The sphere has a constant potential. It is the electric field.
In the sphere, then
Outside the sphere, then
The elements of the electric field include
Which becomes,
<h3>
In a consistent electric field, is force constant?</h3>
Similar to an ordinary object in the uniform gravitational field near the Earth's surface, a charged item in a uniform electric field experiences a constant force and consequently experiences a uniform acceleration. The vector cross product of p and E determines the torque's direction.
If the charge is positive, the force either moves in the same direction as E or in the opposite direction (if charge is negative).
A torque is experienced by an electric dipole (p) in an even electric field (E). The vector cross product of p and E determines the torque's direction.
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Answer:
He traveled 9km
Explanation:
To do this problem you need to use the equation which is Speed= distance/time and this problem gives you the speed which is 18 km/h and it gives you the time 1/2 hour so you write the equation 18= d/ 1/2 which his distance is 9km
Answer:
C
Explanation:
Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces
I hope this helps a little bit
Answer:
F = 11 N
Explanation:
Given,
Mass of a block, m = 5 kg
Acceleration of the block, a = 2.2 m/s²
We need to find the force on the person's hand. Let it is F. We know that force is given by the product of mass and acceleration as follows :
F = ma
F = 5 kg × 2.2 m/s²
F = 11 N
So, the force on a person's hand is 11 N.
Answer:
0.3 m
Explanation:
Initially, the package has both gravitational potential energy and kinetic energy. The spring has elastic energy. After the package is brought to rest, all the energy is stored in the spring.
Initial energy = final energy
mgh + ½ mv² + ½ kx₁² = ½ kx₂²
Given:
m = 50 kg
g = 9.8 m/s²
h = 8 sin 20º m
v = 2 m/s
k = 30000 N/m
x₁ = 0.05 m
(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²
x₂ ≈ 0.314 m
So the spring is compressed 0.314 m from it's natural length. However, we're asked to find the additional deformation from the original 50mm.
x₂ − x₁
0.314 m − 0.05 m
0.264 m
Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.
