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Sedbober [7]
2 years ago
13

A 75,000-watt radio station transmits at 88 MHz. Determine the number of joules transmitted per second. 88 J/s 8800 J/s 2,000,00

0 J/s 75,000 J/s
Physics
1 answer:
Free_Kalibri [48]2 years ago
6 0

Answer:

the first one is the correct answer

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What traits did Sir William Gilbert have that made him a good scientist?
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Gave him good advice to others
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2 years ago
The speed of a nerve impulse in the human body is about 100 m/s. If you accidentally stub you toe in the dark, estimate the time
vodomira [7]

Answer:

0.02 s

Explanation:

Take the (+x) direction to be up.  

The average velocity v during a time interval Δt is the displacement Δx divided by Δt.  

v=Δx/Δt

 =x_f-x_i/t_f-t_i                 (1)

We assume that your height is 1.6m  

Solving [1]

Δt=Δx/v

  = 0.02 s

4 0
3 years ago
What is the resulting direction of a surface wave?​
german

surface wave is a wave that travels along the surface of a medium. The medium is the matter through which the wave travels. Ocean waves are the best-known examples of surface waves. They travel on the surface of the water between the ocean and the air.

HOPE IT HELPS

7 0
3 years ago
Read 2 more answers
An uniform electric field of magnitude E = 100 N/C is oriented along the positive y-axis. What is the magnitude of the flux of t
Ede4ka [16]

Answer:

The magnitude of the flux of electric field through a square of surface area is zero.

Explanation:

E=100 NC^{-1}\\\\A=2 m^2\\\\Electic\,\,flux\,\,flux\,\,is\,\,given\,\,as:\\\\\phi_E=E.A\,cos\,\theta

It is given that square box is parallel to yz-plane which has normal vector perpendicular to plane in x-direction. Angle between normal vector of area and electric field is 90°. Substituting in (1)

\phi_E=E.A\,cos\,(90^o)\\\\\phi_E=0

4 0
3 years ago
A square coil ℓ = 2cm on a side with 30 turns rotates in a uniform magnetic field, B~ = B0zˆ = 0.1Tˆz, such that the normal of t
kow [346]

Answer:

a) 1.2*10^{-3}cos(1.25t)

b) 0.49mV

Explanation:

a) The coil rotates periodically with period T. Hence, we can write the variation of the magnetic flux with a sinusoidal function, and with max flux NAB. Thus, we have that:

\Phi_B(t)=NABcos(\omega t)\\\\\omega=\frac{2\pi}{T}=1.25\frac{rad}{s}\\\\A=l^2=(0.02m)^2=4*10^{-4}m^2\\\\B=0.1T\\\\\Phi_B(t)=1.2*10^{-3}cos(1.25 t) W

where we have used the values given by the information of the problem for N B and A.

b)

the emf is given by:

emf=-\frac{d\Phi_B}{dt}=-NBA\omega sin(\omega t)\\\\emf(t=12.5s)=-(30)(0.1T)(4*10^{-4})(1.25\frac{rad}{s})sin(1.25*12.5)=1.49*10^{-4}V=0.49mV

hope this helps!!

5 0
3 years ago
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