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Ludmilka [50]
3 years ago
15

Ultraviolent radiation is one of many types of energy that comes to Earth from the Sun.

Physics
2 answers:
Leni [432]3 years ago
7 0
True. UV radiation is a form of energy that travels through space. <span>The sun is a major source of ultraviolet radiation but also emits a number of different types of electro magnetic radiation.</span>
motikmotik3 years ago
7 0
True. It is the cause of some types of skin cancer, unlike the heat and light waves.
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A 59.31 kg rock is sitting at the top of a cliff that is 300 m tall. What is the gravitational potential energy of that rock?
Diano4ka-milaya [45]

Answer:

The gravitational potential energy of that rock is 174371.4 J.

Explanation:

Given

  • Mass m = 59.31 kg
  • Height h = 300 m

To determine

We need to find the gravitational potential energy of the rock

We know that the potential energy of a body is termed as the stored energy due to its position.

One kind of energy comes from Earth's gravity — Gravitational potential energy (GPE).

Gravitational potential energy (GPE) can be determined using the formula

GPE = mgh

where

  • m is the mass
  • g is the gravitational acceleration which is equal to g = 9.8 m/s²
  • h is the height
  • GPE is the Gravitational potential energy

now substituting m = 59.31, h = 300 and g = 9.8

GPE = mgh

         =59.31\times 9.8\times 300

         =174371.4 J

Therefore, the gravitational potential energy of that rock is 174371.4 J.

4 0
3 years ago
A weight suspended from a spring is seen to bob up and down over a distance of 20 cm triply each second. What is the period? Wha
Kisachek [45]

Answer:

1) Hence, the period is 0.33 s.

2) The amplitude is 10 cm.

Explanation:

1) The period is given by:

T = \frac{1}{f}

Where:

f: is the frequency = 3 bob up and down each second = 3 s⁻¹ = 3 Hz

T = \frac{1}{f} = \frac{1}{3 Hz} = 0.33 s

Hence, the period is 0.33 s.

2) The amplitude is the distance between the equilibrium position and the maximum position traveled by the spring. Since the spring is moving up and down over a distance of 20 cm, then the amplitude is:          

A = \frac{20 cm}{2} = 10 cm  

Therefore, the amplitude is 10 cm.          

I hope it helps you!                    

5 0
3 years ago
The electron's velocity at that instant is purely horizontal with a magnitude of 2 \times 10^5 ~\text{m/s}2×10 ​5 ​​ m/s then ho
Lesechka [4]

Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10​⁵​​ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]

Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

F = qvB

Also;

F = \frac{MV}{t}

solving this two equations together;

\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}

where;

m is the mass of electron = 9.11 x 10⁻³¹ kg

q is the charge of electron = 1.602 x 10⁻¹⁹ C

B is the strength of the magnetic field = 3.47 x 10⁻³ T

substitute these values and solve for t

t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9}  \ s \ = 1.639 \ ns

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.

5 0
3 years ago
What is the reflectivity of a glass surface (n =1.5) in air (n = 1) at an 45° for (a) S-polarized light and (b) P-polarized ligh
Goshia [24]

Answer:

a) R_s = 0.092

b) R_p = 0.085

Explanation:

given,

n =1.5 for glass surface

n = 1 for air

incidence angle = 45°

using Fresnel equation of reflectivity of S and P polarized light

R_s=\left | \dfrac{n_1cos\theta_i-n_2cos\theta_t}{n_1cos\theta_i+n_2cos\theta_t} \right |^2\\R_p=\left | \dfrac{n_1cos\theta_t-n_2cos\theta_i}{n_1cos\theta_t+n_2cos\theta_i} \right |^2

using snell's law to calculate θ t

sin \theta_t = \dfrac{n_1sin\theta_i}{n_2}=\dfrac{sin45^0}{1.5}=\dfrac{\sqrt{2}}{3}

cos \theta_t =\sqrt{1-sin^2\theta_t} = \dfrac{sqrt{7}}{3}

a) R_s=\left | \dfrac{\dfrac{1}{\sqrt{2}}-\dfrac{1.5\sqrt{7}}{3}}{\dfrac{1}{\sqrt{2}}+\dfrac{1.5\sqrt{7}}{3}} \right |^2

R_s = 0.092

b) R_p=\left | \dfrac{\dfrac{\sqrt{7}}{3}-\dfrac{1.5}{\sqrt{2}}}{\dfrac{\sqrt{7}}{3}+\dfrac{1.5}{\sqrt{2}}} \right |^2

R_p = 0.085

3 0
3 years ago
Is this right? Please help me
morpeh [17]

Answer:

yes

Explanation:

4 0
3 years ago
Read 2 more answers
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