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3 years ago

How many light years away is the sun from the middle of the Millky way

Physics

1 answer:

julsineya [31]3 years ago

4 0

Answer:

The Milky Way is about 1,000,000,000,000,000,000 km (about 100,000 light years or about 30 kpc) across. The Sun does not lie near the center of our Galaxy. It lies about 8 kpc from the center on what is known as the Orion Arm of the Milky Way

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A train travels 4200 km in 90 hours. What is its speed?

denis23 [38]

Answer: I think its 120

Explanation: thx for the free points :)

6 0

3 years ago

When landing after a spectacular somersault, a 35.0 kg gymnast decelerates by pushing straight down on the mat. calculate the fo

Sliva [168]

The deceleration experienced by the gymnast is the 9 times of the acceleration due to gravity.

Now from Newton`s first law, the net force on gymnast,

$F_{net} =F-W=ma$

Here, W is the weight of the gymnast and a is the acceleration experienced by the gymnast ( $9\times g$ acceleration due to gravity)

Therefore,

$F= ma+W$ OR $F=ma+mg=m(g+a)$

Given $m = 30 kg$ and $a=9\times g=9\times 9.8 m/s^{2} =88.2 m/s^{2}$

Substituting these values in above formula and calculate the force exerted by the gymnast,

$F=(40 kg) (88.2 m/s^{2} +9.8 m/s^{2} )$

$F=3.537\times10^{3}N$

6 0

3 years ago

A student gives a 5.0 kg box a brief push causing the box to move with an initial speed of 8.0 m/s along a rough surface. The bo

WITCHER [35]

Answer:

The time taken to stop the box equals 1.33 seconds.

Explanation:

Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:

$F=mass\times acceleration\\\\\therefore acceleration=\frac{Force}{mass}$

Given mass of box = 5.0 kg

Frictional force = 30 N

thus

$acceleration=\frac{30}{5}=6m/s^{2}$

Now to find the time that the box requires to stop can be calculated by first equation of kinematics

The box will stop when it's final velocity becomes zero

$v=u+at\\\\0=8-6\times t\\\\\therefore t=\frac{8}{6}=4/3seconds$

Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.

5 0

4 years ago

An air-track glider attached to a spring oscillates between the 10.0 cm mark and the 57.0 cm mark on the track. The glider compl

pickupchik [31]

Answer:

a. 2.1 s

b.0.48 Hz

c. A=24cm

d. 72cm/s

Explanation:

An air-track glider attached to a spring oscillates between the 10.0 cm mark and the 57.0 cm mark on the track. The glider completes 15.0 oscillations in 31.0 s.What are the (a) period, (b) frequency, (c) amplitude, and (d) maximum speed of the glider?

What are the period,

period is the time taken for a wave particle to make one complete oscillation

a) 31 / 15 = 2.066 seconds

= 2.1 s

(b) frequency : this the number of oscillation made in one seconds.

it is also the inverse of the period.

= oscillations / time

= 15/31= 0.48 Hz

amplitude = 1/2 of the difference of oscillation marks

= 1/2(57-10) = 47/2cm

23.5cm

A=24cm

(d) maximum speed of the glider?

V=ωA

angular frequency *Amplitude

V=a*pi*f*amplitude

2π x frequency x amplitude = maximum speed

= 2π x .48 x 24

=72.38 cm/s

72cm/s

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4 years ago

What radiant energy is sensed by human skin as warmth

almond37 [142]

Infrared radiation <span>is sensed by human skin as warmth.</span>

4 0

3 years ago

How many light years away is the sun from the middle of the Millky way​

How many light years away is the sun from the middle of the Millky way