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3 years ago

Extreme tides are

Physics

1 answer:

nasty-shy [4]3 years ago

7 0

This are tides that are very high and reach up the beach

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5. Describe the shape of the waveform in the secondary coil for a sine, square and triangle wave in the primary coil. How does t

Volgvan

Answer:

When primary coil is exited by sin wave,this will result in sin wave in secondary coil as well.According to law,flux induced in the secondary coil will have same waveform as in the primary coil.

5 0

3 years ago

Which of these is a benefit of replacing a coal-burning power plant with a

Scilla [17]

A. Reduced greenhouse gas emissions.

5 0

3 years ago

Does space have plenty of oxygen

xz_007 [3.2K]

Answer:

I don't think so because we can't breathe in space

Explanation:

6 0

3 years ago

Read 2 more answers

The membrane of a living cell can be approximated by a parallel-plate capacitor with plates of area 4.50×10−9 m2 , a plate separ

Marizza181 [45]

The energy stored in the membrane is $6.44\cdot 10^{-14} J$

Explanation:

The capacitance of a parallel-plate capacitor is given by

$C=\frac{k\epsilon_0 A}{d}$

where

k is the dielectric constant of the material

$\epsilon_0$ is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

For the membrane in this problem, we have

k = 4.6

$A=4.50\cdot 10^{-9} m^2$

$d=8.1\cdot 10^{-9} m$

Substituting, we find its capacitance:

$C=\frac{(4.6)(8.85\cdot 10^{-12})(4.50\cdot 10^{-9})}{8.1\cdot 10^{-9}}=2.26\cdot 10^{-11} F$

Now we can find the energy stored: for a capacitor, it is given by

$U=\frac{1}{2}CV^2$

where

$C=2.26\cdot 10^{-11} F$ is the capacitance

$V=7.55\cdot 10^{-2} V$ is the potential difference

Substituting,

$U=\frac{1}{2}(2.26\cdot 10^{-11} F)(7.55\cdot 10^{-2})^2=6.44\cdot 10^{-14} J$

Learn more about capacitors:

brainly.com/question/10427437

brainly.com/question/8892837

brainly.com/question/9617400

#LearnwithBrainly

6 0

3 years ago

A bullet is fired horizontally from a handgun at a target 100.0 m away. If the initial speed of the bullet as it leaves the gun

Lera25 [3.4K]

Answer:

The distance is 0.53 m.

Explanation:

Given that,

Target distance = 100.0 m

Speed of bullet = 300 m/s

We need to calculate the total time

Using formula of time

$t=\dfrac{d}{v}$

Put the value into the formula

$t=\dfrac{100.0}{300}$

$t=0.33\ sec$

Now, consider vertical motion of bullet.

Initial velocity of bullet in vertical direction = 0 m/s

We need to calculate the vertically distance

Using equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Put the value in the equation

$s=0+\dfrac{1}{2}\times9.8\times(0.33)^2$

$s=0.53\ m$

Hence, The distance is 0.53 m.

5 0

3 years ago