The larger the push, the larger the change in velocity. This is an example of Newton's Second Law of Motion which states that the acceleration an object experiences is

Answer:

a.     F = 2.32*10^-18 N

b.     The force F is 2.59*10^11 times the weight of the electron

Explanation:

a. In order to calculate the magnitude of the force exerted on the electron you first calculate the acceleration of the electron, by using the following formula:

[tex]v^2=v_o^2+2ax[/tex]         (1)

v: final speed of the electron = 6.60*10^5 m/s

vo: initial speed of the electron = 4.00*10^5 m/s

a: acceleration of the electron = ?

x: distance traveled by the electron = 5.40cm = 0.054m

you solve the equation (2) for a and replace the values of the parameters:

[tex]a=\frac{v^2-v_o^2}{2x}=\frac{(6.60*10^5m/s)^2-(4.00*10^5m/s)^2}{2(0.054m)}\\\\a=2.55*10^{12}\frac{m}{s^2}[/tex]

Next, you use the second Newton law to calculate the force:

[tex]F=ma[/tex]

m: mass of the electron = 9.11*10^-31kg

[tex]F=(9.11*10^{-31}kg)(2.55*10^{12}m/s^2)=2.32*10^{-18}N[/tex]

The magnitude of the force exerted on the electron is 2.32*10^-18 N

b. The weight of the electron is given by:

[tex]F_g=mg=(9.11*10^{-31}kg)(9.8m/s^2)=8.92*10^{-30}N[/tex]

The quotient between the weight of the electron and the force F is:

[tex]\frac{F}{F_g}=\frac{2.32*10^{-18}N}{8.92*10^{-30}N}=2.59*10^{11}[/tex]

The force F is 2.59*10^11 times the weight of the electron

Answer: No, we can have a displacement equal to 0 while the distance traveled is different than zero.

Explanation:

Ok, let's write the definitions:

Displacement: The displacement is equal to the difference between the final position and the initial position.

Distance traveled: Total distance that you moved.

So, for example, if at t = 0s, you are in your house, then you go to the store, and then you return to your house, we have:

The displacement is equal to zero, because the initial position is your house and the final position is also your house, so the displacement is zero.

But the distance traveled is not zero, because you went from you traveled the distance from your house to the store two times.

So no, we can have a displacement equal to zero, but a distance traveled different than zero.

Answer:

The answer is electromagnetic

Answer:

electromagnetic

Explanation:

edge 2021

Answer:

t = 0.414s

Explanation:

In order to calculate the time that the ball takes to reach home plate, you assume that the speed of the ball is constant, and you use the following formula:

[tex]t=\frac{d}{v}[/tex]         (1)

d: distance to the plate = 18.4m

v: speed of the ball = 160.0km/h

You first convert the units of the sped of the ball to appropriate units (m/s)

[tex]160.0\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=44.44\frac{m}{s}[/tex]

Then, you replace the values of the speed v and distance s in the equation (1):

[tex]t=\frac{18.4m}{44.44m/s}=0.414s[/tex]

THe ball takes 0.414s to reach the home plate

Answer:

Φ = 361872 N.m^2 / C

Explanation:

Given:-

- The area of the two plates, [tex]A_p = 180 cm^2[/tex]

- The charge on each plate, [tex]q = 17 * 10^-^6 C[/tex]

- Permittivity of free space, [tex]e_o = 8.85 * 10^-^1^2 \frac{C^2}{N.m^2}[/tex]

- The radius for the flux region, [tex]r = 3.3 cm[/tex]

- The angle between normal to region and perpendicular to plates, θ = 4°

Find:-

Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates.

Solution:-

- First we will determine the area of the region ( Ar ) by using the formula for the area of a circle as follows. The region has a radius of r = 3.3 cm:

                             [tex]A_r = \pi *r^2\\\\A_r = \pi *(0.033)^2\\\\A_r = 0.00342 m^2[/tex]

- The charge density ( σ ) would be considered to be uniform for both plates. It is expressed as the ratio of the charge ( q ) on each plate and its area ( A_p ):

                           σ = [tex]\frac{q}{A_p} = \frac{17*10^-^6}{0.018} \\[/tex]

                           σ = 0.00094 C / m^2

- We will assume the electric field due to the positive charged plate ( E+ ) / negative charged plate ( E- ) to be equivalent to the electric field ( E ) of an infinitely large charged plate with uniform charge density.

