A firecracker in a coconut blows the coconut into three pieces. Twopieces of equal mass fly off south and west, perpendicular to
1 answer:
Answer:
V = 13m/s
North-East (i.e. 45 degree from North)
Explanation:
This question deals with the idea of momentum.
Since the directions of a compass are fixed, we can take south as horizontal and west as vertical. Since both pieces of coconut are of equal mass, then the resultant of the two pieces would be in between South and West and the third piece would be opposite this direction and be in the North-East (i.e. 45 degree from North) direction
To find the speed of the third piece, we first find the speed on the two pieces of the coconut in terms of (horizontal component of velocity) and
(vertical component of velocity)
We have
= Velocity in South direction = 18m/s
= Mass of piece in South direction = m
= Velocity in West direction = 18m/s
= Mass of piece in West direction = m
= Velocity of Third Piece in North direction = unknown
= Velocity of Third Piece in East direction = unknown
= Mass of third piece = 2m
Since we now know the horizontal and vertical component of the velocity of the third piece of coconut we find the resultant velocity. Since we know the direction is North-East, we can imagine a right angled triangle with base of 9 and height 9 and the hypotenuse equal to the resultant velocity .
We can simply apply the Pythagoras theorem and find the hypotenuse
V = 13m/s
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Here,
f = Focal Length
R = Radius
Rearranging to find the radius we have,
Replacing with our values,
R = 2(13.8cm)
R = 27.6cm
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Complete Question
Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field
E⃗ 1 inside the cell at a distance of 3.05 μm from the center?
E⃗ 2 Just inside the surface of the cell
E⃗ 3 Just outside the surface of the cell
E⃗ 4 At a point outside the cell 3.05 μm from the surface
Answer:
E⃗ 1
0 V/m
E⃗ 2
0 V/m
E⃗ 3
E⃗ 4
Explanation:
From the question we are told that
The diameter is
The charge is
The permittivity of free space is
The distance considered is
Generally the electric field inside the cell at a distance of 3.05 μm from the center is
0 V/m
This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge
Generally the electric field just inside the surface of the cell is 0 V/m
This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge
Generally the electric field just outside the cell is mathematically represented as
Here is the coulomb constant with value
r is the radius of the sphere which is mathematically as
Generally the electric field at a point outside the cell 3.05 μm from the surface is mathematically represented as
Here R is mathematically represented as
=>
So