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azamat
3 years ago
9

Which of the following frequencies falls in the range of RF waves used by commercial radio broadcasting stations?

Physics
2 answers:
dybincka [34]3 years ago
8 0
<span>  A. 6,000,000 Hz =3 they were correct</span>
shusha [124]3 years ago
8 0

It's D. 600,000 Hz

I just took the test :)

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Give a real life example in which two objects are moving at a constant speed but have different velocities.
Bad White [126]

Answer:

The planets and moons that orbit in the solar system.

Explanation:

For example the earth moves at 67,000 mph (107,000 km/h), and is constant from the gravitational pull of the sun. The moon orbits at about  2,288 mph (3,683 km/h).  these are both traveling at different velocities but at a constant speed.

6 0
2 years ago
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5. Graph A plots a race car's speed for 5 seconds. The car's
Nastasia [14]
The answer for the cars speed is tue
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2 years ago
You attach a 3.10 kg weight to a horizontal spring that is fixed at one end. You pull the weight until the spring is stretched b
dolphi86 [110]

Answer:

0.785 m/s

Explanation:

Hi!

To solve this problem we will use the equation of motion of the harmonic oscillator, <em>i.e.</em>

x(t) = A cos(\omega t )+B sin(\omega t) - (1)

\frac{dx}{dt}(t) = \omega (B cos(\omega t )- A sin(\omega t) - (1)

The problem say us that the spring is released from rest when the spring is stretched by 0.100 m, this condition is given as:

x(0) = 0.100

\frac{dx}{dt}(0) = 0

Since cos(0)=1 and sin(0) = 0:

x(0)=A

\frac{dx}{dt}(0) = -B\omega

We get

A =0.100\\B = 0

Now it say that after 0.4s the weigth reaches zero speed. This will happen when the sping shrinks by 0.100. This condition is written as:

x(0.4) = - 0.100

Since

x(t) = 0.100 cos(\omega t)\\ -0.100=x(0.4)=0.100cos(\omega 0.4)

This is the same as:

-1 = cos(0.4\omega)

We know that cosine equals to -1 when its argument is equal to:

(2n+1)π

With n an integer

The first time should happen when n=0

Therefore:

π = 0.4ω

or

ω = π/0.4  -- (2)

Now, the maximum speed will be reached when the potential energy is zero, <em>i.e. </em>when the sping is not stretched, that is when x = 0

With this info we will know at what time it happens:

0 = x(t) = 0.100cos(\omega t)

The first time that the cosine is equal to zero is when its argument is equal to π/2

<em>i.e.</em>

t_{maxV}=\pi /(2\omega)

And the velocity at that time is:

\frac{dx}{dt}(t_{maxV} ) =- 0.100\omega sin(\omega t_{maxV})\\\frac{dx}{dt}(t_{maxV} ) =- 0.100\omega sin(\pi/2)\\

But sin(π/2) = 1.

Therefore, using eq(2):

\frac{dx}{dt}(t_{maxV} ) = 0.100*\omega = 0.100\frac{\pi}{0.400} = \pi/4

And so:

V_{max} = \pi / 4 =0.785

6 0
2 years ago
A 50kg girl is playing soccer. She runs towards the ball at a speed of 7 m/s. How much kinetic energy does she have as she runs
stepladder [879]
The calculation for kinetic energy is this

KE = 1/2mv^2

KE = 1/2(50)(7^2)

KE = 1/2(49•50)

KE = 1225 kgm^2/s^2.

Or simply 1225 J.

She possess this much energy when she runs.
3 0
3 years ago
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A 1.11 kg piece of aluminum at 78.3 c is put into a glass with 0.210 kg of water at 15.0
ValentinkaMS [17]

M = mass of aluminium = 1.11 kg

c_{a} = specific heat of aluminium = 900

T_{ai} = initial temperature of aluminium = 78.3 c

m = mass of water = 0.210 kg

c_{w} = specific heat of water = 4186

T_{wi} = initial temperature of water = 15 c

T = final equilibrium temperature = ?

using conservation of heat

Heat lost by aluminium = heat gained by water

M c_{a} (T_{ai} - T) = m c_{w} (T - T_{wi} )

(1.11) (900) (78.3 - T) = (0.210) (4186) (T - 15)

T = 48.7 c

7 0
3 years ago
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