Answer:
θ’ = θ₀ / 2
we see that the resolution angle is reduced by half
Explanation:
The resolving power of a radar is given by diffraction, for which we will use the Rayleigh criterion for the resolution of two point sources, they are considered resolved if the maximum of diffraction of one coincides with the first minimum of the other.
The first minimum occurs for m = 1, so the diffraction equation of a slit remains
a sin θ = λ
in general, the diffraction patterns occur at very small angles, so
sin θ = θ
θ = λ / a
in the case of radar we have a circular aperture and the equation must be solved in polar coordinates, which introduces a numerical constant.
θ = 1.22 λ /a
In this exercise we are told that the opening changes
a’ = 2 a
we substitute
θ ‘= 1.22 λ / 2a
θ' = (1.22 λ / a) 1/2
θ’ = θ₀ / 2
we see that the resolution angle is reduced by half
22. a - (vf^2 - vi^2)/(2d)
a = (0 - 23^2)/(170)
a = -3.1 m/s^2
23. Find the time (t) to reach 33 m/s at 3 m/s^2
33-0/t = 3
33 = 3t
t = 11 sec to reach 33 m/s^2
Find the av velocuty: 33+0/2 = 16.5 m/s
Dist = 16.5 * 11 = 181.5 meters to each 33m/s speed. Runway has to be at least this long.
24. The sprinter starts from rest. The average acceleration is found from:
(Vf)^2 = (Vi)^2 = 2as ---> a = (Vf)^2 - (Vi)^2/2s = (11.5m/s)^2-0/2(15.0m) = 4.408m/s^2 estimated: 4.41m/s^2
The elapsed time is found by solving
Vf = Vi + at ----> t = vf-vi/a = 11.5m/s-0/4.408m/s^2 = 2.61s
25. Acceleration of car = v-u/t = 0ms^-1-21.0ms^-1/6.00s = -3.50ms^-2
S = v^2 - u^2/2a = (0ms^-1)^2-(21.0ms^-1)^2/2*-3.50ms^-2 = 63.0m
26. Assuming a constant deceleration of 7.00 m/s^2
final velocity, v = 0m/s
acceleration, a = -7.00m/s^2
displacement, s - 92m
Using v^2 = u^2 - 2as
0^2 - u^2 + 2 (-7.00) (92)
initial velocity, u = sqrt (1288) = 35.9 m/s
This is the speed pf the car just bore braking.
I hope this helps!!
Answer:
TRUE
Explanation:
Balance forces are usually defined as the two distinct force that acts on an object but in opposite directions. These two acting forces are equal in size or magnitude. When this type of force is applied on any object, it signifies that the object is stationary or it is moving at a constant speed and in the same direction.
This force is comprised of two most important properties namely the strength and direction. When any of the two forces is higher then it result in the motion of the object.
Thus, the above given statement is TRUE.
The velocity of the cannonball is 150 m/s, the right option is B. 150 m/s.
The question can be solved, using Newton's second law of motion.
Note: Momentum of the cannon = momentum of the cannonball.
<h3>
Formula:</h3>
- MV = mv................. Equation 1
<h3>Where:</h3>
- M = mass of the cannon
- m = mass of the cannonball
- V = velocity of the cannon
- v = velocity of the cannonball
Make v the subject of the equation.
- v = MV/m................ Equation 2
From the question,
<h3>Given: </h3>
- M = 500 kg
- V = 3 m/s
- m = 10 kg.
Substitute these values into equation 2.
- v = (500×3)/10
- v = 150 m/s.
Hence, The velocity of the cannonball is 150 m/s, the right option is B. 150 m/s.
Learn more about Newton's second law here: brainly.com/question/25545050
I think it is 500 cm. Hope I helped!
