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Alenkinab [10]
2 years ago
7

What is the hottest layer in the Sun's atmosphere?

Physics
1 answer:
erica [24]2 years ago
7 0
I believe it is the core
You might be interested in
Which does not describe the relationship between the nucleus and the electron cloud of an atom?
Margaret [11]
Choices 'a', 'c', and 'd' are true.

In choice 'b', I'm not sure what it means when it says that masses
are 'balanced'.  To me, masses are only balanced when they're on
a see-saw, or on opposite ends of a rope that goes over a pulley.

Maybe the statement means that the mass of the nucleus and the
mass of the electron cloud are equal.  This is way false.  It takes
more than 1,800 electrons to make the mass of ONE proton or
neutron, and the most complex atom in nature only has 92 electrons
in it.  So there's no way that the masses of the nucleus and the electrons
in one atom could ever be anywhere near equal.
4 0
3 years ago
During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be lau
Viktor [21]

Answer:

6.75 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration = 16 m/s²

g = Acceleration due to gravity = 9.81 m/s²

Let y be the distance the rocket is accelerating

960-y is the distance traveled in free fall

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 16\times y+0^2}\\\Rightarrow v^2=32y\ m/s

In free fall

v^2-u^2=2g(960-y)\\\Rightarrow 0-32y=2g(960-y)\\\Rightarrow -32y=2\times -9.81(960-y)\\\Rightarrow 960-y=\dfrac{-32}{2\times -9.81}y\\\Rightarrow 960-y=1.63098878695y\\\Rightarrow 960=2.63098878695y\\\Rightarrow y=\dfrac{960}{2.63098878695}\\\Rightarrow y=364.881828749\ m

The distance the rocket will keep accelerating is 364.881828749 m

After which it will travel 960-364.881828749 = 595.118171251 m in free fall

s=ut+\frac{1}{2}at^2\\\Rightarrow 364.881828749=0t+\frac{1}{2}\times 16\times t^2\\\Rightarrow t=\sqrt{\frac{364.881828749\times 2}{16}}\\\Rightarrow t=6.75353452598\ s

The time the rocket is accelerating is 6.75 seconds

5 0
3 years ago
Calculate the constant acceleration a in g’s which the catapult of an aircraft carrier must provide to produce a launch velocity
goblinko [34]
The solution to the problem is as follows:

<span>First, I'd convert 188 mi/hr to ft/s. You should end up with about ~275.7 ft/s.
 
So now write down all the values you know:

Vfinal = 275.7 ft/s

Vinitial = 0 ft/s

distance = 299ft
</span>
<span>Now just plug in Vf, Vi and d to solve
</span>
<span>Vf^2 = Vi^2 + 2 a d 

</span><span>BTW: That will give you the acceleration in ft/s^2. You can convert that to "g"s by dividing it by 32 since 1 g is 32 ft/s^2.</span>
5 0
2 years ago
A proton is initially moving west at a speed of 1.10 106 m/s in a uniform magnetic field of magnitude 0.281 T directed verticall
kifflom [539]

Answer:

so here it will move in circle with radius 4.06 cm

Explanation:

As we know that proton is moving towards west while the magnetic field is vertically upwards

So here the force on the proton must be perpendicular to the velocity

So here we have

F = q(\vec v \times \vec B)

so here we have

F = qvB sin90

since force is perpendicular to the velocity so here it must be centripetal force

here we have

\frac{mv^2}{R} = qvB

so we have

R = \frac{mv}{qB}

R = \frac{(1.66 \times 10^{-27})(1.10 \times 10^6)}{(1.6 \times 10^{-19})(0.281)}

R = 4.06 cm

so here it will move in circle with radius 4.06 cm

8 0
3 years ago
Why is the physics of hssc2 so difficult?
Ivan
Could be easy for some people and hard for some people.
8 0
3 years ago
Read 2 more answers
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