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Viefleur [7K]
3 years ago
7

A Projectile attains its maximum height when the angle of projection is?​

Physics
1 answer:
FrozenT [24]3 years ago
7 0

Answer:A projectile attains its maximum height when the angle of projection is. Question. A projectile attains its maximum height when the angle of projection is.

Explanation:

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A 0.500-nm x-ray photon is deected through 134 in a Compton scattering event. At what angle (with respect to the incident beam)
natita [175]

Answer:

The angle of recoil electron with respect to incident beam of photon is 22.90°.

Explanation:

Compton Scattering is the process of scattering of X-rays by a charge particle like electron.

The angle of the recoiling electron with respect to the incident beam is determine by the relation :

\cot\phi = (1+\frac{hf}{m_{e}c^{2}  })\tan\frac{\theta }{2}      ....(1)

Here ∅ is angle of recoil electron, θ is the scattered angle, h is Planck's constant, m_{e} is mass of electron, c is speed of light and f is the frequency of the x-ray photon.

We know that, f = c/λ      ......(2)

Here λ is wavelength of x-ray photon.

Rearrange equation (1) with the help of equation (1) in terms of  λ .

\cot\phi = (1+\frac{h}{m_{e}c\lambda  })\tan\frac{\theta }{2}

Substitute 6.6 x 10⁻³⁴ m² kg s⁻¹ for h, 9.1 x 10⁻³¹ kg for m_{e}, 3 x 10⁸ m/s for c, 0.500 x 10⁻⁹ m for λ  and 134° for θ in the above equation.

\cot\phi = (1+\frac{6.6\times10^{-34} }{9.1\times10^{-31}\times3\times10^{8}\times0.5\times10^{-9}  })\tan\frac{134 }{2}

\cot\phi=2.37

\phi = 22.90°

8 0
4 years ago
Select all that apply. There MIGHT be more than one.
Stolb23 [73]
I believe the answer to your question is “Lithosphere plate boundaries”

The planet Earth is covered by a layer formed by land and rocks called the earth's crust or lithosphere. This crust is not smooth and uniform, but rather irregular and composed of tectonic plates, also called lithosphere plates. These plates are not fixed as they are under the magma (high temperature molten rock).

Hope this helps!:)
4 0
3 years ago
What is the device used to measure a potential difference in a circuit called?
kvasek [131]
A voltmeter is the instrument used to measure a potential difference between two points in an electric circuit
8 0
3 years ago
Read 2 more answers
A bare helium nucleus has two positive charges and a mass of 6.64×10−27kg(a) Calculate its kinetic energy in joules at 2.00% of
grandymaker [24]

To solve this problem we will apply the concepts related to kinetic energy and energy conservation. The kinetic energy will be expressed in terms of mass and speed, as well as load and voltage. From this last expression we will find the charges by electron and by Helium nucleus.

a ) Kinetic Energy is given as

KE = \frac{1}{2} mv^2

Replacing with our values we have that

KE = \frac{1}{2} (6.64*10^{-27})(2.0\% (3.00*10^8))^2

KE = 1.1935*10^{-13}J

Therefore the kinetic energy of the helium nucleus is 1.1935*10^{-13}J

PART B) Now for calculate the electron volts we use the kinetic energy as a expression between the charge and the voltage, that is

KE = qV

Here,

q = Charge of an electron

V = Voltage

Rearranging to find the potential we have,

V = \frac{KE}{q}

V = \frac{1.19*10^{-13}}{1.6*10^{-19}}

V = 743750eV

Therefore the kinetic energy in electron vols is 743750eV

PART C) Applying the same relationship but now using the Helium core load, we will have to

KE = QV

Here,

Q = Charge of a helium nucleus

V = Voltage

Rearranging to find the potential we have

V = \frac{KE}{Q}

But we need to note that the charge is equal to the number of charge for the unit charge, then

Q = \text{No. Charge} \times \text{Unit Charge}

Q = (2)(1.6*10^{-19}C)

Q = 3.2*10^{-19}C

Now replacing we have that

V= \frac{1.19*10^{-13}}{32*10^{-19}}

V = 371875V

Therefore the voltage applied is  371875V

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3 years ago
We are on an asteroid with thrusters pointed in 4 directions; Go Left, GO Right, Go Up, and Go Down. We start at coordinate (3,3
GalinKa [24]
we would be at (4,0)
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3 years ago
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