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Lerok [7]
2 years ago
12

The magnetic field of an electromagnetic wave in a vacuum is Bz =(2.4μT)sin((1.05×107)x−ωt), where x is in m and t is in s. You

may want to review (Pages 889 - 892) . For help with math skills, you may want to review: Rearrangement of Equations Involving Multiplication and Division Part A What is the wave's wavelength? Express your answer to three significant figures and include the appropriate units.
Physics
1 answer:
tatiyna2 years ago
7 0

Answer:

Explanation:

Given

B_z=(2.4\mu T)\sin (1.05\times 10^7x-\omega t)

Em wave is in the form of

B=B_0\sin (kx-\omega t)

where \omega =frequency\ of\ oscillation

k=wave\ constant

B_0=Maximum\ value\ of\ Magnetic\ Field

Wave constant for EM wave k is

k=1.05\times 10^7 m^{-1}

Wavelength of wave \lambda =\frac{2\pi }{k}

\lambda =\frac{2\pi }{1.05\times 10^7}

\lambda =5.98\times 10^{-7} m

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To calculate the acceleration of the wooden block, we use the expression F=ma where F is the force applied, m is the mass of the object and a is the acceleration. We calculate as follows:

F = ma
4.9 = 0.5a
a = 9.8 

Hope this answers the question. Have a nice day.

4 0
2 years ago
Read 2 more answers
Consider a steel guitar string of initial length l=1.00m and cross-sectional area a=0.500mm2. the young's modulus of the steel i
laiz [17]
L = 1.00 m, the original length
A = 0.5 mm² = 0.5 x 10⁻⁶ m², the cross sectional area
E = 2.0 x 10¹¹ n/m², Young's modulus
P = 1500 N, the applied tension

Calculate the stress.
σ = P/A = (1500 N)/(0.5 x 10⁻⁶ m²) = 3 x 10⁹ N/m²

Let δ =  the stretch of the string.
Then the strain is
ε = δ/L

By definition, the strain is
ε = σ/E = (3 x 10⁹ N/m²)/(2 x 10¹¹ N/m²) = 0.015
Therefore
δ/(1 m) = 0.015
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Answer:  15 mm
4 0
2 years ago
Calculate the acceleration of the body of mass 3kg on which a force of 42N has been applied for 5s
____ [38]
F=ma

42/3 = a
a = 14m/s^2
5 0
2 years ago
The magnetic field of an electromagnetic wave in a vacuum is Bz =(2.4μT)sin((1.05×107)x−ωt), where x is in m and t is in s. You
tatiyna

Answer:

Explanation:

Given

B_z=(2.4\mu T)\sin (1.05\times 10^7x-\omega t)

Em wave is in the form of

B=B_0\sin (kx-\omega t)

where \omega =frequency\ of\ oscillation

k=wave\ constant

B_0=Maximum\ value\ of\ Magnetic\ Field

Wave constant for EM wave k is

k=1.05\times 10^7 m^{-1}

Wavelength of wave \lambda =\frac{2\pi }{k}

\lambda =\frac{2\pi }{1.05\times 10^7}

\lambda =5.98\times 10^{-7} m

7 0
2 years ago
An object with a heat capacity of 345J∘C experiences a temperature change from 88.0∘C to 45.0∘C. How much heat is released in th
pogonyaev

Answer:

There is 148.35 Joules of heat is released in the process.

Explanation:

Given that,

Heat capacity of the object, c=345J/^oC

Initial temperature, T_i=88^{\circ}C

Final temperature, T_f=45^{\circ}C

We need to find the amount of heat released in the process. It is a concept of heat capacity. The heat released in the process is given by :

Q=mc\Delta T

Let the mass of the object is 10 g or 0.01 kg

So,

Q=0.01\times 345\times (88-45)

Q = 148.35 Joules

So, there is 148.35 Joules of heat is released in the process. Hence, this is the required solution.                                                

5 0
2 years ago
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