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2 years ago

In the aftermath of an intense earthquake, the earth as a whole "rings" with a period of 54 minutes.

Physics

1 answer:

lawyer [7]2 years ago

6 0

The frequency of the oscillation in hertz is calculated to be 0.00031 Hz.

The frequency of a wave is defined as the number of cycles completed per second while the period refers to the time taken to complete a cycle. The frequency is the inverse of period.

So;

Period(T) = 54 minutes or 3240 seconds

Frequency (f) = T-1 = 1/T = 1/3240 seconds = 0.00031 Hz

Learn more: brainly.com/question/14588679

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b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul

Nat2105 [25]

Answer:

b) Betelgeuse would be $\approx 1.43 \cdot 10^{6}$ times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is $\approx 1.37 \cdot 10^{-5} }$ the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

$B = \displaystyle{\frac{L}{4\pi d^2}}$ (1)

where $B$ is the brightness, $L$ is the star luminosity and $d$ , the distance from the star to the point where the brightness is calculated (measured). Thus:

b) $B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}}$ and $B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}}$ where $L_{Sun}$ is the Sun luminosity ( $3.9 x 10^{26} W$ ) but we don't need to know this value for solving the problem. $ly$ is light years.

Finding the ratio between the two brightness we get:

$\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }$

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is $1.581 \cdot 10^{-5}\ ly$ . Then

$\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }$

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

4 0

3 years ago

Optical tweezers use light from a laser to move single atoms and molecules around. Suppose the intensity of light from the tweez

WITCHER [35]

Answer:

Explanation:

a )

If it is totally absorbed pressure is calculated as follows .

Pressure = I / c where I is intensity of light falling .

= 1000 / 3 x 10⁸

= 3.33 x 10⁻⁶ N / m²

b ) weight of tritium atom

= 3 x 1.67 x 10⁻²⁷ kg

acceleration = force / mass

= 3.33x 10⁻⁶ / 3 x 1.67 x 10⁻²⁷

= .6646 x 10²¹ m /s²

= 66.46 x 10¹⁹ m / s²

7 0

3 years ago

A stone is dropped into a river from a bridge 41.7 m above the water. Another stone is thrown vertically down 1.80 s after the f

hram777 [196]

Answer:

31.75 m/s

Explanation:

h = 41.7 m

Let the initial velocity of the second stone is u

Let the time taken to reach to the bottom by the first stone is t then the time taken by the second stone to reach the ground is t - 1.8.

For first stone:

Use second equation of motion

$h=ut+\frac{1}{2}gt^2$

Here, u = 0, g = 9.8 m/s^2 and t be the time and h = 41.7

So, 41.7= 0 + 0.5 x 9.8 x t^2

41.7 = 4.9 t^2

t = 2.92 s ..... (1)

For second stone:

Use second equation of motion

$h=ut+\frac{1}{2}gt^2$

Here, g = 9.8 m/s^2 and time taken is t - 1.8 = 2.92 - 1.8 = 1.12 s, h = 41.7 m and u be the initial velocity

$h=u\left ( t-1.8 \right )+4.9\left ( t-1.8 \right )^2$ .... (2)

By equation the equation (1) and (2), we get

$41.7=1.12 u +4.9 \times 1.12^{2}$

u = 31.75 m/s

5 0

3 years ago

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 60 cm long and has a mass of 3.8 kg, with the center of

Serggg [28]

Answer:

(a) τ = 26.58 Nm

(b) τ = 18.79 Nm

Explanation:

(a)

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

d₁ = perpendicular distance between ball and shoulder = 60 cm = 0.6 m

τ₁ = (29.4 N)(0.6 m)

τ₁ = 17.64 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of 60 cm = (0.4)(60 cm) = 24 cm = 0.24 m

τ₂ = (37.24 N)(0.24 m)

τ₂ = 8.94 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 17.64 Nm + 8.94 Nm

τ = 26.58 Nm

(b)

Now, the arm is at 45° below horizontal line.

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

42.42 cm = 0.4242 m

τ₁ = (29.4 N)(0.4242 m)

τ₁ = 12.47 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of (60 cm)(Cos 45°) = (0.4)(42.42 cm) = 16.96 cm = 0.1696 m

τ₂ = (37.24 N)(0.1696 m)

τ₂ = 6.32 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 12.47 Nm + 6.32 Nm

τ = 18.79 Nm

3 0

2 years ago

How did Orsted discover Electromagnetism?

Liula [17]

Answer:

Hans Christian Oersted began a new scientific epoch when he discovered that electricity and magnetism are linked. He showed by experiment that an electric current flowing through a wire could move a nearby magnet. The discovery of electromagnetism set the stage for the eventual development of our modern technology-based world.

Explanation:

7 0

3 years ago