                         [tex]E+ = E- = \frac{sigma}{2*e_o} \\\\[/tex]

- The electric field experienced by a region between two infinitely long charged plates with uniform charge density is the resultant effect of both plates. So from the principle of super-position we have the following net uniform electric field ( E_net ) between the two plates:

                        [tex]E_n_e_t = (E+) + ( E-)\\\\E_n_e_t = \frac{0.00094}{8.85*10^-^1^2} \\\\E_n_e_t = 106214689.26553 \frac{N}{C} \\[/tex]

- From the Gauss-Law the flux ( Φ ) through a region under uniform electric field ( E_net ) at an angle of ( θ ) is:

                        Φ = E_net * Ar * cos ( θ )

                        Φ = (106214689.26553) * (0.00342) * cos ( 5 )

                        Φ = 361872 N.m^2 / C

Answer:

Yes we can induce current in the coil by moving the magnet in and out of the coil steadily.

Explanation:

A current can be induced there using the magnetic field and the coil of wire. Moving the bar magnet around the coil can induce a current and this is called electromagnetic induction.

The generation of an electromotive force  across an electrical conductor in a fluctuating magnetic field is known as electromagnetic or magnetic induction.

Induction was first observed in 1831 by Michael Faraday, and James Clerk Maxwell mathematically named it Faraday's law of induction. The induced field's direction is described by Lenz's law.

Electrical equipment like electric motors and generators as well as parts like inductors and transformers have all found uses for electromagnetic induction.

Here, moving the bar magnet around the coil generates the electronic movement followed by a generation of electric current.

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Answer:

The distance is  [tex]d = 1.5 *10^{15} \ km[/tex]

Explanation:

From the question we are told that

        The smallest shift is [tex]d = 0.2 \ grid \ units[/tex]

Generally a grid unit is  [tex]\frac{1}{10}[/tex] of  an arcsec

  This implies that  0.2 grid unit is  [tex]k = \frac{0.2}{10} = 0.02 \ arc sec[/tex]

The maximum distance at which a star can be located and still have a measurable parallax is mathematically represented as

           [tex]d = \frac{1}{k}[/tex]

substituting values

           [tex]d = \frac{1}{0.02}[/tex]

           [tex]d = 50 \ parsec[/tex]

Note  [tex]1 \ parsec \ \to 3.26 \ light \ year \ \to 3.086*10^{13} \ km[/tex]

So  [tex]d = 50 * 3.08 *10^{13}[/tex]

     [tex]d = 1.5 *10^{15} \ km[/tex]

Answer:

20 J

Explanation:

From the question, since there is a lost in kinetic energy, Then the collision is an inelastic collision.

m'u'+mu = V(m+m')........... Equation 1

Where m' = mass of the moving object, m = mass of the stick, u' = initial velocity of the moving object, initial velocity of the stick, V = common velocity after collision.

make V the subject of the equation above

V = (m'u'+mu)/(m+m')............. Equation 2

Given: m' = 2 kg, m = 8 kg, u' = 5 m/s, u = 0 m/s (at rest).

Substitute into equation 2

V = [(2×5)+(8×0)]/(2+8)

V = 10/10

V = 1 m/s.

Lost in kinetic energy = Total kinetic energy before collision- total kinetic energy after collision

Total kinetic energy before collision = 1/2(2)(5²) = 25 J

Total kinetic energy after collision = 1/2(2)(1²) +1/2(8)(1²) = 1+4 = 5 J

Lost in kinetic energy = 25-5 = 20 J

The collision is inelastic collision. As a result of collision the kinetic energy lost by the given system is 20 J.

Since there is a lost in kinetic energy, the collision is inelastic collision.  

m'u'+mu = V(m+m')

[tex]\bold {V =\dfrac { (m'u'+mu)}{(m+m')} }[/tex]  

Where

m' = mass of the moving object = 2 kg

m = mass of the stick = 8 kg,

u' = initial velocity of the moving object = 5 m/s

V = common velocity after collision= ?    

u = 0 m/s (at rest).

put the values in the formula,  

[tex]\bold {V = \dfrac {(2\times 5)+(8\times 0)}{(2+8)}}\\\\\bold {V = \dfrac {10}{10}}\\\\\bold {V = 1\\ m/s.}[/tex]

  kinetic energy before collision

[tex]\bold { = \dfrac 1{2} (2)(5^2) = 25 J}[/tex]  

kinetic energy after collision

[tex]\bold { = \dfrac 12(2)(1^2) + \dfrac 12(8)(1^2) = 5\ J}[/tex]  

Lost in kinetic energy = 25-5 = 20 J

Therefore, As a result of collision the kinetic energy lost by the given system is 20 J.

