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Aliun [14]
3 years ago
14

Why is the majority of Earth's freshwater not readily available for our use?

Physics
2 answers:
lara [203]3 years ago
8 0
The answer is B. It is locked up in glaciers and ice caps.

Hope this helped. Good luck! Please give me a thanks and Brainliest.
Simora [160]3 years ago
7 0

Answer:

B.It is locked up in glaciers and ice caps.

Explanation:

Earth is the only planet which we know where water is present is present in all the three forms. Seventy percent of the Earth is filled with water but only 2.5% is freshwater. Rest of it is saline water in oceans. Now, out of this 2.5% freshwater, only 1% is in usable form.  This is because rest of the freshwater is locked up in ice caps and glaciers and cannot be used.

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A transverse wave is a moving wave in which the current is perpendicular to the direction of the wave or path of propagation. A longitudinal wave are waves in which the displacement of the median is in the direction of the propagation.
Example:
Transverse- pond ripple
Longitudinal- crest and troff
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Why are latitude and longitude included on maps
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Answer:

to locate places on earth

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How many new facilities does an average facility manager build during his or her career?
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An average facility manager can build one new facility during his or her career.

 

<span>A </span>facilities manager<span> is the ultimate organiser, making sure that a workplace meets the needs of employees by managing all of the required services. In this job, you will be responsible for the </span>management<span> of services and processes that support the core business of an organisation.</span>

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Two identical small metal spheres with q1 &gt; 0 and |q1| &gt; |q2| attract each other with a force of magnitude 72.1 mN when se
Brrunno [24]

1) +2.19\mu C

The electrostatic force between two charges is given by

F=k\frac{q_1 q_2}{r^2} (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

\frac{Q}{2}

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

F=k\frac{(\frac{Q}{2})^2}{r^2}

And since we know that

r = 1.41 m (distance between the spheres)

F= 21.63 mN = 0.02163 N

(the sign is positive since the charges repel each other)

We can solve the equation for Q:

Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C

So, the final charge on the sphere on the right is

\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C

2) q_1 = +6.70 \mu C

Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

F = -72.1 mN = -0.0721 N (we put a negative sign since the force is attractive, which means that the charges have opposite signs)

r = 1.41 m is the separation between the charges

And also,

q_2 = Q-q_1

So we can rewrite eq.(1) as

F=k \frac{q_1 (Q-q_1)}{r^2}

Solving for q1,

Fr^2=k (q_1 Q-q_1^2})\\kq_1^2 -kQ q_1 +Fr^2 = 0

Since Q=4.37\cdot 10^{-6} C, we can substituting all numbers into the equation:

8.98755\cdot 10^9 q_1^2 -3.93\cdot 10^4 q_1 -0.141 = 0

which gives two solutions:

q_1 = 6.70\cdot 10^{-6} C\\q_2 = -2.34\cdot 10^{-6} C

Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

q_1 = +6.70 \mu C

8 0
3 years ago
The tuning fork has a frequency of 426.7 is found to cause resonance in a closed column of air measuring 0.186 (the diameter of
lidiya [134]

Answer:

The velocity of the tuning fork sound is 317.4648 m/s

Explanation:

The given parameters are;

The frequency of the fork, f = 426.7 Hz

The length of the closed air column, L = 0.186 m

The diameter of the tube, d = 0.15 m

At the fundamental frequency in a closed tube, we have;

λ = 4·L

Where;

λ = The wavelength of the wave

L = The length of the tube

From the equation for the velocity of a wave, we have;

v = f·λ

Where;

v = The velocity of the (sound) wave

f = The frequency of the wave = 426.7 Hz

λ = 4·L = 4 × 0.186 m = 0.744 m

∴ v = 426.7 Hz × 0.744 m = 317.4648 m/s

Therefore, the velocity of the sound produced by the tuning fork, v = 317.4648 m/s

3 0
3 years ago
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