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denis23 [38]
3 years ago
11

A hungry rat is placed in a maze. It walks the following path to find a piece of cheese. 4.0m N, 7.5 m E, 6.8 m S, 3.7 m E, 3.6

m S, 5.3 m W, 3.7 m N, 5.6 m W, 4.4 m S, 4.9 m W. Calculate the distance the rat walked. Find the rat’s displacement and direction.
Physics
2 answers:
storchak [24]3 years ago
6 0

Answer:

Explanation:

We shall take the help of vector form of displacement . Taking east as i and north as j

4.0m N = 4 j

7.5 m E = 7.5 i

6.8 m S = - 6.8 j

3.7 m E, = 3.7 i

3.6 m S = - 3.6 j

5.3 m W = - 5.3 i

3.7 m N, = 3.7 j

5.6 m W = - 5.6 i

4.4 m S = - 4.4 j

4.9 m W = -  4.9 i

Total displacement = 4j +7.5 i -6.8j+3.7i-3.6j-5.3i+3.7j-5.6i-4.4j-4.9i

= -4.6 i -7.1 j

magnitude of displacement = \sqrt{(4.6^2+7.1^2)}

= 8.46 m

Direction

Tanθ = 7.1/ 4.6

θ = 57⁰ south of west .

distance walked = 4+7.5 +6.8+3.7+3.6+5.3+3.7+5.6+4.4+4.9

= 49.5 m

ankoles [38]3 years ago
6 0

Answer:

equal to

Explanation:

i got it right hope it helped :)

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Determine the binding energy per nucleon of an mg-24 nucleus. the mg-24 nucleus has a mass of 24.30506. a proton has a mass of 1
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The mass of Mg-24 is 24.30506 amu, it contains 12 protons and 12 neutrons.

Theoretical mass of Mg-24:

The theoretical mass of Mg-24 is:

Hydrogen atom mass = 12 × 1.00728 amu = 12.0874 amu

Neutron mass = 12 x 1.008665 amu = 12.104 amu

Theoretical mass = Hydrogen atom mass + Neutron mass = 24.1913 amu

Note that the mass defect is:

Mass defect = Actual mass - Theoretical mass : 24.30506 amu- 24.1913 amu= 0.11376 amu

Calculating the binding energy per nucleon:

\frac{B.E.}{nucleon}=\frac{(0.11376amu)(931Mev/amu}{24nucleons}  = 4.41294 Mev/nucleon

So approximately 4.41294 Mev/necleon


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An atom that has an excess positive or negative electrical charge caused by the loss or addition of an electron is called a(n)
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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
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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
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