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Answer:

P = 5.97 × 10^(5) Pa

Explanation:

We are given;

Mass of balloon;m_b = 4.3 kg

Radius;r = 1.54 m

Temperature;T = 289 K

Density;ρ = 1.19 kg/m³

We know that, density = mass/volume

So, mass = Volume x Density

We also know that Force = mg

Thus;

F = mg = Vρg

Where m = mass of balloon(m_b) + mass of helium (m_he)

So,

(m_b + m_he)g = Vρg

g will cancel out to give;

(m_b + m_he) = Vρ - - - eq1

Since a sphere shaped balloon, Volume(V) = (4/3)πr³

V = (4/3)π(1.54)³

V = 15.3 m³

Plugging relevant values into equation 1,we have;

(3 + m_he) = 15.3 × 1.19

m_he = 18.207 - 3

m_he = 15.207 kg = 15207 g

Molecular weight of helium gas is 4 g/mol

Thus, Number of moles of helium gas is ; no. of moles = 15207/4 ≈ 3802 moles

From ideal gas equation, we know that;

P = nRT/V

Where,

P is absolute pressure

n is number of moles

R is the gas constant and has a value lf 8.314 J/mol.k

T is temperature

V is volume

Plugging in the relevant values, we have;

P = (3802 × 8.314 × 289)/15.3

P = 597074.53 Pa

P = 5.97 × 10^(5) Pa

Answer:

Explanation:

Given that

The electric fields of strengths E = 187,500 V/m and

and The magnetic  fields of strengths B = 0.1250 T

The diameter d is 25.05 cm which is converted to 0.2505m

The radius is (d/2)

= 0.2505m / 2 = 0.12525m

The given formula to find the magnetic force is [tex]F_{ma}=BqV---(i)[/tex]

The given formula to find the electric force is [tex]F_{el}=qE---(ii)[/tex]

The velocity of electric field and magnetic field is said to be perpendicular

Electric field is equal to magnectic field

Equate equation (i) and equation (ii)

[tex]Bqv=qE\\\\v=\frac{E}{B}[/tex]

[tex]v=\frac{187500}{0.125} \\\\v=15\times10^5m/s[/tex]

It is said that the particles moves in semi circle, so we are going to consider using centripetal force

[tex]F_{ce}=\frac{mv^2}{r}---(iii)[/tex]

magnectic field is equal to centripetal force

Lets equate equation (i) and (iii)

[tex]Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br} \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg[/tex]

Therefore,  the particle's charge-to-mass ratio is [tex]958.1\times10^5C/kg[/tex]

b)

To identify the particle

Then 1/ 958.1 × 10⁵ C/kg

The charge to mass ratio is very close to that of a proton, which is about 1*10^8 C/kg

Therefore the particle is proton.

Answer:

a) v₁ = 3.92 m / s , b)     ΔP =  = 9.0 10⁴ Pa, c)  t = 0.0297 s  

Explanation:

This is a fluid mechanics exercise

a) let's use the continuity equation

let's use index 1 for the hose and index 2 for the nozzle

        A₁ v₁ = A₂v₂

in area of ​​a circle is

       A = π r² = π d² / 4

we substitute in the continuity equation

        π d₁² / 4 v₁ = π d₂² / 4 v₂

        d₁² v₁ = d₂² v₂

the speed of the water in the hose is v1

       v₁ = v₂ d₂² / d₁²

       v₁ = 14 (1.8 / 3.4)²

        v₁ = 3.92 m / s

b) they ask us for the pressure difference, for this we use Bernoulli's equation

       P₁ + ½ ρ v₁² + m g y₁ = P₂ + ½ ρ v₂² + mg y2

as the hose is horizontal y₁ = y₂

       P₁ - P₂ = ½ ρ (v₂² - v₁²)

      ΔP = ½ 1000 (14² - 3.92²)

       ΔP = 90316.8 Pa = 9.0 10⁴ Pa

c) how long does a tub take to flat

the continuity equation is equal to the system flow

        Q = A₁v₁

        Q = V t

where V is the volume, let's equalize the equations

         V t = A₁ v₁

         t = A₁ v₁ / V

A₁ = π d₁² / 4

let's reduce it to SI units

         V = 120 l (1 m³ / 1000 l) = 0.120 m³

          d1 = 3.4 cm (1 m / 100cm) = 3.4 10⁻² m

let's substitute and calculate

         t = π d₁²/4   v1 / V

         t = π (3.4 10⁻²)²/4 3.92 / 0.120

         t = 0.0297 s

Given that,

Sun exposure = 30%, 45%, 60%, 75%, 90%

Stem mass (g) = 275, 415, 563, 815, 954

Stem volume (ml) = 1100, 1215, 1425, 1610, 1742

(a). We need to convert the mass measurements to kilograms (kg) and the volume measurements to cubic meters

