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3 years ago

This graph depicts an object that has _____. positive acceleration constant velocity negative acceleration no acceleration

Physics

2 answers:

fredd [130]3 years ago

3 0

The object has a positive acceleration

Explanation:

The graph in the question is missing; find it in attachment.

The acceleration of an object is defined as the rate of change in velocity of the object:

$a=\frac{\Delta v}{\Delta t}$

where

$\Delta v$ is the change in velocity

$\Delta t$ is the change in time

We can verify that in a velocity-time graph, the acceleration corresponds to the slope of the graph. In fact, the slope is given by:

$m=\frac{\Delta y}{\Delta x}$

where

$\Delta y$ is the change in the y-variable, so it is the change in velocity, $\Delta v$

$\Delta x$ is the change in the x-variable, so it is the change in time, $\Delta t$

Which means that the slope is equal to the acceleration:

$m=\frac{\Delta y}{\Delta x}=\frac{\Delta v}{\Delta t}=a$

In this problem, the slope of the curve shown in the velocity-time graph in the figure is positive: therefore, this means that the object has a positive acceleration.

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

LekaFEV [45]3 years ago

3 0

Answer: It's "positive acceleration" I got the same question and got it right.

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OLEGan [10]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

Calculate the broadcast wavelength of the radio station 99.90 fm. seconds

taurus [48]

The radio waves are electromagnetic wave, so it travels with velocity of light i.e $3\times 10^{8} \ m/s$ .

We can use the relation between frequency, wavelength and speed as

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

Here c is speed of light, $\lambda$ is wavelength and f is frequency and its value is given 99.90 FM, it is actually in megahertz (i.e 99.90 MHz).

Therefore,

$\lambda =\frac{3 \times 10^{8}\ m/s }{99.90 \times 10^{6} \ Hz} = 3.003 \ m$ .

Thus, the broadcast wavelength of the given radio station is 3.003 m.

8 0

4 years ago

A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent

Julli [10]

Answer:

The kinetic energy of the merry-go-round is $\bf{475.47~J}$ .

Explanation:

Given:

Weight of the merry-go-round, $W_{g} = 826~N$

Radius of the merry-go-round, $r = 1.17~m$

the force on the merry-go-round, $F = 57.8~N$

Acceleration due to gravity, $g= 9.8~m.s^{-2}$

Time given, $t=3.47~s$

Mass of the merry-go-round is given by

$m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg$

Moment of inertial of the merry-go-round is given by

$I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}$

Torque on the merry-go-round is given by

$\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m$

The angular acceleration is given by

$\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}$

The angular velocity is given by

$\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}$

The kinetic energy of the merry-go-round is given by

$E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J$

5 0

4 years ago

Two people stand across from one another at the top edges of identical buildings, 50 meters above the ground. One person throws

cricket20 [7]

Answer:

Explanation:

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6 0

3 years ago

What is the frequency of a wave having a period equal to 18 seconds <br>

Ivanshal [37]

Explanation:

The time taken by a wave crest to travel a distance equal to the length of wave is known as wave period.

The relation between wave period and frequency is as follows.

T = \frac{1}{f}T=

where, T = time period

f = frequency

It is given that wave period is 18 seconds. Therefore, calculate the wave period as follows.

T = \frac{1}{f}T=

or, f = \frac{1}{T}f=

= \frac{1}{18 sec}

18sec

= 0.055 per second (1cycle per second = 1 Hertz)

or, f = 5.5 \times 10^{-2} hertz5.5×10 −2 hertz

<h3>Thus, we can conclude that the frequency of the wave is 5.5 \times 10^{-2} hertz5.5×10 −2 hertz .</h3>

3 0

3 years ago