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evablogger [386]
3 years ago
9

How does an an electron microscope work?

Physics
2 answers:
Eddi Din [679]3 years ago
8 0
With electricity it work..
EastWind [94]3 years ago
5 0
<em>it works on the following principals:

</em>

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Dry air is primarily composed of nitrogen. In a classroom demonstration, a physics instructor pours 3.6 L of liquid nitrogen int
neonofarm [45]

Answer:

The  value is  V_n  =  2.2498 \  m^3

Explanation:

From the question we are told that

   The volume of  liquid nitrogen is  V_n  =  3.6 \  L=  3.6 *10^{-3} \ m^3

   The  density of  nitrogen at gaseous form   is  \rho_n =  1.2929 \  kg/m^3  =  The dry air at sea level

   

Generally the density of nitrogen at liquid form is  

         \rho _l = 808 \  kg/m^3

And this is mathematically represented as

      \rho_l  =  \frac{m}{V_l }

=>   m  =  \rho_l  *  V_l

Now the density of  gaseous nitrogen is

       \rho_n  =  \frac{m}{V_n }

=>   m  =  \rho_n  *  V_n

Given that the mass is constant

       \rho_n  *  V_n  =   \rho_l  *  V_l

        1.2929*  V_n  =   808  *  3.6*10^{-3}

=>   V_n  =  2.2498 \  m^3

       

3 0
3 years ago
A 2kg toy car traveling at 2m/s strikes a second toy car at rest. The mass of the second toy car is 3. The two stock together an
Nutka1998 [239]
Mass of first car = Initial mass (Mi) = 2 kg
Initial velocity (Vi) = 2 m/s

Mass of both cars together = Final mass (Mf) = 2 + 3 kg = 5 kg
Final Velocity (Vf) = ?

Applying law of conservation of momentum,

Mi x Vi = Mf x Vf
2 x 2 = 5 x Vf
Vf = 4/5 = 0.8 m/s

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Se um corpo 1 está em movimento em relação a um corpo 2, e se o corpo 2 está em repouso em relação a um corpo 3, pode-se afirmar
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Yo habla ingles no espanol
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Read 2 more answers
Inductors in parallel. Two inductors L1 = 1.31 H and L2 = 2.24 H are connected in parallel and separated by a large distance so
mote1985 [20]

Answer:

a) 0.83H

b) 3.22/N Henry

Explanation:

Given two inductors L1 = 1.31 H and L2 = 2.24 H connected in parallel, their equivalent inductance derivative is similar to that of resistance in parallel.

Derivation:

If the voltage across an inductor

VL = IXL

I is the current

XL is the inductive reactance

XL = 2πfL

VL = I(2πfL)

L is the inductance.

From the formula, I = V/2πfL

Given two inductors in parallel, different current will flow through them.

I1 = V/2πfL1 (current in L1)

I2 = V/2πfL2 (current in L2)

Total current I = I1+I2

I = V/2πfL1 + V/2πfL2

I = V/2πf{1/L1+1/L2}

V/2πfL = V/2πf{1/L1+1/L2}

1/L = 1/L1+1/L2 (equivalent inductance in parallel)

Given L1 = 1.31 H and L2 = 2.24

1/L = 1/1.31 + 1/2.24

1/L = 0.763 + 0.446

1/L = 1.209

L = 1/1.209

L = 0.83H

The equivalent inductance is 0.83H

b) Given similar inductors L = 3.22H in parallel, the equivalent inductance will be:

1/L = 1/3.22+1/3.22+1/3.22+1/3.22+1/3.22

+1/3.22+1/3.22+1/3.22+1/3.22+1/3.22

1/L = 10/3.22 (since that all have the same denominator)

L = 3.22/10

If N = 10, the generalization of 10 similar inductors in parallel will be:

L = 3.22/N Henry

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3 years ago
What happens to myosin and actin as sarcomeres relax?
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The myosin heads pull on the actin, bringing them closer together

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