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klasskru [66]
3 years ago
8

Help!! Help!! Help!! Help!!

Physics
1 answer:
Mice21 [21]3 years ago
3 0

I think it's Quadratic. With all the rapid increase of the values of y.

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Glass absorbs ultraviolet (UV) rays from the Sun. Would a fraction of the incident UV light be reflected from the air/glass boun
fiasKO [112]

Answer:

Yes

Explanation:

Any transparent surface in practical is neither a perfect absorber of electromagnetic waves neither a perfect reflector. Generally all the transparent surfaces reflect some amount of irradiation and the other parts are absorbed and transmitted.

<u>That is given by as relation:</u>

\alpha+\rho+\tau=1

where:

\alpha= absorptivity which is defined as the ratio of the absorbed radiation to the total irradiation

\rho= reflectivity is defined as the ratio of reflected radiation to the total irradiation

\tau= transmittivity is defined as the ratio of total transmitted radiation to the total irradiation

6 0
3 years ago
Select all that apply. Which of the following are characteristics of acids? contain hydroxide ion or produce it in a solution ta
julia-pushkina [17]
These are the characteristics that apply:

 - In a solution taste sour: which is consequence of the H+ concentration.

- Corrode metals: the H+ ion reacts with the metal producing a salt and water

-Produce hydronium ion in solution: as per the Bronsted - Lowry definition an acid is a substance that donates a proton, H+. This proton will react with H2O to form H3O+ (hydronium), as per this scheme: 

HA + H2O --> A(-) + H3O(+)
5 0
3 years ago
When a piece of metal mass of 72.17 g is
nordsb [41]

Answer:

Explanation:

The trick is in finding the volume.

Final Volume = 26.64

Initial Volume=<u>20.92</u>                   Subtract

Metal Volume  5.72  cm^3

Density = mass / volume

Density = 72.17 / 5.72

Density = 12.617

7 0
2 years ago
The magnetic field in a cyclotron is 1.25 T, and the maximum orbital radius of the circulating protons is 0.40 m. (a) What is th
Darya [45]

Answer:

1.92 x 10⁻¹²J

Explanation:

The magnetic force from the magnetic field gives the circulating protons gives the particle the necessary centripetal acceleration to keep it orbiting round the circular path. And from Newton's second law of motion, the force(F) is equal to the product of the mass(m) of the proton and the centripetal acceleration(a). i.e

F = ma

Where;

a = \frac{v^2}{r}             [v = linear velocity, r = radius of circular path]

=> F = m\frac{v^2}{r}           ------------(i)

We also know that the magnitude of this magnetic force experienced by the moving charge (proton) in a magnetic field is given by;

F = q v B sin θ       ----------(ii)

Where;

q = charge of the particle

v = velocity of the particle

B = magnetic field

θ = the angle between the velocity and the magnetic field.

Combining equations (i) and (ii) gives

m\frac{v^2}{r} = q v B sin θ           [θ = 90° since the proton is orbiting at the maximum orbital radius]

=> m\frac{v^2}{r} = q v B sin 90°

=> m\frac{v^2}{r} = q v B

Divide both side by v;

=> m\frac{v}{r} = qB

Make v subject of the formula

v = \frac{qBr}{m}

From the question;

B = 1.25T

m = mass of proton = 1.67 x 10⁻²⁷kg

r = 0.40m

q = charge of a proton = 1.6 x 10⁻¹⁹C

Substitute these values into equation(iii) as follows;

v = \frac{(1.6*10^{-19})(1.25)(0.4)}{(1.67*10^{-27})}

v = 4.79 x 10⁷m/s

Now, the kinetic energy, K, is given by;

K = \frac{1}{2}mv²

m = mass of proton

v = velocity of the proton as calculated above

K = \frac{1}{2}(1.67*10^{-27} * (4.79 * 10^7)^2 )

K = 1.92 x 10⁻¹²J

The kinetic energy is 1.92 x 10⁻¹²J

8 0
3 years ago
An exoplanet with one half of Earth's mass and 50% of Earth's radius is discovered.
Georgia [21]

Answer:

The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet

Explanation:

The given parameters are;

The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M

The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R

The weight of the cadet on Earth = 800 N

The \ weight, W  =G\dfrac{M \times m}{R^{2}} = 800 \ N

Therefore, for the weight of the cadet on the exoplanet, W₁, we have;

W_1   =G\dfrac{\dfrac{M}{2}  \times m}{ \left ( \dfrac{R}{2} \right ) ^{2}} = G\dfrac{\dfrac{M}{2}  \times m \times 4}{ R ^{2}} = 2 \times G \times  \dfrac{M \times m}{R^{2}} = 2 \times 800 \, N = 1,600 \, N

The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.

7 0
3 years ago
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