physics approach to study macromoelcues at nanoscales
in detail plx

Answer:F = 255 N

Explanation:

It is given that,

Mass of the boy, m = 25 kg

Acceleration of the elevator,  

The elevator is accelerating in upward direction. The net force acting on the boy is given by :

g is the acceleration due to gravity

F = 255 N

The scale reading is 255 N as it begins to accelerate upward. hence, this is the required solution.

Answer:

C. Mass

Explanation:

Answer:

use light of the same wavelength but decrease it's intensity

Answer:

Explanation:

150/5  = 30

30mph per 1 second

I assume friction is the only force acting on the book as it slides.

(A) By the work-energy theorem, the total work performed on the book as it slides is equal to the change in its kinetic energy:

W = ∆K

W = 1/2 (1.50 kg) (1.25 m/s)² - 1/2 (1.50 kg) (3.21 m/s)²

W ≈ -6.56 J

(B) Using the work-energy theorem again, the speed v of the book at point C is such that

-0.750 J = 1/2 (1.50 kg) v ² - 1/2 (1.50 kg) (1.25 m/s)²

==>   v = 0.750 m/s

(C) Take the left side to be positive, then solve again for v.

0.750 J = 1/2 (1.50 kg) v ² - 1/2 (1.50 kg) (1.25 m/s)²

==>   v ≈ 1.60 m/s

Answer:

A solenoid is a type of electromagnet, the purpose of which is to generate a controlled magnetic field through a coil wound into a tightly packed helix. The coil can be arranged to produce a uniform magnetic field in a volume of space when an electric current is passed through it.

The car cannot continue to travel in the circular path, if the radius of the circular track is less than 2.57 m.

Centripetal force is described as the force applied to a body that is travelling in a circular motion and is pointed in the direction towards the center of the circular path.

Here,

Mass of the car, m = 0.5 kg

Velocity of the car, v = 6 m/s

Radial friction between the car and the track, f = 7 N

The necessary centripetal force for the car to execute the circular motion is provided by the maximum radial frictional force between the car and the track.

So, the condition that the car cannot continue to travel in the circular path is that the centripetal force required is greater than the maximum radial friction.

So,

mv²/r > f

0.5 x 6²/r > 7

Therefore, the radius of the circular track,

r < 18/7

r < 2.57 m

Hence,

The car cannot continue to travel in the circular path, if the radius of the circular track is less than 2.57 m.

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Answer:

Explanation:

       Kirchhoff's Current Law, often shortened to KCL, states that “The algebraic sum of all currents entering and exiting a node must equal zero.

           #I AM ILLITERATE

Answer:

[tex]5/27[/tex]

Explanation:

wavelengths for Lyman series

[tex]\lambda=\frac{1}{R(1-\frac{1}{4} })=\frac{4}{3R}[/tex]

wavelengths for Balmer series

[tex]\lambda_B=\frac{1}{R(\frac{1}{4}-\frac{1}{9}) } =\frac{1}{R(\frac{5}{36}) } =\frac{36}{5R}[/tex]

[tex]\frac{ \lambda_L}{ \lambda_B} =\frac{4}{3R} \times\frac{5R}{36} =5/27[/tex]

OAmalOHopeO

The ratio of longest wavelengths corresponding to the Lyman and Balmer series in the hydrogen spectrum is 5/27. The correct option is 3.

Lyman and Balmer series are sets of spectral lines in the emission spectrum of hydrogen, which result from the transitions of the electron from higher energy levels to lower energy levels.

The Lyman series consists of spectral lines that are produced by transitions of the electron from higher energy levels to the n=1 energy level. These transitions release energy in the form of ultraviolet photons. The lowest energy level in hydrogen is the n=1 energy level, which is also called the ground state. Therefore, the Lyman series includes the transition of the electron from any energy level greater than or equal to n=2 to the ground state.

