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3 years ago

What is the mass of basswood wing with these dimensions: 26.9 cm x 5.5 cm x 0.15 cm?

Physics

2 answers:

Softa [21]3 years ago

6 0

The answer is 22.1925 I hope you get right let me know I got my answer because I multiply it all together.

MatroZZZ [7]3 years ago

4 0

Some said it already but the answer is 22.1925

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The energy of a wave will remain constant if which of the following changes are made to it?AThe wavelength is cut in half, and t

Makovka662 [10]

The energy of a wave will remain constant if which of the following changes are made to it is given below

Explanation:

1.In electromagnetic waves, energy is transferred through vibrations of electric and magnetic fields. In sound waves, energy is transferred through vibration of air particles or particles of a solid through which the sound travels. In water waves, energy is transferred through the vibration of the water particles.

2.First of all for light which is a electromagnetic wave c=frequency*wavelength. As the wavelength increases the frequency decreases. This can be physically understood as the increase in red shift of the light. Also energy =h*frequency,hence increasing wavelength decreases the energy carried by the photon.

3.Wave frequency is related to wave energy. the more energy in a wave, the higher its frequency. The lower the frequency is, the less energy in the wave.

4 0

3 years ago

Two skaters collide and grab on to each other on frictionless ice. One of them, of mass 77.0 kg, is moving to the right at 4.00

Alex777 [14]

Answer:3.31m/s², to the right

Explanation:

According to the law of conservation of momentum of a body, change in momentum of bodies before collision is equal to the change in momentum after collision.

Momentum = mass × velocity

M1 and M2 be the masses of the first and second skaters respectively

Let u1 and u2 be the velocities of the first and second skaters respectively.

v be their common velocity after collision

M1 = 77kg M2 = 66kg u1 = 4m/s² u2 = 2.5m/s²

According to the law we have

M1u1 + M2u2 = (M1+M2)v

77(4) + 66(2.5) = (77+66)v

308 + 165 = 143v

V = 473/143

V = 3.31m/s²

Their velocity after collision will become 3.31m/s²

They will both move towards the right after collision because the mass of the body moving to the right is higher than the other mass and the mass is also moving at a higher velocity than the other.

5 0

2 years ago

Explain relative velocity briefly

fomenos

Answer:

Explanation:

Relative velocity is defined as the velocity of an object B in the rest frame of another object A.

4 0

3 years ago

X-rays with an energy of 400 keV undergo Compton scattering with a target. If the scattered X-rays are detected at \theta = 30^{

dedylja [7]

<h2>Answer: 37.937 keV</h2>

Explanation:

<u>Photons have momentum</u>, this was proved by he American physicist Arthur H. Compton after his experiments related to the <u>scattering of photons from electrons</u> (Compton Effect or Compton Shift). In addition, energy and momentum are conserved in the process.

In this context, the Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}.c}$ , being $h=4.136(10)^{-15}eV.s$ the Planck constant, $m_{e}$ the mass of the electron and $c=3(10)^{8}m/s$ the speed of light in vacuum.

$\theta=30\°$ the angle between incident phhoton and the scatered photon.

We are told the scattered X-rays (photons) are detected at $30\°$ :

$\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos(30\°))$ (2)

$\Delta \lambda=\lambda' - \lambda_{o}=3.2502(10)^{-13}m$ (3)

Now, the initial energy $E_{o}=400keV=400(10)^{3}eV$ of the photon is given by:

$E_{o}=\frac{h.c}{\lambda_{o}}$ (4)

From this equation (4) we can find the value of $\lambda_{o}$ :

$\lambda_{o}=\frac{h.c}{E_{o}}$ (5)

$\lambda_{o}=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{400(10)^{3}eV}$

$\lambda_{o}=3.102(10)^{-12}m$ (6)

Knowing the value of $\Delta \lambda$ and $\lambda_{o}$ , let's find $\lambda'$ :

$\Delta \lambda=\lambda' - \lambda_{o}$

Then:

$\lambda'=\Delta \lambda+\lambda_{o}$ (7)

$\lambda'=3.2502(10)^{-13}m+3.102(10)^{-12}m$

$\lambda'=3.427(10)^{-12}m$ (8)

Knowing the wavelength of the scattered photon $\lambda'$ , we can find its energy $E'$ :

$E'=\frac{h.c}{\lambda'}$ (9)

$E'=\frac{(4.136(10)^{-15}eV.s)(3(10)^{8}m/s)}{3.427(10)^{-12}m}$

$E'=362.063keV$ (10) This is the energy of the scattered photon

So, if we want to know the energy of the recoiling electron $E_{e}$ , we have to calculate all the energy lost by the photon, which is:

$E_{e}=E_{o}-E'$ (11)

$E_{e}=400keV-362.063keV$

Finally we obtain the energy of the recoiling electron:

$E_{e}=37.937keV$

5 0

3 years ago

Students are experimenting with circuits in their physics class and they build the two working circuits pictured below. The batt

bija089 [108]

The complete observation about adding bulb 3 is the brightness of the bulbs has to do with power which considers both the voltage and the current: less voltage x less current = dimmer bulbs. In circuit A, the voltage is divided across the resistors and the current decreases as resistance increases. In circuit B, the voltage is the same in each parallel section of the circuit and the current through that section of the circuit only depends on the resistor in that section.

<h3>What is power of the circuit?</h3>

The power of the bulb or any resistor is equal to the product of voltage and current flowing through it.

P = VI

Circuit A has bulbs in series while the circuit B has bulbs in parallel.

When bulb 3 added to circuit A, the brightness of all the bulbs dimmed but when bulb 3 (R3) added to circuit B, nothing changed in the brightness of the bulb.

The brightness is depended on the power of the circuit. When both the voltage and current are less, the bulb will be dimmed. In circuit A, series resistors divide the voltage across them. In circuit B, voltage is equal for all the resistors.

Thus, the last option is correct.

Learn more about power.

brainly.com/question/2933971

#SPJ1

4 0

1 year ago