The number of protons in the nucleus of an atom determines the species of the atom, i.e., the element to which the atom belongs. An atom has the same number of protons and neutrons. But the electron number cannot be used instead because
A. electrons are not within the nucleus
B. electrons are negatively charged
C. electrons can be removed from or added to an atom
D.electrons are lighter than protons

Answer:

U = 5.37*10^33 J

Explanation:

The gravitational potential energy between two bodies is given by:

[tex]U_{1,2}=-G\frac{m_1m_2}{r_{1,2}}[/tex]

G: Cavendish's constant = 6.67*10^-11 m^3/kg.s

For three bodies the total gravitational potential energy is:

[tex]U_{T}=U_{1,2}+U_{1,3}+U_{2,3}\\\\U_{T}=-G[\frac{m_1m_2}{r_{1,2}}+\frac{m_1m_3}{r_{1,3}}+\frac{m_2m_3}{r_{2,3}}][/tex]

BY replacing the values of the parameters for 1->earth, 2->moon and 3->sun you obtain:

[tex]U_{T}=-(6.67*10^{-11}m^3/kg.s)[\frac{(5.98*10^{24}kg)(7.36*10^{22}kg)}{3.84*10^{8}m}+\\\\\frac{(5.98*10^{24}kg)(1.99*10^{30}kg)}{1.496*10^{11}m}+\frac{(7.36*10^{22}kg)(1.99*10^{30}kg)}{1.496*10^{11}m-3.84*10^8m}]\\\\U_{T}=5.37*10^{33}J[/tex]

hence, the total gravitational energy is 5.37*10^33 J

Answer:

$2018.016

Explanation:

Applying,

P = I²R.............. Equation 1

Where P = power, drawn by the oven, I = current drawn by the oven, R = resistance of the oven.

Given: I = 10 A, R = 2002 Ω

Substitute this values into equation 1

P = 10²(2002)

P = 100(2002)

P = 200200 W.

P = 200.2 kW

If the electric oven works 3 hours daily,

Then the total time it works in a week = 3×7 = 21 hours.

E(kWh) = P(kW)×t(h)

E = 200.2×21

E = 4204.2 kWh.

If 1 kWh of energy cost $0.48,

Then, 4204.2 kWh =  $4204.2(0.48) =  $2018.016

Answer:

speed after one riveting operation = 230.5 rpm

mass moment of inertia = 9.90 kg

Explanation:

given data

torque energy  = 2.5 kW = 2500 W

mass = 125 kg

radius = 700 mm

energy = 1000 J

Speed of the flywheel N = 240 rpm

solution

we know here Speed of the flywheel N so here angular speed ω

ω = [tex]\frac{2\pi N}{60}[/tex]   .....................1

ω = [tex]\frac{2\pi 240}{60}[/tex]

ω = 25.13 rad/s

so here

change in energy is

ΔE = E1 - E2    ..............2

ΔE = 2500 - 1000

ΔE = 1500 J/sec

and

I = mr²    .........3

I = 125 × 0.7²

I = 61.25 kg-m²

and ΔE is express as here

ΔE = 0.5 × I × (ω² - ω1² )      ........4

put here value and we get

1500 =  0.5 × 61.25 × (25.13² - ω1² )

ω1 = 24.13 rad /s

and

reduction in speed after one riveting operation will be

N = [tex]\frac{24.13\times 60 }{2\pi }[/tex]

speed after one riveting operation = 230.5 rpm

and

for 35 rpm

ω1 = [tex]\frac{2\pi 35}{60}[/tex]

ω1 = 3.65

so ΔE  will be here

ΔE = 0.5 × mr² × (ω² - ω1² )    ....................5

put here value and we get m

1500 = 0.5 × m (0.7)² × (25.13² - 3.65² )

solve it we get

mass moment of inertia = 9.90 kg

Answer:

The period is  [tex]T = 8.27 \ sec[/tex]

Explanation:

