Which of the following statements describes an anticyclone?

A person observes a firework display for A safe distance of .750 km. Assuming that sound travels at 340 m/s in air what is the t

WINSTONCH [101]

Answer:

t = 2.2 s

Explanation:

Given that,

A person observes a firework display for A safe distance of 0.750 km.

d = 750 m

The speed of sound in air, v = 340 m/s

We need to find the between the person see and hear a firework explosion. let it is t. So, using the formula of speed.

$v=\dfrac{d}{t}\\\\t=\dfrac{d}{v}\\\\t=\dfrac{750\ m}{340\ m/s}\\\\t=2.2\ s$

So, the required time is 2.2 seconds.

3 0

2 years ago

Five identical quintuplets leave earth when they reach the age of 21, in the year 2121. Each quintuplet goes on a spaceship jour

Elena-2011 [213]

Answer:

Explanation:

This is a problem based on time dilation , a theory given by Albert Einstein .

The formula of time dilation is as follows .

t₁ = $\frac{t}{\sqrt{1-\frac{v^2}{c^2} } }$

t is time measured on the earth and t₁ is time measured by man on ship .

A ) Given t = 20 years , t₁ = ? v = .4c

$\frac{20}{\sqrt{1-\frac{.16c^2}{c^2} } }$

=1.09 x 20

t₁= 21.82 years

B ) Given t = 5 years , t₁ = ? v = .2c

$\frac{5}{\sqrt{1-\frac{.04c^2}{c^2} } }$

=1.02 x 5

t₁= 5.1 years

C ) Given t = 10 years , t₁ = ? v = .8c

$\frac{10}{\sqrt{1-\frac{.64c^2}{c^2} } }$

=1.67 x 10

t₁= 16.7 years

D ) Given t = 10 years , t₁ = ? v = .4c

$\frac{10}{\sqrt{1-\frac{.16c^2}{c^2} } }$

=1.09 x 10

t₁= 10.9 years

E ) Given t = 20 years , t₁ = ? v = .8c

$\frac{20}{\sqrt{1-\frac{.64c^2}{c^2} } }$

=1.67 x 20

t₁= 33.4 years

7 0

2 years ago

The cytoplasm of a cell consists mainly of water.

ankoles [38]

Deal! The answer is true

5 0

3 years ago

Read 2 more answers

Review. As an astronaut, you observe a small planet to be spherical. After landing on the planet, you set off, walking always st

anyanavicka [17]

To find the mass of the planet we will apply the relationship of the given circumference of the planet with the given data and thus find the radius of the planet. From the kinematic equations of motion we will find the gravitational acceleration of the planet, and under the description of this value by Newton's laws the mass of the planet, that is,

The circumference of the planet is,

$\phi = 25.1m$

Under the mathematical value the radius would be

$\phi = 2\pi r$

$r = \frac{25}{2\pi}$

$r = 3.9788km$

Using second equation of motion

$x = \frac{1}{2} at^2$

Replacing the values given,

$1.4 = \frac{1}{2} a (29.2)^2$

Rearranging and solving for 'a' we have,

$a = 0.003283m/s^2$

Using the value of acceleration due to gravity from Newton's law we have that

$a = \frac{GM}{r^2}$

Here,

r = Radius of the planet

G = Gravitational Universal constant

M = Mass of the Planet

$\frac{(6.67*10^{-11})*M}{(3.9788*10^3)^2} = 0.003283$

$M = 7.79201*10^{14}kg$

Therefore the mass of this planet is $7.79201*10^{14}kg$

5 0

3 years ago

07. How do scientists use spectroscopy (the study of light frequencies released

Vera_Pavlovna [14]

Every element is able to be recognized individually in many different ways. A very easy and common way is using light absorption also known as spectroscopy. Every atom has electrons, and these electrons like to stay in their lowest-energy configuration. However, when photons collide with an electron it can increase it to a higher energy level.. This is absorption, and each element’s electrons absorb light at specific wavelengths related to the difference between energy levels in that atom. But the electrons want to return to their original levels, so they don’t hold onto the energy for long. When they emit the energy, they release photons with exactly the same wavelengths of light that were absorbed in the first place. An electron can release this light in any direction, so most of the light is emitted in directions away from our line of sight. Therefore, a dark line appears in the spectrum at that particular wavelength.

Because the wavelengths at which absorption lines occur are unique for each element, astronomers can measure the position of the lines to determine which elements are present in a target. The amount of light that is absorbed can also provide information about how much of each element is present.

8 0

3 years ago