The Event Horizon Telescope needs a 22 micro-arcsecond resolution to view the event horizon regions around black holes. If the a
1 answer:
Answer:
14869817.395 m
Explanation:
=22 microarcsecond
λ = Wavelength = 1.3 mm
Converting to radians we get
From Rayleigh Criterion
Diameter of the effective primary objective is 14869817.395 m
It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.
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Answer:
k = 40 N/m
Explanation:
A spring's energy is given:
U is the energy in the spring, k is the spring constant and x is the spring displacement.
We are told that the spring stores 5J of energy, therefore, U = 5J. We are also told that the spring is compressed by 0.5m, so the spring x = 0.5m
k = 40 N/m
Hope this helps!
Answer:
His launching angle was 14.72°
Explanation:
Please, see the figure for a graphic representation of the problem.
In a parabolic movement, the velocity and displacement vectors are two-component vectors because the object moves along the horizontal and vertical axis.
The horizontal component of the velocity is constant, while the vertical component has a negative acceleration due to gravity. Then, the velocity can be written as follows:
v = (vx, vy)
where vx is the component of v in the horizontal and vy is the component of v in the vertical.
In terms of the launch angle, each component of the initial velocity can be written using the trigonometric rules of a right triangle (see attached figure):
sin angle = opposite / hypotenuse
cos angle = adjacent / hypotenuse
In our case, the side opposite the angle is the module of v0y and the side adjacent to the angle is the module of vx. The hypotenuse is the module of the initial velocity (v0). Then:
sin angle = v0y / v0 then: v0y = v0 * sin angle
In the same way for vx:
vx = v0 * cos angle
Using the equation for velocity in the x-axis we can find the equation for the horizontal position:
dx / dt = v0 * cos angle
dx = (v0 * cos angle) dt (integrating from initial position, x0, to position at time t and from t = 0 and t = t)
x - x0 = v0 t cos angle
x = x0 + v0 t cos angle
For the displacement in the y-axis, the velocity is not constant because the acceleration of the gravity:
dvy / dt = g ( separating variables and integrating from v0y and vy and from t = 0 and t)
vy -v0y = g t
vy = v0y + g t
vy = v0 * sin angle + g t
The position will be:
dy/dt = v0 * sin angle + g t
dy = v0 sin angle dt + g t dt (integrating from y = y0 and y and from t = 0 and t)
y = y0 + v0 t sin angle + 1/2 g t²
The displacement vector at a time "t" will be:
r = (x0 + v0 t cos angle, y0 + v0 t sin angle + 1/2 g t²)
If the launching and landing positions are at the same height, then the displacement vector, when the object lands, will be (see figure)
r = (x0 + v0 t cos angle, 0)
The module of this vector will be the the total displacement (65 m)
module of r =
65 m = x0 + v0 t cos angle ( x0 = 0)
65 m / v0 cos angle = t
Then, using the equation for the position in the y-axis:
y = y0 + v0 t sin angle + 1/2 g t²
0 = y0 + v0 t sin angle + 1/2 g t²
replacing t = 65 m / v0 cos angle and y0 = 0
0 = 65m (v0 sin angle / v0 cos angle) + 1/2 g (65m / v0 cos angle)²
cancelating v0:
0 = 65m (sin angle / cos angle) + 1/2 g * (65m)² / (v0² cos² angle)
-65m (sin angle / cos angle) = 1/2 g * (65m)² / (v0² cos² angle)
using g = -9.8 m/s²
-(sin angle / cos angle) * (cos² angle) = -318.5 m²/ s² / v0²
sin angle * cos angle = 318.5 m²/ s² / (36 m/s)²
(using trigonometric identity: sin x cos x = sin (2x) / 2
sin (2* angle) /2 = 0.25
sin (2* angle) = 0.49
2 * angle = 29.44
<u>angle = 14.72°</u>
