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3 years ago

15

Why is the study of black holes important?

1 answer:

mariarad [96]3 years ago

7 0

It's important because first, we might be able to understand further with how gravity works. And if we can harness the power that black hole possesses, which pulls everything in, then we might be able to use that power for a huge amount of energy, thus possibly replacing solar energy, fossil fuels, etc.,.

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The law of conservation of mass states that

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The law of conservation of mass states that <em>Matter/substances/energy can not be created or destroyed, it can only be transferred from state to state.</em>

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3 years ago

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Guitar string has an overall length of 1.22 m and a total mass of 3.5 g before being strung on a guitar. Once it is used on the

ikadub [295]

Answer:

Fundamental frequency= 174.5 hz

Explanation:

We know

fundamental frequency= $\frac{velocity}{2 *length}$

velocity = $\sqrt{\frac{tension}{mass per unit length} }$

mass per unit length= $\frac{3.5}{1000*1.22}$ =0.00427 $\frac{kg}{m}$

Now calculating velocity v= $\sqrt{\frac{255}{0.00427} }$

=244.3 $\frac{m}{sec}$

Distance between two nodes is 0.7 m.

Plugging these values into to calculate frequency

f = $\frac{244.3}{2 *0.7}$ =174.5 hz

6 0

3 years ago

A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth

andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

$F_{D}=\frac{kq^{2}}{DC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

$F_{B}=\frac{kq^{2}}{BC^{2}}$

$F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

$F_{A}=\frac{kq^{2}}{AC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N$

The direction of FA is along A to C.

The net force along +X axis

$F_{x}=F_{A}Cos45-F_{D}$

$F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N$

The net force along +Y axis

$F_{y}=F_{B}-F_{A}Sin45$

$F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N$

The resultant force is given by

$F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}$

$F = 4.03\times10^{7}N$

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

5 0

3 years ago