Answer:
GFCI outlets are found in wet areas.
GFCI outlets prevent electrocution if you are touching a wet appliance.
To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.
PART A) We will begin by finding the two net distances.

And the distance 'd' is



Through the free-body diagram the tension components are given by


Here we can watch that,

Dividing both expression we have that,

Replacing the values,


PART B) Using the vertical component we can find the tension,




(50 gal / 5 min) x (.0037854 m³/gal) x (1 min / 60 sec)
= (50 · 0.0037854 · 1) / (5 · 60) m³/sec
= 0.000631 m³/sec
Answer:
a=2500J,b=1000K,c=1000J,d=14.142m/s
Explanation:
V²=U²+2gh
V²=0 + 2×10×10=200m/s
a).kinetic energy=(1/2)mv²=(1/2)25×200=2500
potential energy=mgh
p.e=25×10×10=2500J
pe+ke=2500+2500=5KJ
b).mgh=25×10×4=1000J
c). V²=U²+2gh
V²=0+2×10×4
V²=80
kinetic energy=(1/2)mv²
=(1/2)25×80
=1KJ
d). From my first paragraph V²=200
V=√200
V=14.142m/s
<h2>Right answer: acceleration due to gravity is always the same </h2><h2 />
According to the experiments done and currently verified, in vacuum (this means there is not air or any fluid), all objects in free fall experience the same acceleration, which is <u>the acceleration of gravity</u>.
Now, in this case we are on Earth, so the gravity value is
Note the objects experience the acceleration of gravity regardless of their mass.
Nevertheless, on Earth we have air, hence <u>air resistance</u>, so the afirmation <em>"Free fall is a situation in which the only force acting upon an object is gravity" </em>is not completely true on Earth, unless the following condition is fulfiled:
If the air resistance is <u>too small</u> that we can approximate it to <u>zero</u> in the calculations, then in free fall the objects will accelerate downwards at
and hit the ground at approximately the same time.