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3 years ago

How can astronomers detect a bianary star if only one of the two stars is visible from earth

Physics

1 answer:

Svet_ta [14]3 years ago

6 0

Answer:

The star would continually shift its position from apparently stationary to

reflect the motion of the invisible star around the visible star.

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Cars driving 15.7 m/s on a flat ground drives off of 25.3 m High cliff how far from the base of the cliff does it land

MA_775_DIABLO [31]

The car lands 35.6 m from the base of the cliff

Explanation:

The motion of the car is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction

We start by analyzing the vertical motion, in order to find the time of flight of the car. We do it by using the following suvat equation:

$s=u_y t+\frac{1}{2}at^2$

where:

s = 25.3 m is the vertical displacement of the car (the height of the cliff)

$u_y=0$ is the initial vertical velocity of the car (because it is moving horizontally)

t is the time of the fall

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t, we find :

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(25.3)}{9.8}}=2.27 s$

Now we can analyze the horizontal motion of the car, which moves with constant horizontal velocity of

$v_x = 15.7 m/s$

Therefore, the horizontal distance covered in a time t is

$d=v_x t$

and by substituting t = 2.27 s, we find the how far from the base of the cliff the car lands:

$d=(15.7)(2.27)=35.6 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

7 0

3 years ago

How deep can an object with 6360N hitting on a sponge get ???

borishaifa [10]

Answer:

The sponge must go $\dfrac{636}{m}\ \text{meter}$ deep

Explanation:

If F = 6360 N, then it is required to find how deep can an object with this force hitting on a sponge get.

We know that, F = mgh

m is mass

g is acceleration due to gravity

$h=\dfrac{F}{mg}\\\\h=\dfrac{6360}{10m}\\\\h=\dfrac{636}{m}\ \text{meter}$

So, the sponge must go $\dfrac{636}{m}\ \text{meter}$ deep.

5 0

4 years ago

Please help in physics

Minchanka [31]

As there is no postive or negative assigned so
Initial velocity= -2.8759
Displacement= 0.5at^2+ut
= 0.5(-1.77)(3.33)^2+(-2.8759)(3.33)=-19.4m

4 0

3 years ago

answer key to Student Exploration: Feed the Monkey (Projectile Motion) page 3 Activity B Question: As the banana flies through s

murzikaleks [220]

As the banana flies through space, up the red arrow showing its path of motion takes a curved or parabolic shape.

<h3>What is a projectile?</h3>

A projectile is an object launched into space at an angle to the horizontal ground and allowed to fall freely under gravity.

The path of motion of a projectile is parabolic.

Examples of objects that are projectiles include:

A thrown banana
A kicked football
An arrow shot from a bow

Therefore, as the banana flies through space, up the red arrow showing its path of motion takes a curved or parabolic shape.

Learn more about projectiles at: brainly.com/question/24216590

6 0

3 years ago

In an experiment, a rectangular block with height h is allowed to float in two separate liquids. In the first liquid, which is w

Amiraneli [1.4K]

Answer:

The relative density of the second liquid is 7.

Explanation:

From archimede's principle we know that the force that a liquid exerts on a object equals to the weight of the liquid that the object displaces.

Let us assume that the volume of the object is 'V'

Thus for the liquid in which the block is completely submerged

The buoyant force should be equal to weight of liquid

Mathematically

$F_{buoyant}=Weight\\\\\rho _{1}\times V\times g=m\times g\\\\\therefore \rho _{1}=\frac{m}{V}...............(i)$

Thus for the liquid in which the block is 1/7 submerged

The buoyant force should be equal to weight of liquid

Mathematically

$F'_{buoyant}=Weight\\\\\rho _{2}\times \frac{V}{7}\times g=m\times g\\\\\therefore \rho _{2}=\frac{7m}{V}.................(ii)$

Comparing equation 'i' and 'ii' we see that

$\rho_{2}=7\times \rho _{1}$

Since the first liquid is water thus $\rho _{1}=1gm/cm^3$

Thus the relative density of the second liquid is 7.

6 0

3 years ago

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How can astronomers detect a bianary star if only one of the two stars is visible from earth￼

How can astronomers detect a bianary star if only one of the two stars is visible from earth