Answer:
so maximum velocity for walk on the surface of europa is 0.950999 m/s
Explanation:
Given data
legs of length r = 0.68 m
diameter = 3100 km
mass = 4.8×10^22 kg
to find out
maximum velocity for walk on the surface of europa
solution
first we calculate radius that is
radius = d/2 = 3100 /2 = 1550 km
radius = 1550 × 10³ m
so we calculate no maximum velocity that is
max velocity = √(gr) ...............1
here r is length of leg
we know g = GM/r² from universal gravitational law
so G we know 6.67 × 
N-m²/kg²
g = 6.67 × ( 4.8×10^22 ) / ( 1550 × 10³ )
g = 1.33 m/s²
now
we put all value in equation 1
max velocity = √(1.33 × 0.68)
max velocity = 0.950999 m/s
so maximum velocity for walk on the surface of europa is 0.950999 m/s
The candle flame releases hot gases, which directly go in upwards directions. Due to which the air near the flame of the candle is very hot and dense. The particles along with vapour move up. And since the sideways, the air is not very dense and hot, we are able to hold the candle. In anti-gravity region, there will be no density differences and also, the convection process wont occur. So, the candle quickly snuffs off.
Answer:
Difference threshold or also Just Noticeable Difference
Explanation:
The above mentioned case between room mates, where one room mate was able to detect a minute change in volume shows an instance of the difference threshold.
Difference threshold can be defined as stimulation at its minimum level that can be detected by an individual almost 50 % of the times.
It is the lowest possible level of sound that is detectable by a person.
That is what happened in the mentioned case that when the volume was increased from 14 to 15, Amber was able to detect it.
Answer:
Explanation:
The strength of the electric field produced by a charge Q is given by
where
Q is the charge
r is the distance from the charge
k is the Coulomb's constant
In this problem, the electric field that can be detected by the fish is
and the fish can detect the electric field at a distance of
Substituting these numbers into the equation and solving for Q, we find the amount of charge needed:
