Answer:
v = 54.2 m/s
Explanation:
We can utilize conservation of energy to solve this issue.
Initial condition Higher
Em₀ = U = m g h
Final condition. Lower
= K = ½ m v²
Em₀ = Em_{f}
m g h = ½ m v²
v² = 2gh
v = √ (2gh)
Now let's perform the calculation
v = √ (2 * 9.8 * 150)
v = 54.2 m/s
Response:
a) 80 V
b) The electric field has a strength of 100 N/C, directed from point B toward point A, where the charge is negative.
Clarification:
Given:
An object with a charge of q = -6.00 x 10^-9 C starts from rest at point A, making its kinetic energy zero ( = 0) and moving to point B at a distance l = 0.500m where its kinetic energy is (
= 5.00 x 10^-7J). Additionally, the electric potential of q at point A is VA = +30.0 v.
Required:
(a) We seek to find the electric potential VB
(b) We need to compute the magnitude and orientation of the electric field E.
Solution
(a) Utilizing the given values for VA, and q, we derive a relationship among the three parameters and VB to compute VB.
.........................................(1)
Where = 0, and the potential energy U of the charge is defined as U = q V
In this equation, V represents the electric potential. Thus, equation (1) can be expressed as:
0+qVA= +qVB (Dividing by q)
VA= /q + VB (Restructuring for VB)
VB=VA- /q.......................................(2)
We now have the relation between VB, VA, and , allowing us to substitute our values for VA,
, and q into equation (2) to obtain VB
VB=VA- /q
=30V-(5.00 x 10^-7J)/(-6.00 x 10^-9)
=80 V
(b) After calculating VB, we may use equation a to derive the electric field E affecting the charge q, where the potential difference between the two points equals the integration of the electric field multiplied by the distance l between these points
VA-VB =...................................(a)
VA-VB=E (Restructuring for E)
E= VA-VB/..................................(3)
Now, substituting our values for VA, Vs, and l into equation (3) allows us to compute the electric field E
E= VA-VB/
=-100 N/C
The electric field's magnitude equals 100 N/C and it directs from point B to point A towards the negative charge.
Answer:
The runner's deceleration is -23.33
Given:
Initial speed = 3.5
Final speed = 0
Time taken = 0.15 s
To determine:
Deceleration of the runner =?
Used Formula:
Using the first equation of motion,
v = u + at
Where, v = final speed
u = initial speed
a = deceleration
t = duration
Solution:
<pusing the="" first="" equation="" of="" motion="">v = u + at
Where, v = final speed
u = initial speed
a = deceleration
t = duration
0 = 3.5 + a (0.15)
-3.5 = 0.15 (a)
a =
a = -23.33
The negative sign indicates that this represents deceleration.
Hence, the deceleration of the runner is -23.33
Explanation:
The formula for the electric field produced by an infinite sheet of charge is outlined below.
E =
where, is the surface charge density
Following this, the formula for the electric force acting on a proton is given as:
F = eE
where, e is the charge of a proton
According to Newton's second law of motion, the overall force on the proton can be expressed as follows.
F = ma
a =
=
=
According to kinematic equations, the proton's speed in the perpendicular direction can be described as follows.
=
=
= 683.974 m/s
Thus, the overall speed of the proton can be calculated as follows.
v' =
=
= 1178.73 m/s
Consequently, we conclude that the proton's speed is 1178.73 m/s.