Answer:
Explanation:
<u>Given:</u>
= uniform electric field in the space = 
- Q = Charge placed in the region =
<u>Assume:</u>
= Electric force on the charge due to electric field
We know that the electric field is the electric force applied on a unit positive charge i.e.,
This means the electric force applied on this additional charge placed in the field is given by:
From the above expression of force, we have the following y-component of force on this additional charge.
Hence, the y-component of the electric force on the this charge is 
.
Answer:
a = 6.1 m / s²
Explanation:
For this kinematics exercise, to solve the exercise we must set a reference system, we place it in the initial position of the fastest vehicle
Let's find the relative initial velocity of the two vehicles
v₀ = v₀₂ - v₀₁
v₀ = 25.4 - 13.6
v₀ = 11.8 m / s
the fastest vehicle
x = v₀ t + ½ a t²
The faster vehicle has an initial speed relative to the slower vehicle, therefore it is as if the slower vehicle were stopped, so the distance that must be traveled in a fast vehicle to reach this position is
x = 11.4 m
let's use the expression
v² = v₀² - 2 a x
how the vehicle stops v = 0
a = v₀² / 2x
a = 
a = 6.1 m / s²
this velocity is directed to the left
Answer:
5..No work is being done.
Explanation:
Hello!
Remember that the concept of work is defined as the force required to move a certain body distance, it is calculated as the product of force by distance.
therefore, the work required to raise the box is 150J.
However, the work required to keep the box lifted without moving is zero since although the box has a force due to the weight it does not move.
