The point on a line through the ends of the meter stick is the electric field equal to zero is 4.7 m from the 0 cm mark.
<h3>What is electric field?</h3>
An electric field is the physical field that surrounds the electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them. It also refers to physical field for a system of charged particles. Electric fields originate from the electric charges and time-varying electric currents. Electric fields and the magnetic fields are both manifestations of the electromagnetic field, one of the four fundamental interactions (also called forces) of nature.
Electric fields are important in many areas of the physics, and are exploited in electrical technology. In atomic physics and chemistry, for the instance, the electric field is the attractive force holding the atomic nucleus and electrons together in atoms. It is also force responsible for chemical bonding between atoms that result in molecules.
The electric field is defined as a vector field that associates to each point in the space the (electrostatic or Coulomb) force per unit of charge exerted on an infinitesimal positive test charge at rest at that point. The derived SI unit for electric field is the volt per meter (V/m), which is equal to the newton per coulomb (N/C)
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The correct question will be:
"A +5.00-μC point charge is placed at 0.0 cm mark of meter stick and a -4.00-μC point charge is placed at 50.0 cm mark. At what point on a line through the ends of meter stick is electric field equal to zero?"
Answer:
1400 N
Explanation:
Change in momentum equals impulse which is a product of force and time
Change in momentum is given by m(v-u)
Equating this to impulse formula then
m(v-u)=Ft
Making F the subject of the formula then
Take upward direction as positive then downwards is negative
Substituting m with 0.3 kg, v with 2 m/s, and u with -5 m/s and t with 0.0015 s then
Answer:
The maximum altitude reached with respect to the ground = 0.5 + 0.510 = 1.01 km
Explanation:
Using the equations of motion,
When the rocket is fired from the ground,
u = initial velocity = 0 m/s (since it was initially at rest)
a = 10 m/s²
The engine cuts off at y = 0.5 km = 500 m
The velocity at that point = v
v² = u² + 2ay
v² = 0² + 2(10)(500) = 10000
v = 100 m/s
The velocity at this point is the initial velocity for the next phase of the motion
u = 100 m/s
v = final velocity = 0 m/s (at maximum height, velocity = 0)
y = vertical distance travelled after the engine shuts off beyond 0.5 km = ?
g = acceleration due to gravity = - 9.8 m/s²
v² = u² + 2gy
0 = 100² + 2(-9.8)(y)
- 19.6 y = - 10000
y = 510.2 m = 0.510 km
So, the maximum altitude reached with respect to the ground = 0.5 + 0.510 = 1.01 km
Hope this helps!!!
The radio because u arent pushing as many watts out and it wont pass through as much
