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Bumek [7]
2 years ago
12

Which of the following is NOT a result of supernova explosions? The neutron core is completely destroyed. Any planets within a f

ew dozen light-years receive a life-threatening dose of radiation. Many of the elements the star fused during its life are blasted out into space. Many new elements, including some heavier than iron, are fused during the supernova explosion.
Physics
1 answer:
DanielleElmas [232]2 years ago
4 0

Answer:

 The neutron core is completely destroyed

Explanation:

 A earth - supernova is an explosion resulting to the death of a star that occurs close enough to the earth but this does not completely destroy a star. Supernovae are the most violent explosions in the universe. But they do not explode like a bomb explodes, blowing away every bit of the original bomb. Rather, when a star explodes into a supernova, its core survives. The reason for this is that the explosion is caused by a gravitational rebound effect and not by a chemical reaction. Stars are so large that the gravitational forces holding them together are strong enough to keep the nuclear reactions from blowing them apart. It is the gravitational rebound that blows apart a star in a supernova.

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A vehicle moving with a uniform a acceleration of 2m/s and has a velocity of 4m/s at a certain time .
Evgesh-ka [11]

Answer:

1 second later the vehicle's velocity will be:

v(1)= 6\,\,\frac{m}{s} \\

5 seconds later the vehicle's velocity will be:

v(5)=14\,\,\frac{m}{s}

Explanation:

Recall the formula for the velocity of an object under constant accelerated motion (with acceleration "a"):

v(t)=v_0+a\,t

Therefore, in this case v_0=4\,\,\frac{m}{s}  and a=2\,\,\frac{m}{s^2}

so we can estimate the velocity of the vehicle at different times just by replacing the requested "t" in the expression:

v(t)=v_0+a\,t\\v(t)=4+2\,\,t\\v(1)=4+2\,(1) = 6\,\,\frac{m}{s} \\v(5)=4+2\,(5)=14\,\,\frac{m}{s}

3 0
3 years ago
Two light bulbs have resistances of 400 Ω and 800 ΩThe two light bulbs are connected in series across a 120- V line. Find the cu
Natasha2012 [34]

1) Current in each bulb: 0.1 A

The two light bulbs are connected in series, this means that their equivalent resistance is just the sum of the two resistances:

R_{eq}=R_1 + R_2 = 400 \Omega + 800 \Omega=1200 \Omega

And so, the current through the circuit is (using Ohm's law):

I=\frac{V}{R_{eq}}=\frac{120 V}{1200 \Omega}=0.1 A

And since the two bulbs are connected in series, the current through each bulb is the same.

2) 4 W and 8 W

The power dissipated by each bulb is given by the formula:

P=I^2 R

where I is the current and R is the resistance.

For the first bulb:

P_1 = (0.1 A)^2 (400 \Omega)=4 W

For the second bulb:

P_1 = (0.1 A)^2 (800 \Omega)=8 W

3) 12 W

The total power dissipated in both bulbs is simply the sum of the power dissipated by each bulb, so:

P_{tot} = P_1 + P_2 = 4 W + 8 W=12 W

3 0
3 years ago
The moment of water from ocean through the atmosphere and back
melisa1 [442]

The water cycle is all about storing water and moving water on, in, and above the Earth. Although the atmosphere may not be a great storehouse of water, it is the superhighway used to move water around the globe. Evaporation and transpiration change liquid water into vapor, which ascends into the atmosphere due to rising air currents. Cooler temperatures aloft allow the vapor to condense into clouds and strong winds move the clouds around the world until the water falls as precipitation to replenish the earthbound parts of the water cycle. About 90 percent of water in the atmosphere is produced by evaporation from water bodies, while the other 10 percent comes from transpiration from plants.

There is always water in the atmosphere. Clouds are, of course, the most visible manifestation of atmospheric water, but even clear air contains water—water in particles that are too small to be seen. One estimate of the volume of water in the atmosphere at any one time is about 3,100 cubic miles (mi3) or 12,900 cubic kilometers (km3). That may sound like a lot, but it is only about 0.001 percent of the total Earth's water volume of about 332,500,000 mi3 (1,385,000,000 km3), If all of the water in the atmosphere rained down at once, it would only cover the globe to a depth of 2.5 centimeters, about 1 inch.

6 0
3 years ago
Read 2 more answers
A bucket of water of mass 10 kg is pulled at constant velocity up to a platform 30 meters above the ground. This takes 10 minute
Rudik [331]

Answer:

W = 2352 J

Explanation:

Given that:

  • mass of the bucket, M = 10 kg
  • velocity of pulling the bucket, v = 3m.min^{-1}
  • height of the platform, h = 30 m
  • time taken, t = 10 min
  • rate of loss of water-mass, m = 0.4 kg.min^{-1}

Here, according to the given situation the bucket moves at the rate,

v=3 m.min^{-1}

The mass varies with the time as,

M=(10-0.4t) kg

Consider the time interval between t and t + ∆t. During this time the bucket moves a distance

∆x =  3∆t meters

So, during this interval change in work done,

∆W = m.g∆x

<u>For work calculation:</u>

W=\int_{0}^{10} [(10-0.4t).g\times 3] dt

W= 3\times 9.8\times [10t-\frac{0.4t^{2}}{2}]^{10}_{0}

W= 2352 J

5 0
3 years ago
A girl throws a pebble into a deep well at 4.0 m/s (downward). It hits the water in 2.0 s. How far below the ground is the water
alexandr402 [8]

Answer:

  • 27.6 m
  • 13.8 m/s

Explanation:

(b) The initial velocity is added to that due to acceleration by gravity. The velocity is increased linearly by gravity at the rate of 9.8 m/s². The average velocity of the pebble will be its velocity halfway through the 2-second time period.* That is, it will be ...

  4 m/s + (9.8 m/s²)(2 s)/2 = 13.8 m/s . . . . average velocity

__

(a) The distance covered in 2 seconds at an average velocity of 13.8 m/s is ...

  d = vt

  d = (13.8 m/s)(2 s) = 27.6 m

The water is about 27.6 m below ground.

_____

* We have chosen to make use of the fact that the velocity curve is linear, so the average velocity is half the sum of initial and final velocities:

  vAvg = (vInit + vFinal)/2 = (vInit + (vInit +at))/2 = vInit +at/2

__

If you work this in a straightforward way, you would find distance as the integral of velocity, then find average velocity from the distance and time.

  \displaystyle d=\int_0^t{(v_0+at)}\,dt=v_0t+\dfrac{1}{2}at^2=t\left(v_0+a\dfrac{t}{2}\right)\\\\v_{avg}=\dfrac{d}{t}=v_0+a\dfrac{t}{2}\qquad\text{the formula we started with}

8 0
3 years ago
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