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valentinak56 [21]
4 years ago
12

Numerical problemsnumerical problems:

Physics
1 answer:
White raven [17]4 years ago
6 0

Answer:

3.4m

Explanation:

As 1 m = 100cm

So 1cm = 1/100 m or 0.01m

To convert 340cm into m simply divide it by 100

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In an experiment, 108 J of work was done on a closed system. During this phase of the experiment, 79 J of heat energy was added
amid [387]

Answer:

187 J

Explanation:

First Law of Thermodynamics :

ΔQ = ΔW + ΔU

ΔQ : Heat. If it added to system then positive and if it is rejected by system then negative.

ΔW : Work. If it done by the system then positive and if it is done on system then negative.

ΔU : Internal Energy. If it positive then temperature of system increased and if it is negative then temperature of system decreased.

ΔQ = 79 J

ΔW = - 108 J

ΔU = ?

substituting the value in the equation:

79 = -108 + ΔU

∴ ΔU = 187 J

7 0
4 years ago
The membrane that surrounds a certain type of living cell has a surface area of 5.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A
kotykmax [81]

Answer:

2.1\times 10^{-12} c

Explanation:

We are given that

Surface area of membrane=5.3\times 10^{-9} m^2

Thickness of membrane=1.1\times 10^{-8} m

Assume that membrane behave like a parallel plate capacitor.

Dielectric constant=5.9

Potential difference between surfaces=85.9 mV

We have to find the charge resides on the outer surface of membrane.

Capacitance between parallel plate capacitor is given by

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

Substitute the values then we get

Capacitance between parallel plate capacitor=\frac{5.9\times 8.85\times 10^{-12}\times 5.3\times 10^{-9}}{1.1\times 10^{-8}}

C=0.25\times 10^{-12}F

V=85.9 mV=85.9\times 10^{-3}

Q=CV

Q=0.25\times 10^{-12}\times 85.9\times 10^{3}=2.1\times 10^{-12} c

Hence, the charge resides on the outer surface=2.1\times 10^{-12} c

5 0
3 years ago
Read 2 more answers
Calculate the wavelengths of the first five members of the Lyman series of spectral lines, providing the result in units Angstro
Oduvanchick [21]

Answer:

Explanation:

The formula for hydrogen atomic  spectrum is as follows

energy of photon due to transition from higher orbit n₂ to n₁

E=13.6 (\frac{1}{n_1^2 } - \frac{1}{n_2^2})eV

For layman series n₁ = 1 and n₂ = 2 , 3 , 4 ,   ...   etc

energy of first line

E_1=13.6 (\frac{1}{1^2 } - \frac{1}{2 ^2})

10.2 eV

wavelength of photon = 12375 / 10.2 = 1213.2 A

energy of 2 nd line

E_2=13.6 (\frac{1}{1^2 } - \frac{1}{3 ^2})

= 12.08 eV

wavelength of photon = 12375 / 12.08 = 1024.4 A

energy of third line

E_3=13.6 (\frac{1}{1^2 } - \frac{1}{4 ^2})

12.75 e V

wavelength of photon = 12375 / 12.75 = 970.6 A

energy of fourth line

E_4=13.6 (\frac{1}{1^2 } - \frac{1}{5 ^2})

= 13.056 eV

wavelength of photon = 12375 / 13.05 = 948.3 A

energy of fifth line

E_5=13.6 (\frac{1}{1^2 } - \frac{1}{6 ^2})

13.22 eV

wavelength of photon = 12375 / 13.22 = 936.1 A

7 0
3 years ago
The maximum allowed leakage of microwave radiation from a microwave oven is 5.0 mw/cm2. if microwave radiation outside an oven h
irina [24]

Answer:

194 V/m

Explanation:

In order to find electric field, we can use the formula of power density

i.e Pd = E^2 / Z

where:

Pd = power density in W/m^2

E = electric field strength in V/m

Z = impedance of free space = 120 * π

E = sqrt(Pd * Z) -----> re-arranging it for E

before solving, convert Pd unit into W/m^2  

Pd= 5mW/cm^2 = 50 W/m^2    

Solving for E:

E= sqrt(50 * 120 * π)

E = 137.3 V/m  

the above value is RMS value

In order to find the peak amplitude of the oscillating field will therefore be 137.3 * sqrt(2) = 194 V/m

8 0
4 years ago
Read 2 more answers
what is the largest and smallest possible resultant force of two force with magnitude of 41N and 14N​
MArishka [77]

Explanation:

6gtvctyvyvyvyvyfcfufy

8 0
3 years ago
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