The Balmer series consists of spectral lines that are produced by transitions of the electron from higher energy levels to the n=2 energy level. These transitions release energy in the form of visible photons. The lowest energy level in the Balmer series is the n=2 energy level. Therefore, the Balmer series includes the transition of the electron from any energy level greater than or equal to n=3 to the n=2 energy level.

Lyman and Balmer's series are named after the scientists who discovered them. The Lyman series is named after Theodore Lyman, an American physicist who discovered the series in 1906. The Balmer series is named after Johann Balmer, a Swiss mathematician who discovered the series in 1885.

Here in the Question,

The longest wavelength in the Lyman series of the hydrogen spectrum corresponds to the transition from the n = 2 energy level to the n = 1 energy level, while the longest wavelength in the Balmer series corresponds to the transition from the n = 3 energy level to the n = 2 energy level.

The wavelengths of these transitions can be calculated using the Rydberg formula:

1/λ = R(1/n1^2 - 1/n2^2)

where λ is the wavelength of the photon emitted, R is the Rydberg constant (1.097 × 10^7 m^-1), and n1 and n2 are the initial and final energy levels of the electron.

For the longest wavelength in the Lyman series, we have n1 = 2 and n2 = 1, so:

1/λ_lyman = R(1/2^2 - 1/1^2) = 3R/4

For the longest wavelength in the Balmer series, we have n1 = 3 and n2 = 2, so:

1/λ_balmer = R(1/3^2 - 1/2^2) = 5R/36

Therefore, the ratio of the longest wavelengths in the Lyman and Balmer series is:

λ_lyman/λ_balmer = (3R/4)/(5R/36) = 27/20

Simplifying this ratio gives:

λ_lyman/λ_balmer = 27/20

Multiplying both the numerator and denominator by 1/3R, we get:

λ_lyman/λ_balmer = (1/2)/(1/3) = 3/2

Therefore, the ratio of the longest wavelengths in the Lyman and Balmer series is 3:2, or 3/5 in fractional form. Simplifying this ratio gives:

λ_lyman/λ_balmer = 5/3

Taking the reciprocal of both sides, we get:

λ_balmer/λ_lyman = 3/5

Therefore, the correct answer is (3) 5/27.

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Answer:

Sweat also contains ammonia and urea, which are produced by the body when it breaks down proteins from the foods you eat.

Hope this helps..

Answer:

It tell you that the velocity is constant, what means that there's no acceleration

Answer:

well its about our thinking but i do believe in ghost a little

Answer:

0.032 [tex]m/s^2[/tex]

Explanation:

Given :

Weight of the astronaut = 69 kg

Weight of the physics book = 28 kg

Acceleration of the astronaut = 0.0130 [tex]m/s^2[/tex]

The force that is applied on the astronaut :

[tex]F=ma[/tex]

   [tex]$=69 \times 0.013$[/tex]

   = 0.897 N

Therefore, by Newton's 3rd law, we know that the force applied on the physics book is also F = 0.897 N

Therefore, the acceleration of the physics book is given by :

[tex]$a = \frac{\text{Force on physics book}}{\text{mass of physics book}}$[/tex]

[tex]$a = \frac{0.897}{28}$[/tex]

a = 0.032 [tex]m/s^2[/tex]

Hence, the acceleration of the physics book is  0.032 [tex]m/s^2[/tex].

Answer:

The acceleration of astronaut is 5.27 x 10^-3 m/s^2.

Explanation:

mass of astronaut, M = 69 kg

Mass of book, m = 28 kg

acceleration of book, a = 0.013 m/s^2

Let the acceleration of astronaut is A.

According to the Newton's third law, for every action there is an equal and opposite reaction.

So, the force acting on the book is same as the force acting on the astronaut but the direction is opposite to each other.

M A = m a

69 x A = 28 x 0.013

A = 5.27 x 10^-3 m/s^2

Answer:

6v

Explanation:

V=IR

V= 2* 3

V= 6 volts

The question is incomplete. The complete question is :

A wire 0.6 m long and with mass m = 11 g is positioned horizontally near the earth's surface and perpendicular to a horizontal magnetic field of magnitude B = 0.4 T. What current I must flow through the wire in order that the wire accelerate neither upwards nor downwards? The magnetic field is directed into the page.

