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djverab [1.8K]
3 years ago
8

The radar system of a navy cruiser transmits at a wavelength of 1.4 cm, from a circular antenna with a diameter of 2.7 m. At a r

ange of 7.8 km, what is the smallest distance in meters that two speedboats can be from each other and still be resolved as two separate objects by the radar system?
Physics
1 answer:
Goshia [24]3 years ago
4 0

Answer:

Distance will be 49.34 m

Explanation:

We have given wavelength \lambda =1.4cm =0.014m

Diameter of the antenna d = 2.7 m

Range L = 7.8 km = 7800 m

We have to find the smallest distance hat two speedboats can be from each other and still be resolved as two separate objects D

We know that distance is given by D=\frac{L1.22\lambda }{d}=\frac{7800\times 1.22\times 0.014}{2.7}=49.3422m

So distance D will be 49.34 m

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Ulleksa [173]
Distance traveled by him = circumference of that circular path = 2πr = 2π(3.5)
= 7π = 7×3.14 = 21.98 m
time = 8.9 s  [ Given ]

Now, Average speed = distance / time
s = 21.98 / 8.9
s = 2.46 m/s

Hope this helps!
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3 years ago
Why was Earth very hot at the beginning of its formation? Select all that apply.
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Answer:

heat came from the supernova that led to the formation of solar system

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A lightning bolt with 13 kA strikes an object for 14 μ s. How much charge is deposited on the object?
pogonyaev

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0.182C

Explanation:

Using Q= It

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3 years ago
The Great Sandini is a 60 kg circus performer who is shotfrom a cannon (actually a spring gun). You don't find many men ofhis ca
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Answer:

V=15.3 m/s

Explanation:

To solve this problem, we have to use the energy conservation theorem:

U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}

the elastic potencial energy is given by:

U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J

The work is defined as:

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this work is negative because is opposite to the movement.

The gravitational potencial energy at 2.5 m aboves is given by:

U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s

3 0
3 years ago
What would the person’s mass be on the earth? Part B pls
Dafna1 [17]

Answer:

I hopes it helps

thank you

4 0
3 years ago
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