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2 years ago

Explain the uses of Electrophorus

Physics

1 answer:

Maslowich2 years ago

8 0

Answer:

Explanation:

Comment

Interesting creature.

These creatures are the only survivors from an era long past, where nothing else has survived. They are eels capable of discharging a large electrical current. A pack of them (they can hunt in groups) can kill just about anything.

Tribes use and tame them to guard the tribe's tame Electrophorus. They are bread and maintained in schools. While the charge is huge, once discharged, these creatures take a very long time to recharge, so they are not entirely invincible.

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3 years ago

15) On a cold day, you take in 4.2 L (i.e., 4.2 x 10-3 m3) of air into your lungs at a temperature of 0°C. If you hold your brea

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Answer:

4.8L ( i.e 4.8 x 10^-3 m3)

Explanation:

Step 1:

Data obtained from the question.

Initial volume (V1) = 4.2L

Initial temperature (T1) = 0°C

Final temperature (T2) = 37°C

Final volume (V2) =?

Step 2:

Conversion of celsius temperature to Kelvin temperature. This is illustrated below

K = °C + 273

T1 = 0°C = 0°C + 273 = 273K

T2 = 37°C = 37°C + 273 = 310K

Step 3:

Determination of the final volume.

Since the pressure is constant,

Charles' Law equation will be applied as shown below:

V1 /T1 = V2/T2

4.2/273 = V2 /310

Cross multiply to express in linear form

273 x V2 = 4.2 x 310

Divide both side by 273

V2 = (4.2 x 310)/273

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Therefore, the volume of the air in the lungs at that point is 4.8L ( i.e 4.8 x 10^-3 m3)

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3 years ago

Cobaltâ’60 is a radioactive isotope used to treat cancers of the brain and other tissues. A gamma ray emitted by an atom of this

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Energy of gamma rays is given by equation

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here we know that

h = Planck's constant

$\nu = frequency$

now energy is given as

$E = 4.70 MeV = 4.70 \times 10^6 \times 1.6 \times 10^{-19}$

$E = 7.52 \times 10^{-13} J$

now by above equation

$E = h\nu$

$7.52 \times 10^{-13} = 6.6 \times 10^{-34} \nu$

$\nu = 1.14 \times 10^{21} Hz$

now for wavelength we can say

$\lambda = \frac{c}{\nu}$

$\lambda = \frac{3\times 10^8}{1.14 \times 10^{21}}$

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2 years ago

A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.68 m/s. The block

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Answer:

a) $v'=-0.227\ m.s^{-1}$

b) $v=1.36\ m.s^{-1}$

Explanation:

Given:

mass of the lighter block, $m'=0.02\kg$

velocity of the lighter block, $u'=0.68\ m.s^{-1}$

mass of the heavier block, $m=0.04\ kg$

velocity of the heavier block, $u=0\ m.s^{-1}$

Using conservation of linear momentum:

$m'.u'+m.u=m'.v'+m.v$

where:

$v'=$ final velocity of the lighter block

$v=$ final velocity of the heavier block

$m'.u'=m'.v'+m.v$

$m'(u'-v')=m.v$ ........................(1)

Since kinetic energy is conserved in elastic collision:

$\frac{1}{2}m'.u'^2=\frac{1}{2}m'.v'^2+\frac{1}{2}m.v^2$

$m'(u'^2-v'^2)=m.v^2$

$m'(u'-v')(u'+v')= m.v^2$

divide the above equation by eq. (1)

$v=u'+v'$ .............................(2)

now we substitute the value of v from eq. (2) in eq. (1)

$m'(u'-v')=m(u'+v')$

$\frac{m'+m}{m'-m} =\frac{u'}{v'}$

$\frac{0.02+0.04}{0.02-0.04} =\frac{0.68}{v'}$

$v'=-0.227\ m.s^{-1}$ (negative sign denotes that the direction is towards left)

now we substitute the value of v' from eq. (2) in eq. (1)

$m'(u'-v+u')=m.v$

$2m'.u'=(m-m')v$

$2\times 0.02\times 0.68=(0.04-0.02)\times v$

$v=1.36\ m.s^{-1}$

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Answer:

The answer is A

Explanation:

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Explain the uses of Electrophorus ​

Explain the uses of Electrophorus