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4 years ago

12

If an object accelerates at 40 m/s per second in four minutes, what will be the final velocity? (assume the object started from

rest)

1 answer:

Sliva [168]4 years ago

5 0

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The Hubble Space Telescope orbits the Earth at approximately 612,000m altitude. Its mass is 11,100 kg and the mass of earth is 5

nexus9112 [7]

Answer:

7.55 km/s

Explanation:

The force of gravity between the Earth and the Hubble Telescope corresponds to the centripetal force that keeps the telescope in uniform circular motion around the Earth:

$G\frac{mM}{R^2}=m\frac{v^2}{R}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2}$ is the gravitational constant

$m=11,100 kg$ is the mass of the telescope

$M=5.97\cdot 10^{24} kg$ is the mass of the Earth

$R=r+h=6.38\cdot 10^6 m+612,000 m=6.99\cdot 10^6 m$ is the distance between the telescope and the Earth's centre (given by the sum of the Earth's radius, r, and the telescope altitude, h)

v = ? is the orbital velocity of the Hubble telescope

Re-arranging the equation and substituting numbers, we find the orbital velocity:

$v=\sqrt{\frac{GM}{R}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.97\cdot 10^{24} kg)}{6.99\cdot 10^6 m}}=7548 m/s=7.55 km/s$

6 0

3 years ago

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 2.0 rev/s in 10.0 s. At th

Arte-miy333 [17]

Answer:

22 revolutions

Explanation:

2 rev/s = 2*(2π rad/rev) = 12.57 rad/s

The angular acceleration when it starting

$\alpha_a = \frac{\Delta \omega}{\Delta t} = \frac{12.57}{10} = 1.257 rad/s^2$

The angular acceleration when it stopping:

$\alpha_o = \frac{\Delta \omega}{\Delta t} = \frac{-12.57}{12} = -1.05 rad/s^2$

The angular distance it covers when starting from rest:

$\omega^2 - 0^2 = 2\alpha_a\theta_a$

$\theta_a = \frac{\omega^2}{2\alpha_a} = \frac{12.57^2}{2*1.257} = 62.8 rad$

The angular distance it covers when coming to complete stop:

$0 - \omega^2 = 2\alpha_o\theta_o$

$\theta_o = \frac{-\omega^2}{2\alpha_o} = \frac{-12.57^2}{2*(-1.05)} = 75.4 rad$

So the total angular distance it covers within 22 s is 62.8 + 75.4 = 138.23 rad or 138.23 / (2π) = 22 revolutions

6 0

3 years ago

In heavy rush-hour traffic you drive in a straight line at 12 m/s for 1.5 milrutes, then you have to stop for 3.5 minutes, and f

g100num [7]

Average velocity between t= 0 and t = 7.5 minutes = 7.4 m / s

<h3>Further explanation </h3>

Regular straight motion is the motion of objects on a straight track that has a fixed speed

Formula used

$\large{\boxed{\boxed{\bold{S\:=\:v\:\times\:t}}}$

S = distance = m

v = speed = m / s

t = time = seconds

The motion of objects can be expressed in graphical form

This relationship graph can be in the form of a graph S-t, v-t or a-t

From the S-t graph, the average speed can be determined by determining the slope distance of the curve line from the distance to time ratio

average velocity = distance traveled / time taken

From the statement of questions there are 3 stages of time that occur

1. first time

move with v = 12 m / s, t = 1.5 min = 90 s

So the distance traveled =

s = v x t

S = 12. 90

S = 1080 m

2. second time

stop (v = 0) so that s = 0 and t = 3.5 min = 210 s

3. third time

move with v = 15 m / s and t = 2.5 min = 150 s

so the distance traveled

s = v x t

s = 15 x 150

s = 2250 m

Total distance = 2250 + 1080 = 3330 m

If we look at the existing chart (attached)

then we can draw a straight line from the starting point to the end point so we get a slash curve that shows the average speed

v average = distance traveled / time taken

v average = 3330/450

v average = 7.4 m / s

<h3>Learn more </h3>

The distance of the elevator

brainly.com/question/8729508

velocity position

brainly.com/question/2005478

resultant velocity

brainly.com/question/4945130

Keywords: average velocity, Motion of objects, s-t graph

6 0

4 years ago

Read 2 more answers

Explain why the student should open the switch after each<br>reading.

vova2212 [387]

Answer:

Maybe to get different readings

5 0

3 years ago

Two guitarists attempt to play the same note of wavelength 6.50 cm at the same time, but one of the instruments is slightly out

PolarNik [594]

Answer:

The two value of the wavelength for the out of tune guitar is

$\lambda _2 = (6.48,6.52) \ cm$

Explanation:

From the question we are told that

The wavelength of the note is $\lambda = 6.50 \ cm = 0.065 \ m$

The difference in beat frequency is $\Delta f = 17.0 \ Hz$

Generally the frequency of the note played by the guitar that is in tune is

$f_1 = \frac{v_s}{\lambda}$

Where $v_s$ is the speed of sound with a constant value $v_s = 343 \ m/s$

$f_1 = \frac{343}{0.0065}$

$f_1 = 5276.9 \ Hz$

The difference in beat is mathematically represented as

$\Delta f = |f_1 - f_2|$

Where $f_2$ is the frequency of the sound from the out of tune guitar

$f_2 =f_1 \pm \Delta f$

substituting values

$f_2 =f_1 + \Delta f$

$f_2 = 5276.9 + 17.0$

$f_2 = 5293.9 \ Hz$

The wavelength for this frequency is

$\lambda_2 = \frac{343 }{5293.9}$

$\lambda_2 = 0.0648 \ m$

$\lambda_2 = 6.48 \ cm$

For the second value of the second frequency

$f_2 = f_1 - \Delta f$

$f_2 = 5276.9 -17$

$f_2 = 5259.9 Hz$

The wavelength for this frequency is

$\lambda _2 = \frac{343}{5259.9}$

$\lambda _2 = 0.0652 \ m$

$\lambda _2 = 6.52 \ cm$

8 0

3 years ago