Using conversion of mass

[tex]1\ g=0.001\ kg[/tex]

Conservation of volume

[tex]1\ Lt=0.001\ m^3[/tex]

[tex]1\ mL=1\times10^{-6}\ m^3[/tex]

So, mass in kg

Stem mass (kg) = 0.275, 0.415, 0.563, 0.815, 0.954

Volume in m³,

Stem volume (m³) = 0.0011, 0.001215, 0.001425, 0.001610, 0.001742

(b). We need to calculate the density of the samples

Using formula of density

[tex]\rho=\dfrac{m}{V}[/tex]

Where, m = mass

V = volume

If the m = 0.275 kg and V = 0.0011 m³

Put the value into the formula

[tex]\rho=\dfrac{0.275}{0.0011}[/tex]

[tex]\rho=250\ kg/m^3[/tex]

If the m = 0.415 kg and V = 0.001215 m³

Put the value into the formula

[tex]\rho=\dfrac{0.415}{0.001215}[/tex]

[tex]\rho=341.56\ kg/m^3[/tex]

[tex]\rho=342\ kg/m^3[/tex]

If the m = 0.563 kg and V = 0.001425 m³

Put the value into the formula

[tex]\rho=\dfrac{0.563}{0.001425}[/tex]

[tex]\rho=395.08\ kg/m^3[/tex]

If the m = 0.815 kg and V = 0.001610 m³

Put the value into the formula

[tex]\rho=\dfrac{0.815}{0.001610}[/tex]

[tex]\rho=506.21\ kg/m^3[/tex]

If the m = 0.954 kg and V = 0.001742 m³

Put the value into the formula

[tex]\rho=\dfrac{0.954}{0.001742}[/tex]

[tex]\rho=547.6\ kg/m^3[/tex]

[tex]\rho=548\ kg/m^3[/tex]

(c). We need to convert the density values to scientific notation

In scientific notation

The densities are

[tex]\rho\ (kg/m^3)= 2.50\times10^{2}, 3.42\times10^{2}, 3.95\times10^{2}, 5.06\times10^{2}, 5.48\times10^{2}[/tex]

Hence, This is required solution.

Answer:

x = 46.54m

Explanation:

In order to find the length of the incline you use the following formula:

[tex]x=v_ot+\frac{1}{2}at^2[/tex]      (1)

vo: initial speed of the soccer ball = 0 m/s

t: time

a: acceleration

You first use the the fact that the ball traveled 22 m in 2.2 s. Whit this information you can calculate the acceleration a from the equation (1):

[tex]22m=\frac{1}{2}a(2.2s)^2\\\\a=9.09\frac{m}{s^2}[/tex]      (2)

Next, you calculate the distance traveled by the ball for t = 3.2 s (one second later respect to t = 2.2s). The values of the distance calculated is the lenght of the incline:

[tex]x=\frac{1}{2}(9.09m/s^2)(3.2s)^2=46.54m[/tex]       (3)

The length of the incline is 46.54 m

Answer:

The fraction of the cooper's electrons that is removed is [tex]8.5222\times 10^{-11}[/tex].

Explanation:

An electron has a mass of [tex]9.1 \times 10^{-31}\,kg[/tex] and a charge of [tex]-1.6 \times 10^{-19}\,C[/tex]. Based on the Principle of Charge Conservation, [tex]-2.10\times 10^{-6}\,C[/tex] in electrons must be removed in order to create a positive net charge. The amount of removed electrons is found after dividing remove charge by the charge of a electron:

[tex]n_{R} = \frac{-2.10\times 10^{-6}\,C}{-1.6 \times 10^{-19}\,C}[/tex]

[tex]n_{R} = 1.3125 \times 10^{13}\,electrons[/tex]

The number of atoms in 56 gram cooper ball is determined by the Avogadro's Law:

[tex]n_A = \frac{m_{ball}}{M_{Cu}}\cdot N_{A}[/tex]

Where:

[tex]m_{ball}[/tex] - Mass of the ball, measured in kilograms.

[tex]M_{Cu}[/tex] - Atomic mass of cooper, measured in grams per mole.

