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kirza4 [7]
3 years ago
8

When a beam of light is incident on a surface, it delivers energy to the surface. The intensity of the beam is defined as the en

ergy delivered per unit area per unit time.
What is the unit of intensity, expressed in SI base units?​
Physics
1 answer:
scoundrel [369]3 years ago
6 0

Answer:

Kg

Explanation:

Given that the definition of intensity is:

The intensity of the beam is defined as the energy delivered per unit area per unit time. That is,

I =( E × t)/A

Which can expressed as

I = J/m^2/s

The S.I Units of

Energy is Joule = kgm^2s^-1

Area A = m^2

Time t = s

Intensity I = kgm^2s^-1 × m^-2 × s

Intensity I = kg

The cubic metres and seconds are cancelled out

Therefore, the S.I units of intensity is kg

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If a proton and an electron are released when they are 2.50×10^-10m apart (typical atomic distances), find the initial accelerat
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To solve this exercise, we will first proceed to calculate the electric force given by the charge between the proton and the electron (it). From the Force we will use Newton's second law that will allow us to find the acceleration of objects. The Coulomb force between two charges is given as

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Here,

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Replacing we have that,

F = (9*10^9)(\frac{(1.602*10^{-19})^2}{2.5*10^{-10}})

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The force between the electron and proton is calculated. From Newton's third law the force exerted by the electron on proton is same as the force exerted by the proton on electron.

The acceleration of the electron is given as

a_e = \frac{F}{m_e}

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The acceleration of the proton is given as,

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3 years ago
To understand how to find the velocities of objects after a collision.
trasher [3.6K]

There are some information missing on Part D: Let the mass of object 1 be m and the mass of object 2 be 3m. If the collision is perfectly inelastic, what are the velocities of the two objects after the collision? Give the velocity v_1 of object one, followed by object v_2 of object two, separated by a comma. Express each velocity in terms of v.

Answer: Part A: v_1 = 0; v_2 = v

Part B: v_1 = v_2 = \frac{v}{2}

Part C: v_1 = \frac{v}{3}; v_2 = \frac{4v}{3}

Part D: v_1 = v_2 = \frac{v}{4}

Explanation: In elastic collisions, there no loss of kinetic energy and momentum is conserved. Momentum is determined as p = m.v and kinetic energy as K = \frac{1}{2}m.v^{2}

Conserved means that the amount of initial momentum is equal to the amount of final momentum:

m_{1}.v_{1i} + m_{2}.v_{2i} = m_{1}.v_{1f} + m_{2}.v_{2f}

No loss of energy means that initial kinietc energy is the same as the final kinetic energy:

\frac{1}{2}(m_{1}.v_{1i} + m_{2}.v_{2i}) = \frac{1}{2} (m_{1}.v_{1f} + m_{2}.v_{2f}  )

To determine the final velocities of each object, there are 2 variables and two equations, so working those equations, the result is:

v_{2f} = \frac{2.m_{1} } {m_{1} + m_{2} }.v_{1i}  + \frac{(m_{2} - m_{1})}{m_{1} + m_{2} } . v_{2i}

v_{1f} = \frac{m_{2} - m_{1} }{m_{1} + m_{2} } . v_{1i} + \frac{2.m_{2} }{m_{1} + m_{2} } .v_{2i}

For all the collisions, object 2 is static, i.e. v_{2i} = 0

<u>Part A</u>: Both objects have the same mass (m), v_{1i} = v and collision is elastic:

v_1 = \frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}

v_1 = 0

v_2 = \frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}

v_2 = \frac{2.m}{m+m}.v

v_2 = v

When the masses are the same and there is an object at rest, the object in movement stops and the object at rest has the same same velocity as the object who hit it.

<u>Part B</u>: Same mass but collision is inelastic: An inelastic collision means that after it happens, the two objects has the same final velocity, then:

m_{1}.v_{1i} + m_{2}.v_{2i} = m_{1}.v_{1f} + m_{2}.v_{2f}

m_{1}.v_{1i} = (m_{1}+m_{2}).v_{f}

v_{f} =  \frac{m_{1}.v_{1i}}{m_{1} + m_{2} }

v_1 = v_2 = \frac{m.v}{m+m}

v_1 = v_2 = \frac{v}{2}

<u>Part C:</u> Object 1 is 2m, object 2 is m and elastic collision:

v_1 = \frac{m_{2} - m_{1}}{m_{1} + m_{2} } . v_{1i}

v_1 = \frac{2m - m}{2m + m } . v

v_1 = \frac{v}{3}

v_2 = \frac{2.m_{1} }{m_{1} + m_{2}}.v_{1i}

v_2 = \frac{2.2m}{2m+m}.v

v_2 = \frac{4v}{3}

<u>Part D</u>: Object 1 is m, object is 3m and collision is inelastic:

v_1 = v_2 = v_{f} =  \frac{m_{1}.v_{1i}}{m_{1} + m_{2} }

v_1 = v_2 = \frac{m}{m+3m}.v

v_1 = v_2 = \frac{v}{4}

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Answer:

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