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3 years ago

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(Math Question) Angle A is a complement of Angle B and the midpoint of Angle A is 36 degrees. Find the midpoint of Angle B.

1 answer:

shusha [124]3 years ago

6 0

Complementary angles are those whose sum is 90 degrees, right?, then let's find A, A=72 degrees, because its mid point is 36 deg. Then, B = 18 degrees and its mid point is 9 degrees! :)

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{WORTH BRAINLIEST WITH AN EXPLANATION} Janine is a police officer patrolling a street. She drives 2.2 km east, then 4.4 km west,

icang [17]

Answer:

Options (3) and (6)

Explanation:

Janine drives 2.2 km East.

Since, distances measured in the East are positive,

Displacement = 2.2 km

Then she drives 4.4 km west

Displacement = -4.4 km

Followed by 1.7 km in the East

Displacement = 1.7 km

Total displacement = 2.2 - 4.4 + 1.7 = -0.5 km

$\overrightarrow{d}=-0.5$ km

Total distance covered by Janine = 2.2 + 4.4 + 1.7 = 8.3 km

d = 8.3 km

Therefore, Options (3) and (6) will be the answer.

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4 years ago

A uniform steel rod has mass 0.400 kg and length 50.0 cm and is horizontal. A uniform sphere with radius

Eddi Din [679]

Explanation:

Center of gravity , Let center of rod is origin

= $\frac{(0.400*0)+(0.800*\frac{50}{2}+10)+ (0.200 *(-(25+5)) }{0.400+0.800+0.200}$

=15.71cm from the center of rod

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3 years ago

The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu

Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

$E=\phi + K$

where

$E=\frac{hc}{\lambda}$ is the energy of the incident light, with h being the Planck constant, c being the speed of light, and $\lambda$ being the wavelength

$\phi$ is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

$\lambda=190 nm=1.9\cdot 10^{-7}m$ , so the energy of the incident light is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J$

Converting in electronvolts,

$E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV$

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

$\phi = E-K=6.5 eV-4.0 eV=2.5 eV$

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

$\lambda=510 nm=5.1\cdot 10^{-7} m$

So its energy is

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J$

Converting in electronvolts,

$E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV$

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: $\phi = 4.5 eV$

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

$E=\phi+K=4.5 eV+2.7 eV=7.2 eV$

Then the copper is replaced with sodium, which has work function of

$\phi = 2.3 eV$

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

$K=E-\phi = 7.2 eV-2.3 eV=4.9 eV$

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valentina_108 [34]

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4 years ago

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lisov135 [29]

Answer:

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Explanation:

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4 years ago