From the question we are told that

   The extension of the spring is  [tex]e_1 = 33 \ m[/tex]

   The  first weight  applied is  [tex]F_1 = 19 N[/tex]

     The second weight applied is  [tex]F_2 = W[/tex]

     The second extension is [tex]e_2 = 17 \ m[/tex]

The spring constant of the spring is mathematically evaluated as

         [tex]k = \frac{F_1}{e_1 }[/tex]

substituting values

       [tex]k = \frac{19}{33 }[/tex]

       [tex]k = 0.576[/tex]

We are told that

         19 N extended the spring to 33 m    

Then W N  will extended it by  17 m

Therefore     [tex]W = \frac{19 * 17}{33}[/tex]

                    [tex]W = 9.788 \ N[/tex]

Generally the period of the oscillation is mathematically represented as

           [tex]T = 2 \pi \sqrt{\frac{M}{K} }[/tex]

where M is the mass of the W which is mathematically evaluated as

         [tex]M = \frac{9.788}{9.8}[/tex]

          [tex]M = 1.0 \ kg[/tex]

substituting values

        [tex]T = 2 \pi \sqrt{\frac{1}{0.576} }[/tex]

       [tex]T = 8.27 \ sec[/tex]

The force required to stretch a spring is directly proportional to the amount the spring is stretched. The force constant can be calculated using the equation F = kx. By substituting the given values into the equation, we can find the unknown weight W.

The force required to stretch a spring is directly proportional to the amount the spring is stretched. This can be represented by the equation F = kx, where F is the force, k is the force constant, and x is the amount the spring is stretched. In this scenario, we are given that a 19 N weight stretches the spring by 33 m at equilibrium.

To find the force constant, we can use the equation:

k = F / x = 19 N / 33 m = 0.58 N/m

Next, we need to find the unknown weight W that will stretch the spring to a new equilibrium position 17 m below the position with no weight attached. Using the force constant we calculated, we can use the equation F = kx to find the force required:

F = kx = (0.58 N/m)(17 m) = 9.86 N

Therefore, the unknown weight W that will stretch the spring to the new equilibrium position is 9.86 N.

Answer:

a) An equation for the x-component of the electric field.

Eₓ = (-15xy³ + 5.32xy⁴z²) N/C

b) An equation for the y-component of the electric field.

Eᵧ = (-22.5x²y² + 10.64x²y³z²) N/C

c) An equation for the z-component of the electric field.

Ez = (5.32x²y⁴z) N/C

d) At (-5.0, 2.0, 1.5) m, the electric field is given as

E = (-357.6î + 2,538ĵ + 3,192ķ) N/C

Magnitude of the electric field = 4,093.7 N/C

Explanation:

The electric field is given by the negative of the gradient of the electric potential,

E = −grad V

E = - ∇V

The electric potential is given as

V(x,y,z) = 3αx²y³ - 2γx²y⁴z²

α = 2.5 V/m⁵ and γ = 1.33 V/m⁸

V(x,y,z) = 7.5x²y³ - 2.66x²y⁴z²

grad = ∇ = (∂/∂x)î + (∂/∂y)ĵ + (∂/∂z)ķ

E = -grad V = -∇V

= -[(∂V/∂x)î + (∂V/∂y)ĵ + (∂V/∂z)ķ

E = -(∂V/∂x)î - (∂V/∂y)ĵ - (∂V/∂z)ķ

E = Eₓî + Eᵧĵ + Ez ķ

a) An equation for the x-component of the electric field.

Eₓ = -(∂V/∂x) = -(∂/∂x)(V)

= -(∂/∂x)(7.5x²y³ - 2.66x²y⁴z²)

= -(15xy³ - 5.32xy⁴z²)

= (-15xy³ + 5.32xy⁴z²)

b) An equation for the y-component of the electric field.

