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3 years ago

Why do astronomers think that the milky way is a spiral galaxy?

1 answer:

5 0

1) When you look toward the Galactic Center with your eye, you see a long, thin strip. This suggests a disk seen edge-on, rather than a ellipsoid or another shape. We can also detect the bulge at the center. Since we see spiral galaxies which are disks with central bulges, this is a bit of a top.

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True or false

ElenaW [278]

If im not mistaken it is true

4 0

3 years ago

Determine the work that is being done by tension in pulling the box 198.0 cm along the table.

Julli [10]

Work, Kinetic Energy and Potential Energy
6.1 The Important Stuff 6.1.1 Kinetic Energy
For an object with mass m and speed v, the kinetic energy is defined as K = 1mv2
2
(6.1)
Kinetic energy is a scalar (it has magnitude but no direction); it is always a positive number; and it has SI units of kg · m2/s2. This new combination of the basic SI units is
known as the joule:
As we will see, the joule is also the unit of work W and potential energy U. Other energy
1joule = 1J = 1 kg·m2 (6.2) s2
units often seen are:
6.1.2 Work
1erg=1g·cm2 =10−7J 1eV=1.60×10−19J s2
When an object moves while a force is being exerted on it, then work is being done on the object by the force.
If an object moves through a displacement d while a constant force F is acting on it, the force does an amount of work equal to
W =F·d=Fdcosφ (6.3)
where φ is the angle between d and F.

5 0

3 years ago

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

5 0

3 years ago

A child slides down the water slide at a swimming pool and enters the water at a final speed of 4.22 m/s. At what final speed wo

zheka24 [161]

Answer:

v'=5.97 m/s

Explanation:

If the friction and fluid friction are ignored, then by law of conservation of mechanical energy, potential energy at the top of the slide must be equal to the kinetic energy at the bottom of the slide. Thus, the height of slide and final speed at the bottom of the slide are related as:

$mgh = \frac {1}{2} mv^2\\v =\sqrt{2gh}$

let the final speed be v' when h' = 2 h

$\frac{v'}{v}=\frac{\sqrt h'}{\sqrt h}\\v'=\sqrt{\frac{2h}{h}}\times v\\v'=\sqrt2 v\\v'=5.97 m/s$

8 0

3 years ago

A 100.0-kg bakery sign hangs from two thin cables as shown.

harina [27]

Answer:

1. T₁ = 500 N

2. T₂ = 866 N

Explanation:

Please see attached photo for the diagram.

Thus, we can obtain obtained the value of T₁ and T₂ as follow:

1. Determination of T₁

Angle θ = 30

Hypothenus = 100 kg

Opposite = T₁ =?

Sine θ = Opposite /Hypothenus

Sine 30 = T₁ / 100

Cross multiply

T₁ = 100 × Sine 30

T₁ = 100 × 0.5

T₁ = 50 Kg

Multiply by 10 to express in Newton

T₁ = 50 × 10

T₁ = 500 N

2. Determination of T₂

Angle θ = 60

Hypothenus = 100 kg

Opposite = T₂ = ?

Sine θ = Opposite /Hypothenus

Sine 60 = T₂ / 100

Cross multiply

T₂ = 100 × Sine 60

T₂ = 100 × 0.8660

T₂ = 86.6 Kg

Multiply by 10 to express in Newton

T₂ = 86.6 × 10

T₂ = 866 N

5 0

3 years ago