You would need to ear defenders (I'm assuming these are ear protectors for use with loud sounds) while firing at a gun range. You could use it for loud construction areas.
C) The longer the line, the greater the magnitude of the vector. As for the direction, just think of a compass.
Answer:
Explanation:
Given
![B_z=(2.4\mu T)\sin (1.05\times 10^7x-\omega t)](https://tex.z-dn.net/?f=B_z%3D%282.4%5Cmu%20T%29%5Csin%20%281.05%5Ctimes%2010%5E7x-%5Comega%20t%29)
Em wave is in the form of
![B=B_0\sin (kx-\omega t)](https://tex.z-dn.net/?f=B%3DB_0%5Csin%20%28kx-%5Comega%20t%29)
where ![\omega =frequency\ of\ oscillation](https://tex.z-dn.net/?f=%5Comega%20%3Dfrequency%5C%20of%5C%20oscillation)
![k=wave\ constant](https://tex.z-dn.net/?f=k%3Dwave%5C%20constant)
![B_0=Maximum\ value\ of\ Magnetic\ Field](https://tex.z-dn.net/?f=B_0%3DMaximum%5C%20value%5C%20of%5C%20Magnetic%5C%20Field)
Wave constant for EM wave k is
![k=1.05\times 10^7 m^{-1}](https://tex.z-dn.net/?f=k%3D1.05%5Ctimes%2010%5E7%20m%5E%7B-1%7D)
Wavelength of wave ![\lambda =\frac{2\pi }{k}](https://tex.z-dn.net/?f=%5Clambda%20%3D%5Cfrac%7B2%5Cpi%20%7D%7Bk%7D)
![\lambda =\frac{2\pi }{1.05\times 10^7}](https://tex.z-dn.net/?f=%5Clambda%20%3D%5Cfrac%7B2%5Cpi%20%7D%7B1.05%5Ctimes%2010%5E7%7D)
![\lambda =5.98\times 10^{-7} m](https://tex.z-dn.net/?f=%5Clambda%20%3D5.98%5Ctimes%2010%5E%7B-7%7D%20m)
The time it would take a 2500 W electric kettle to boil away 1.5 Kg of water is 2400 seconds
<h3>How to calculate the time</h3>
Use the formula:
Power × time = mass × specific heat
Given mass = 1. 5kg
Specific latent heat of vaporization = 4000000 J/ Kg
Power = 2500 W
Substitute the values into the formula
Power × time = mass × specific heat
2500 × time = 1. 5 × 4000000
Make 'time' the subject
time = 1. 5 × 4000000 ÷ 2500 = 6000000 ÷ 2500 = 2400 seconds
Therefore, the time it would take a 2500 W electric kettle to boil away 1.5 Kg of water is 2400 seconds.
Learn more about specific latent heat of vaporization:
https://brainly.in/question/1580957
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Answer:
A) 0.50 mV
Explanation:
In this problem, we can think the wings of the bird as a metal rod moving across a magnetic field. So, and emf will be induced into the wings of the bird, according to the formula:
![\epsilon = BvL sin \theta](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%20BvL%20sin%20%5Ctheta)
where
is the strength of the magnetic field
v = 13 m/s is the speed of the bird
L = 1.2 m is the wingspan of the bird
is the angle between the direction of motion and the direction of the magnetic field
Substituting numbers into the formula, we find
![\epsilon = (5.0\cdot 10^{-5} T)(13 m/s)(1.2 m) sin 40^{\circ}=0.00050 V = 0.50 mV](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%20%285.0%5Ccdot%2010%5E%7B-5%7D%20T%29%2813%20m%2Fs%29%281.2%20m%29%20sin%2040%5E%7B%5Ccirc%7D%3D0.00050%20V%20%3D%200.50%20mV)