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xz_007 [3.2K]
4 years ago
10

Which type of radiation travels at the speed of light and penetrates matter easily?

Physics
2 answers:
goblinko [34]4 years ago
7 0
The radiation is ultra voilet or Gamma radiation , because their wave length is very short i e 1..0 to 2.5 (angstrom)Ao.
Alla [95]4 years ago
6 0

Answer:

D. gamma rays

Explanation:

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Read 2 more answers
A long, current-carrying solenoid with an air core has 1800 turns per meter of length and a radius of 0.0165 m. A coil of 210 tu
Hoochie [10]

Answer:

The mutual inductance is  M  =  0.000406 \  H

Explanation:

From the question we  are told that

    The  number of turns per unit length  is  N =  1800

    The radius is  r = 0.0165 \ m

     The  number of turns of the solenoid is  N_s  =  210 \ turns

   

Generally the mutual inductance of the  system is mathematically represented as

       M  =  \mu_o  *  N *  N_s  *  A

Where A is the cross-sectional area of the system which is mathematically represented as

       A  =  \pi  *  r^2

substituting values

      A  =  3.142 *  (0.0165)^2

       A  =  0.0008554 \ m^2

also   \mu_o is the permeability of free space with the value  \mu_o  =   4\pi * 10^{-7} N/A^2

So  

      M  =  4\pi * 10^{-7}   *1800 *  210  *  0.0008554

      M  =  0.000406 \  H

3 0
3 years ago
The parallel plates in a capacitor, with a plate area of 8.00 cm2 and an air-filled separation of 2.70 mm, are charged by a 8.70
MrRa [10]

Answer:

a)  ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J,  d)  W = 150 10⁻¹² J

Explanation:

Let's find the capacitance of the capacitor

         C = \epsilon_o \frac{A}{d}

         C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³

         C = 2.62 10⁻¹² F

for the initial data let's look for the accumulated charge on the plates

          C = \frac{Q}{\Delta V}

          Q₀ = C ΔV

           Q₀ = 2.62 10⁻¹² 8.70

           Q₀ = 22.8 10⁻¹² C

a) we look for the capacity for the new distance

          C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³

          C₁ = 1.04 10⁻¹² F

       

          C₁ = Q₀ / ΔV₁

          ΔV₁ = Q₀ / C₁

          ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²

          ΔV₁ = 21.9 V

b) initial stored energy

          U₀ = \frac{Q_o}{ 2C}

          U₀ = (22.8 10⁻¹²)²/(2  2.62 10⁻¹²)

          U₀ = 99.2 10⁻¹² J

c) final stored energy

          U_f = (22.8 10⁻¹²) ² /(2  1.04 10⁻⁻¹²)

          U_f = 249.9 10⁻¹² J

d) the work of separating the plates

as energy is conserved work must be equal to energy change

          W = U_f - U₀

          W = (249.2 - 99.2) 10⁻¹²

          W = 150 10⁻¹² J

note that as the energy increases the work must be supplied to the system

6 0
3 years ago
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