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Tju [1.3M]
3 years ago
11

Radar uses radio waves of a wavelength of 2.2 m . The time interval for one radiation pulse is 100 times larger than the time of

one oscillation; the time between pulses is 10 times larger than the time of one pulse. What is the shortest distance to an object that this radar can detect?
Physics
1 answer:
liq [111]3 years ago
8 0

Answer:

<em>The shortest distance to an object that this radar can detect would be </em>

<em>111 m</em>

Explanation:

The shortest distance is the minimum distance that would be detected by the radar, The time of oscillation can be obtained thus;

T = 1/f   ..........1

but f= v/λ substituting f into equation 1 we have;

T = λ/v...............2

Where λ is the wavelength = 2.2 m

v is the velocity of  light  since the radar is an electromagnetic wave

   = 3 x10^{8} m/s

T = 2.2 m / 3 x10^{8} m/s

T = 7.33 x10^{-9} s

<u>Calculating the time interval required by the pulse</u>

Since the interval of the pulse (t) is 100 times greater than the period of oscillation, the time of the pulse is expressed as;

t = 100 x T

t = 100 x 7.33 x10^{-9} s

t = 7.33 x 10^{-7} s

Therefor the time of the pulse is 7.33 x 10^{-7} s

<u>Calculating for the shortest distance of the radar</u>

The shortest distance of the radar can be obtained using the equation below;

λ_s = (t/2) x v ............3

Substituting into equation 3 we have

λ_s = (7.33 x 10^{-7} /2) x 3 x10^{8} m/s

λ_s  =  111 m

Therefore the shortest distance to an object that this radar can detect would be 111 m

   

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T₂ = 10.19N

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• The mass of the ball, m = 1.8kg

First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,

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\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}

The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,

\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}

We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,

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Solve for T₁,

T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N

Now, we use the second equation to find the tension in the horizontal string,

T_2-T_1\sin 30\degree=0

Solve for T₂,

T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N

Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.

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