Current is inversely proportional to the resistance of the resistor and directly to the potential difference across it.
I = V/R = 6/12 = 0.5 A
Answer:
She can swing 1.0 m high.
Explanation:
Hi there!
The mechanical energy of Jane (ME) can be calculated by adding her gravitational potential (PE) plus her kinetic energy (KE).
The kinetic energy is calculated as follows:
KE = 1/2 · m · v²
And the potential energy:
PE = m · g · h
Where:
m = mass of Jane.
v = velocity.
g = acceleration due to gravity (9.8 m/s²).
h = height.
Then:
ME = KE + PE
Initially, Jane is running on the surface on which we assume that the gravitational potential energy of Jane is zero (the height is zero). Then:
ME = KE + PE (PE = 0)
ME = KE
ME = 1/2 · m · (4.5 m/s)²
ME = m · 10.125 m²/s²
When Jane reaches the maximum height, its velocity is zero (all the kinetic energy was converted into potential energy). Then, the mechanical energy will be:
ME = KE + PE (KE = 0)
ME = PE
ME = m · 9.8 m/s² · h
Then, equallizing both expressions of ME and solving for h:
m · 10.125 m²/s² = m · 9.8 m/s² · h
10.125 m²/s² / 9.8 m/s² = h
h = 1.0 m
She can swing 1.0 m high (if we neglect dissipative forces such as air resistance).
Answer:
radius comes out to be 3 m
height of the cylinder comes out to be 3m
Explanation:
given
volume of cylinder = 27π m³
π r² h = 27π
r² h = 27.............(1)
surface area of cylinder open at the top
S = 2πrh + π r²




for least amount of material requirement.

hence radius comes out to be 3 m
for height put the value in the equation 1
so, height of the cylinder comes out to be 3m
Desireurban has given a good discussion in the first answer,
but be careful with it:
-- Yes, skimming instead of reading carefully can be non-productive.
-- Using credible sources is MORE productive, and using less reliable sources
can be LESS productive.
Another nonproductive learning behavior ... one that I worry about a lot ...
is slapping the question up on Brainly and waiting for someone to give
you the answer, instead of taking some time to think about it.