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3 years ago

The buildup of charges on an object is called

Physics

2 answers:

In-s [12.5K]3 years ago

7 0

what whitey said he is right because he is smart

Nastasia [14]3 years ago

4 0

Static electricity.

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Who was the man who lived from 460B.C. 370B.C. and was among the first to suggest the idea of atoms?

Jlenok [28]

C Democritus

That's the Answer

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3 years ago

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If 5 mm of rain falls in a 100 m2 field, what volume of rain, in m3, fell in the field?

Nina [5.8K]

The volume of rain that fells in the field is simply given by the area of the field, which is

$A=100 mm^2$

multiplied by the height of rain that fell, which is

$h=5.0 mm$

Therefore, the volume is

$V=hA=(5 mm)(100 mm^2)=500 mm^3$

7 0

3 years ago

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:

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

3 0

3 years ago

You are operating a pwc in an area where swimmers are in the water. When must you slow your pwc to "slow, no wake speed"?.

Lostsunrise [7]

You should slow your pwc to "slow, no wake speed" when within 100 feet of anchored vessels or non-motorized craft.

This is the process of operating a personal watercraft at the slowest possible speed.

This helps to maintain steerage which prevents different forms of accident or risks when in motion in the water.

Read more about Slow-no-wake here brainly.com/question/10410716

#SPJ1

4 0

2 years ago

When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa

MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

$\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2$

In parallel

$\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}$

$\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0$

Solving the above quadratic equation

$\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}$

$\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega$

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

6 0

3 years ago