[tex]N_{A}[/tex] - Avogradro's Number, measured in atoms per mole.

If [tex]m_{ball} = 56\,g[/tex], [tex]M_{Cu} = 63.5\,\frac{g}{mol}[/tex] and [tex]N_{A} = 6.022\times 10^{23}\,\frac{atoms}{mol}[/tex], the number of atoms is:

[tex]n_{A} = \left(\frac{56\,g}{63.5\,\frac{g}{mol} } \right)\cdot \left(6.022\times 10^{23}\,\frac{atoms}{mol} \right)[/tex]

[tex]n_{A} = 5.3107\times 10^{23}\,atoms[/tex]

As there are 29 protons per each atom of cooper, there are 29 electrons per atom. Hence, the number of electrons in cooper is:

[tex]n_{E} = \left(29\,\frac{electrons}{atom} \right)\cdot (5.3107\times 10^{23}\,atoms)[/tex]

[tex]n_{E} = 1.5401\times 10^{23}\,electrons[/tex]

The fraction of the cooper's electrons that is removed is the ratio of removed electrons to total amount of electrons when net charge is zero:

[tex]x = \frac{n_{R}}{n_{E}}[/tex]

[tex]x = \frac{1.3125\times 10^{13}\,electrons}{1.5401\times 10^{23}\,electrons}[/tex]

[tex]x = 8.5222 \times 10^{-11}[/tex]

The fraction of the cooper's electrons that is removed is [tex]8.5222\times 10^{-11}[/tex].

Answer:

a) [tex]E = \frac{1}{2} \cdot k \cdot x^{2} + \frac{1}{2} \cdot m \cdot v^{2}[/tex], b) Amplitude of the motion is [tex]A = \sqrt{\frac{2\cdot E}{k} }[/tex], c) The maximum speed attained by the object during its motion is [tex]v_{max} = \sqrt{\frac{2\cdot E}{m} }[/tex].

Explanation:

a) The total energy of the object is equal to the sum of potential and kinetic energies. That is:

[tex]E = K + U[/tex]

Where:

[tex]K[/tex] - Kinetic energy, dimensionless.

[tex]U[/tex] - Potential energy, dimensionless.

After replacing each term, the total energy of the object at any point in its motion is:

[tex]E = \frac{1}{2} \cdot k \cdot x^{2} + \frac{1}{2} \cdot m \cdot v^{2}[/tex]

b) The amplitude of the motion occurs when total energy is equal to potential energy, that is, when objects reaches maximum or minimum position with respect to position of equilibrium. That is:

[tex]E = U[/tex]

[tex]E = \frac{1}{2} \cdot k \cdot A^{2}[/tex]

Amplitude is finally cleared:

[tex]A = \sqrt{\frac{2\cdot E}{k} }[/tex]

Amplitude of the motion is [tex]A = \sqrt{\frac{2\cdot E}{k} }[/tex].

c) The maximum speed of the motion when total energy is equal to kinetic energy. That is to say:

[tex]E = K[/tex]

[tex]E = \frac{1}{2}\cdot m \cdot v_{max}^{2}[/tex]

Maximum speed is now cleared:

[tex]v_{max} = \sqrt{\frac{2\cdot E}{m} }[/tex]

The maximum speed attained by the object during its motion is [tex]v_{max} = \sqrt{\frac{2\cdot E}{m} }[/tex].

Answer:

The total internal energy change for the entire process is  -0.94 kJ

Explanation:

Process A to B is an isothermal process, therefore, [tex]u_A[/tex] - [tex]u_B[/tex] = 0

Process B to C

P = -mV + C

When P = 12, V = 0.12

When P = 4, V = 0.135

Therefore, we have;

12 = -m·0.12 + C

4 = -m·0.135 + C

Solving gives

m = 533.33

C = 76

[tex]T = \dfrac{1}{nR} \times (-533.33 \times V^2 + 76 \times V)[/tex]

p₂ = p₁V₁/V₂ = 12*0.1/0.12 = 10 kPa

The work done = 0.5*(0.135 - 0.12)*(4 - 10.0) = -0.045 kJ = -45 J

For heat supplied

Assuming an approximate polytropic process, we have;

Work done = (p₃×v₃ - p₂×v₂)/(n - 1)

Which gives;

-45 = (4*0.135 - 10*0.12)/(n -1)

∴ n -1 = (4*0.135 - 10*0.12)/-45 =   14.67

n = 15.67

Q = W×(n - γ)/(γ - 1)