Solution :

Given :

Length of the wire, L = 0.6 m

Mass of the wire length, m = 11 g

                                             = [tex]11 \times 10^{-3}[/tex] kg

Magnetic field , B = 0.4 T

Know we know that :

ILB = mg

or [tex]$I=\frac{mg}{BL}$[/tex]

 [tex]$I= \frac{(11 \times 10^{-3})(9.81)}{(0.4)(0.6)}$[/tex]

 [tex]I=0.44963\ A[/tex]

 [tex]I = 449.63 \ mA[/tex]

Answer:

there are 3 photos attached. so check

Explanation:

Explanation:

I=2.5 Ampere ; V=100V ;t = 1 hour=60secs

We know Heat = VIt

H=100×2.5×60=15,000J

Complete Question

(c) It takes you 5 hours to to bring the turkey from 10.0°C to 75.0 °C. During that time, the electrical grid transfers a constant 2500.0 Watts of power into the the oven. Take the turkey and the air in the oven to be your system. What was the thermal transfer of energy between the system and the surroundings?

Answer:

[tex]Q=4.50 *10^7J[/tex]

Explanation:

From the question we are told that:

Time [tex]t=5hours[/tex]

Temperature rise [tex]dT= 65\textdegree[/tex]

Power [tex]P=2500.0 Watts[/tex]

Generally, the equation for Power is mathematically given by

[tex]P=\frac{Q}{t}[/tex]

Therefore

[tex]Q=2500*5*360[/tex]

[tex]Q=4.50 *10^7J[/tex]

The frequency is 4,6E14 Hz.

What is the frequency?

Frequency is the fee at which modern changes direction in step with 2nd. it's far measured in hertz (Hz), a worldwide unit of degree wherein 1 hertz is identical to 1 cycle in line with 2d. Hertz (Hz) = One hertz is the same as 1 cycle in step with the second. Cycle = One entire wave of alternating present-day voltage.

Frequency describes the number of waves that pass a hard and fast place in a given quantity of time. So if the time it takes for a wave to skip is half of 2d, the frequency is 2 per 2nd. If it takes 1/one hundred of an hour, the frequency is a hundred in step with hour.

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The question is incomplete. The complete question is :

Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.

What is the frequency of the sound?

Solution :

Given :

The distance between the two loud speakers, [tex]d = 1.8 \ m[/tex]

The speaker are in phase and so the path difference is zero constructive interference occurs.

At the point [tex]D[/tex], the speakers are out of phase and so the path difference is [tex]$=\frac{\lambda}{2}$[/tex]

Therefore,

[tex]$AD-BD = \frac{\lambda}{2}[/tex]

[tex]$\sqrt{(1.8)^2+(3)^2-3} =\frac{\lambda}{2}$[/tex]

[tex]$\lambda = 2 \times 0.4985$[/tex]

[tex]$\lambda = 0.99714 \ m$[/tex]

Thus the frequency is :

[tex]$f=\frac{v}{\lambda}$[/tex]

[tex]$f=\frac{340}{0.99714}$[/tex]

[tex]f=340.9744[/tex] Hz

Answer:

Here , mass of bucket ,m = 3.2 Kg

Now , let the tension in upper rope is T1

the tension in the middle rope is T2

a)

For lower bucket, balancing forces in vertical direction

T2 - mg = 0

T2 = mg

T2 = 3.2 *9.8

T2 = 31.36 N

tension in the middle rope is 31.36 N

For the upper bucket , balancing forces in vertical direction

T1 - T2 - mg = 0

T1 = T2 + 3.2 *9.8

T1 = 62.72 N

the tension in the upper rope is 62.72 N

B)

for a = 1.25 m/s^2

Using second law of motion ,for both the buckets

Fnet = ma

T1 - 2mg = 2m*a

T1 = 2*3.2*(9.8 +1.25)

