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Oliga [24]
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).
Physics
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.

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if two gliders of equal mass m and equal and opposite initial velocity v collide perfectly elastically, using both the momentum
Inga [223]

Answer:

initial kinetic energy = final kinetic energy = m · v²

Explanation:

Hi there!

Since the gliders collide elastically, the kinetic energy and momentum of the system after the collision are the same as before the collision.

Then, the initial kinetic energy of the system will be equal to the final kinetic energy of the system.

The equation of kinetic energy for each glider is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the glider.

v =velocity of the glider.

The sum of the kinetic energies of each glider is the kinetic energy of the system. Then, the initial kinetic energy of the system will be:

initial KE = 1/2 · m · v² + 1/2 · m · v²

initial KE = m · v²

Since the initial kinetic energy of the system is equal to the final kinetic energy of the system:

final KE = m · v²

Using the equation of momentum of the system:

initial momentum = m · v + m · (-v) = m (v-v) = 0

The initial and final momentum of the system is zero because both vectors cancel each other for being of the same magnitude but of opposite direction.

4 0
4 years ago
a person, sunbathing on a warm day, is lying horizontally on the deck of a boat. her mass is 59.5 kg, and the coefficient of sta
Setler [38]

a) since a=0 therefore, friction force =0

b) friction = 59.5 × 1 =59.5 N.

c)maximum acceleration = 0.566 × g =5.54 m/s².

<h3>What is Friction force?</h3>

The force preventing sliding against one another of solid surfaces, fluid layers, and material components is known as friction. There are various kinds of friction: Two solid surfaces in touch are opposed to one another's a relative lateral motion by dry friction.

<h3>What is acceleration?</h3>

Acceleration is the rate at which an object's velocity with respect to time changes. They are vector quantities. The direction of the net force acting on an object determines the direction of its acceleration.

Hence, (a) since a=0 therefore, friction force =0, (b) friction = 59.5 × 1 =59.5 N, (c)maximum acceleration = 0.566 × g =5.54 m/s².

To know more about Friction, check out:

brainly.com/question/24338873

#SPJ4

4 0
1 year ago
A scientist is examining an unknown solid. which procedure would most likely help determine a chemical property of the substance
asambeis [7]

Answer:

exposing it to a flame to see if it catches on fire

Explanation:

A chemical property is when the matter changes into a different substance. This is true when you expose it to a flame because if a substance is burned, it will change into a new substance. All the other options are examples of physical changes.

-I also took the quiz and got 100%

Hope this helps!

7 0
3 years ago
Un reloj de péndulo de largo L y período T, aumenta su largo en ΔL (ΔL &lt;&lt; L). Demuestre que su período aumenta en: ΔT = π
Kruka [31]

Answer:

 ΔT = \pi \ \frac{\Delta L}{\sqrt{Lg} }

Explanation:

In a simple harmonic motion, specifically in the simple pendulum, the angular velocity

          w = \sqrt{\frac{g}{L} }

angular velocity and period are related

          w = 2π / T

we substitute

          2π / T = \sqrt{\frac{g}{L} }

          T = 2\pi  \ \sqrt{\frac{L}{g} }

In this exercise indicate that for a long Lo the period is To, then and increase the long

          L = L₀ + ΔL

we substitute

           T = 2\pi  \ \sqrt{\frac{L + \Delta L}{g} }

            T = 2\pi  \ \sqrt{\frac{L}{g} } \ \sqrt{1+ \frac{\Delta L}{L} }

in general the length increments are small ΔL/L «1, let's use a series expansion

           \sqrt{1+ \ \frac{\Delta L}{L} } = 1 + \frac{1}{2} \frac{\Delta L}{L} + ...  

we keep the linear term, let's substitute

           T = 2\pi  \ \sqrt{\frac{L}{g} } \ ( 1 + \frac{1}{2} \frac{\Delta L}{L}  )  

if we do

           T = T₀ + ΔT

           

           T₀ + ΔT = 2\pi  \sqrt{\frac{\Delta L}{g} }  + \pi  \ \sqrt{\frac{L}{g} } \ \frac{\Delta L}{L}

            T₀ + ΔT = T₀ + \pi  \sqrt{\frac{1}{Lg} } \ \Delta L

            ΔT = \pi \ \frac{\Delta L}{\sqrt{Lg} }

4 0
3 years ago
What is the kinetic energy of a cat running 5 m/s
dusya [7]

The cat's kinetic energy is

                   (1/2) x (the cat's mass in kg) x (25 m²/sec²).

The unit is [joules] .

4 0
3 years ago
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