Q = -45*(15.67 - 1.4)/(1.4 - 1) = -1,605.375 J

u₃ - u₂ = Q + W = -1,605.375 J - 45 J = -1650 J = -1.65 kJ

For the constant pressure process D to C, we have;

[tex]Q = c_p \times \dfrac{p}{R} \times (V_4 -V_3) = \dfrac{5}{2} \times p \times (V_4 -V_3)[/tex]

Q₄₋₃ = (0.1 - 0.135) * 4*5/2 = -0.35 kJ

W₄₋₃ = 4*(0.1 - 0.135) = -0.14 kJ

u₄ - u₃ = Q₄₋₃ + W₄₋₃ = -0.14 kJ + -0.35 kJ = -0.49 kJ

For the process D to A, we have a constant volume process

[tex]Q_{1-4} = \dfrac{c_v}{R} \times V \times (p_1 - p_4) = \dfrac{3}{2} \times 0.1 \times (12 - 4) = 1.2 \ kJ[/tex]

W₁₋₄ = 0 for constant volume process, therefore, u₁ - u₄ = 1.2 kJ

The total internal energy change Δ[tex]u_{process}[/tex] for the entire process is therefore;

Δ[tex]u_{process}[/tex] = u₂ - u₁ + u₃ - u₂ + u₄ - u₃ + u₁ - u₄ = 0  - 1.65 - 0.49 + 1.2 = -0.94 kJ.

Answer: According to Newtons second Law of motion ;

F = ma (Force  equals  mass multiplied by acceleration.)

The acceleration is directly proportional to the net force; the net force equals mass times acceleration; the acceleration in the same direction as the net force; an acceleration is produced by a net force

Explanation:

Answer:

Cost per year = $131.4

Explanation:

We are given;

Power rating of computer with monitor;P = 300 W = 0.3 KW

Cost of power per KWh = 15 cents = $0.15

Time used per day by the computer with monitor = 8 hours

Thus; amount of power consumed per 8 hours each day = 0.3 × 8 = 2.4 KWh per day

Thus, for 365 days in a year, total amount amount of power = 2.4 × 365 = 876 KWh

Now, since cost of power per KWh is $0.15, then cost for 365 days would be;

876 × 0.15 = $131.4

Answer:

(b) electrons will not be ejected

Explanation:

Determine the number of photons ejected by 10 W red laser pointer.

The wavelength (λ) of red light is  700 nm = 700 x 10⁻⁹ m

Energy of a photon is given as;

[tex]E = \frac{hc}{\lambda}[/tex]

where;

h is Planck's constant, = 6.626 x 10⁻³⁴ J/s

c is speed of light, = 3 x 10⁸ m/s

[tex]E = \frac{6.626*10^{-34} *3*10^8}{700 X 10^{-9}} \\\\E = 2.8397 *10^{-19} \ J/photon[/tex]

The number of photons emitted by 10 W red laser pointer

10 W = 10 J/s

[tex]Number \ of \ photons = 10(\frac{ J}{s}) * \frac{1}{2.8397*10^{-19}} (\frac{photon}{J} ) = 3.522 *10^{19} \ photons/s[/tex]

Determine the number of photons ejected by 5 W red green pointer

The wavelength (λ) of green light is  500 nm = 500 x 10⁻⁹ m

[tex]E = \frac{hc}{\lambda} = \frac{6.626*10^{-34} *3*10^8}{500*10^{-9}} = 3.9756 *10^{-19} \ J/photon[/tex]

The number of photons emitted by 5 W green laser pointer

5 W = 5 J/s

[tex]Number \ of \ photons = \frac{5J}{s} *\frac{photon}{3.9756*10^{-19}J} = 1.258 *10^{19} \ Photons/s[/tex]

The number of photons emitted by 10 W red laser pointer is greater than the number of photons emitted by 5 W green laser pointer.

Thus, 5 W green laser pointer will not be able to eject electron from the same metal.

The correct option is "(b) electrons will not be ejected"

Answer:

the acceleration [tex]a^{\to} = (0.0159 \ \ m/s^2 )i[/tex]

Explanation:

Given that:

the initial speed v₁ = 0 m/s i.e starting from rest ; since the car accelerates at a distance Δx = 6 miles in order to teach that final speed v₂ of 63.15 km/h.

