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3 years ago

Question 29 (1 point)

Physics

1 answer:

bulgar [2K]3 years ago

8 0

The answer is true. fluorescence demonstrates a sparkly like light beam and also can bounce off of other beams

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Solnce55 [7]

Answer:

47c

Explanation:

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3 years ago

một động cơ nhiệt khi hoạt động nhiệt lượng mà nó nhận vào gấp bốn lần công mà nó thực hiện. Hiệu suất của động cơ là?

OlgaM077 [116]

$\frac{1}{4} = 0.25 = 25\%$

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3 years ago

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laila [671]

What? Do you have a picture of the problem ?

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3 years ago

A wire carrying a 32.0 A current passes between the poles of a strong magnet such that the wire is perpendicular to the magnet's

marin [14]

Answer:

2.24 T

Explanation:

From Electromagnetic Field,

F = BILsin∅................ Equation 1

Where F = Force on the wire, B = Field strength, I = current flowing in the conductor, L = length of the conductor, ∅ = The angle the conductor makes with the magnetic field.

Making B the subject of the equation,

B = F/ILsin∅..................... Equation 2

Given: F = 2.15 N, I = 32 A, L = 3.00 cm = 0.03 m, ∅ = 90° ( the wire is perpendicular to the magnetic field)

Substitute into equation 2

B = 2.15/(32×0.03×sin90°)

B = 2.15/0.96

B = 2.24 T.

Hence the Field strength = 2.24 T

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4 years ago

A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this

wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

$r=r_o\times A^{\frac{1}{3}}$

$r_o=1.25 \times 10^{-15} m$ = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

$r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}$

$r=4.6656\times 10^{-15} m=4.6656 fm$

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) $F=k\times \frac{q_1q_2}{a^2}$

k= $9\times 10^9 N m^2/C^2$ = Coulombs constant

$q_1,q_2$ = charges kept at distance 'a' from each other

F = electrostatic force between charges

$q_1=+1.602\times 10^{-19} C$

$q_2=+1.602\times 10^{-19} C$

Force of repulsion between two protons on opposite sides of the diameter

$a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m$

$F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}$

$F=2.6527 N$

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

6 0

3 years ago