A physical property is a characteristic of a substance is one that ____

imagine we wanted to use telescopes to look for synestias in the same general part of the milky way galaxy where our solar syste

Naya [18.7K]

We might have a trouble viewing the general part of galaxy because : the Milky Way's center is so brilliant and filled with stars, it has been particularly challenging for astronomers to examine it because it is impossible to distinguish individual stars and clusters.

<h3>What is a Milky Way ?</h3>

There are many stars, grains of dust, and gas in the Milky Way. It is known as a spiral galaxy because, from the top or bottom, it would appear to be whirling like a pinwheel. About 25,000 light-years from the galaxy's nucleus, the Sun is situated on one of the spiral arms.

The Milky Way galaxy is made up of billions of stars, as well as gas and dust, which are all drawn to one another by gravitational pull, as well as a significant amount of dark matter. Our galaxy is approximately 100,000 light years [e1] across.

To know more about milky way galaxy you may visit the link :

brainly.com/question/2905713

#SPJ4

6 0

1 year ago

What force must the deltoid muscle provide to keep the arm in this position?

ruslelena [56]

Answer:

Deltoid Force, $F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}$

Additional Information:

Some numerical information are missing from the question. However, I will derive the formula to calculate the force of the deltoid muscle. All you need to do is insert the necessary information and calculate.

Explanation:

The deltoid muscle is the one keeping the hand arm in position. We have two torques that apply to the rotating of the arm.

1. The torque about the point in the shoulder for the deltoid muscle, $T_{Deltoid}$

2. The torque of the arm, $T_{arm}$

Assuming the arm is just being stretched and there is no rotation going on,

$T_{Deltoid}$ = 0

$T_{arm}$ = 0

⇒ $T_{Deltoid} = T_{arm}$

$r_{d}F_{d}sin\alpha_{d} = r_{a}F_{a}sin\alpha_{a}$

Where,

$r_{d}$ is radius of the deltoid

$F_{d}$ is the force of the deltiod

$\alpha_{d}$ is the angle of the deltiod

$r_{a}$ is the radius of the arm

$F_{a}$ is the force of the arm , $F_{a} = mg$ which is the mass of the arm and acceleration due to gravity

$\alpha_{a}$ is the angle of the arm

The force of the deltoid muscle is,

$F_{d} = \frac {r_{a}F_{a}sin\alpha_{a}}{r_{d}sin\alpha_{d}}$

but $F_{a} = mg$ ,

∴ $F_{d} = \frac {r_{a}mgsin\alpha_{a}}{r_{d}sin\alpha_{d}}$

7 0

3 years ago

A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each

ra1l [238]

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

$E_{1}+E_{3}-E_{2}=0$

We know that the electric field is:

$E=k\frac{q}{r^{2}}$

Where:

k is the Coulomb constant
q is the charge
r is the distance from the charge to the point

So, we have:

$k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0$

Let's solve it for r(3).

$\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0$

$r_{3}=0.0743\:$

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

3 0

3 years ago

A spherical balloon has a radius of 6.55 m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of a

Ksju [112]

Answer:

$M_1 = 317.7 kg$

Explanation:

Mass of the helium gas filled inside the volume of balloon is given as

$m = \rho V$

$m = 0.179(\frac{4}{3}\pi R^3)$

$m = 0.179(\frac{4}{3}\pi 6.55^3)$

$m = 210.7 kg$

now total mass of balloon + helium inside balloon is given as

$M = 210.7 + 990$

$M = 1200.7 kg$

now we know that total weight of balloon + cargo = buoyancy force on the balloon

so we will have

$(M + M_1)g = \rho_{air} V g$

$(1200.7 + M_1) = (\frac{4}{3}\pi 6.55^3) (1.29)$

$1200.7 + M_1 = 1518.4$

$M_1 = 317.7 kg$

3 0

3 years ago

Which of the following would cause the greatest decrease in gravitational force between the earth and the moon?

Marat540 [252]

Answer:

A decrease in the distance between the earth and the moon.

7 0

3 years ago

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A physical property is a characteristic of a substance is one that _____.