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3 years ago

How are stars important to the Milky Way

2 answers:

6 0

Stars are a source of light and heat. They recycle all the matter, gas, and dust and process them into new material. They were created as a major driving force of evolution of the universe.

PolarNik [594]3 years ago

4 0

Stars are the light and heat of the milky way the stars work with the milky way like the sun works with the earth the stars keep the milky way lit up and semi warm like how the sun (which is a star of the milky way ) keeps the earth warm. Just remember that stars r just very big balls of gas burning in the sky :)

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A train Traveled from hong kong to beijing. It traveled at an average speed of 160 km/h in the first four hours. After that, it

4vir4ik [10]

The answer is 7 hours

3 0

3 years ago

Two particles with oppositely signed charges are held a fixed distance apart. The charges are equal in magnitude and they exert

damaskus [11]

Answer:

the force will decrease to 3/4 of its original value.

Explanation:

The initial electric force between the two charges is:

$F = k \frac{q\cdot q}{r^2}$

where

k is the Coulomb's constant

q is the magnitude of each charge

r is their separation

Later, half of one charge is transferred to the other charge; this means that one charge will have a charge of

$q+\frac{q}{2}=\frac{3}{2}q$

while the other charge will be

$q-\frac{q}{2}=\frac{q}{2}$

So, the new force will be

$F' = k \frac{(\frac{q}{2})\cdot (\frac{3}{2}q)}{r^2}=\frac{3}{4} (k\frac{q\cdot q}{r^2})=\frac{3}{4}F$

So, the force will decrease to 3/4 of its original value.

6 0

3 years ago

A subway train starts from rest at a station and accelerates at a rate of 1.68 m/s2 for 14.2 s. It runs at constant speed for 68

liubo4ka [24]

Answer:

total distance = 1868.478 m

Explanation:

given data

accelerate = 1.68 m/s²

time = 14.2 s

constant time = 68 s

speed = 3.70 m/s²

to find out

total distance

solution

we know train start at rest so final velocity will be after 14 .2 s is

velocity final = acceleration × time ..............1

final velocity = 1.68 × 14.2

final velocity = 23.856 m/s²

and for stop train we need time that is

final velocity = u + at

23.856 = 0 + 3.70(t)

t = 6.44 s

and

distance = ut + 1/2 × at² ...........2

here u is initial velocity and t is time for 14.2 sec

distance 1 = 0 + 1/2 × 1.68 (14.2)²

distance 1 = 169.37 m

and

distance for 68 sec

distance 2= final velocity × time

distance 2= 23.856 × 68

distance 2 = 1622.208 m

and

distance for 6.44 sec

distance 3 = ut + 1/2 × at²

distance 3 = 23.856(6.44) - 0.5 (3.70) (6.44)²

distance 3 = 76.90 m

total distance = distance 1 + distance 2 + distance 3

total distance = 169.37 + 1622.208 + 76.90

total distance = 1868.478 m

7 0

3 years ago

Read 2 more answers

A 20-newton weight is attached to a spring.

Makovka662 [10]

Answer:

40 N/m

Explanation:

The diagram attached is used to answer the question

We know from Hooke's law that extension is directly proportional to the applied force hence

F=kx where x is extension, F is applied force and k is the spring constant. Making k the subject of the formula then

$k=\frac {F}{k}$

From the attached diagram extension is given by subtracting unstretched spring from stretched spring hence extension, x=1-0.5=0.5m

Substituting 20 N for F and 0.5 m for x then

$k=\frac {20}{0.5}=40 N/m$

8 0

3 years ago

At what distance from the wire is the magnitude of the electric field equal to 2.41 n/cn/c ?

Sladkaya [172]

The correct answer is 1.07m.

The area surrounding an electric charge where its impact may be felt is known as the electric field. When another charge enters the field, the presence of an electric field may be felt. The electric field will either attract or repel the charge depending on its makeup. Any electric charge has a property known as the electric field. The charge and electrical force working in the field determine the strength or intensity of the electric field.

Here, is the charge per unit length, r is the distance from the wire, and

is the free space permittivity ε_0. Electric field due to the long straight wire is,

E= λ/2πε_0r

Rearrange the equation for r.

r=λ/2πε_0E

Substitute 2.41 N/C for E,

E=1.44×10^-10C/m

λ=8.85×10^-12C^2/Nm^2

r=(1.44×10^-10C/m)/(2(3.14)(8.85×10^-12C^2/Nm^2)(2.41N/C))

r=1.07m

At a distance of 1.07 m the magnitude of electric field is 2.41 N/C.

To learn more about electric field refer the link:

brainly.com/question/12821750

#SPJ4

5 0

1 year ago