T1 = 70.72 N

the tension in the upper rope is 70.7 N

Now , the lower bucket

Using second law of motion,

T2 - mg = ma

T2 = 3.2 * (9.8 + 1.25)

T2 = 35.36 N

the tension in the lower rope is 35.36 N

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

a) The mass flow rate through the nozzle can be calculated with the following equation:

[tex] \dot{m_{i}} = \rho_{i} v_{i}A_{i} [/tex]

Where:

[tex]v_{i}[/tex]: is the initial velocity = 20 m/s

[tex]A_{i}[/tex]: is the inlet area of the nozzle = 60 cm²  

[tex]\rho_{i}[/tex]: is the density of entrance = 2.21 kg/m³

[tex] \dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s [/tex]  

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

[tex] \rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f} [/tex]

[tex] 0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f} [/tex]

[tex] A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2} [/tex]

Therefore, the exit area of the nozzle is 23.6 cm².

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a) Mass flow rate through the nozzle: 0.265 kilograms per second, b) Exit area of the nozzle: 23.202 square centimeters.

We determine the Mass Flow Rate through the nozzle and the Exit Area of the nozzle by means of the Principle of Mass Conservation. A nozzle is a device that works at Steady State, so that Mass Balance can be reduced into this form:

[tex]\dot m_{in} = \dot m_{out}[/tex] (1)

Where:

[tex]\dot m_{in}[/tex] - Inlet mass flow, in kilograms per second.

[tex]\dot m_{out}[/tex] - Outlet mass flow, in kilograms per second.

Given that air flows at constant rate, we expand (1) by dimensional analysis:

[tex]\rho_{in} \cdot A_{in}\cdot v_{in} = \rho_{out}\cdot A_{out}\cdot v_{out}[/tex] (2)

Where:

[tex]\rho_{in}, \rho_{out}[/tex] - Air density at inlet and outlet, in kilograms per cubic meter.

[tex]A_{in}, A_{out}[/tex] - Inlet and outlet area, in square meters.

[tex]v_{in}, v_{out}[/tex] - Inlet and outlet velocity, in meters per second.

a) If we know that [tex]\rho_{in} = 2.21\,\frac{kg}{m^{3}}[/tex], [tex]A_{in} = 60\times 10^{-4}\,m^{2}[/tex] and [tex]v_{in} = 20\,\frac{m}{s}[/tex], then the mass flow rate through the nozzle is:

[tex]\dot m = \rho_{in}\cdot A_{in}\cdot v_{in}[/tex]

[tex]\dot m = \left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)[/tex]

[tex]\dot m = 0.265\,\frac{kg}{s}[/tex]

The mass flow rate through the nozzle is 0.265 kilograms per second.

b) If we know that [tex]\rho_{in} = 2.21\,\frac{kg}{m^{3}}[/tex], [tex]A_{in} = 60\times 10^{-4}\,m^{2}[/tex], [tex]v_{in} = 20\,\frac{m}{s}[/tex], [tex]\rho_{out} = 0.762\,\frac{kg}{m^{3}}[/tex] and [tex]v_{out} = 150\,\frac{m}{s}[/tex], then the exit area of the nozzle is:

[tex]\rho_{in} \cdot A_{in}\cdot v_{in} = \rho_{out}\cdot A_{out}\cdot v_{out}[/tex]

[tex]A_{out} = \frac{\rho_{in}\cdot A_{in}\cdot v_{in}}{\rho_{out}\cdot v_{out}}[/tex]

[tex]A_{out} = \frac{\left(2.21\,\frac{kg}{m^{3}} \right)\cdot (60\times 10^{-4}\,m^{2})\cdot \left(20\,\frac{m}{s} \right)}{\left(0.762\,\frac{kg}{m^{3}} \right)\cdot \left(150\,\frac{m}{s} \right)}[/tex]

[tex]A_{out} = 2.320\times 10^{-3}\,m^{2}[/tex]

[tex]A_{out} = 23.202\,cm^{2}[/tex]

The exit area of the nozzle is 23.202 square centimeters.

