In what ways is the earth constantly being worn down

In order to effectively radiate a radio signal, an antenna must be at least one-half the wavelength of transmission in length, i

Virty [35]

<span>The equation that relates Wavelength & frequency of electromagnetic waves is : Velocity (c) = Wavelength (λ) * frequency (f) ------ (1) Electromagnetic wave velocity (c) = Speed of Light = 3 * 10^8 m/s From (1), Wavelength , λ = c / f λ= (3*10^8)/(980*10^3) λ = 306.12 m Minimum height of the antenna for effective transmission = λ * 0.5 = 306.12*0.5 Answer : 153 m (rounded off)</span>

6 0

3 years ago

Imagine you could travel to the moon where the acceleration due to gravity is 1.6 m/s^2. What would be the period of

VladimirAG [237]

Answer:

4.9612 s

Explanation:

Applying,

T = 2π√(L/g)............... Equation 1

Where T = period of the pendulum, L = Lenght of the pendulum, g = acceleration due to gravity of the moon, π = pie.

From the question,

Given: L = 1 m, g = 1.6 m/s²

Constant: π = 3.14

Substitute these values into equation 1

T = 2×3.14×√(1/1.6)

T = 6.28√(0.625)

T = 6.28×0.79

T = 4.9612 s

6 0

3 years ago

What statement best describes the kind of change that could have taken place

Anarel [89]

In order to answer the question, we need to know what was there before
AND what was there after. Is there a reason why you're concealing that
information ?

While you're at it, you might pass along the answer choices too.

5 0

4 years ago

A player kicks a football from ground level with a velocity of 26.2m/s at an angle of 34.2° above the horizontal. How far back f

Amanda [17]

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.

For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

<h3>Explanation</h3>

How long does it take for the ball to reach the goal?

Let the distance between the kicker and the goal be $x$ meters.

Horizontal velocity of the ball will always be $26.2\times\cos{34.2\textdegree}$ until it lands if there's no air resistance.

The ball will arrive at the goal in $\displaystyle \frac{x}{26.2\times\cos{34.2\textdegree}}$ seconds after it leaves the kicker.

What will be the height of the ball when it reaches the goal?

Consider the equation

$\displaystyle h(t) = -\frac{1}{2}\cdot g\cdot t^{2} + v_{0,\;\text{vertical}} \cdot t + h_0$ .

For this soccer ball:

$g = 9.81\;\text{m}\cdot\text{s}^{-2}$ ,
$v_{0,\;\text{vertical}} = 26.2\times \sin{34.2\textdegree{}}\;\text{m}\cdot\text{s}^{-2}$ ,
$h_0 = 0$ since the player kicks the ball "from ground level."

$\displaystyle t=\frac{x}{26.2\times\cos{34.2\textdegree}}$

when the ball reaches the goal.

$\displaystyle h= - 9.81 \times \frac{x^2}{(26.2\times\cos{34.2\textdegree})^2} + (26.2 \times \sin{34.2\textdegree})\times\frac{x}{26.2\times\cos{34.2\textdegree}} \\\phantom{h} = -\frac{9.81}{(26.2\times\cos{34.2\textdegree})^2}\cdot x^{2} + \frac{\sin{34.2\textdegree}}{\cos{34.2\textdegree}}\cdot x$ .

Solve this quadratic equation for $x$ , $x > 0$ .

$x = 65.1$ meters when $h = 0$ meters.
$x = 6.54$ or $58.5$ meters when $h = 4$ meters.

In other words,

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.
For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

3 0

4 years ago

The frequency of a wave is 200 Hz. The wavelength is 0.1 m. What is the period of the wave?

Aleks04 [339]

The formula for the period of wave is: wave period is equals to 1 over the frequency. $waveperiod=\frac{1}{frequency}$
To get the value of period of wave you need to divide 1 by 200 Hz. However, beforehand, you have to convert 200 Hz to cycles per second. So that would be, 200 cyles per second or 200/s.
By then, you can start the computation by dividing 1 by 200/s. Since 200/s is in fractional form, you have to find its reciprocal form and multiply it to one which would give you 1 (one) second over 200. This would then lead us to the value 0.005 seconds as the wave period.

wave period= 1/200 Hz
Convert Hz to cycles per second first
200 Hz x 1/s= 200/second
Make 200/second as your divisor, so:

wave period= 1/ 200/s

get the reciprocal form of 200/s which is s/200

then you can start the actual computation:

wave period= 1 x s divided by 200

this would give us an answer of 0.005 s.

6 0

3 years ago