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3 years ago

What type of front usually brings thunder clouds and storms

Physics

2 answers:

a_sh-v [17]3 years ago

7 0

The type of front that brings storms and thunder clouds is called a cold front

Iteru [2.4K]3 years ago

7 0

Cold front forms Thunderclouds and storms.

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If a positively charged particle moves into a magnetic field traveling in a straight line, how would you expect it's motion to c

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It would begin to curve toward the magnetic pull

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3 years ago

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A solution is prepared by dissolving 49.3 g of KBr in enough water to form 473 mL of solution. Calculate the mass % of KBr in th

son4ous [18]

Answer:

9.31%

Explanation:

We are given that

Mass of KBr=49.3 g

Volume of solution=473 mL

Density of solution =1.12g/mL

We have to find the mass% of KBr.

Mass = $volume\times density$

Using the formula

Mass of solution= $1.12\times 473=529.76 g$

Mass % of KBr= $\frac{mass\;of\;KBr}{Total\;mass\;of\;solution}\times 100$

Mass % of KBr= $\frac{49.3}{529.76}\times 100$

Mass % of KBr=9.31%

Hence, the mass% of KBr=9.31%

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3 years ago

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I think it’s A protons .. hope this helps

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2 years ago

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A mass is oscillating with amplitude A at the end of a spring.

Dmitry_Shevchenko [17]

A) $x=\pm \frac{A}{2\sqrt{2}}$

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

$E=\frac{1}{2}kA^2$ (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

$E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2$ (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

$U=\frac{1}{3}K$

Using (2) we can rewrite this as

$U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}$

And using (1), we find

$U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2$

Substituting $U=\frac{1}{2}kx^2$ into the last equation, we find the value of x:

$\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}$

B) $x=\pm \frac{3}{\sqrt{10}}A$

In this case, the kinetic energy is 1/10 of the total energy:

$K=\frac{1}{10}E$

Since we have

$K=E-U$

we can write

$E-U=\frac{1}{10}E\\U=\frac{9}{10}E$

And so we find:

$\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A$

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3 years ago

What force is needed to move a 5 kg mass with an acceleration of 5 m/s²?

ElenaW [278]

Answer:

b)25N

Explanation:

F=ma

F=(5kg)(5m/s^2)

F=25N

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2 years ago

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