Answer:
Radio stations have dipole type antennas
this field increases in intensity and propagates outwards,
Explanation:
Radio stations have dipole type antennas, that is, all sides are isolated from each other, when the AC signal from the radio station arrives, the lcharge begins at times and by the Lens law a field appears that opposes this movement, this field increases in intensity and propagates outwards, when the voltage reaches a maximum, the generated wave also reaches the maximum, now the incident wave begins to decrease, an electric hand appears to oppose this prisoner, and in this way a cap is created. electric .
Answer:
The answer is the 1st one
Answer: false
Explanation:
Hi, The Michelson-Morley experiment was designed to measure the extra time it took a light beam to travel "there-and-back" against the ether wind, compared to a light beam travelling "sideways across" the ether wind.
The experiment failed, there was no variation, and years later Einstein explained that there is no ether. Light is a special sort of wave which requires no medium to carry it.
Answer:
(a) v = 1.71 m/s
(b) μ = 0.005
Explanation:
(a)
Using the law of conservation of the momentum:
![m_1u_1+m_2u_2=m_1v_1+m_2v_2\\](https://tex.z-dn.net/?f=m_1u_1%2Bm_2u_2%3Dm_1v_1%2Bm_2v_2%5C%5C)
where,
m₁ = mass of person = 61.1 kg
m₂ = mass of sled = 16.1 kg
u₁ = initial speed of the person = 2.16 m/s
u₂ = initial speed of the sled = 0 m/s
v₁ = v₂ = v = final speeds of both the person and the sled = ?
Therefore,
![(61.1\ kg)(2.16\ m/s)+(16.1\ kg)(0\ m/s)=(61.1\ kg)(v)+(16.1\ kg)(v)\\\\v = \frac{131.976\ kgm/s}{77.2\ kg}](https://tex.z-dn.net/?f=%2861.1%5C%20kg%29%282.16%5C%20m%2Fs%29%2B%2816.1%5C%20kg%29%280%5C%20m%2Fs%29%3D%2861.1%5C%20kg%29%28v%29%2B%2816.1%5C%20kg%29%28v%29%5C%5C%5C%5Cv%20%3D%20%5Cfrac%7B131.976%5C%20kgm%2Fs%7D%7B77.2%5C%20kg%7D)
<u>v = 1.71 m/s</u>
<u></u>
(b)
The kinetic energy lost by the sled must be equal to the frictional energy:
K.E = fd
![\frac{1}{2}mv^2=\mu Rd = \mu Wd\\\\\frac{1}{2}mv^2=\mu mgd\\\\\frac{1}{2}v^2=\mu g\\\\\mu = \frac{v^2}{2gd}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2%3D%5Cmu%20Rd%20%3D%20%5Cmu%20Wd%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7Dmv%5E2%3D%5Cmu%20mgd%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7Dv%5E2%3D%5Cmu%20g%5C%5C%5C%5C%5Cmu%20%3D%20%5Cfrac%7Bv%5E2%7D%7B2gd%7D)
where,
μ = coefficient of kinetic friction = ?
d = distance covered = 30 m
g = acceleration due to gravity = 9.81 m/s²
Therefore,
![\mu = \frac{(1.71\ m/s)^2}{(2)(9.81\ m/s^2)(30\ m)}](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%5Cfrac%7B%281.71%5C%20m%2Fs%29%5E2%7D%7B%282%29%289.81%5C%20m%2Fs%5E2%29%2830%5C%20m%29%7D)
<u>μ = 0.005</u>
Answer:
10.36m/s
Explanation:
From the question we are given;
Distance S = 5.40m
Initial velocity v= 1.17m/s
Required
Final velocity of the diver
Using the equation of motion
v² = u²+2gS
v²= 1.17²+2(9.8)(5.40)
v² = 1.3689+105.84
v² = 107.2089
v = √107.2089
v= 10.36m/s
Hence the required velocity is 10.36m/s