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Angelina_Jolie [31]

3 years ago

11

An object moves with simple harmonic motion. If the amplitude and the period are both doubled, the object's maximum speed is 1.

quartered 2. halved 3. quadrupled. 4. doubled 5. unchanged

1 answer:

Kisachek [45]3 years ago

6 0

Answer:

The object's maximum speed remains unchanged.

Explanation:

The speed of a particle in SHM is given by :

$v(t)=A\omega\ sin(\omega t)$

Maximum speed is, $v_{max}=A\omega$

If A' = 2A and T' = 2T

$v'_{max}=(2A)\dfrac{2\pi}{2T}$

$v'_{max}=(A)\dfrac{2\pi}{T}$

$v'_{max}=v_{max}$

So, the maximum speed of the object remains the same i.e. it remains unchanged. Hence, this is the required solution.

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mihalych1998 [28]

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Hope this helps

8 0

4 years ago

A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.

Elanso [62]

The tension in the cable is 169.43 N and the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

<h3>Tension in the cable</h3>

Apply the principle of moment and calculate the tension in the cable;

Clockwise torque = TL sinθ

Anticlockwise torque = ¹/₂WL

TL sinθ = ¹/₂WL

T sinθ = ¹/₂W

T = (W)/(2 sinθ)

T = (29 x 9.8)/(2 x sin57)

T = 169.43 N

<h3>Vertical component of the force</h3>

T + F = W

F = W - T

F = (9.8 x 29) - 169.43

F = 114.77 N

Thus, the tension in the cable is 169.43 N and the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

Learn more about tension here: brainly.com/question/24994188

#SPJ1

6 0

2 years ago

A 1-kilogram ball has a kinetic energy of 50 joules the speed of the ball is

sveta [45]

Answer:

10 m/s

Explanation:

3 0

3 years ago

A wave has a speed of 360 m/s. It has a frequency of 20hz what is its wavelength (include correct unit)

abruzzese [7]

Answer:

18m

Explanation:

v=frequency × wavelenght

wavelength=v/f

wavelength=360÷20

=18m

7 0

3 years ago

Read 2 more answers

A 70.0-kg person throws a 0.0430-kg snowball forward with a ground speed of 32.0 m/s. A second person, with a mass of 58.5 kg, c

Aleks04 [339]

Answer:

The velocities of the skaters are $v_{1} = 3.280\,\frac{m}{s}$ and $v_{2} = 0.024\,\frac{m}{s}$ , respectively.

Explanation:

Each skater is not under the influence of external forces during process, so that Principle of Momentum Conservation can be used on each skater:

First skater

$m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}$ (1)

Second skater

$m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}$ (2)

Where:

$m_{1}$ - Mass of the first skater, in kilograms.

$m_{2}$ - Mass of the second skater, in kilograms.

$v_{1,o}$ - Initial velocity of the first skater, in meters per second.

$v_{1}$ - Final velocity of the first skater, in meters per second.

$v_{b}$ - Launch velocity of the meter, in meters per second.

$v_{2}$ - Final velocity of the second skater, in meters per second.

If we know that $m_{1} = 70\,kg$ , $m_{b} = 0.043\,kg$ , $v_{b} = 32\,\frac{m}{s}$ , $m_{2} = 58.5\,kg$ and $v_{1,o} = 3.30\,\frac{m}{s}$ , then the velocities of the two people after the snowball is exchanged is:

By (1):

$m_{1} \cdot v_{1, o} = m_{1} \cdot v_{1} + m_{b}\cdot v_{b}$

$m_{1}\cdot v_{1,o} - m_{b}\cdot v_{b} = m_{1}\cdot v_{1}$

$v_{1} = v_{1,o} - \left(\frac{m_{b}}{m_{1}} \right)\cdot v_{b}$

$v_{1} = 3.30\,\frac{m}{s} - \left(\frac{0.043\,kg}{70\,kg}\right)\cdot \left(32\,\frac{m}{s} \right)$

$v_{1} = 3.280\,\frac{m}{s}$

By (2):

$m_{b}\cdot v_{b} = (m_{2}+m_{b})\cdot v_{2}$

$v_{2} = \frac{m_{b}\cdot v_{b}}{m_{2}+m_{b}}$

$v_{2} = \frac{(0.043\,kg)\cdot \left(32\,\frac{m}{s} \right)}{58.5\,kg + 0.043\,kg}$

$v_{2} = 0.024\,\frac{m}{s}$

5 0

3 years ago