Answer:
a) 
b)
parallel to the earth surface.
- In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.
Explanation:
Given:
mass of the bee, 
charge acquired by the bee, 
a.
Electrical field near the earth surface, 
Now the electric force on the bee:
we know:




The weight of the bee:



Therefore the ratio :


b.
The condition for the bee to hang is its weight must get balanced by the electric force acing equally in the opposite direction.
So,



parallel to the earth surface.
- In this case according to the Fleming's left hand rule the direction of movement of bee must be in a direction parallel to the earth surface and perpendicular to the electric field at the same time.
Jsisisusuagahannananana this is worth 22 points g that’s crazy 25
<h2>
Spring constant is 14.72 N/m</h2>
Explanation:
We have for a spring
Force = Spring constant x Elongation
F = kx
Here force is weight of mass
F = W = mg = 0.54 x 9.81 = 5.3 N
Elongation, x = 36 cm = 0.36 m
Substituting
F = kx
5.3 = k x 0.36
k = 14.72 N/m
Spring constant is 14.72 N/m
Answer:
a = 8.06 m/s²
Explanation:
The acceleration of this car can be found using the first equation of motion:

where,
a = acceleration = ?
vf = final speed = 26.8 m/s
vi = initial speed = 0 m/s
t = time = 3.323 s
Therefore,

<u>a = 8.06 m/s²</u>
Answer:
The value is 
Explanation:
From the question we are told that
The magnitude of the horizontal force is 
The mass of the crate is 
The acceleration of the crate is 
Generally the net force acting on the crate is mathematically represented as

Here
is force of kinetic friction (in N) acting on the crate
So

=> 