Eᵧ = -(∂V/∂y) = -(∂/∂x)(V)

= -(∂/∂y)(7.5x²y³ - 2.66x²y⁴z²)

= -(22.5x²y² - 10.64x²y³z²)

= (-22.5x²y² + 10.64x²y³z²)

c) An equation for the z-component of the electric field.

Ez = -(∂V/∂z) = -(∂/∂x)(V)

= -(∂/∂z)(7.5x²y³ - 2.66x²y⁴z²)

= -(0 - 5.32x²y⁴z)

= (5.32x²y⁴z)

d) E = Eₓî + Eᵧĵ + Ez ķ

E = (-15xy³ + 5.32xy⁴z²)î + (-22.5x²y² + 10.64x²y³z²)ĵ + (5.32x²y⁴z) ķ

At (-5.0, 2.0, 1.5) m

x = -5 m

y = 2 m

z = 1.5 m

Eₓ = (-15xy³ + 5.32xy⁴z²)

= (-15×-5×2³) + (5.32×-5×2⁴×1.5²)

= 600 - 957.6 = -357.6

Eᵧ = (-22.5x²y² + 10.64x²y³z²)

= (-22.5×(-5)²×2²) + (10.64×(-5)²×2³×1.5²)

= -2250 + 4788 = 2538

Ez = (5.32x²y⁴z) = (5.32×(-5)²×2⁴×1.5)

= 3192

E = -357.6î + 2,538ĵ + 3,192ķ

Magnitude = /E/ = √[(-357.6)² + 2538² + 3192²]

= 4,093.6763135353 = 4,093.7 N/C

Hope this Helps!!!!

The correct answers are as follows:

(a) The equation for the x-component of the electric field is[tex]\[ E_x = -15xy^3 + 5.32xy^4z^2 \][/tex]

(b) The equation for the y-component of the electric field is [tex]\[ E_y = -22.5x^2y^2 + 10.64x^2y^3z^2 \][/tex]

(c) The equation for the z-component of the electric field is [tex]\[ E_z = 5.32x^2y^4z \][/tex]

(d) the magnitude of the electric field at the point x13-(-5.0, 2.0, 1.5) m in units of newtons per coulomb is 14907.5 N/C.

To find the components of the electric field, we need to take the negative gradient of the electric potential function V(x,y,z). The gradient of a function is a vector field whose components are the partial derivatives of the function with respect to each variable. The electric field [tex]\( \vec{E} \)[/tex] is related to the electric potential [tex]\( V \)[/tex] by the equation:

[tex]\[ \vec{E} = -\nabla V \][/tex]

where [tex]\( \nabla \)[/tex] is the gradient operator.

(a) The x-component of the electric field [tex]\( E_x \)[/tex] is given by the negative partial derivative of [tex]\( V \)[/tex] with respect to [tex]\( x \)[/tex] :

[tex]\[ E_x = -\frac{\partial V}{\partial x} = -\frac{\partial}{\partial x}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex]

[tex]\[ E_x = -3α(2xy^3) + 2γ(2xy^4z^2) \][/tex]

[tex]\[ E_x = -6αxy^3 + 4γxy^4z^2 \][/tex]

Substituting the given values of [tex]\( α \)[/tex] and [tex]\( γ \)[/tex] :

[tex]\[ E_x = -6(2.5)xy^3 + 4(1.33)xy^4z^2 \][/tex]

[tex]\[ E_x = -15xy^3 + 5.32xy^4z^2 \][/tex]

(b) The y-component of the electric field [tex]\( E_y \)[/tex] is given by the negative partial derivative of [tex]\( V \)[/tex] with respect to [tex]\( y \)[/tex] :

[tex]\[ E_y = -\frac{\partial V}{\partial y} = -\frac{\partial}{\partial y}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex] [tex]\[ E_y = -3αx^2(3y^2) + 2γx^2(4y^3z^2) \][/tex]

[tex]\[ E_y = -9αx^2y^2 + 8γx^2y^3z^2 \][/tex]

