A 4.5-m-long wooden board with a 24-kg mass is supported in two places. One support is directly under the center of the board, a

<span>Let Q be the charge, thus Q = -20.0 µC.</span>
Define D as the distance between the center of the rod and the specified point. Therefore,
D=0.32 - 0.12 = 0.2 m
<span>L = 0.12 m, which represents the length of the rod
</span><span>To find the magnitude and direction of the electric field along the axis of the rod at a point 32.0 cm from its center, use the formula:
</span><span>E = K·Q/r²
</span>or<span>E = kQ/D(D+L), where k</span> is a constant equal to 8.99 x 10<span>9</span> N m
2/C2.<span>Consequently,[TAG_21]]E=(</span>8.99 x 109 N m2/C2.* (-20.0 µC))/(<span>0.2 m*0.32m)</span><span>

</span>

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.

The work performed to elevate the box equals 515.03 J.

Answer:2.53*10^-10F

Explanation:

C=£o£r*A/d

Where £ represents the permittivity constant

£o= 8.85*10^-12f/m

£r=6.3

A=150mm^2=0.015m^2

d=3.3mm= 0.0033m

C=8.85*10^-12*6.3*0.015/0.0033

C=8.85*6.3*10^-12*0.015/0.0033

C=55.755*0.015^-12/0.003

C=8.36/3.3*10^-13+3

C=2.53*10^-10F

Response:

y= 240/901 cos 2t+ 8/901 sin 2t

Clarification:

To determine mass m=weight/g

  m=8/32=0.25

To calculate the spring constant

Kx=mg    (with c=6 inches and mg=8 pounds)

K(0.5)=8               (6 inches converts to 0.5 feet)

K=16 lb/ft

The governing equation for the spring-mass system is

my''+Cy'+Ky=F  

Inserting the known values yields

0.25 y"+0.25 y'+16 y=4 cos 20 t  ----(1) (given C=0.25 lb.s/ft)

Assuming the steady state equation for y is

y=A cos 2t+ B sin 2t

To determine constants A and B, we must equate this with equation 1.

Next, we find y' and y" by differentiating with respect to t.

y'= -2A sin 2t+2B cos 2t

y"=-4A cos 2t-4B sin 2t

Now, substitute the values of y", y' and y into equation 1

0.25 (-4A cos 2t-4B sin 2t)+0.25(-2A sin 2t+2B cos 2t)+16(A cos 2t+ B sin 2t)=4 cos 20 t

By comparing coefficients on both sides

30 A+ B=8

A-30 B=0

From this, we find

A=240/901 and B=8/901

Thus, the steady state response

y= 240/901 cos 2t+ 8/901 sin 2t