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2 years ago

If arts gravitational pool just got three times stronger what would happen to our weight

Physics

1 answer:

IRINA_888 [86]2 years ago

4 0

We would just get three times heavier.

weight = mass x gravity (9.8 m/s/s)

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A manufacturer of CD-ROM drives claims that the player can spin the disc as frequently as 1200 revolutions per minute.

Leya [2.2K]

a. 7.0 m/s

First of all, we need to convert the angular speed (1200 rpm) from rpm to rad/s:

$\omega = 1200 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=125.6 rad/s$

Now we know that the row is located 5.6 cm from the centre of the disc:

r = 5.6 cm = 0.056 m

So we can find the tangential speed of the row as the product between the angular speed and the distance of the row from the centre of the circle:

$v=\omega r = (125.6 rad/s)(0.056 m)=7.0 m/s$

b. $875 m/s^2$

The acceleration of the row of data (centripetal acceleration) is given by

$a=\frac{v^2}{r}$

where we have

v = 7.0 m/s is the tangential speed

r = 0.056 m is the distance of the row from the centre of the trajectory

Substituting numbers into the formula, we find

$a=\frac{(7.0 m/s)^2}{0.056 m}=875 m/s^2$

3 0

2 years ago

In a compoumd are atoms physically or chemically combined

Andrei [34K]

Answer:

They are...if I'm correct Chemically combined, sorry if I'm wrong.

3 0

2 years ago

The frequency of a certain sound is 440 Mz. What is the wavelength of this sound when the temperature of the air is (a) 20°C; (b

Serggg [28]

Answer:

Explanation:

We know the frequency and the velocity, both of which have good units. All we have to do is rearrange the equation and solve for

Let's plug in our given values and see what we get!

340

440

−

0.773

3 0

1 year ago

a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the

tatuchka [14]

The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
$W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N$
the force of the wind F, acting horizontally, with intensity
$F=12 N$
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
$T \cos \alpha -F=0$
$T \sin \alpha -W=$
By dividing the second equation by the first one, we get
$\tan \alpha = \frac{W}{F}= \frac{24.5 N}{12 N}=2.04$
From which we find
$\alpha = 63.8 ^{\circ}$
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
$T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N$