed of the observer and c is the speed of light.

Given :

Source wavelength, $\lambda_{s}$ = 660 nm = 660 x 10⁻⁹ m

Observed wavelength, $\lambda_{0}$ = 555 nm = 555 x 10⁻⁹ m

Substitute these values in the equation (1).

$555\times10^{-9} } = 660\times10^{-9} \sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } }$

$\sqrt{\frac{1-\frac{v}{c} }{1+\frac{v}{c} } } = 0.84$

${\frac{1-\frac{v}{c} }{1+\frac{v}{c} } } = (0.84)^{2} = 0.7056$

$1-\frac{v}{c}=0.7056+0.7056\frac{v}{c}$

$\frac{v}{c}=\frac{0.2944}{8.056}$

$v = 0.036c=0.036\times3\times10^{8}$

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8 0

2 years ago

REMARKS The speed found in part (a) is the same as if the woman fell vertically through a distance of 21.9 m. The result of part

sasho [114]

Answer:

Yes, if the system has friction, the final result is affected by the loss of energy.

Explanation:

The result that you are showing is the conservation of mechanical energy between two points in the upper one, the energy is only potential and the lower one is only kinetic.

In the case of some type of friction, the change in energy between the same points is equal to the work of the friction forces

$W_{fr}$ = ΔEm

$W_{fr}$ = $Em_{f}$ -Em₀

As we can see now there is another quantity and for which the final energy is lower and therefore the final speed would be less than what you found in the case without friction.

$Em_{f}$ = $W_{fr}$ + Em₀

Remember that the work of the rubbing force is negative, let's write the work of the rubbing force explicitly, to make it clearer

½ m v² = -fr d + mgh

v = √(-fr d 2/m + 2 gh)

v = √ (2gh - 2fr d/m)

Now it is clear that there is a decrease in the final body speed.

Consequently, if the system has friction, the final result is affected by the loss of energy.

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2 years ago