The ratio between the force of sliding friction and the normal force of an object is called?

Light with a wavelength of 395 nm illuminates a metal cathode. The maximum kinetic energy of the emitted electrons is 0.76 eV .

lutik1710 [3]

Answer:

520.8 nm

Explanation:

We are given that

$\lambda=395 nm=395\times 10^{-9} m$

1 nm= $10^{-9} m$

Maximum kinetic energy, $K_{max}=0.76 eV=0.76\times 1.6\times 10^{-19}$ V

$1 e=1.6\times 10^{-19} C$

We have to find the maximum wavelength of light.

We know that

$h\nu=\frac{hc}{\lambda_0}+K_{max}$

Where $c=3\times 10^8 m/s$

$h=6.625\times 10^{-34}$

$6.625\times 10^{-34}\times \frac{3\times 10^8}{395\times 10^{-9}}=\frac{6.625\times 10^{-34}\times 3\times 10^8}{\lambda_0}+0.76\times 1.6\times10^{-19}$

$6.625\times 10^{-34}\times \frac{3\times 10^8}{395\times 10^{-9}}- 0.76\times 1.6\times10^{-19}=\frac{6.625\times 10^{-34}\times 3\times 10^8}{\lambda_0}$

$\lambda_0=\frac{6.625\times 10^{-34}\times 3\times 10^8}{6.625\times 10^{-34}\times \frac{3\times 10^8}{395\times 10^{-9}}- 0.76\times 1.6\times10^{-19}}$

$\lambda_0=5.208\times 10^{-7} m$ =520.8 nm

6 0

3 years ago

In a vacuum, which has a greater acceleration while in free fall: a7 kg bowling ball or a 0.007 kg feather?

Klio2033 [76]

Answer:the will have the same acceleration

Explanation:according to fállelos theory they land at the same time

6 0

4 years ago

A book weighing 5 N rests on top of a table. 1) A downward force of magnitude 5 N is exerted on the book by the force of

andreev551 [17]

Answer:

1) A downward force of magnitude 5 N is exerted on the book by the force of of gravity

2) An upward force of magnitude 5 N is exerted on the book by the table

Explanation:

First of all, any object near the Earth's surface experiences the forces of gravity, which is also called weight of the object. This force always acts downward.

For the book in the problem, the magnitude of the weight is 5 N.

We also know that the book is at rest: this means that the net force acting on it is zero, and there must be another force balancing the weight, in order to give a zero net force. This other force is the reaction force exerted by the table on the book: the magnitude of this force must be equal to the force of gravity (so, 5 N) and its direction is opposite to the weight, therefore upward.

7 0

3 years ago

SHOW ADEQUATE WORKINGS IN THIS SECTION

cluponka [151]

Answer:

12 i. The work done by Wale = 107.910 kJ

The work done by Lekan = 117.720 kJ

Total work done = 225.36 kJ

ii. Wale's power = 4.3164 kW

Lekan's power = 3.924 kW

Wale has more power and is more powerful than Lekan

13. 313.92 N

Explanation:

i. The work done, W = Force, F × Distance moved by the force, D

The given parameters are

The mass of Wale = 55 kg

The mass of Lekan = 60 kg

The acceleration due to gravity, g =9.81 m/s²

The motion force of Wale and Lekan are;

Motion force of Wale = 9.81 × 55 = 539.55 N

Motion force of Lekan = 9.81 × 60 = 588.6 N

The work done by Wale = 539.55 × 200 = 107910 J = 107.910 kJ

The work done by Lekan= 588.6 × 200 = 117720 J = 117.720 kJ

107910 + 117720 =225630 J = 225.36 kJ

ii. Power = Work done/time

Wale finished the race in 25 s, therefore, his power = 107910/25 = 4316.4 W

Lekan finished the race in 30 s, therefore, his power = 117720/30 = 3924 W

Wale has more power and is more powerful than Lekan

13. The velocity ratio = 5

V. R. = Distance moved by effort/(Distance moved by load)

Efficiency = 80%

Work done by effort = x

Work done by machine = Efficiency × Work done by effort = 0.8 × x

Distance moved by effort, E = V. R. × Distance moved by load, D = 5 × D

Work done by effort = Force × Distance moved = 200×9.81× E

Work done by effort = 1962×E = 1962×E = 1962×5×D

Work done by machine = 1962 × D, when D = 1, we have;

0.8 × 1962×1 = 1569.6 J

Work done by effort = Force × Distance moved

Work done by effort = Force × 5×D = Force × 5 (D = 1)

From the principle of conservation of energy, we have;

Energy is neither created nor destroyed

Therefore

Work done by effort = Force × 5 = 1569.6 J

Force = 1569.6 /5 = 313.92 N.

3 0

3 years ago

PLEASSE HELP WILL MARK BRAINLIEST!!!!!!!!!!

faust18 [17]

They are halogen elements, or nonmetallic elements in the same GROUP, specifically group 17

6 0

3 years ago