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larisa [96]
2 years ago
9

Hoosier Manufacturing operates a production shop that is designed to have the lowest unit production cost at an output rate of 1

15 units per hour. In the month of July, the company operated the production line for a total of 205 hours and produced 21,400 units of output. What was its capacity utilization rate for the month?
Physics
1 answer:
abruzzese [7]2 years ago
3 0

Answer:

90.77%

its capacity utilization rate for the month is 90.77%

Explanation:

The capacity utilisation rate can be expressed mathematically as;

Capacity utilisation rate = capacity used/Best operating level × 100%

Given;

Total Number of production time = 205hours

Production output/capacity used = 21400 units

Best operation rate = 115units/hour

Best operation output for the month of July( at best operation level )

=115units/hour × 205 hours = 23575 units

Capacity utilisation rate = 21400/23575 × 100%

= 90.77%

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4.A photon of green light strikes an unknown metal and an electron is emitted. The voltage is set to 2 volts. The electron canno
Anarel [89]
4) The correct answer is:
<span>B. An electron will be emitted in the second experiment, but it cannot be determined whether it will reach the second plate. 

In fact, violet light has higher frequency than green light. This means that photons of violet light carry more energy than photons of green light (remember that the energy of a photon is proportional to its frequency: </span>E=hf)<span>, so when they hit the surface of the metal, more energy is transferred to the electrons. The electron was already emitted with green light, so it must be emitted also with violet light, given the more energy transferred. The electron will also have more kinetic energy when hit by violet light, however, we cannot determine if it will reach the second plate, since we don't know how much energy has been used to extract the electron from the metal (in fact, we don't know the work function of the metal, i.e. the energy needed to extract the electron)


3)  The correct answer is
</span><span>A. Violet light will cause electrons to be emitted at greater velocities than those removed by green light.

In fact, </span>violet light has higher frequency than green light. This means that photons of violet light carry more energy than photons of green light (remember that the energy of a photon is proportional to its frequency: E=hf), so when they hit the surface of the metal, more energy is transferred to the electrons. Therefore, the emitted electrons will have on average greater energy (and so, greater velocity) than those removed by green light.
3 0
2 years ago
A spherical drop of water carrying a charge of 30 pC has a potential of 500 V at its surface (with V 0 at infinity). (a) What is
Delicious77 [7]

Answer:

A. 5.4 * 10^(-4) m

B. 500V

Explanation:

A. Electric potential, V is given as:

V = kq/r

This means that radius, r is

r = kq/V

r = (9 * 10^9 * 30 * 10^(-12))/500

r = (270 * 10^(-3))/500

r = 5.4 * 10^(-4) m

B. Now the radius is doubled and the charge is doubled,

V = (9 * 10^9 * 2 * 30 * 10^(-12))/(2 * 5.4 * 10^(-4) * 2)

V = 500V

7 0
2 years ago
Two 5.0 mm × 5.0 mm electrodes are held 0.10 mm apart and are attached to 7.5 V battery. Without disconnecting the battery, a 0.
Musya8 [376]

Answer:

A) V = 7.5 V

B) E = 75,000 V/m

C) Q = 16.6 pC

D) V = 7.5 V

E) E = 24,000 V/m

F) Q = 52 pC

Explanation:

Given:

- The Area of plate A = ( 5 x 5 ) mm^2

- The distance between plates d = 0.10 mm

- The thickness of Mylar added t = 0.10 mm

- Voltage supplied by battery V = 7.5 V

Solution:

A) What is the capacitor's potential difference before the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

B) What is the capacitor's electric field before the Mylar is inserted?

- The Electric Field E between the capacitor plates is given by:

                                E = V / k*d

k = 1 (air)                  E = 7.5 / 0.10*10^-3

                                E = 75,000 V/m

C) What is the capacitor's charge Q before the Mylar is inserted?

                                C = k*A*ε / d

k = 1 (air)                   C = ( 0.005^2 * 8.85*10^-12 ) / 0.0001

                                C = 2.213 pF

                                Q = C*V

                                Q = 7.5*(2.213)

                                Q = 16.6 pC

D) What is the capacitor's potential difference after the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

E) What is the capacitor's electric field after the Mylar is inserted?    

- The Electric Field E between the capacitor plates is given by:

                                E = V / k*d

k = 3.13                     E = 7.5 / (3.13)0.10*10^-3

                                E = 24,000 V/m              

F) What is the capacitor's charge after the Mylar is inserted?      

                                C = k*A*ε / d

k = 3.13                    C = 3.13*( 0.005^2 * 8.85*10^-12 ) / 0.0001

                                C = 6.927 pF

                                Q = C*V

                                Q = 7.5*(6.927)

                                Q = 52 pC                                      

6 0
2 years ago
The table below shows data of sprints of animals that traveled 75 meters. At each distance marker, the animals' times were recor
mr_godi [17]

Answer:

Explanation:

Animal 1 because it takes 3s to go 25 meters 3.5s to go 50 meters and 5s to go 75 meters while the others take longer.

5 0
2 years ago
Read 2 more answers
A particle moves along a straight line with a velocity in meters per second given by v = 12 - 3t + 8t, where t is in seconds. Wh
elena-14-01-66 [18.8K]

Answer:

Explanation:

Given the velocity of a particle modeled by the equation

v = 13-3t+8t² where t is in seconds

Given t = 0 and position s = +5m

A) To get the position as a function of time, we will integrate the function with respect to t ad shown;

v = 13-3t+8t²

S = ∫13-3t+8t² dt

S = 13t-13t²/2+8t³/3 + C

at t = 0 and S = +5m

5 = 13(0)-13(0)²/2+8(0)³/3+C

5 = 0-0+0+C

C = 5

Substituting c = 5 into the displacement function

S = 13t-13t²/2+8t³/3 + C

S = 13t-13t²/2+8t³/3 + 5

B) acceleration is the change in velocity with respect to time.

a = dv/dt

Given v = 13-3t+8t²

a= dv/dt = -3+16t

a = 16-3t

C) acceleration at t = 6s is derived by plugging in t = 6 into the resulting equations in (B)

a = 16-3t

a = 16-3(6)

a = 16-18

a = -2m/s²

D) net displacement from t = 0 to t = 6s

At t = 0:

S(0) = 13(0)-13(0)²/2+8(0)³/3 + 5

S(0) = 0+5

S(0) = 5m

At t = 6s

S(6) = 13(6)-13(6)²/2+8(6)³/3 + 5

S(6) = 78-234+576+5

S(6) = 425m

Net displacement from t = 0s to t = 6s is s(6)-s(0)

= 425-5

= 420m

E) Total distance travelled D = S(6)+S(0)

= 425+5

= 430m

F) Average velocity = ∆S/∆t

Average velocity = S(6)-S(0)/6-0

Average velocity = 425-5/6

Average velocity = 420/6

Average velocity = 70m/s

4 0
2 years ago
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