Answer:
The asteroid's acceleration at this point is 
Explanation:
The equation that governs the trajectory of asteroid is given by :
The velocity of asteroid is given by :
At some point during the trip across the screen, the asteroid is at rest. It means, v = 0
So,
Acceleration,
Put t = 0.971 s
So, the asteroid's acceleration at this point is and it is decelerating.
Well, first of all, there's no such thing as "fully charged" for a capacitor.
A capacitor has a "maximum working voltage", because of mechanical
or chemical reasons, just like a car has a maximum safe speed. But
anywhere below that, cars and capacitors do their jobs just fine, without
any risk of failing.
So we have a capacitor that has some charge on it, and therefore some
voltage across it. From the list of choices above . . .
<span>-- Both plates have the same amount of charge.
Yes. And both plates have opposite TYPES of charge.
One plate is loaded with electrons and is negatively charged.
The other plate is missing electrons and is positively charged.
-- There is a potential difference between the plates.
Yes. That's the "voltage" mentioned earlier.
It's a measure of how badly the extra electrons want to jump
from the negative plate to the positive plate.
-- Electric potential energy is stored.
Yes. It's the energy that had to be put into the capacitor
to move electrons away from one plate and cram them
onto the other plate.
</span>
Answer:
final velocity will be44.72m/s
Explanation:
HEIGHT=h=100m
vi=0m/s
vf=?
g=10m/s²
by using third equation of motion for bodies under gravity
2gh=(vf)²-(vi)²
evaluating the formula
2(10m/s²)(100m)=vf²-(0m/s)²
2000m²/s²=vf²
√2000m²/s²=√vf²
44.72m/s=vf
Answer:
=170kcal
Explanation:
We first calculate the amount of energy required to melt the alcohol using the formula: MLf, where Lf is the latent heat of fussion
We then calculate amount of heat required to raise the temperature of liquid alcohol to -14° C using MC∅. We then add the two.
Thus ΔH=MLf+MC∅
ΔH=2kg×25kcal/kg+ 2kg×(0.6kcal/kg.K×(-14-⁻114)
=50kcal+120kcal
=170kcal
A seesaw remains stationary when two students of equal weight sit on the ends
c