Answer:

5.71×10¹⁴ Hz

Explanation:

Applying,

v = λf................. Equation 1

Where v = speed of the electromagnetic radiation, λ = wavelength of the electromagnetic radiation, f = frequency

make f the subject of the equation

f = v/λ............. Equation 2

From the question,

Given: λ = 525 nm = 5.25×10⁻⁷ m,

Constant: Speed of electromagnetic wave (v) = 3.0×10⁸ m/s

Substitute these values into equation 2

f = (3.0×10⁸)/(5.25×10⁻⁷)

f = 5.71×10¹⁴ Hz

Hence the frequency of light is 5.71×10¹⁴ Hz

Answer:

No ejection of photo electron takes place.

Explanation:

When a photon of suitable energy falls on cathode, then the photoelectrons is emitted from the cathode. This phenomenon is called photo electric effect.

The minimum energy required to just  eject an electron is called work function.

The photo electric equation is

E = W + KE

where, E is the incident energy, W is the work function and KE is the kinetic energy.

W = h f

where. h is the Plank's constant and f is the threshold frequency.

Now, when the violet light is falling, no electrons is ejected. When the red light is falling, whose frequency is less than the violet light, then again no photo electron is ejected from the metal surface.

Answer:

21.55 m

Explanation:

Answer:

B. -40 kgm/s is the answer

Answer:

I think 9.5

Explanation:

............

Answer:

The SI unit of power is watt

The vector sum is the algebraic sum if the two vectors have the same direction.

The sum vector is zero if the two vectors have the same magnitude and opposite direction

Vector addition is a process that can be performed graphically using the parallelogram method, see  attached, where the second vector is placed at the tip of the first and the vector sum goes from the origin of the first vector to the tip of the second.

There are two special cases where the vector sum can be reduced to the algebraic sum if the vectors are parallel

case 1. if the two vectors are parallel, the sum vector has the magnitude of the sum of the magnitudes of each vector

case 2. If the two vectors are antiparallel and the magnitude of the two vectors is the same, the sum gives zero.

In summary in the sum of vectors If the vectors are parallel it is reduced to the algebraic sum, also in the case of equal magnitude and opposite direction the sum is the null vector

a) Magnitudes: [tex]\| \vec A\| \ge 0[/tex], [tex]\|\vec B\| \ge 0[/tex], [tex]\|\vec C\| \ge 0[/tex]; Directions: [tex]\theta_{A} \in (-\infty, +\infty)[/tex] for [tex]\|\vec A\|\ne 0[/tex]. Undefined for [tex]\|\vec A\| = 0[/tex], [tex]\theta_{B} \in (-\infty, +\infty)[/tex] for [tex]\|\vec B\|\ne 0[/tex]. Undefined for [tex]\|\vec B\| = 0[/tex], [tex]\theta_{C} \in (-\infty, +\infty)[/tex] for [tex]\|\vec C\|\ne 0[/tex]. Undefined for [tex]\|\vec C\| = 0[/tex].

b) Magnitudes: [tex]\|\vec A\| \ge 0[/tex], [tex]\|\vec B\| \ge 0[/tex], [tex]\|\vec C\| = 0[/tex]; Directions: [tex]|\theta_{A}-\theta_{B}| = 180^{\circ}[/tex], [tex]\theta_{C}[/tex] is undefined.

a) Let suppose that [tex]\vec A \ne \vec O[/tex], [tex]\vec B \ne \vec O[/tex] and [tex]\vec C \ne \vec O[/tex], where [tex]\vec O[/tex] is known as Vector Zero. By definitions of Dot Product and Inverse Trigonometric Functions we derive expression for the magnitude and directions of [tex]\vec A[/tex], [tex]\vec B[/tex] and [tex]\vec C[/tex]:

Magnitude ([tex]\vec A[/tex])

[tex]\|\vec A\| = \sqrt{\vec A\,\bullet\,\vec A}[/tex]

