What are the 4 spheres that make up the earth

Convert 13.1 miles to feet. Using one step conversion

Marat540 [252]

Answer:

69,168 ft

Explanation:

6 0

3 years ago

A laser beam is incident on two slits with a separation of 0.195 mm, and a screen is placed 5.30 m from the slits. If the bright

kumpel [21]

Answer:

λ = 596 nm.

Explanation:

Fringe width = λ D / d

λ is wave length , D is screen distance and d is slit separation.

Putting the values

1.62 x 10⁻² =( λ x 5.3 ) / .195 x 10⁻³

$\lambda=\frac{1.62\times10^{-2}\times195\times10^{-6}}{5.3}$

λ = 596 nm.

8 0

2 years ago

Sirius, the brightest star in the sky, is 2.6 parsecs (8.6 light-years) from Earth, giving it a parallax of 0.379 arcseconds. An

Vikentia [17]

Answer: Sirius, the brightest star in the sky, is 2.6 parsecs (8.6 light-years) from Earth, giving it a parallax of 0.379 arcseconds. Another bright star, Regulus, has a parallax of 0.042 arcseconds. Then, the distance in parsecs will be,23.46.

Explanation: To find the answer, we have to know more about the relation between the distance in parsecs and the parallax.

<h3>What is the relation between the distance in parsecs and the parallax?</h3>

Let's consider a star in the sky, is d parsec distance from the earth, and which has some parallax of P amount.
Then, the equation connecting parallax and the distance in parsec can be written as,

$d=\frac{1}{P}$

We can say that,

$dP=constant.\\thus,\\d_1P_1=d_2P_2$

<h3>How to solve the problem?</h3>

We have given that,

$d_1=2.6 parsecs.\\P_1=0.379arcseconds.\\P_2=0.042 arcseconds.\\d_2=?$

Thus, we can find the distance in parsecs as,

$d_2=\frac{d_1P_1}{P_2} =23.46 parsecs$

Thus, we can conclude that, the distance in parsecs will be, 23.46.

Learn more about the relation connecting distance in parsecs and the parallax here: brainly.com/question/28044776

#SPJ4

6 0

1 year ago

Was the cannonball able to hit its target of 50 meters when the initial velocity was 20 m/s? Why?/Why Not?

frosja888 [35]

Answer:

The cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

Explanation:

From the question given above, the following data were obtained:

Height to which the target is located = 50 m

Initial velocity (u) = 20 m/s

To know whether or not the cannon ball is able to hit the target, we shall determine the maximum height to which the cannon ball attained. This can be obtained as follow:

Initial velocity (u) = 20 m/s

Final velocity (v) = 0 (at maximum height)

Acceleration due to gravity (g) = 10 m/s²

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 20² – (2 × 10 × h)

0 = 400 – 20h

Collect like terms

0 – 400 = – 20h

– 400 = – 20h

Divide both side by – 20

h = – 400 / – 20

h = 20 m

Thus, the the maximum height to which the cannon ball attained is 20 m.

From the calculations made above, we can conclude that the cannon ball was not able to hit the target because the target is located at a height of 50 m whereas the cannon ball was only above to get to a height of 20 m.

3 0

2 years ago

Alice kicks a soccer ball with a mass of 2 kg. The ball leaves the ground moving 30 meters per second. What is the kinetic energ

sleet_krkn [62]

By using the formula:

• K.E = ½mv²

mass, m = 2 kg
Velocity, v = 30 m/s

→ K.E = ½ × 2 × (30)² J

→ K.E = (30)² J

→ K.E = 900 J (Answer)

5 0

1 year ago