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Bad White [126]
2 years ago
14

How does the gravitational pull of the blue supergiants impact the direction of your star?

Physics
1 answer:
Gennadij [26K]2 years ago
8 0

Answer:

Blue supergiants represent a slower burning phase in the death of a massive star. Due to core nuclear reactions being slightly slower, the star contracts and since very similar energy is coming from a much smaller area (photosphere) then the star's surface becomes much hotter.

Explanation:

I know this may not be the answer youre looking for, but hopefully this can help somehow!

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Consider an embedded system which uses a battery with a 17.15-Amp-Hour capacity. What is the maximum average current draw (in mi
Marianna [84]

Answer:

<em>85.12 μAmp</em>

<em></em>

Explanation:

The battery power output = 17.15 Amp-hr

If the battery is to last 23 years, we have to calculate how many hours there are in 23 years

in one year there are 24 hours x 365 day = 8760 hrs

in 23 years there are 23 x 8760 = 201480 hours

maximum current to be drawn from the battery = (17.15 Amp-hr) ÷ (201480 hours) = 85.12 x 10^-6 Amp = <em>85.12 μA</em>

7 0
3 years ago
A _______ is a repeating disturbance or vibration that transfers or moves energy from place to place without transporting mass.
cupoosta [38]

the answer is a wave


7 0
3 years ago
Read 2 more answers
Three observers watch a train pull away from a station toward the right of the platform. Observer A is in one of the train’s car
juin [17]

Observer A is moving inside the train

so here observer A will not be able to see the change in position of train as he is standing in the same reference frame

So here as per observer A the train will remain at rest and its not moving at all

Observer B is standing on the platform so here it is a stationary reference frame which is outside the moving body

So here observer B will see the actual motion of train which is moving in forward direction away from the platform

Observer C is inside other train which is moving in opposite direction on parallel track. So as per observer C the train is coming nearer to him at faster speed then the actual speed because they are moving in opposite direction

So the distance between them will decrease at faster rate

Now as per Newton's II law

F = ma

Now if train apply the brakes the net force on it will be opposite to its motion

So we can say

- F = ma

a = \frac{-F}{m}

so here acceleration negative will show that train will get slower and its distance with respect to us is now increasing with less rate

It is not affected by the gravity  because the gravity will cause the weight of train and this weight is always counterbalanced by normal force on the train

So there is no effect on train motion



5 0
3 years ago
Why you cannot use an elastic measuring tape to measure distance. What problem you may face if you use it
Stells [14]

The problem you would encounter is measuring the height of two different people, a tall one and a short one, and getting the same answer for both of them.

No matter WHAT we're hearing out of the White House these days, you CAN'T bend and stretch your standard measuring devices, or any other 'facts', to make them fit the thing that you're measuring.  This does not work.  You're always entitled to your own opinions, but you're not entitled to your own facts.

3 0
3 years ago
To determine the muzzle velocity of a bullet fired from a rifle, you shoot the 2.47-g bullet into a 2.43-kg wooden block. The bl
Elza [17]

Answer:

The velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

Explanation:

Given;

mass of the bullet, m₁ = 2.47 g = 0.00247 kg

mass of the wooden block, m₂ = 2.43 kg

initial velocity of the wooden block, u₂ = 0

height reached by the bullet-block system after collision = 0.295 cm = 0.00295 m

let the initial velocity of the bullet on leaving the gun's barrel = v₁

let final velocity of the bullet-wooden block system after collision = v₂

Apply the principle of conservation of linear momentum;

Total initial momentum = Total final momentum

m₁v₁ + m₂u₂ = v₂(m₁ + m₂)

0.00247v₁  + 2.43 x 0  =  v₂(2.43 + 0.00247)

0.00247v₁ = 2.4325v₂ -------(1)

The kinetic energy of the bullet-block system after collision;

K.E = ¹/₂(m₁ + m₂)v₂²

K.E = ¹/₂ (2.4325)v₂²

The potential energy of the bullet-block system after collision;

P.E = mgh

P.E = (2.4325)(9.8)(0.00295)

P.E = 0.07032

Apply the principle of conservation of mechanical energy;

K.E = P.E

¹/₂ (2.4325)v₂² = 0.07032

1.21625 v₂²  = 0.07032

v₂²  = 0.07032  / 1.21625

v₂² = 0.0578

v₂ = √0.0578

v₂ = 0.24 m/s

Substitute v₂ in equation (1), to obtain the initial velocity of the bullet;

0.00247v₁ = 2.4325v₂

0.00247v₁ = 2.4325 (0.24)

0.00247v₁ = 0.5838

v₁ = 0.5838 / 0.00247

v₁ = 236.36 m/s

Therefore, the velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

5 0
3 years ago
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