What is the difference between potential and kinetic types of energy? 1 pois
2 answers:
You might be interested in
Answer:
d) 1.2 mT
Explanation:
Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.
First of all, we observe that:
- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is
I = 15 A
- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).
Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by
where
is the vacuum permeability
I = 15 A is the current in the conductor
r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field
Substituting, we find:
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
Answer:
The work done by the drag force is given by 29.96 J
Explanation:
Given :
Thrust force N
Displacement m
Mass of rocket Kg
From work energy theorem,
Where thrust work
gravitational work
After cutoff kinetic energy is converted into potential energy,
Put value of KE
Work done by drag force is given by,
J
Therefore, the work done by the drag force is given by 29.96 J