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3 years ago

9

A metal foil has a threshold frequency of 5.45×1014 hz . Which of the colors of visible light have enough energy to eject electr

ons from this metal?

1 answer:

trapecia [35]3 years ago

8 0

Answer:

Green, blue, indigo and violet.

Explanation:

The velocity of a wave is given by

$v=f\lambda$

where

f = Frequency = $5.45\times 10^{14}\ Hz$

$\lambda$ = Wavelength

$\lambda=\frac{v}{f}\\\Rightarrow \lambda=\frac{3\times 10^8}{5.45\times 10^{14}}\\\Rightarrow \lambda=5.50459\times 10^{-7}\ m=550.459\times 10^{-9}\ m\\ =550.459\ nm$

The wavelength of the wave is 550.459 nm

Green light has wavelength which ranges from 500-565 nm.

So, the light which observed is green light and lower in wavelength than green light i.e., green, blue, indigo and violet.

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you are traveling in a convertible with the top down. the car is moving at a constant velocty of 25 m/s due east along falt grou

velikii [3]

Answer:

Teh velocity of ball with respect to the ground is 29.2 m/s.

Explanation:

Velocity of car, vc = 25 due m/s east

velocity of ball with respect to car, v(b,c) = 15 m/s due Z axis

Write the velocities in the vector form

$\overrightarrow{v}_{c}=25 \widehat{i}\\\\\overrightarrow{v}_{(b,c)}=15 \widehat{k}$

The velocity of ball with respect to ground is

$\overrightarrow{v}_{(b,c)}= \overrightarrow{v}_{(b,g)}-\overrightarrow{v}_{(c,g)}\\\\15 \widehat{k} = \overrightarrow{v}_{(b,g)} - 25 \widehat{i}\\\\\overrightarrow{v}_{(b,g)} = 25 \widehat{i} + 15 \widehat{k}\\\\v_{(b,g)}=\sqrt{25^2 + 15^2}=29.2 m/s$

3 0

3 years ago

The Chartered Institute of Management Accountants defines activity-based cost management as follows: "An approach to the costing

egoroff_w [7]

Answer:

ABC is the activity based accounting. It is the costing done for each separate activity which maybe unit level, batch level, product sustaining or facility sustaining.

Explanation:

The basic difference between ABC and traditional costing systems can be explained with the help of the following diagram.

Traditional Costing System

Overhead Cost Accounts (For each individual expense e.g. tax)

First Stage Allocation

Cost Centers ( normally Departments)
Cost Centers ( normally Departments)

N. Cost Centers ( normally Departments)

Second Stage Allocation ( Direct Labor Or Machine Hours)

Cost Objects ( products, services and customers)

ACTIVITY BASED COSTING SYSTEMS

Overhead Cost Accounts (For each individual expense e.g. tax)

First Stage Allocation ( resource cost drivers)

Activity Cost Centers
Activity Cost Centers

N. Activity Cost Centers

Second Stage Allocation ( activity cost drivers)

Cost Objects ( products, services and customers) (Direct Costs)

Four steps are involved in the design of ABC systems.

identifying the major activities that take place in an organization
assigning cost to cost pools / cost centers for each activity
determining the cost for each major activity
assigning the cost of activities to products according to the product's demand for activities.

The first two steps relate to the first stage and final two steps to the second stage of the two - stage allocation process shown above.

5 0

3 years ago

S Consider a ring of radius R with the total charge Q spread uniformly over its perimeter. What is the potential difference betw

WITCHER [35]

There the potential difference is -0. 553 keQ/R.

What is Electric potential ?

The amount of labor required to convey a unit of electric charge from a reference point to a given place in an electric field is known as the electric potential (also known as the electric field potential, potential drop, or the electrostatic potential).

Electric potential at point 0,

V1 = ke Q/R

Electric potential at x= 2R

V2= keQ/ root5. R

Therefore potential difference is

dV = V2 - V1

Putting the values we get ,

dV = -0. 553 keQ/R

To learn more about electric potential click on the link below:

brainly.com/question/23980243

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6 0

1 year ago

Two capacitors, C1 = 25.0 μF and C2 = 31.0 μF, are connected in series, and a 6.0-V battery is connected across them.a. Find the

Brrunno [24]

Answer:

$13.83928\times 10^{-6}\ F$

$249.10704\times 10^{-6}\ J$

$137.89848\times 10^{-6}\ J$

$111.20842\times 10^{-6}\ J$

2.98273 V

Explanation:

$C_1=25\ mu F$

$C_2=31\ mu F$

V = Voltage = 6 V

Equivalent capacitance is given by

$\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}\\\Rightarrow C=\dfrac{C_1C_2}{C_1+C_2}\\\Rightarrow C=\dfrac{25\times 10^{-6}\times 31\times 10^{-6}}{(25+31)\times 10^{-6}}\\\Rightarrow C=13.83928\times 10^{-6}\ F$

Equivalent capacitance is $13.83928\times 10^{-6}\ F$

Energy stored is given by

$E=\dfrac{1}{2}CV^2\\\Rightarrow E=\dfrac{1}{2}\times 13.83928\times 10^{-6}\times 6^2\\\Rightarrow E=249.10704\times 10^{-6}\ J$

Total energy stored is $249.10704\times 10^{-6}\ J$

Charge is given by

$Q=CV\\\Rightarrow Q=13.83928\times 10^{-6}\times 6\\\Rightarrow Q=83.03568\times 10^{-6}\ C$

Voltage is given by

$V_1=\dfrac{Q}{C_1}\\\Rightarrow V_1=\dfrac{83.03568\times 10^{-6}}{25\times 10^{-6}}\\\Rightarrow V_1=3.3214272\ V$

$E_1=\dfrac{1}{2}C_1V_1^2\\\Rightarrow E_1=\dfrac{1}{2}\times 25\times 10^{-6}\times 3.3214272^2\\\Rightarrow E_1=137.89848\times 10^{-6}\ J$

Energy strored in C1 is $137.89848\times 10^{-6}\ J$

$V_2=\dfrac{Q}{C_2}\\\Rightarrow V_2=\dfrac{83.03568\times 10^{-6}}{31\times 10^{-6}}\\\Rightarrow V_2=2.67857\ V$

$E_2=\dfrac{1}{2}C_2V_2^2\\\Rightarrow E_2=\dfrac{1}{2}\times 31\times 10^{-6}\times2.67857^2\\\Rightarrow E_2=111.20842\times 10^{-6}\ J$

Energy stored in C2 is $111.20842\times 10^{-6}\ J$

$E=E_1+E_2\\\Rightarrow E=137.89848\times 10^{-6}+111.20842\times 10^{-6}\\\Rightarrow E=249.107\times 10^{-6}\ J$

So, the energy is equivalent

Equivalent capacitance

$C=C_1+C_2\\\Rightarrow C=25+31\\\Rightarrow C=56\times 10^{-6}\ F$

$E=\dfrac{1}{2}CV^2\\\Rightarrow V=\sqrt{\dfrac{2E}{C}}\\\Rightarrow V=\sqrt{\dfrac{2\times 249.107\times 10^{-6}}{56\times 10^{-6}}}\\\Rightarrow V=2.98273\ V$

The voltage would be 2.98273 V

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4 years ago

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Agata [3.3K]

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