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dezoksy [38]
2 years ago
13

Which characterstic is related to kinetic enegry but not potential energy.

Physics
1 answer:
V125BC [204]2 years ago
8 0

Answer:

A

Explanation:

Kinetic energy is the energy of motion

KE=.5mv^2

>m= mass

>v= velocity (m/s)

PE=mgh

>m= mass

>g= acceleration due to graviry

>h= height

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Which of the following describes a situation that would violate the law of entropy?
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Answer:

B. An Iron Bar place in a room becomes cooler than its surroundings.

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How many laws does newton have?
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Newton has 3 Laws specifically The Three Laws of Motion
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When UV light of wavelength 248 nm is shone on aluminum metal, electrons are ejected withmaximum kinetic energy 0.92 eV. What ma
Lina20 [59]

Answer:

The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm

Explanation:

Given;

wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m

maximum kinetic energy of the ejected electron, K.E = 0.92 eV

let the work function of the aluminum metal = Ф

Apply photoelectric equation:

E = K.E + Ф

Where;

Ф is the minimum energy needed to eject electron the aluminum metal

E is the energy of the incident light

The energy of the incident light is calculated as follows;

E = hf = h\frac{c}{\lambda} \\\\where;\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\c \ is \ speed \ of \ light = 3 \times 10^{8} \ m/s\\\\E = \frac{(6.626\times 10^{-34})\times (3\times 10^8)}{248\times 10^{-9}} \\\\E = 8.02 \times 10^{-19} \ J

The work function of the aluminum metal is calculated as;

Ф = E - K.E

Ф = 8.02 x 10⁻¹⁹  -  (0.92 x 1.602 x 10⁻¹⁹)

Ф =  8.02 x 10⁻¹⁹ J   -  1.474 x 10⁻¹⁹ J

Ф = 6.546 x 10⁻¹⁹ J

The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;

\phi = hf = \frac{hc}{\lambda_{max}} \\\\\lambda_{max} = \frac{hc}{\phi} \\\\\lambda_{max} = \frac{(6.626\times 10^{-34}) \times (3 \times 10^8) }{6.546 \times 10^{-19}} \\\\\lambda_{max} = 3.037 \times 10^{-7} m\\\\\lambda_{max} = 303.7 \ nm

3 0
3 years ago
A cylindrical rod of length L is connected across a fixed potential difference, creating a current I through the rod. What would
Inessa05 [86]

Answer:

The value of current through the rod becomes half

i' = \frac{i}{2}

Explanation:

As per Ohm's law we know that the current through a resistor is given as

i = \frac{V}{R}

here we know that

R = \rho \frac{L}{A}

here we know that the length of the cylinder is L and area is A so the value of current through the rod is given as

i = \frac{V A}{\rho L}

now we have change the length of the conductor to twice of initial value and rest all parameters will remain the same

so we will have

i' = \frac{VA}{\rho (2L)}

now from above two equations we have

\frac{i}{i'} = 2

so new current will become

i' = \frac{i}{2}

3 0
3 years ago
hummingbird beat its wings up and down with a frequency of 65 hz. what is the period of the hummingbirds flaps?
kakasveta [241]

Answer:

T ≈ 15.4 ms

Explanation:

T = 1/f = 1/65 = 0.01538

7 0
3 years ago
Read 2 more answers
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