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3 years ago

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A particular radio station broadcast radio waves at 100 MHz (100 million Hz). If the radio waves travel at the speed of light (3

00 million m/s), then what is the wavelength of the radio waves that the station is broadcasting?

1 answer:

Svetlanka [38]3 years ago

4 0

Wavelength = (speed) / (frequency)

= (3 x 10⁸ m/s) / (1 x 10⁸ /s) = 3 meters

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Scientific _____ must be supported by observations and results many investigations and are not absolute

n200080 [17]

The correct answer you're looking for would be Theories.

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3 years ago

Salmon often jump waterfalls to reach their

natta225 [31]

Answer:

5.0 m/s

Explanation:

The horizontal motion of the salmon is uniform, so the horizontal component of the salmon's velocity is constant and it is

$v_x = u cos \theta$

where u is the initial speed and $\theta=37.7^{\circ}$ . The horizontal distance travelled by the salmon is

$d=v_x t = (ucos \theta)t$

where d = 1.95 m and t is the time needed to reach the final point.

Re-arranging for t,

$t=\frac{d}{v_x}=\frac{d}{u cos \theta}$ (1)

Along the vertical direction, the equation of motion is

$y=h+u_y t -\frac{1}{2}gt^2$

where:

y = 0.311 m is the final height reached by the salmon

h = 0 is the initial height

$u_y = u sin \theta$ is the vertical component of the initial velocity of the salmon

$g=9.81 m/s^2$ is the acceleration of gravity

t is the time

Substituting t as found in eq.(1), we get the equation

$y=(u sin \theta) \frac{d}{u cos \theta}- \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}$

and we can solve this formula for u, the initial speed of the salmon:

$y=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}\\\\u=\sqrt{\frac{gd^2}{2(dtan \theta -y)cos^2 \theta}}=\sqrt{\frac{(9.81)(1.95)^2}{2((1.95)(tan 37.7^{\circ}) -0.311)cos^2 37.7^{\circ}}}=5.0 m/s$

5 0

3 years ago

A projectile is launched from the ground with an initial velocity of 12ms at an angle of 30° above the horizontal. The projectil

netineya [11]

$vi^{2}sin2thita/g =12^{2}sin2[30]/9.8=12.7$ Answer:

Explanation:

range is given as

6 0

3 years ago

Read 2 more answers

Projectile Motion—A tennis ball is thrown out a window 28 m above the ground at an initial velocity of 15.0 m/s and 20.0° below

NNADVOKAT [17]

Answer:

The distance will be x = 41.7 [m]

Explanation:

We must first find the components in the x & y axes of the initial velocity.

$(v_{o})_{x} = 15*cos(20)= 14.09[m/s]\\(v_{o})_{y} = 15*sin(20)= 5.13[m/s]$

The acceleration is the gravity acceleration therefore.

g = 9.81 [m/s^2]

Now we can calculate how long it takes to fall.

$y=(v_{o})_{y}*t-0.5*g*t^2\\-28 = 5.13*t-0.5*9.81*t^2\\-28=-4.905*t^2+5.13*t\\4.905*t^2-5.13*t=28\\t = 2.96[s]$

With this time we can find the horizontal distance that runs the projectile.

$x=(v_{o})_{x}*t\\x=14.09*2.96\\x=41.7[m]$

5 0

3 years ago

Someone help me haha;(

Nikitich [7]

Answer: vf = 51 m/s

d = 112 m

Explanation: Solution attached:

To find vf we use acceleration equation:

a = vf - vi / t

Derive to find vf

vf = at + vi

Substitute the values

vf = 3.5 m/s² ( 8.0 s) + 23 m/s

= 51 m/s

To solve for distance we use

d = (∆v)² / 2a

= (51 m/s - 23 m/s )² / 2 ( 3.5 m/s²)

= (28 m/s)² / 7 m/s²

= 784 m/s / 7 m/s²

= 112 m

6 0

3 years ago