So;  the acceleration for the first 6 miles can be calculated by using the formula:

v₂² = v₁² + 2a (Δx)

Making acceleration  a the subject of the formula in the above expression ; we have:

v₂² - v₁² = 2a (Δx)

[tex]a = \dfrac{v_2^2 - v_1^2 }{2 \Delta x}[/tex]

[tex]a = \dfrac{(63.15 \ km/s)^2 - (0 \ m/s)^2 }{2 (6 \ miles)}[/tex]

[tex]a = \dfrac{(17.54 \ m/s)^2 - (0 \ m/s)^2 }{2 (9.65*10^3 \ m)}[/tex]

[tex]a =0.0159 \ m/s^2[/tex]

Thus;

Assume the car moves in the +x direction;

the acceleration [tex]a^{\to} = (0.0159 \ \ m/s^2 )i[/tex]

Answer:

E = (1 / 4π ε₀ )  q r / R³

Explanation:

Thomson's stable model that the negative charge is mobile within the atom and the positive charge is uniformly distributed, to calculate the force we can use Coulomb's law

       F = K q₁ q₂ / r²

we used law Gauss

Ф = ∫ E .dA = q_{int} /ε₀

E 4π r² = q_{int} /ε₀  

E = q_{int} / 4π ε₀ r²

we replace the charge inside  

E = (1 / 4π ε₀ r²) ρ 4/3 π r³  

E = ρ r / 3 ε₀

the density for the entire atom is  

ρ = Q / V  

V = 4/3 π R³  

we substitute  

E = (r / 3ε₀ ) Q 3/4π R³  

E = (1 / 4π ε₀ ) q r / R³

Answer:

0.218

Explanation:

Given that

Total vibrations completed by the wave is 43 vibrations

Time taken to complete the vibrations is 33 seconds

Length of the wave is 424 cm = 4.24 m

to solve this problem, we first find the frequency.

Frequency, F = 43 / 33 hz

Frequency, F = 1.3 hz

Also, we find the wave velocity. Which is gotten using the relation,

Wave velocity = 4.24 / 15

Wave velocity = 0.283 m/s

Now, to get our answer, we use the formula.

Frequency * Wavelength = Wave Velocity

Wavelength = Wave Velocity / Frequency

Wavelength = 0.283 / 1.3

Wavelength = 0.218

Answer:

Force

Explanation:

Force is simply known as pull or push of an object

Answer:

The found acceleration in terms of h and t is:

[tex]a=\frac{h}{5(t_1)^2}[/tex]

Explanation:

(The complete question is given in the attached picture. We need to find the acceleration in terms of h and t in this question)

We are given 3 stages of movement of elevator. We'll first model them each of the stage one by one to find the height covered in each stage. After that we'll find the total height covered by adding heights covered in each stage, and equate it to Total height h. From that we can find the formula for acceleration.

Constant acceleration, starts from rest.

Distance = [tex]y = \frac{1}{2}a(t_1)^2[/tex]

Velocity = [tex]v_1=at_1[/tex]

Constant velocity where

Velocity = [tex]v_o=v_1=at_1[/tex]

Distance =

Constant deceleration where

Velocity = [tex]v_0=v_1=at_1[/tex]

Distance =

[tex]y_3=v_1t_3-\frac{1}{2}a(t_3)^2\\\text{Where}~t_3=t_1\\y_3=v_1t_1-\frac{1}{2}a(t_1)^2\\\text{Where}~ v_1t_1=a(t_1)^2\\y_3=a(t_1)^2-\frac{1}{2}a(t_1)^2\\\text{Subtracting both terms:}\\y_3=\frac{1}{2}a(t_1)^2[/tex]

Total height = y₁ + y₂ + y₃

Total height = [tex]\frac{1}{2}a(t_1)^2+4a(t_1)^2+\frac{1}{2}a(t_1)^2 = 5a(t_1)^2[/tex]

Find acceleration by rearranging the found equation of total height.

Total Height = h

h = 5a(t₁)²

[tex]a=\frac{h}{5(t_1)^2}[/tex]

Answer:

force on larger piston = [tex]\frac{fA}{a}[/tex]

Explanation:

we label the pistons as piston A and piston B

small piston A:

area = a

force = f

large piston B:

area = A

force  = ?

Pascal's law of pressure state that the pressure delivered to a liquid is transmitted undiminished to every portion of the fluid.

we know that pressure = force ÷ area

pressure of piston A = [tex]\frac{f}{a}[/tex]

pressure of piston B = [tex]\frac{(force on piston B)}{A}[/tex]

obeying Pascal's law, the system pressures must be equal. Therefore

[tex]\frac{f}{a} = \frac{(force on piston B)}{A}[/tex]

force on large piston (B) = F = [tex]\frac{fA}{a}[/tex]

Answer:

F₂ = 925.92 N

Explanation:

In a hydraulic lift the normal stress applied to one arm must be equally transmitted to the other arm. Therefore,

σ₁ = σ₂

F₁/A₁ = F₂/A₂

F₂ = F₁ A₂/A₁

where,

F₂ = Initial force that must be applied to narrow arm = ?

