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3 years ago

Which graph shows uniform motion.

Physics

2 answers:

bagirrra123 [75]3 years ago

5 0

Answer:

1. shows uniform motion

Explanation:

prisoha [69]3 years ago

3 0

Option number: 3
Because it’s constant

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Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

DiKsa [7]

Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

<u>Given:</u>

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

$\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.$

It is the displacement of the particle in 2 seconds.

7 0

3 years ago

An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic

malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

$\omega_f = \omega_0 + \alpha t$

Where,

$\omega_f =$ Final Angular Velocity

$\omega_0 =$ Initial Angular velocity

$\alpha =$ Angular acceleration

t = time

The relation between the tangential acceleration is given as,

$a = \alpha r$

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

$\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s$

$\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s$

$t = 11s$

Replacing the previous equation with our values we have,

$\omega_f = \omega_0 + \alpha t$

$9.8436 = 6.5973 + \alpha (11)$

$\alpha = \frac{9.8436- 6.5973}{11}$

$\alpha = 0.295rad/s^2$

The tangential velocity then would be,

$a = \alpha r$

$a = (0.295)(0.2)$

$a = 0.059m/s^2$

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

$\omega_f^2=\omega_0^2+2\alpha\theta$

Replacing with our values and re-arrange to find $\theta,$

$\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}$

$\theta = \frac{9.8436^2-6.5973^2}{2*0.295}$

$\theta = 90.461rad$

That is equal in revolution to

$\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev$

The linear displacement of the system is,

$x = \theta*(2\pi*r)$

$x = 14.397*(2\pi*\frac{0.25}{2})$

$x = 11.3m$

5 0

3 years ago

The speed of sound in aluminum is 5200 m/s. Can you hear a sound with a

romanna [79]

Answer:

v = wavelength * frequency

frequency = 5200 m/s / .2 m = 26000 / sec

20,000 / sec is optimistic for the upper frequency of human hearing

So 26,000 is above the hearing range for human ears

5 0

2 years ago

HELP PLZ "25 meters per second Northeast" is a description of

Hoochie [10]

Answer:

ion no

Explanation:

cuz i dont

7 0

3 years ago

Read 2 more answers

The graph represents velocity over time...

Nady [450]

Answer:

where is the graph I can't see it how can I solve the problem if I don't see the graph can you show the graph please

8 0

2 years ago

Which graph shows uniform motion.​

Which graph shows uniform motion.