[tex]\| \vec A\| \ge 0[/tex]

Magnitude ([tex]\vec B[/tex])

[tex]\|\vec B\| = \sqrt{\vec B\,\bullet\,\vec B}[/tex]

[tex]\|\vec B\| \ge 0[/tex]

Magnitude ([tex]\vec C[/tex])

[tex]\|\vec C\| = \sqrt{\vec C\,\bullet \,\vec C}[/tex]

[tex]\|\vec C\| \ge 0[/tex]

Direction ([tex]\vec A[/tex])

[tex]\vec A \,\bullet \,\vec u = \|\vec A\|\cdot \|u\|\cdot \cos \theta_{A}[/tex]

[tex]\theta_{A} = \cos^{-1} \frac{\vec A\,\bullet\,\vec u}{\|\vec A\|\cdot \|u\|}[/tex]

[tex]\theta_{A} = \cos^{-1} \frac{\vec A\,\bullet\,\vec u}{\|\vec A\|}[/tex]

[tex]\theta_{A} \in (-\infty, +\infty)[/tex] for [tex]\|\vec A\|\ne 0[/tex]. Undefined for [tex]\|\vec A\| = 0[/tex].

Direction ([tex]\vec B[/tex])

[tex]\vec B\,\bullet \, \vec u = \|\vec B\|\cdot \|\vec u\| \cdot \cos \theta_{B}[/tex]

[tex]\theta_{B} = \cos^{-1} \frac{\vec B\,\bullet\,\vec u}{\|\vec B\|\cdot \|\vec u\|}[/tex]

[tex]\theta_{B} = \cos^{-1} \frac{\vec B\,\bullet\,\vec u}{\|\vec B\|}[/tex]

[tex]\theta_{B} \in (-\infty, +\infty)[/tex] for [tex]\|\vec B\|\ne 0[/tex]. Undefined for [tex]\|\vec B\| = 0[/tex].

Direction ([tex]\vec C[/tex])

[tex]\vec C \,\bullet\,\vec u = \|\vec C\|\cdot\|\vec u\|\cdot \cos \theta_{C}[/tex]

[tex]\theta_{C} = \cos^{-1}\frac{\vec C\,\bullet\,\vec u}{\|\vec C\|\cdot\|\vec u\|}[/tex]

[tex]\theta_{C} = \cos^{-1} \frac{\vec C\,\bullet\,\vec u}{\|\vec C\|}[/tex]

[tex]\theta_{C} \in (-\infty, +\infty)[/tex] for [tex]\|\vec C\|\ne 0[/tex]. Undefined for [tex]\|\vec C\| = 0[/tex].

Please notice that [tex]\vec u[/tex] is the Vector Unit.

b) Let suppose that [tex]\vec A \ne \vec O[/tex] and [tex]\vec B \ne \vec O[/tex] and [tex]\vec C = \vec O[/tex]. Hence, [tex]\vec A = -\vec B[/tex]. In other words, we find that both vectors are antiparallel to each other, that is, that angle between [tex]\vec A[/tex] and [tex]\vec B[/tex] is 180°. From a) we understand that [tex]\|\vec A\| \ge 0[/tex], [tex]\|\vec B\| \ge 0[/tex], but [tex]\|\vec C\| = 0[/tex].

Then, we have the following conclusions:

Magnitude ([tex]\vec A[/tex])

[tex]\|\vec A\| \ge 0[/tex]

Magnitude ([tex]\vec B[/tex])

[tex]\|\vec B\| \ge 0[/tex]

Magnitude ([tex]\vec C[/tex])

[tex]\|\vec C\| = 0[/tex]

Directions ([tex]\vec A[/tex], [tex]\vec B[/tex]):

[tex]|\theta_{A}-\theta_{B}| = 180^{\circ}[/tex]

Direction ([tex]\vec C[/tex]):

Undefined

Answer:

C. A force must be acting on the object.

Explanation:

This is due to the action of its momentum direction.

[tex].[/tex]