A pendulum oscillates 12 times in 4 seconds. what is the length of the pendulum?

During metamorphism, what is the major effect of chemically active fluids?

xz_007 [3.2K]

Answer:

Option b

Explanation:

Metamorphism is the process where the variation of the geological texture resulting from the different arrangement of the minerals or the variation of minerals in protoliths, i.e., pre- existing rocks take place such that there occurs no change in state of the protolith, i.e., it does not melt into magma.

The change takes place as a result of the presence of chemically active fluids, heat and pressure.

There is a reaction between the chemically active fluid and the rock through which it passes and promotes the movement of the dissolved ions of silicate and promotes the growth of the mineral grains.

7 0

3 years ago

How large is the area of water found on Mars?

ehidna [41]

More than five million cubic kilometers of ice have been identified.

8 0

3 years ago

Read 2 more answers

Find the speed of a satellite in a circular orbit around the Earth with a radius 3.57 times the mean radius of the Earth. (Radiu

Gnesinka [82]

The orbiting velocity of the satellite is 4.2km/s.

To find the answer, we need to know about the orbital velocity of a satellite.

<h3>What's the expression of orbital velocity of a satellite?</h3>

Mathematically, orbital velocity= √(GM/r)
r = radius of the orbital, M = mass of earth

<h3>What's the orbital velocity of a satellite orbiting earth with a radius 3.57 times the earth radius?</h3>

M= 5.98×10²⁴ kg, r= 3.57× 6.37×10³ km = 22.7×10⁶m
Orbital velocity= √(6.67×10^(-11)×5.98×10²⁴/22.7×10⁶)

=4.2km/s

Thus, we can conclude that the orbiting velocity of the satellite is 4.2km/s.

Learn more about the orbital velocity here:

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8 0

1 year ago

When exercising in the heat, which of the following hydration strategies is best for temperature regulation during an event (e.g

Olegator [25]

Answer:

Your question was incomplete so here is the complete question and answer.

Q. When exercising in the heat, which of the following hydration strategies is best for temperature regulation during an event (e.g., 10K race)

a) plain water

b) 5-7 percent glucose solution

c) Glucose polymer solution of 6-8 percent

d) There appears to be no difference among these different forms of hydration techniques relative to temperature regulation.

Ans. d) There appears to be no difference among these different forms of hydration techniques relative to temperature regulation.

Explanation:

Temperature Regulation is an important phenomenon for the person exposed to extreme hot conditions or weather. Exercising in hot conditions increase the body temperature. Greater and intense exercise, greater the production of heat. Then the heat dissipation takes place in the form of excessive sweating which results in dehydration. That was just the brief overview of temperature regulation. Above mentioned techniques are equally good hydration techniques so there is no difference at all. You can have a plain water or glucose solutions of above mentioned percentages.

3 0

3 years ago

Two large rectangular aluminum plates of area 180 cm2 face each other with a separation of 3 mm between them. The plates are cha

Westkost [7]

Answer:

Φ = 361872 N.m^2 / C

Explanation:

Given:-

- The area of the two plates, $A_p = 180 cm^2$

- The charge on each plate, $q = 17 * 10^-^6 C$

- Permittivity of free space, $e_o = 8.85 * 10^-^1^2 \frac{C^2}{N.m^2}$

- The radius for the flux region, $r = 3.3 cm$

- The angle between normal to region and perpendicular to plates, θ = 4°

Find:-

Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates.

Solution:-

- First we will determine the area of the region ( Ar ) by using the formula for the area of a circle as follows. The region has a radius of r = 3.3 cm:

$A_r = \pi *r^2\\\\A_r = \pi *(0.033)^2\\\\A_r = 0.00342 m^2$

- The charge density ( σ ) would be considered to be uniform for both plates. It is expressed as the ratio of the charge ( q ) on each plate and its area ( A_p ):

σ = $\frac{q}{A_p} = \frac{17*10^-^6}{0.018} \\$

σ = 0.00094 C / m^2

- We will assume the electric field due to the positive charged plate ( E+ ) / negative charged plate ( E- ) to be equivalent to the electric field ( E ) of an infinitely large charged plate with uniform charge density.

$E+ = E- = \frac{sigma}{2*e_o} \\\\$

- The electric field experienced by a region between two infinitely long charged plates with uniform charge density is the resultant effect of both plates. So from the principle of super-position we have the following net uniform electric field ( E_net ) between the two plates:

$E_n_e_t = (E+) + ( E-)\\\\E_n_e_t = \frac{0.00094}{8.85*10^-^1^2} \\\\E_n_e_t = 106214689.26553 \frac{N}{C} \\$

- From the Gauss-Law the flux ( Φ ) through a region under uniform electric field ( E_net ) at an angle of ( θ ) is:

Φ = E_net * Ar * cos ( θ )

Φ = (106214689.26553) * (0.00342) * cos ( 5 )

Φ = 361872 N.m^2 / C

5 0

2 years ago