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3 years ago

If you're not satisfied with your fitness evaluation score, there's really little you can do about it.

Physics

3 answers:

Doss [256]3 years ago

4 0

Answer:

false

Explanation:

you can improve your score next time

BARSIC [14]3 years ago

3 0

Answer:

False

Explanation:

Alizaha2 years ago

0 0

Answer:False :>

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A 40-cm-long tube has a 40-cm-long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. As th

Umnica [9.8K]

Answer:

The frequency of the tuning is 1.065 kHz

Explanation:

Given that,

Length of tube = 40 cm

We need to calculate the difference between each of the lengths

Using formula for length

$\Delta L=L_{2}-L_{1}$

$\Delta L=74.7-58.6$

$\Delta L=16.1\ m$

For an open-open tube,

We need to calculate the fundamental wavelength

Using formula of wavelength

$\lambda=2\Delta L$

Put the value into the formula

$\lambda=2\times16.1$

$\lambda=32.2\ cm$

We need to calculate the frequency of the tuning

Using formula of frequency

$f=\dfrac{v}{\lambda}$

Put the value into the formula

$f=\dfrac{343}{32.2\times10^{-2}}$

$f=1065.2\ Hz$

$f=1.065\ kHz$

Hence, The frequency of the tuning is 1.065 kHz

3 0

3 years ago

In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used inst

Archy [21]

Answer:

16 cm

Explanation:

For protons:

Energy, E = 300 keV

radius of orbit, r1 = 16 cm

the relation for the energy and velocity is given by

$E = \frac{1}{2}mv^{2}$

So, $v = \sqrt{\frac{2E}{m}}$ .... (1)

Now,

$r = \frac{mv}{Bq}$

Substitute the value of v from equation (1), we get

$r = \frac{\sqrt{2mE}}{Bq}$

Let the radius of the alpha particle is r2.

For proton

So, $r_{1} = \frac{\sqrt{2m_{1}E}}{Bq_{1}}$ ... (2)

Where, m1 is the mass of proton, q1 is the charge of proton

For alpha particle

So, $r_{2} = \frac{\sqrt{2m_{2}E}}{Bq_{2}}$ ... (3)

Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle

Divide equation (2) by equation (3), we get

$\frac{r_{1}}{r_{2}}=\frac{q_{2}}{q_{1}}\times \sqrt{\frac{m_{1}}{m_{2}}}$

q1 = q

q2 = 2q

m1 = m

m2 = 4m

By substituting the values

$\frac{r_{1}}{r_{2}}=\frac{2q}}{q}}\times \sqrt{\frac{m}}{4m}}=1$

So, r2 = r1 = 16 cm

Thus, the radius of the alpha particle is 16 cm.

8 0

3 years ago

Read 2 more answers

How long will it take to go 150 km [E] traveling at 50 km/hr?

pantera1 [17]

It will take 3 hours

7 0

4 years ago

Read 2 more answers

A body with a mass of 2,000 kg is lifted to a height of 15 m within a time of 15 s. Which one of the following statements concer

Aleksandr [31]

<span>So we want to know which statement is true for the body of mass m=2000kg that is lifted to a height of h=15m in t=15 s. Lets calculate each of the following: Gravity force on the body is F=m*g=2000*9.81=19620 N so a is FALSE. Potential energy of the body when it is lifted to the height of 15 m is Ep=m*g*h=2000*9.81*15=294300 J so b is FALSE. Work to lift the body is: W=Fg*h=2000*9.81*15= Ep=294300 J so c is FALSE. Power P=W/t=294300/15=19620 W So d is TRUE. </span>

8 0

3 years ago

A large helium filled balloon is used as the center piece for a graduation party. The balloon alone has a mass of 222 kg and it

inn [45]

Answer:

The buoyant force is 3778.8 N in upward.

Explanation:

Given that,

Mass of balloon = 222 Kg

Volume = 328 m³

Density of air = 1.20 kg/m³

Density of helium = 0.179 kg/m³

We need to calculate the buoyant force acting

Using formula of buoyant force

$F_{b}=\rho_{air}\times V_{b}\times g$

Where, $\rho_{air}$ = density of air

V = Volume of balloon

g = acceleration due to gravity

Put the value into the formula

$F_{b}=1.20\times321\times9.81$

$F_{b}=3778.8\ N$

This buoyant force is in upward direction.

Hence, The buoyant force is 3778.8 N in upward.

4 0

3 years ago