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3 years ago

7. A golfer hits a golf ball a through a horizontal displacement of 80m. The time of flight for the ball was 1.8

Physics

1 answer:

Darya [45]3 years ago

6 0

Answer:

44.4 m/s

Explanation:

d = 80 m, t = 1.8 s, find v

d = v*t

v = d/t = 80/1.8 = 44.4 m/s

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The polar coordinates of the collar A are given as functions of time in seconds by r = 2+ 0.7 t2 ft and ????= 3.5t rad. What are

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Answer with explanation:

Part a)

$v_{radial}=\frac{dr}{dt}=\frac{d(2+0.7t^{2})}{dt}\\\\v_{radial}=1.4t\\\\\therefore v_{radial}|_{t=4}=1.4\times 4=5.6ft/s\\\\v_{angular}=r|_{t=4}\times \frac{d\theta }{dt}=13.2\frac{3.5t}{dt}=46.2fts^{-1}\\\\\therefore v=\sqrt{v_{radial}^{2}+v_{angular}^{2}}\\\\v=46.53ft/s$

Part b)

$a_{radial}=\frac{d^{2}r}{dt^{2}}=\frac{d^{2}(2+0.7t^{2})}{dt^{2}}\\\\a_{radial}=1.4ft/s^{2}\\\\a_{angular}=r\times \frac{d^{2}\theta }{dt^{2}}\\\\a_{angular}=r\times \frac{d^{2}(3.5t) }{dt^{2}}\\\\\therefore a_{angular}=0\\\\\therefore Accleration=1.4ft/s^{2}$

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3 years ago

Read 2 more answers

A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7

aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

$f'=(\frac{v+v_r}{v+v_s})f$

where

f is the original frequency

f is the apparent frequency

$v$ is the velocity of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

$v_s$ is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

$v_s = v_A = -28.7 m/s$ (surface A is the source, which is moving towards the receiver)

$v_r = +62.2 m/s$ (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

$f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz$

(b) 0.232 m

The wavelength of a wave is given by

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

$\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m$

(c) 1481.2 Hz

Again, we can use the same formula

$f'=(\frac{v+v_r}{v+v_s})f$

In the source frame (= on surface A), we have

$v_s = v_B = -62.2 m/s$ (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

$v_r = +28.7 m/s$ (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

$f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz$

(d) 0.225 m

The wavelength of the wave is given by

$\lambda=\frac{v}{f}$

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

$\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m$

8 0

3 years ago

A 55.6-kg skateboarder starts out with a speed of 2.44 m/s. He does 80.4 J of work on himself by pushing with his feet against t

Vikentia [17]

(a) -1620.8 J

The initial kinetic energy of the skateboarder is:

$K_i = \frac{1}{2}mu^2 = \frac{1}{2}(55.6 kg)(2.44 m/s)^2=165.5 J$

where m is the skateboarder's mass and u his initial speed;

While the final kinetic energy is

$K_f = \frac{1}{2}mv^2 = \frac{1}{2}(55.6 kg)(7.24 m/s)^2=1457.2 J$

where v is his final speed.

So the change in kinetic energy is

$\Delta K=K_f - K_i = 1457.2 J -165.6 J = 1291.6 J$

According to the work-energy theorem, the change in mechanical energy (kinetic+potential) of the skateboarder is equal to the work done on it:

$\Delta K + \Delta E_p = W + W_f$

where

$W = 80.4 J$ is the work done by the skateboarder on himself

$W_f = -244 J$ is the work done by friction

$\Delta E_p = E_p_f - E_p_i$ is the change in gravitational potential energy

Solving for $\Delta E_p$ ,

$\Delta E_p = W+W_f - \Delta K=80.4 J - 244 J - 1457.2 J = -1620.8 J$

(b) 2.97 m

The change in potential energy of the skateboarder can be written as

$\Delta E_p = mg \Delta h$

where

m = 55.6 kg is the mass

g = 9.8 m/s^2 is the acceleration of gravity

$\Delta h$ is the change in vertical height of the skateboarder

Solving for $\Delta h$ ,

$\Delta h = \frac{\Delta E_p}{mg}=\frac{-1620.8 J}{(55.6 kg)(9.8 m/s^2)}=-2.97 m$

Where the negative sign means the skateboarder has moved downwards. Since we are interested only in the absolute value, the answer is

h = 2.97 m

7 0

3 years ago