All categories

3 years ago

1) How can astronomers determine the chemical composition of stars?

2 answers:

8 0

1. The astronomers can determine the chemical composition of stars by analyzing their emission and absorption as the spectra of the sun and the stars exhibits bright and dark lines are these can be caused by the elements absorbing and emitting light at the specific wavelengths,

2. All objects emit radiation as a result of the growing temperature of the charged particles within the objects as the amount of radiation emitted depends on the temperature of the object.

3. The main components of the Milky Way are:

Disk, bulge and halo.

4. The main difference between the atmospheres of Jupiter and Neptune is that Jupiter has methane clouds.

5. The techniques used to find the exosolar planets are based on An analysis of light.

tatiyna3 years ago

6 0

1). B
2). none of those choices satisfies me
3). B
4). A
5). D

You might be interested in

When should scientific claims be questioned?

Kryger [21]

Answer:

C) if they are based on a very small sample

4 0

2 years ago

True or false: You can calculate a three variable equation knowing only one of the variables.

makvit [3.9K]

Answer:

false

Explanation:

you cant calculate a three variable equation knowing only one of the variables

3 0

3 years ago

A projectile of mass 2.0 kg is fired in the air at an angle of 40.0 ° to the horizon at a speed of 50.0 m/s. At the highest poin

tekilochka [14]

Answer:

a) The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

0.7 kg is 109.4 ms on the x axis

b) Y = 109.3 m

Explanation:

This is a moment and projectile launch exercise.

a) Let's start by finding the initial velocity of the projectile

sin 40 = voy / v₀

$v_{oy}$ = v₀ sin 40

$v_{oy}$ = 50.0 sin40

$v_{oy}$ = 32.14 m / s

cos 40 = v₀ₓ / V₀

v₀ₓ = v₀ cos 40

v₀ₓ = 50.0 cos 40

v₀ₓ = 38.3 m / s

Let us define the system as the projectile formed t all fragments, for this system the moment is conserved in each axis

Let's write the amounts

Initial mass of the projectile M = 2.0 kg

Fragment mass 1 m₁ = 1.0 kg and its velocity is vₓ = 0 and $v_{y}$ = -10.0 m / s

Fragment mass 2 m₂ = 0.7 kg moves in the x direction

Fragment mass 3 m₃ = 0.3 kg moves up (y axis)

Moment before the break

X axis

p₀ₓ = m v₀ₓ

Y Axis y

$p_{oy}$ = 0

After the break

X axis

$p_{fx}$ = m₂ v₂

Axis y

$p_{fy}$ = m₁ v₁ + m₃ v₃

Let's write the conservation of the moment and calculate

Y Axis

0 = m₁ v₁ + m₃ v₃

Let's clear the speed of fragment 3

v₃ = - m₁ v₁ / m₃

v₃ = - (-10) 1 / 0.3

v₃ = 33.3 m / s

X axis

M v₀ₓ = m₂ v₂

v₂ = v₀ₓ M / m₂

v₂ = 38.3 2 / 0.7

v₂ = 109.4 m / s

The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

0.7 kg is 109.4 ms on the x axis

b) The speed of the fragment is 33.3 m / s and has a starting height of where the fragmentation occurred, let's calculate with kinematics

$v_{fy}$ ² = $v_{oy}$ ² - 2 gy

0 = $v_{oy}$ ²-2gy

y = $v_{oy}$ ² / 2g

y = 32.14² / 2 9.8

y = 52.7 m

This is the height where the break occurs, which is the initial height for body movement of 0.3 kg

$v_{f}$ ² = $v_{y}$ ² - 2 g y₂

0 = $v_{y}$ ² - 2 g y₂

y₂ = $v_{y}$ ² / 2g

y₂ = 33.3²/2 9.8

y₂ = 56.58 m

Total body height is

Y = y + y₂

Y = 52.7 + 56.58

Y = 109.3 m

8 0

2 years ago

A block with mass ma = 14.0 kg on a smooth horizontal surface is connected by a thin cord that passes over a pulley to a second

Drupady [299]

<span>We can assume that the horizontal surface has no friction and the pulley is massless. We can use Newton's second law to set up an equation. F = Ma F is the net force M is the total mass of the system a is the acceleration a = F / M a = (mb)(g) / (ma + mb) a = (6.0 kg)(9.80 m/s^2) / (6.0 kg + 14.0 kg) a = 58.8 N / 20 kg a = 2.94 m/s^2 The magnitude of the acceleration of the system is 2.94 m/s^2</span>

7 0

3 years ago

A force is applied to a block sliding along a surface (Figure 2). The magnitude of the force is 15 N, and the horizontal compone

Lemur [1.5K]

Answer:

Fy = 14.3 [N]

Explanation:

To be able to solve this problem we must know that the force is a vector and has magnitude and direction, therefore it can be decomposed into the force in the X & y components:

When we have the components on the horizontal and vertical axes we must use the Pythagorean theorem.

$F = \sqrt{F_{x}^{2} +F_{y}^{2} }$

where:

F = 15 [N]

Fx = horizontal component = 4.5 [N]

Fy = vertical component [N]

$15=\sqrt{4.5^{2}+F_{y}^{2}}\\ 15^{2}= (\sqrt{4.5^{2}+F_{y}^{2}})^{2} \\225 = 4.5^{2}+F_{y} ^{2}\\ F_{y}^{2} =225 -4.5^{2}\\ F_{y}^{2}=204.75\\F_{y}=\sqrt{204.75}\\ F_{y}=14.3 [N]$

7 0

3 years ago