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1 year ago

Dashun Mack: Attempt

Physics

1 answer:

Montano1993 [528]1 year ago

4 0

The time elapsed when the ball reaches the window is 1.55s

As per the question:

Final velocity, v = 11 m/s

Height, h = 29m

Acceleration, a = g = -9.8 m/s

From the third equation of motion:

v² = u² +2as

u² = v² - 2as

u² = (11)² - 2 × (-9.8) × (29)

u² = 689.4

u = √689.4

u = 26.25 m/s

Now, from the first equation of motion:

v = u + at

$t=\frac{v-u}{-g}$

$t = \frac{11 -26.25}{-9.8}$

t = 1.55 s

Therefore, the time elapsed when the ball reaches the window is 1.55s.

Read more about Equations of Motion:

brainly.com/question/13436432

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3 years ago

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3 years ago

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The blades in a blender rotate at a rate of 6100 rpm. When the motor is turned off during operation, the blades slow to rest in

MissTica

Answer:

Explanation:

Using the equation of motion to find the angular acceleration:

$\omega_f = \omega_i + \alpha t$

$\omega_f$ is the final angular velocity in rad/s

$\omega_i$ is the initial angular velocity in rad/s

$\alpha$ is the angular acceleration

t is the time taken

Given the following

$\omega_f = 6100rpm$

Time = 4.1secs

Convert the angular velocity to rad/s

1rpm = 0.10472rad/s

6100rpm = x

x = 6100 * 0.10472

x = 638.792rad/s

$\omega_f = 638.792rad/s\\$

Get the angular acceleration:

Recall that:

$\omega_f = \omega_i + \alpha t$

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2 years ago

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

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

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3 years ago