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8_murik_8 [283]

2 years ago

6

What is the wavelength of a 22.75×109Hz radar signal in free space? The speed of light is 2.9979×108m/s. Express your answer to

four significant figures and include the appropriate units.

2 answers:

Romashka-Z-Leto [24]2 years ago

8 0

Answer:

1.318 * 10^(-2) m

Explanation:

Parameters given:

Frequency, f = 22.75 * 10^9 Hz

Velocity, v = 2.9979 * 10^8 m/s

Wavelength is given as:

Wavelength = v/f

Wavelength = (2.9979 * 10^8) / (22.75 * 10^9)

Wavelength = 0.01318 m = 1.318 * 10^(-2) m

krek1111 [17]2 years ago

8 0

Answer:

wave length = 13.2 × 10⁻²m or 13.2cm

Explanation:

wavelength = speed of light / frequency

speed of light = 2.9979×10^8m/s

Frequency = 22.75×10^9Hz

wavelength= speed / frequency

$\frac{2.9979 \times10^8}{22.75 \times 10^9}\\\\= 13.2 \times10^-^2 m$

wave length = 13.2 × 10⁻²m or 13.2cm

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3 years ago

A motorcycle daredevil plans to ride up a 2.0-m-high, 20° ramp, sail across a 10-m-wide pool filled with hungry crocodiles, and

il63 [147K]

Answer:

He becomes a croco-dile food

Explanation:

From the question we are told that

The height is h = 2.0 m

The angle is $\theta = 20^o$

The distance is $w = 10m$

The speed is $u = 11 m/s$

The coefficient of static friction is $\mu = 0.02$

At equilibrium the forces acting on the motorcycle are mathematically represented as

$ma = mgsin \theta + F_f$

where $F_f$ is the frictional force mathematically represented as

$F_f =\mu F_x =\mu mgcos \theta$

where $F_x$ is the horizontal component of the force

substituting into the equation

$ma = mgsin \theta + \mu mg cos \theta$

$ma =mg (sin \theta + \mu cos \theta )$

making a the subject of the formula

$a = g(sin \theta = \mu cos \theta )$

substituting values

$a = 9.8 (sin(20) + (0.02 ) cos (20 ))$

$= 3.54 m/s^2$

Applying " SOHCAHTOA" rule we mathematically evaluate that length of the ramp as

$sin \theta = \frac{h}{l}$

making $l$ the subject

$l = \frac{h}{sin \theta }$

substituting values

$l = \frac{2}{sin (20)}$

$l = 5.85m$

Apply Newton equation of motion we can mathematically evaluate the final velocity at the end of the ramp as

$v^2 =u^2 + 2 (-a)l$

The negative a means it is moving against gravity

substituting values

$v^2 = (11)^2 - 2(3.54) (5.85)$

$v= \sqrt{79.582}$

$= 8.92m/s$

The initial velocity at the beginning of the pool (end of ramp) is composed of two component which is

Initial velocity along the x-axis which is mathematically evaluated as

$v_x = vcos 20^o$

substituting values

$v_x = 8.92 * cos (20)$

$= 8.38 m/s$

Initial velocity along the y-axis which is mathematically evaluated as

$v_y = vsin\theta$

substituting values

$v_y = 8.90 sin (20)$

$= 3.05 m/s$

Now the motion through the pool in the vertical direction can mathematically modeled as

$y = y_o + u_yt + \frac{1}{2} a_y t^2$

where $y_o$ is the initial height,

$u_y$ is the initial velocity in the y-axis

$a_y$ is the initial acceleration in the y axis with a constant value of ( $g = 9.8 m/s^2$ )

at the y= 0 which is when the height above ground is zero

Substituting values

$0 = 2 + (3.05)t - 0.5 (9.8)t^2$

The negative sign is because the acceleration is moving against the motion

$-(4.9)t^2 + (2.79)t + 2m = 0$

Solving using quadratic formula

$\frac{-b \pm \sqrt{b^2 -4ac} }{2a}$

substituting values

$\frac{-3.05 \pm \sqrt{(3.05)^2 - 4(-4.9) * 2} }{2 *( -4.9)}$

$t = \frac{-3.05 + 6.9}{-9.8} \ or t = \frac{-3.05 - 6.9}{-9.8}$

$t = -0.39s \ or \ t = 1.02s$

since in this case time cannot be negative

$t = 1.02s$

At this time the position the motorcycle along the x-axis is mathematically evaluated as

$x = u_x t$

x $=8.38 *1.02$

$x =8.54m$

So from this value we can see that the motorcycle would not cross the pool as the position is less that the length of the pool

7 0

3 years ago