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2 years ago

Which planet is known as evening star

Physics

1 answer:

Verdich [7]2 years ago

8 0

Answer:

Mercury is a metal that is used in

Mercury can be seen as an evening "star" near the sun's setting point or as a morning "star" near the sun's rising point. The evening star was given the name Hermes, and the morning star was given the name Apollo, since the ancient Greeks thought they were two separate things. Mercury, the Roman god's messenger, is the planet's name.

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What kind of waves are present during an earthquake? ...

Brrunno [24]

Answer:

1. The two main types of waves are body waves and surface waves. Body waves can travel through the earth's inner layers, but surface waves can only move along the surface of the planet like ripples on water. Earthquakes radiate seismic energy as both body and surface waves.

2. potential energy

3. Newton's second law of motion is F = ma, or force is equal to mass times acceleration.

4. Refraction is the bending of light

5. Density uses the formula p=m/V, or density (p) is equal to mass (m) divided by volume (V). Density is defined as mass per unit volume.

Explanation:

6 0

3 years ago

Light of wavelength 630 nm is incident on a long, narrow slit. Determine the angular deflection of the first diffraction minimum

asambeis [7]

Answer:

a) 1.8°

b) 0.18°

c) 0.018°

Explanation:

Wavelength (λ) = 630nm = 630 *10^-9m

The equation that describes the angular deflection of a dark band is

Wsin(βm) = mλ

w = width of the single slit

λ = wavelength of the light

βm = angular deflection of the mth dark band.

a) In order to get the angular deflection of the first dark band for a slit with 0.02mm width, substitute w = 0.02mm = 0.02*10^-3 , m = 1 , λ= 630*10^-9

0.02*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.02*10^-3

Sin(β1) = 0.0315

β1 = Sin^-1(0.0315)

= 1.8°

b) substitute w = 0.2mm = 0.2*10^-3 , m = 1 , λ= 630*10^-9

0.2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.2*10^-3

Sin(β1) = 0.00315

β1 = Sin^-1(0.00315)

= 0.18°

c) substitute w = 2mm = 2*10^-3 , m = 1 , λ= 630*10^-9

2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 2*10^-3

Sin(β1) = 3.15*10^-4

β1 = Sin^-1(3.15*10^-4)

= 0.018°

6 0

3 years ago

A long string is wrapped around a 6.6-cm-diameter cylinder, initially at rest, that is free to rotate on an axle. The string is

lys-0071 [83]

Answer:

$\omega_f=571.42\ rpm$

Explanation:

It is given that,

Diameter of cylinder, d = 6.6 cm

Radius of cylinder, r = 3.3 cm = 0.033 m

Acceleration of the string, $a=1.5\ m/s^2$

Displacement, d = 1.3 m

The angular acceleration is given by :

$\alpha =\dfrac{a}{r}$

$\alpha =\dfrac{1.5}{0.033}$

$\alpha =45.46\ rad/s^2$

The angular displacement is given by :

$\theta=\dfrac{d}{r}$

$\theta=\dfrac{1.3}{0.033}$

$\theta=39.39\ rad$

Using the third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta$

Here, $\omega_i=0$

$\omega_f=\sqrt{2\alpha \theta}$

$\omega_f=\sqrt{2\times 45.46\times 39.39}$

$\omega_f=59.84\ rad/s$

Since, 1 rad/s = 9.54 rpm

So,

$\omega_f=571.42\ rpm$

So, the angular speed of the cylinder is 571.42 rpm. Hence, this is the required solution.

5 0

3 years ago

A person hears a siren as a fire truck approaches and passes by. The frequency varies from 480Hz on approach to 400Hz going away

alekssr [168]

Answer:

31.2 m/s

Explanation:

$f_{app}$ = Frequency of approach = 480 Hz

$f_{aw}$ = Frequency of going away = 400 Hz

$V$ = Speed of sound in air = 343 m/s

$v$ = Speed of truck

Frequency of approach is given as

$f_{app} = \frac{Vf}{V - v}$ eq-1

Frequency of moving awayy is given as

$f_{aw} = \frac{Vf}{V + v}$ eq-2

Dividing eq-1 by eq-2

$\frac{f_{app}}{f_{aw}} = \frac{V + v}{V - v}$

$\frac{480}{400} = \frac{343 + v}{343 - v}$

$v$ = 31.2 m/s

7 0

3 years ago

Standing waves can ruin the acoustics of a concert hall if there is excessive reflection of the sound waves that the performers

Dmitrij [34]

Answer:

The answer to the questions is;

In terms of standing waves, the listener moves from a location with high amplitude to one with lower amplitude or vibration (anti-node to node)

The distance 4.1 cm is equivalent to λ/4

Explanation:

For standing waves we have is a stationary wave comprising of two opposite direction moving waves that have equal amplitude and frequency, resulting in the superimposition of the waves. As such certain points are fixed along the wave path that is the peaks amplitude of the wave oscillation is constant at a particular point. A node occurring at a point and an anti-node occurring at another fixed point

When the listener moves 4.1 cm he or she has left the anti-node to the node hence the faintness of the sound

The distance from the node to the anti-node is 1/4 wavelength, or 1/4×λ

Therefore 4.1 cm is λ/4

6 0

3 years ago

Which planet is known as evening star​

Which planet is known as evening star