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4 years ago

The outer most layer of the sun's atmosphere that extends into space and can be seen during an eclipse is called the

1 answer:

8 0

The outer most layer of the sun's atmosphere that extends into space that can be seen during an eclipse is called the SOLAR ECLIPSE

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tamaranim1 [39]

Answer:

I’m so sorry I tried solving it but I don’t understand it can you explain the question a little bit more ty

Explanation:

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4 years ago

In which state of matter has the least kinetic energy

Savatey [412]

Solids have the least kinetic energy

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4 years ago

Read 2 more answers

What are the two main categories of ecosystems?

snow_lady [41]

It would be autotroph and hetrotroph

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3 years ago

A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a forc

____ [38]

Answer:

Part a)

$a = 0.36 m/s^2$

Part b)

$a = -1.29 m/s^2$

Explanation:

Force applied by the student on the box is 80 N at an angle of 25 degree

so here two components of the force on the box is given as

$F_x = Fcos25$

$F_x = 80 cos25 = 72.5 N$

$F_y = Fsin25$

$F_y = 80 sin25 = 33.8 N$

now in vertical direction we can use force balance for the box to find the normal force on it

$F_n + F_y = mg$

$F_n = (25)(9.81) - 33.8$

$F_n = 211.45 N$

now kinetic friction on the box opposite to applied force due to rough floor is given as

$F_k = \mu F_n$

$F_k = (0.300)(211.45) = 63.44 N$

now the net force on the box in forward direction is given as

$F_{net} = F_x - F_k$

$F_{net} = 72.5 - 63.44 = 9.065 N$

now the acceleration of the box is given as

$a = \frac{F_{net}}{m}$

$a = \frac{9.065}{25} = 0.36 m/s^2$

Part b)

when box is pulled up along the inclined surface of angle 10 degree

now the two components of the force will be same along the inclined and perpendicular to inclined plane

$F_x = 72.5 N$

$F_y = 33.8 N$

now force balance perpendicular to inclined plane is given as

$F_n + F_y = mgcos\theta$

$F_n = (25)(9.81)cos10 - 33.8 = 207.7 N$

now the friction force opposite to the motion on the box is given as

$f_k = \mu F_n$

$f_k = (0.300)(207.7) = 62.3 N$

now the net pulling force along the inclined plane is given as

$F_{net} = F_x - F_k - mgsin10$

$F_{net} = 72.5 - 62.3 - (25)(9.81)sin10$

$F_{net} = -32.38 N$

now the box will decelerate and it is given as

$a = \frac{F_{net}}{m}$

$a = \frac{-32.38}{25} = -1.29 m/s^2$

7 0

4 years ago

Your grandfather clock's pendulum has a length of 0.9930 m.

posledela

We will start from the period definition to find the relationship between the perioricity and the length. After finding the relationship between the two lengths and periods we will proceed to calculate the length two with the loss of the period, that is

$T = 2\pi \sqrt{\frac{l}{g}}$

Therefore the period is proportional to the square root of the length

$T \propto l^{1/2}$

Then

$\frac{T_2}{T_1} = \frac{\sqrt{L_2}}{\sqrt{L_1}}$

$(\frac{T_2}{T_1})^2 = \frac{L_2}{L_1}$

$L_2 = (\frac{T_2}{T_1})^2(L_1)$

The period $T_1$ is equivalent to the seconds that a day has, that is 86400 seconds while period two will be the seconds that have one day less the loss of 16 consecutive announced in the statement therefore,