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3 years ago

1) A speaker vibrates at a frequency of 2500 Hz. What is its period?

Physics

1 answer:

yarga [219]3 years ago

7 0

It's easier for you to solve these than to try and read my solutions if I solve them.

Use this magic formula:

(period) · (frequency) = 1

If you handle the magic formula carefully and correctly, you can get these facts out of it:

-- Period = 1 / frequency

-- Frequency = 1 / period

Use the first one to solve #1.

Use the second one to solve #2.

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A shell is fired with 500km/h on a target half kilometer away.In what time in which shell will hit the target.

Drupady [299]

It depends on the direction in which the shell is launched. The time can be anything from 3.6 seconds to never.

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3 years ago

n a level football field a football is projected from ground level. It has speed 9.0 m/sm/s when it is at its maximum height. It

Pani-rosa [81]

Answer:

T = 0.225 s

Explanation:

The speed of a projectile at the highest point of its motion is the horizontal speed of the projectile. Considering the horizontal motion with negligible air resistance, we can use the following formula:

$v_x = RT\\\\T = \frac{v_x}{R}$

where,

T = Total time of ball in air = ?

R = Horizontal distance covered = 40 m

$v_x$ = horizontal speed = 9 m/s

Therefore,

$T = \frac{9\ m/s}{40\ m}$

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3 years ago

A 1400 kg car is traveling east on the highway at 31 m/s and collides into the rear of a slower moving pickup truck of 2400 kg,

zloy xaker [14]

Answer: 31 m/s due east

Explanation: this question can be solved using the law of conservation of linear momentum.

This law states that in a closed or isolated system, during collision, the vector sum of momentum before collision equals the vector sum of momentum after collision.

Momentum = mass × velocity

From our question, our parameters before collision are given below as

Mass of car = mc = 1400kg

Speed of car =vc = 31 m/s (due east)

Mass of truck = mt = 2400kg

Velocity of truck = vt = 25 m/s ( due east )

After collision

Velocity of car = ?

Velocity of truck = 34 m/s ( due east )

Vector sum of momentum before collision is given as

1400 (31) + 2400 (25) = 43400 + 60000 = 103400 kgm/s

After collision the truck is seen to move faster (v = 34 m/s) which implies that the car also moves due east .

1400 (v) + 2400(25) .... A positive value is between both momenta because they are in the same direction.

After collision, we have that

1400v + 60000

Vector sum of momentum before collision = vector sum of momentum after collision

103400 = 1400v + 60000

103400 - 60000 = 1400v

43400 = 1400v

v = 43400/ 1400

v = 31 m/s due east

4 0

4 years ago

X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr

mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

$\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)$

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

$\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m$

b) The change in photon energy is given by:

$\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm$

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

$P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\$

$E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV$

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4 years ago

-09utydrbfshjklo<br><br>iuohyfdgxfncj

adoni [48]

Answer:

thanks

Explanation:

5 0

3 years ago