The main equation we are using to measure the e/m ratio is:

Charged beads are placed at the corners of a square in the various configurations shown in

Akimi4 [234]

Answer:

The consecutive charge configuration has a more intense field than alternating

Explanation:

In each corner we place a different account there are only two different settings, see attached.

In the case of alternating charging (+ - + -) see diagram 1, the electric field in the center is canceled in pairs, resulting in a zero field

In the case of consecutive loads (+ + - -) in this case we have a result between the two charges, therefore the total field is

E = 2 k q / ra2 a cos 45

The consecutive charge configuration has a more intense field than alternating

8 0

2 years ago

A diffusion couple composed of two silver– gold alloys is formed; these alloys have compositions of 98 wt% Ag–2 wt% Au and 95 wt

jeyben [28]

Answer:

for this problem, 2.5 = (5+2/2)-(5-2/2)erf (50×10-6m/2Dt)

It now becomes necessary to compute the diffusion coefficient at 750°C (1023 K) given that D0= 8.5 ×10-5m2/s and Qd= 202,100 J/mol.

we have D= D0exp( -Qd/RT)

=(8.5×105m2/s)exp(-202,100/8.31×1023)

= 4.03 ×10-15m2/s

4 0

2 years ago

A child of mass m is at the edge of a merry-go-round of diameter d. When the merry-go-round is rotating with angular acceleratio

dem82 [27]

Answer:

The torque on the child is now the same, τ.

Explanation:

It can be showed that the external torque applied by a net force on a rigid body, is equal to the product of the moment of inertia of the body with respect to the axis of rotation, times the angular acceleration.
In this case, as the movement of the child doesn't create an external torque, the torque must remain the same.
The moment of inertia is the sum of the moment of inertia of the merry-go-round (the same that for a solid disk) plus the product of the mass of the child times the square of the distance to the center.
When the child is standing at the edge of the merry-go-round, the moment of inertia is as follows:

$I_{to} = I_{d} + m*r^{2} = m*\frac{r^{2}}{2} + m*r^{2} = \frac{3}{2}* m*r^{2} (1)$

So, τ = 3/2*m*r²*α (2)

When the child moves to a position half way between the center and the edge of the merry-go-round, the moment of inertia of the child decreases, as the distance to the center is less than before, as follows:

$I_{t} = I_{d} + m*\frac{r^{2}}{4} = m*\frac{r^{2}}{2} + m*\frac{r^{2}}{4} = \frac{3}{4}* m*r^{2} (3)$

Since the angular acceleration increases from α to 2*α, we can write the torque expression as follows:

τ = 3/4*m*r² * (2α) = 3/2*m*r²

same result than in (2), so the torque remains the same.

7 0

2 years ago

The following represents a process used to assemble a chair with an upholstered seat. stationsa, b, and c make the seat; station

Mekhanik [1.2K]

There are three questions here:

A. The possible daily output of this process if there is 8 processing time each day?

the time it takes to assemble a chair in seconds = A + B + C + J + K + L + X + Y + Z

which is equal to = 38 + 34 + 35 + 32 + 30 + 34 + 22 + 18 + 20 =263 per chair

Processing hours = 8 hours x 60 minutes x 60 seconds

= 8 x 60 x 60

= 480 x 60 = 28,800 seconds is available in an 8 hr day.

28,800 / 263 =109.5057034220532

Therefore, It is possible to make 109 chairs in an 8 hour day.

B. Given your output rate in above, what is the efficiency of the process?

The time it takes to assemble a chair in seconds = A+ B + C + J + K + L + X + Y + Z.

= 34 + 34 + 34 + 30 + 30 + 30 + 22 + 18 + 20 = 252.6315789473684 per chair

A total of 252 per chair

Processing hours = 8 hours x 60 minutes x 60 seconds

= 8 x 60 x 60

= 480 x 60 = 28,800 seconds is available in an 8 hour day

= 28,000 / 252 = 114.2857142857143114 chairs

= 109/114 x 100 = 95.6140350877193

Therefore, the efficiency of making the chair is 95.61%

C. What is the flow time of the process?

Flow time is the period that it takes a completed chair to flow through the process from the beginning assemblage step to the last step. Take note that ABC and JKL are parallel legs in the process, and as a result both do not include to flow time to the procedure. In addition, Flow time comprises both pass time and run time at each position in the procedure.

7 0

2 years ago

You, Archimedes, suspect that the king’s crown is not solid gold but is instead gold-plated lead. To test your theory, you weigh

hjlf

Answer:

a) 16675.75 Kg/m³ b) 77.6%

Explanation:

the weight of the crown = 60 N, density of gold = 19300 Kg/m^3, density of lead = 11340 kg/m^3, density of water = 1000kg/m^3 and acceleration due to gravity = 9.8 m/s^2

upthrust on the crown = weight in air - weight when fully submerged in water = 60 - 56.4 = 3.6 N

mass of water displaced = 3.6 / 9.8 since weight = mass × g

mass of water displaced = 0.367 Kg

density of water = mass / volume

1000 = 0.367 / volume

cross multiply and find volume

volume of the crown = 0.367 / 1000 = 0.000367 m³ since the crown will displace water of equal volume according to Archimedes principle

Let V1 represent the volume of Gold and let V2 represent the volume of lead

Total volume of the crown = V1 + V2

also

density of gold = mass of gold / V1 and density of lead = mass of lead / V2

19300 = mass of gold in the crown / V1 and 11340 = mass of lead in the crown / V2

19300 V1 = mass of gold and 11340 V2 = mass of lead

add the two together

19300 V1 + 11340 V2 = weigth of the crown / 9.8

19300 V1 + 11340 V2 = 6.12 also

V1 + V2 = 0.000367

make V1 subject of the formula in equation 2

V1 = 0.000367 - V2

substitute for V1 in equation 1

19300 (0.000367 - V2) + 11340 V2 = 6.12

open the bracket

7.083 - 19300 V2 + 11340 V2 = 6.12

rearrange the equation

-7960 V2 = 6.12 - 7.083

-7960 V2 = -0.963

V2 = -0.963 / -7960 = 0.000121 (volume of lead in the crown)

substitute V2 into equation 2

V1 + 0.000121 = 0.000367m³

V1 = 0.000367 - 0.000121 = 0.000246m³ (volume of gold in the crown)

so mass of gold in the crown = 19300 × 0.000246 = 4.748 kg

and mass of lead = 11340 × 0.000121 = 1.372 kg

average density of the crown = (mass of gold + mass of lead) / total volume = 6.12 / 0.000367 = 16675.75 kg/ m³

b) percentage make of gold = mass of gold / total mass × 100 = 77.6 % approx

4 0

2 years ago