Ask question

Ask question

All categories

3 years ago

9

____ is the detection of halo objects like brown dwarfs when their gravitational fields concentrate the light of distant sources

(like galaxies).

1 answer:

maksim [4K]3 years ago

7 0

Answer:

Microlensing.

Explanation:

This techniques is called Microlensing.

Microlensing is a method of gravitational lensing where light from a backdrop point of origin is curved to develop distorted, numerous and/or lightened images by the gravity field of a foreground lens.

This method is very effective in discovering planets that are far-far from earth.It is actually an astronomical effect that was predicted by Albert Einstein's general theory of relativity.

You might be interested in

A car starts from rest and travels for 5.8 s with a uniform acceleration of 1.6 m/s² in the negative direction. What is the fina

elena-s [515]

Answer:

Final velocity of the car will be -9.28 m/sec

Explanation:

We have given that the car starts from the rest so initial velocity of the car u = 0 m /sec

Acceleration of the car $a=1.6m/sec^2$ in negative direction so acceleration will be $a=-1.6m/sec^2$

From first equation of motion we know that

v = u+at

So $v=0+(-1.6)\times 5.8=-9.28m/sec$

So final velocity will be -9.28 m/sec

8 0

3 years ago

1.)A tank travels at a rate of 10.0 km/hr for 12.00 minutes, then at 15.0 km/hr for 8.00

rosijanka [135]

12.00 min = 0.2 hr

8.00 min = 0.15 hr

Total distance:

(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) + (20.0 km/hr) (0.2 hr)

= 8.25 km

Average speed:

(10.0 km/hr + 15.0 km/hr + 20.0 km/hr) / 3

= 15 km/hr

Change in position:

(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) - (20.0 km/hr) (0.2 hr)

= 0.25 km

Average velocity:

(10.0 km/hr + 15.0 km/hr - 20.0 km/hr) / 3

≈ 1.67 m/s

8 0

2 years ago

Unpolarized light is incident upon two ideal polarizing filters that do not have their transmission axes aligned. If 19% of the

Zepler [3.9K]

Answer:

51.94°

Explanation:

$I_0$ = Unpolarized light

$I_2$ = Light after passing though second filter = $0.19I_0$

Polarized light passing through first filter

$I_1=\frac{I_0}{2}$

Polarized light passing through second filter

$I_2=\frac{I_0}{2}cos^2\theta\\\Rightarrow 0.19I_0=\frac{I_0}{2}cos^2\theta\\\Rightarrow cos^2\theta=\frac{0.19I_0}{\frac{I_0}{2}}\\\Rightarrow cos\theta=\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{0.19\times 2}\\\Rightarrow \theta=cos^{-1}\sqrt{0.38}\\\Rightarrow \theta=51.94^{\circ}$

The angle between the two filters is 51.94°

5 0

2 years ago

The potential energy stored in the compressed spring of a dart gun, with a spring constant of 32.50 N/m, is 0.640 J. Find by how

liraira [26]

Answer:

A

$x = 0.198456 \ m$

B

$h = 1.3061 \ m$

C

$v = 5.06 \ m/s$

D

$d = 4.0273 \ m$

Explanation:

Considering the first question

From the question we are told that

The spring constant is $k = 32.50 N/m$

The potential energy is $PE = 0.640 \ J$

Generally the potential energy stored in spring is mathematically represented as $PE = \frac{1}{2} * k * x^2$

=> $0.640= \frac{1}{2} * 32.50 * x^2$

=> $x = \sqrt{0.03938}$

=> $x = 0.198456 \ m$

Considering the second question

From the question we are told that

The mass of the dart is m = 0.050 kg

Generally from the law of energy conservation

$PE = mgh$

=> $0.640 = 0.050 * 9.8 * h$

=> $h = 1.3061 \ m$

Considering the third question

The height at which the dart was fired horizontally is $H = 3.90\ m$

Generally from the law of energy conservation

$PE = KE$

Here KE is kinetic energy of the dart which is mathematical represented as

$KE = \frac{1}{2} * mv^2$

=> $0.640 = \frac{1}{2} * 0.050 * v^2$

=> $v^2 = 25.6$

=> $v = 5.06 \ m/s$

Considering the fourth question

Generally the total time of flight of the dart is mathematically represented as

$t = \frac{ 2 * H }{g}$

=> $t = \frac{ 2 * 3.90 }{9.8 }$

=> $t = 0.7959 \ s$

Generally the horizontal distance from the equilibrium position to the ground is mathematically represented as

$d = v * t$

=> $d = 5.06 * 0.7959$

=> $d = 4.0273 \ m$

5 0

2 years ago

The car travels 52m in 21 seconds before it comes to a complete stop after the driver applies the brakes. What was the car's ini

prisoha [69]

Answer:

2.48 m/s

Explanation:

D=vt

52 = V 21

V= 52/21=2.47m/s

5 0

2 years ago