An airplane has an average speed of 740 mph

A bicycle wheel rotates uniformly through 2.0 revolutions in 4.0 s. what is the frequency of the wheel's rotation?

muminat

The frequency is defined as the number of cycles done in one second:
 $f= \frac{N}{t}$ 
where N is the number of cycles and t is the time taken to complete N cycles. The wheel in this problem does N=2 revolutions in a time of t=4.0s, therefore its frequency of rotation is:
 $f= \frac{N}{t} = \frac{2.0}{4.0s} = 0.5 Hz$

8 0

3 years ago

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A proton (charge +e, mass mp), a deuteron (charge +e, mass 2mp), and an alpha particle (charge +2e, mass 4mp) are accelerated fr

icang [17]

Answer:

(a) $r_{d}=\sqrt{2}\times r_{P}$

(b) $r_{\alpha}=\sqrt{2}\times r_{P}$

Explanation:

charge on proton, q = e

mass pf proton, m = mp

charge on deuteron, q' = e

mass pf deuteron, m' = 2mp

charge on alpha particle, q'' = 2e

mass pf alpha particle, m'' = 4mp

Potential difference = V

Magnetic field = B

The kinetic energy is given by eV.

So, 1/2 mv^2 = e V

where, v be the velocity of the particle.

$v=\sqrt{\frac{2eV}{m}}$

The formula for the radius of circular path is given by

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

By substituting the value of v

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

$r\alpha \frac{\sqrt{m}}{q}$

The radius of path of proton is given by

$r_{p}\alpha \frac{\sqrt{m_{p}}}{e}$ .... (1)

(a) radius of deuteron

$r_{d}\alpha \frac{\sqrt{2m_{p}}}{e}$

By comparing equation (1), we get

$r_{d}=\sqrt{2}\times r_{P}$

(b) radius of alpha particle

$r_{\alpha}\alpha \frac{\sqrt{4m_{p}}}{2e}$

By comparing equation (1), we get

$r_{\alpha}=\sqrt{2}\times r_{P}$

5 0

3 years ago

The energy of a wave depends on its ___________. Fill in the blank.

Trava [24]

It mainly depends on its wavelength

8 0

3 years ago

Which statement best describes the two reactions?

Oksana_A [137]

Answer:

b

Explanation:

i guessed annd got it right

3 0

2 years ago

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A rifle with a weight of 25 N fires a 4.5-g bullet with a speed of 240 m/s. (a) Find the recoil speed of the rifle. m/s (b) If a

asambeis [7]

Answer:

The recoil speed of the man and rifle is $v_{man}=0.016 ms^{-1}$ .

Explanation:

The expression for the force in terms of mg is as follows;

F=mg

Here, m is the mass and acceleration due to gravity.

Rearrange the expression for mass.

$m=\frac{F}{g}$

Calculate the combined mass of the man and rifle.

$m_{man,rifle}=\frac{650+25}{g}$

Put $g=9.8 ms^{-2}$ .

$m_{man,rifle}=\frac{650+25}{9.8}$

$m_{man,rifle}=68.88 kg$

The expression for the conservation of momentum is as follows as;

$m_{man}u_{man}+m_{bullet}u_{bullet}=m_{man}v_{man}+m_{rifle}v_{man,rifle}$

Here, $m_{man,rifle}$ is the mass of the man and rifle, $m_{rifle}$ is the mass of the rifle, $u_{man},u{bullet}$ are the initial velocities of the man and bullet and $v_{man},v{man,rifle}$ are the final velocities of the man and rifle and rifle.