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Galina-37 [17]
3 years ago
13

Your destroyer has a RADAR antenna height of 40 m. Using RADAR, what is the maximum detection range of a patrol boat with a mast

height of 11 m above the water?
Physics
1 answer:
Lisa [10]3 years ago
6 0

Answer:

The maximum detection range is 39.75 km

Explanation:

Given that;

Antenna height h1 = 40 m

Target height  ( patrol boat mast ) h2 = 11 m

Using RADAR, what is the maximum detection range = ?

Using RADAR

we know that; Maximum detection range = (√17h1 + √17h2) km

where h1 and h2 are heights of the antenna and target height in meters

so we substitute in our values

Maximum detection range = (√(17 × 40) + √(17 × 11)) km

Maximum detection range = (√680 + √187) km

Maximum detection range = (26.0768 + 13.6747) km

Maximum detection range = (26.0768 + 13.6747) km

Maximum detection range = 39.75 km

Therefore, The maximum detection range is 39.75 km

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A kangaroo jumps straight up to a vertical height of 1.66 m. How long was it in the air before returning to Earth? Express your
Nataly [62]

Answer:

The kangaroo was 1.164s in the air before returning to Earth

Explanation:

For this we are going to use the equation of distance for an uniformly accelerated movement, that is:

x = x_{0} + V_{0}t + \frac{1}{2}at^2

Where:

x = Final distance

xo = Initial point

Vo = Initial velocity

a = Acceleration

t = time

We have the following values:

x = 1.66m      

xo = 0m (the kangaroo starts from the floor)

Vo = 0 m/s (each jump starts from the floor and from a resting position)

a = 9.8 m/s^2 (the acceleration is the one generated by the gravity of earth)

t =This is just the time it takes to the kangaoo reach the 1.66m, we don't know the value.

Now replace the values in the equation

x = x_{0} + V_{0}t + \frac{1}{2}at^2

1.66 = 0 + 0t + \frac{1}{2}9.8t^2

1.66 = 4.9t^2

\frac{1.66}{4.9}  = t^2

\sqrt{0.339} = t\\ t = 0.582s

It takes to the kangaroo 0.582s to go up and the same time to go down then the total time it is in the air before returning to earth is

t = 0.582s + 0.582s

t = 1.164s

The kangaroo was 1.164s in the air before returning to Earth

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