Answer:
the average force 11226 N
Explanation:
Let's analyze the problem we are asked for the average force, during the crash, we can find this from the impulse-momentum equation, but this equation needs the speeds and times of the crash that we could look for by kinematics.
Let's start looking for the stack speeds, it has a free fall, from rest (Vo=0)
Vf² = Vo² - 2gY
Vf² = 0 - 2 9.8 7.69 = 150.7
Vf = 12.3 m / s
This is the speed that the battery likes when it touches the beam. They also give us the distance it travels before stopping, let's calculate the time
Vf = Vo - g t
0 = Vo - g t
t = Vo / g
t = 12.3 / 9.8
t = 1.26 s
This is the time to stop
Now let's use the equation that relates the impulse to the amount of movement
I = Δp
F t = pf-po
The amount of final movement is zero because the system stops
F = - po / t
F = - mv / t
F = - 1150 12.3 / 1.26
F = -11226 N
This is the average force exerted by the stack on the vean
Answer:
D) equal to the flux of electric field through the Gaussian surface B.
Explanation:
Flux through S(A) = Flux through S (B ) = Charge inside/ ∈₀
Answer:
θ = (7π / 3) rad
Explanation:
given,
displacement of simple harmonic motion along x-axis
equation is given as
x = 5 sin (π t + π/3 )
general equation of simple harmonic motion
x = A sin θ
θ is the phase angle
θ = π t + π/3
at t = 2 s


Phase of the motion at t =2 s is θ = (7π / 3) rad
Answer:
The x-component of the electric field at the origin = -11.74 N/C.
The y-component of the electric field at the origin = 97.41 N/C.
Explanation:
<u>Given:</u>
- Charge on first charged particle,

- Charge on the second charged particle,

- Position of the first charge =

- Position of the second charge =

The electric field at a point due to a charge
at a point
distance away is given by

where,
= Coulomb's constant, having value 
= position vector of the point where the electric field is to be found with respect to the position of the charge
.
= unit vector along
.
The electric field at the origin due to first charge is given by

is the position vector of the origin with respect to the position of the first charge.
Assuming,
are the units vectors along x and y axes respectively.

Using these values,

The electric field at the origin due to the second charge is given by

is the position vector of the origin with respect to the position of the second charge.

Using these values,

The net electric field at the origin due to both the charges is given by

Thus,
x-component of the electric field at the origin = -11.74 N/C.
y-component of the electric field at the origin = 97.41 N/C.
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