Answer:
5.92×10⁷ J
Explanation:
We'll begin by converting 15 tons to Newton. This can be obtained as follow:
1 ton = 9806.65 N
Therefore,
15 ton = 15 ton × 9806.65 N / 1 ton
15 ton = 147099.75 N
Next, we shall convert one-quarter (¼) or 0.25 mile to metre. This can be obtained as follow:
100 mi = 160934 m
0.25 mi = 0.25 mi × 160934 / 100 mi
0.25 mi = 402.335 m
Finally, we shall determine the Workdone. This can be obtained as follow:
Force (F) = 147099.75 N
Distance (d) = 402.335 m
Workdone (Wd) =?
Wd = F × d
Wd = 147099.75 × 402.335
Wd = 5.92×10⁷ J
Thus, the Workdone is 5.92×10⁷ J
According to all the given requirements
stratosphere x starts at an altitude of 16 km
that means
<span>16≤x
</span> and,as given ,stratosphere ends at an altitude of 30 km:
which means
<span>x≤30</span>
combining both of them we get
<span>16≤x≤30</span>
So the range of the stratosphere is from 16 to 30 km.
The right half will be a new bar magnet of 2cm with north pole on the right side and south pole on teh left.
1. Amperes, is the SI unit (also a fundamental unit) responsible for current.
2.
Δq over Δt technically
Rearrange for Δq
I x Δt = Δq
1.5mA x 5 = Δq
Δq = 0.0075
Divide this by the fundamental charge "e"
Electrons: 0.0075 / 1.60 x 10^-19
Electrons: 4.6875 x 10^16 or 4.7 x 10^16
3. So we know that the end resistances will be equal so:
ρ = RA/L
ρL = RA
ρL/A = R
Now we can set up two equations one for the resistance of the aluminum bar and one for the copper: Where 1 represents aluminum and 2 represents copper

We are looking for L2 so we can isolate using algebra to get:

If you fill in those values you get 0.0205
or 2.05 cm