If a volcano epulses massive amounts of dust into the atmosphere, those two things will/can happen.
The events will last until the dust lays down on the earth.
Answer: 6250 joules
Explanation:
The work needed to lift an object of mass M by a height H is equal to:
w = M*g*H
where h = 10m/s^2
then the total work that he did is equal to the sum of the work for every stone:
W = (100kg*g*H) + (120kg*g*H) + (140kg*g*H) + (160kg*g*H) + (180kg*g*H)
= (100kg + 120kg + 140kg + 160kg + 180kg)*g*H
= (500kg)*g*H
and now we can repalce g by 10m/s^2 and H by 125cm
But you can notice that we have two different units of distance, so knowing that 100cm = 1m
we can write H = 125cm = (125/100) m = 1.25 m
Then we have:
H = 500kg*10m/s^2*1.25m = 6250 J
They have some but not very much, the particles in the ice are still vibrating just not as much as in water. the only time a substance would have 0 kinetic energy is when that substance is at 0 degrees kelvin(absolute zero) so far no place in the universe has been recorded at absolute zero though
According to Newton's 3rd law, there will be equal and opposite force on the astronaut which is -6048 N
<h3>
What does Newton's third law say ?</h3>
The law state that in every action, there will be equal and opposite reaction.
Given that a rocket takes off from Earth's surface, accelerating straight up at 69.2 m/s2. We are to calculate the normal force (in N) acting on an astronaut of mass 87.4 kg, including his space suit.
Let us first calculate the force involved in the acceleration of the rocket by using the formula
F = ma
Where mass m = 87.4 kg, acceleration a = 69.2 m/s2
Substitute the two parameters into the formula
F = 87.4 x 69.2
F = 6048.08 N
According to the Newton's 3rd law, there will be equal and opposite force on the astronaut.
Therefore, the normal force acting on the astronaut is -6048 N approximately
Learn more about forces here: brainly.com/question/12970081
#SPJ1
Answer:
0.0109 m ≈ 10.9 mm
Explanation:
proton speed = 1 * 10^6 m/s
radius in which the proton moves = 20 m
<u>determine the radius of the circle in which an electron would move </u>
we will apply the formula for calculating the centripetal force for both proton and electron ( Lorentz force formula)
For proton :
Mp*V^2 / rp = qp *VB ∴ rp = Mp*V / qP*B ---------- ( 1 )
For electron:
re = Me*V/ qE * B -------- ( 2 )
Next: take the ratio of equations 1 and 2
re / rp = Me / Mp ( note: qE = qP = 1.6 * 10^-19 C )
∴ re ( radius of the electron orbit )
= ( Me / Mp ) rp
= ( 9.1 * 10^-31 / 1.67 * 10^-27 ) 20
= ( 5.45 * 10^-4 ) * 20
= 0.0109 m ≈ 10.9 mm