Substituting the given values of [tex]\( α \)[/tex] and [tex]\( γ \)[/tex] :

[tex]\[ E_y = -9(2.5)x^2y^2 + 8(1.33)x^2y^3z^2 \][/tex]

[tex]\[ E_y = -22.5x^2y^2 + 10.64x^2y^3z^2 \][/tex]

(c) The z-component of the electric field

[tex]\[ E_z = -\frac{\partial V}{\partial z} = -\frac{\partial}{\partial z}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex]

[tex]\[ E_z = -\frac{\partial V}{\partial z} = -\frac{\partial}{\partial z}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex]

is given by the negative partial derivative of [tex]\( V \)[/tex] with respect to [tex]\( z \)[/tex] :

[tex]\[ E_z = -\frac{\partial V}{\partial z} = -\frac{\partial}{\partial z}(3αx^2y^3 - 2γx^2y^4z^2) \][/tex]

[tex]\[ E_z = -2γx^2y^4(-2z) \][/tex]

[tex]\[ E_z = 4γx^2y^4z \][/tex]

Substituting the given value of[tex]\( γ \)[/tex] :

[tex]\[ E_z = 4(1.33)x^2y^4z \]\\ E_z = 5.32x^2y^4z \][/tex]

(d) To calculate the magnitude of the electric field at the point[tex]\( P(-5.0, 2.0, 1.5) \)[/tex] m, we first substitute [tex]\( x = -5.0 \)[/tex] m, [tex]\( y = 2.0 \)[/tex] m, and [tex]\( z = 1.5 \)[/tex] m into the equations for[tex]\( E_x \)[/tex], [tex]\( E_y \)[/tex] , and [tex]\( E_z \)[/tex] :

[tex]E_x = -15(-5.0)(2.0)^3 + 5.32(-5.0)(2.0)^4(1.5)^2 \]\\ E_x = 1500 - 5.32(-5.0)(16)(2.25) \]\\ E_x = 1500 - (-851.88) \]\\ E_x = 1500 + 851.88 \]\\ E_x = 2351.88 \text{ N/C} \][/tex]

[tex]\[ E_y = -22.5(-5.0)^2(2.0)^2 + 10.64(-5.0)^2(2.0)^3(1.5)^2 \]\\ E_y = -22.5(25)(4) + 10.64(25)(8)(2.25) \]\\ E_y = -2200 + 10.64(25)(8)(2.25) \]\\ E_y = -2200 + 4752 \]\\ E_y = 2552 \text{ N/C} \][/tex]

[tex]\[ E_z = 5.32(-5.0)^2(2.0)^4(1.5) \]\\ E_z = 5.32(25)(16)(1.5) \]\\ E_z = 5.32(600) \]\\ E_z = 3192 \text{ N/C} \][/tex]

Now, the magnitude of the electric field \( \vec{E} \) is given by:

[tex]\[ |\vec{E}| = \sqrt{E_x^2 + E_y^2 + E_z^2} \][/tex]

[tex]\[ |\vec{E}| = \sqrt{(2351.88)^2 + (2552)^2 + (3192)^2} \][/tex]

[tex]\[ |\vec{E}| = \sqrt{5527641.64 + 6512644 + 10180416} \][/tex]

[tex]\[ |\vec{E}| = \sqrt{22220502.64} \][/tex]

[tex]\[ |\vec{E}| \approx 14907.5 \text{ N/C} \][/tex]

Therefore, the magnitude of the electric field at the point [tex]\( P(-5.0, 2.0, 1.5) \)[/tex]m is approximately [tex]\( 14907.5 \)[/tex] N/C.

