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natta225 [31]
3 years ago
9

Literally don't know how to do this.

Physics
1 answer:
yulyashka [42]3 years ago
4 0

Answer:

a

c

b

c

c

b

a

a

Explanation:

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PLEASE HELP ANSWER ASAP. 10 points. I will give brainliest.
alina1380 [7]

The answer is D. I know because I already answered the question.

3 0
3 years ago
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 brick of mass 4 kg hangs from the end of a spring. When the brick is at rest, the spring is stretched by 3 cm. The spring is t
jeka57 [31]

Answer:

Explanation:

Let s be displacement from equilibrium position . Restoring force

m d²s / dt² = - k s

d²s / dt² = - k /m  s

Put k /m  = ω

d²s / dt² + ω² s = 0

The solution of this differential equation

= s = A cosωt

Now when t = 0 ,  s = 2 cm

A =  2 cm

Putting the values we have

2 = A cos 0

A = 2 cm

s ( t) = 2 cos ωt

3 0
3 years ago
Low birth rates, death rates are constant
Mariana [72]

1 is b high birth rate and death rate 2 is a low birth rates death are constant and 3 is c high birth rate and low death rate

Explanation:

4 0
3 years ago
At a particular instant the magnitude of the gravitational force exerted by a planet on one of its moons is 7 × 1021 N. Collapse
9966 [12]

Answer:

Fg = 4.2*10²² N

Explanation:

The gravitational force between any two masses, provided that can be approximated by point masses (comparing their diameters with the distance between them), obeys the Newton's Universal Law of Gravitation, which states that the force (always attractive) is proportional to the product of  the masses and inversely proportional to the square of the distance between them (this as a consequence of our Universe being three-dimensional), as follows:

Fg =\frac{G*m1*m2}{r^{2}}

So, if one of the masses increases 6 times, the force between them will be directly 6 times larger, so the new magnitude of the force will be as follows:

Fg₂ = Fg₁*6 = 7*10²¹ N* 6 = 4.2*10²² N

7 0
3 years ago
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