F₁ = Load on Wider Arm to be raised = 12000 N

A₁ = Area of wider arm = πr₁² = π(18 cm)² = 324π cm²

A₂ = Area of narrow arm = πr₂² = π(5 cm)² = 25π cm²

Therefore,

F₂ = (12000 N)(25π cm²)/(324π cm²)

F₂ = 925.92 N

Answer:

4.1 seconds

Explanation:

The height of the football is given by the equation:

[tex]H = -16t^2 + V*t + S[/tex]

Using the inicial position S = 4 and the inicial velocity V = 64, we can find the time when the football hits the ground (H = 0):

[tex]0 = -16t^2 + 64*t + 4[/tex]

[tex]4t^2 - 16t - 1 = 0[/tex]

Using Bhaskara's formula, we have:

[tex]\Delta = b^2 - 4ac = (-16)^2 - 4*4*(-1) = 272[/tex]

[tex]t_1 = (-b + \sqrt{\Delta})/2a[/tex]

[tex]t_1 = (16 + 16.49)/8 = 4.06\ seconds[/tex]

[tex]t_2 = (-b - \sqrt{\Delta})/2a[/tex]

[tex]t_2 = (16 - 16.49)/8 = -0.06\ seconds[/tex]

A negative time is not a valid result for this problem, so the amount of time the football is in the air before hitting the ground is 4.1 seconds.

The amount of time the football spent in air before it hits the ground is 4.1 s.

The given parameters;

To find:

Using the vertical model equation given as;

[tex]H = -16t^2 + Vt + S\\\\[/tex]

the final height when the ball hits the ground, H = 0

[tex]0 = -16t^2 + 64t + 4\\\\16t^2 - 64t - 4 = 0\\\\divide \ through \ by\ 4\\\\4t^2 - 16t - 1= 0\\\\solve \ the \ quadratic \ equation \ using \ the \ formula \ method;\\\\\\a = 4, \ b = -16, \ c = - 1\\\\t = \frac{-b \ \ + /- \ \ \ \sqrt{b^2 - 4ac} }{2a} \\\\[/tex]

[tex]t = \frac{-(-16) \ \ + /- \ \ \ \sqrt{(-16^2 )- 4(4\times -1)} }{2\times 4}\\\\t = \frac{16 \ \ + /- \ \ \sqrt{272} }{8} \\\\t = \frac{16 \ \ +/- \ \ 16.49}{8} \\\\t = \frac{16 - 16.49}{8} \ \ \ \ or \ \ \ \frac{16 + 16.49}{8} \\\\t = -0.61 \ s \ \ or \ \ \ 4.06 \ s\\\\t\approx 4.1 \ s[/tex]

Thus, the amount of time the football spent in air before it hits the ground is 4.1 s.

Learn more here: https://brainly.com/question/2018532

Answer:

The radius of the sphere is 4.05 m

Explanation:

Given;

potential at surface, [tex]V_s[/tex] = 450 V

potential at radial distance, [tex]V_r[/tex] = 150

radial distance, l = 8.1 m

Apply Coulomb's law of electrostatic force;

[tex]V = \frac{KQ}{r} \\\\V_s = \frac{KQ}{r} \\\\V_r = \frac{KQ}{r+ l}[/tex]

[tex]450 = \frac{KQ}{r} ------equation (i)\\\\150 = \frac{KQ}{r+8.1} ------equation (ii)\\\\divide \ equation (i)\ by \ equation \ (ii)\\\\\frac{450}{150} = (\frac{KQ}{r} )*(\frac{r+8.1}{KQ} )\\\\3 = \frac{r+8.1}{r} \\\\3r = r + 8.1\\\\2r = 8.1\\\\r = \frac{8.1}{2} \\\\r = 4.05 \ m[/tex]

Therefore, the radius of the sphere is 4.05 m

Answer:

It is the force that opposes motion between two surfaces touching each other. ( OR ) The force between two surfaces that are sliding or trying to slide across each other.

Explanation:

Answer:

a constant force that acts on objects that rub together

Explanation:

a constant force that acts on objects that rub together