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borishaifa [10]
3 years ago
9

What is the energy in joules and ev of a photon in a radio wave from an am station that has a 1500 khz broadcast frequency?

Physics
1 answer:
LekaFEV [45]3 years ago
4 0
E=hf
E=6.626e-34[Js]•1500e3[Hz]=9.939e-28[J]
For eV divide this by electron charge
9.939e-28[J]/1.6022e-19[C/e]=
6.203e-9eV
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Complete the statement below with the correct term.
dimaraw [331]
Two or three i believe to be the answer
7 0
3 years ago
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In your own words, explain the effects of changes in Earth's magnetic field over time.
viva [34]
It's an interesting fact that scientists don't fully understand how it works. But it seems to be to do with molten metal circulating in the core. Given that it's just liquid metal sloshing around, it seems understandable that it won't always circulate perfectly - imagine the cloud bands in Jupiter's atmosphere - they are reasonably stable but change from time to time. When the liquid changes its speed or direction, however slowly it does so, the resulting magnetic field will move or switch direction.

<span> As scientists try to build better mathematical models of how the core works, they should be able to learn more about the magnetic field it produces. Hope this helps</span>
3 0
3 years ago
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A spring-loaded gun has a spring for which
e-lub [12.9K]

Answer:

7.5 m

Explanation:

k = Spring constant = 180 N/m

x = Displacement of spring = 14 cm

m = Mass of projectile = 0.024 kg

a = g = Acceleration due to gravity = 9.81\ \text{m/s}^2

s = Displacement of projectile

v = Final velocity = 0

u = Initial velocity

The potential energy of the spring will be equal to the kinetic energy of the object

\dfrac{1}{2}kx^2=\dfrac{1}{2}mu^2\\\Rightarrow u=\sqrt{\dfrac{kx^2}{m}}\\\Rightarrow u=\sqrt{\dfrac{180\times 0.14^2}{0.024}}\\\Rightarrow u=12.12\ \text{m/s}

v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0-12.12^2}{2\times -9.81}\\\Rightarrow s=7.5\ \text{m}

The maximum height reached above the initial position is 7.5 m

5 0
3 years ago
A golfer tees off from the top of a rise, giving the golf ball an initial velocity of 44 m/s at an angle of 25° above the horizo
mars1129 [50]

Answer:

a) y=17.6m

b) v_{x}=39.87m/s\\v_{y}=-28.05m/s

Explanation:

From the exercise we got our <u>initial data</u>

v_{o}=44m/s\\\beta =25\\x=190m

a) To find <u><em>maximum height</em></u> we know that at that point v_{y}=0

v_{y} ^{2} =v_{oy} ^{2} +2a(y-y_{o} )

0=(44sin25)^{2} -2(9.8)y

Solving for y

y=\frac{(44sin25)^{2} }{2(9.8)} =17.6 m/s

b) Since we know that the ball strikes the fairway 190 m away

x=v_{ox}t\\ 190=44cos25t

Solving for t

t=4.76s

Now, we can calculate the speed of the ball in both axes

v_{x}=44cos25=39.87m/s

v_{y}=v_{oy}+at

v_{y}=44sin25-(9.8)(4.76)=-28.05m/s

The <em>negative</em> sign means the direction of the ball at that point.

5 0
4 years ago
Help me to solve it . It’s urgent
Artist 52 [7]

Answer: 0°

Explanation:

Step 1: Squaring the given equation and simplifying it

Let θ be the angle between a and b.

Given: a+b=c

Squaring on both sides:

... (a+b) . (a+b) = c.c

> |a|² + |b|² + 2(a.b) = |c|²

> |a|² + |b|² + 2|a| |b| cos 0 = |c|²

a.b = |a| |b| cos 0]

We are also given;

|a+|b| = |c|

Squaring above equation

> |a|² + |b|² + 2|a| |b| = |c|²

Step 2: Comparing the equations:

Comparing eq( insert: small n)(1) and (2)

We get, cos 0 = 1

> 0 = 0°

Final answer: 0°

[Reminders: every letters in here has an arrow above on it]

7 0
3 years ago
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