Answer:
a) The maximum potential difference the wet human body can take = 5.88 V
b) This value is very much lower than the normal voltage of household outlets (120 V) and isn't in the same range at all.
This means one should be very careful operating electrical appliances (even at home) while the body is wet.
Explanation:
From Ohm's law, V = IR
The resistance of the body when the body is wet = 1.2 kΩ = 1200 Ω
Maximum safe current that the body can withstand = 4.90 mA = 0.0049 A
Maximum voltage the body can withstand while wet = (Maximum safe current that the body can withstand) × (Resistance of the body when the body is wet) = 0.0049 × 1200 = 5.88 V
b) This value is still very much lower than the normal voltage of household outlets (120 V) and isn't in the same range at all.
This means one should be very careful operating electrical appliances (even at home) while the body is wet.
Answer:
Negative 9.8 meters per second squared
Explanation:
The negative is for the direction (down, towards the center of the earth). Often this can be estimated as -10 m/s^2 to make calculations easier.
1/2 x b x h if the question is askinf u to make a shape and it comes out a triangle
if not triangle then 1/2 x (x x b) x h
if none of these questions are there then multiply time by position
hope this helped
Answer:
In physics, the kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes.
Explanation:
Answer:
Time taken, t = 3.53 seconds
Explanation:
It is given that,
The formula for calculating the time t is given by :
...... (1)
Where
s is the distance in feets
We need to find the time taken by the stone to drop a distance of 200 feets, s = 200 feets
Put the value of s in equation (1) such that,

t = 3.53 seconds
So, the time taken by the object is 3.53 seconds. Hence, this is the required solution.