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Katen [24]
3 years ago
12

Some radar systems detect the size and shape of objects such as aircraft and geological terrain. What is the frequency of such a

system which can detect objects as small as 19.1 cm?
Physics
1 answer:
vova2212 [387]3 years ago
4 0

Answer:

f=1.57\times 10^9\ Hz

Explanation:

Given that,

A system can detect objects as small as 19.1 cm i.e. 0.191 m. It is the wavelength.

We know that,

Frequency, f=\dfrac{c}{\lambda}

So,

f=\dfrac{3\times 10^8}{0.191}\\\\=1.57\times 10^9\ Hz

So, the frequency of such a system is equal to1.57\times 10^9\ Hz.

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If you are 5.00ft and 10.0in tall, what is your height in meters
Stolb23 [73]

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Hope this helped!

Nate

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If someone weighs 80n on Mars and comes back on earth will they weigh the same
Alexandra [31]

Answer:

No

Explanation:

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8 0
2 years ago
Read 2 more answers
A giraffe, standing 6.00 m tall, bites a branch off a tree to chew on the leaves
Lemur [1.5K]

Answer:

1.11 s.

Explanation:

From the question given above, the following data were obtained:

Height (H) = 6 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =.?

The time taken for the branch to hit the ground can be obtained as follow:

H = ½gt²

6 = ½ × 9.8 × t²

6 = 4.9 × t²

Divide both side by 4.9

t² = 6/4.9

Take the square root of both side

t = √(6/4.9)

t = 1.11 s

Therefore, it will take 1.11 s for the branch to hit the ground.

6 0
3 years ago
What is the product of mass and velocity called?
Crank
The product of mass and velocity is momentum. P=mv. The 'P' in the equation stands for momentum.
3 0
3 years ago
Male Rana catesbeiana bullfrogs are known for their loud mating call. The call is emitted not by the frog's mouth but by its ear
Sidana [21]

Answer:

The amplitude of the eardrum's oscillation is 6.65×10^-13 m.

Explanation:

Given data:

The sound has a frequency of 262 Hz

The sound level is 84 dB

The air density is 1.21 kg/m^3

The speed of sound is 346 m/s

Solution:

As, Intensity of sound is given by,

I = Io×10^(s/10 db)

I = 2×π^2×ρ×v×f^2×Sm^2

Thus,

Sm = √(Io×10^(s/10 db)) / √( 2×π^2×ρ×v×f^2)

Now, put the values,

Sm = √( 10^-12 × 10^(84/10) ) / √( 2×(3.14)^2×1.21×346×(262)^2 )

Sm = √(2.51×10^-4 / 5.66×10^8)

Sm = √0.443×10^-12

Sm = 6.65×10^-13 m.

8 0
3 years ago
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