Answer:

a) [tex]\ddot n = -1.563\,\frac{rev}{min^{2}}[/tex], b) [tex]\Delta n = 7197.697\,rev[/tex], c) [tex]a_{t} = 1.009\times 10^{-3}\,\frac{m}{s^{2}}[/tex], d) [tex]a = 22.823\,\frac{m}{s^{2}}[/tex]

Explanation:

a) Constant angular acceleration is:

[tex]\ddot n = \frac{\dot n - \dot n_{o}}{\Delta t}[/tex]

[tex]\ddot n = \frac{0\,\frac{rev}{min} - 150\,\frac{rev}{min}}{(1.6\,h)\cdot \left(60\,\frac{min}{h} \right)}[/tex]

[tex]\ddot n = -1.563\,\frac{rev}{min^{2}}[/tex]

b) The amount of revolutions required to stop the flywheel is:

[tex]\Delta n = \frac{\dot n^{2}-\dot n_{o}^{2}}{2\cdot \ddot n}[/tex]

[tex]\Delta n = \frac{\left(0\,\frac{rev}{min} \right)^{2}-\left(150\,\frac{rev}{min} \right)^{2}}{2\cdot \left(-1.563\,\frac{rev}{min^{2}} \right)}[/tex]

[tex]\Delta n = 7197.697\,rev[/tex]

c) The tangential acceleration of the particle is:

[tex]a_{t} = \left(1.563\,\frac{rev}{min^{2}} \right)\cdot \left(\frac{1}{3600}\,\frac{min^{2}}{s^{2}}\right)\cdot \left(2\pi\,\frac{rad}{rev}\right)\cdot (0.37\,m)[/tex]

[tex]a_{t} = 1.009\times 10^{-3}\,\frac{m}{s^{2}}[/tex]

d) The radial acceleration of the particle is:

[tex]a_{r} = \left[\left(75\,\frac{rev}{min} \right)\cdot \left(\frac{1}{60}\,\frac{min}{s} \right)\cdot \left(2\pi\,\frac{rad}{rev} \right)\right]^{2}\cdot (0.37\,m)[/tex]

[tex]a_{r} = 22.823\,\frac{m}{s}[/tex]

The net linear acceleration is:

[tex]a = \sqrt{a_{r}^{2}+a_{t}^{2}}[/tex]

[tex]a = \sqrt{\left(22.823\,\frac{m}{s^{2}} \right)^{2}+\left(1.009\times 10^{-3}\,\frac{m}{s^{2}} \right)^{2}}[/tex]

[tex]a = 22.823\,\frac{m}{s^{2}}[/tex]

Answer:

99 dB

Explanation:

To find the new sound intensity level you calculate first the initial intensity by using the following formula:

[tex]\beta=10log(\frac{I}{I_o})\\\\10^{\frac{\beta}{10}}=10^{log(\frac{I}{I_o})}\\\\10^{\frac{\beta}{10}}=\frac{I}{I_o}\\\\I=I_o10^{\frac{\beta}{10}}[/tex]

where β is the sound level of 79dB and Io is the hearing threshold of 10^-12 W/m^2. By replacing you obtain:

[tex]I=(10^{-12}W/m^2)10^{\frac{79}{10}}=7.94*10^{-5}W/m^2[/tex]

The new sound intensity level is given by:

[tex]\beta'=10log(\frac{100I}{I_o})=10log(\frac{100(7.94*10^{-5}W/m^2)}{10^{-12}W/m^2})\\\\\beta'=99\ dB[/tex]

hence, the answer is 99 dB

Answer:

The amount of electrons that flow in the given time is 3.0 C.

Explanation:

An electric current is defined as the ratio of the quantity of charge flowing through a conductor to the time taken.

i.e           I = [tex]\frac{Q}{t}[/tex] ...................(1)

It is measure in Amperes and can be measured in the laboratory by the use of an ammeter.

In the given question, I = 1.5A, t = 2s, find Q.

From equation 1,

            Q = I × t

                = 1.5 × 2

               = 3.0 Coulombs

The amount of electrons that flow in the given time is 3.0 C.

Final answer:

The accurate statement about magnetic field lines is that they form closed loops and never intersect, originating at the north pole and terminating at the south pole of a magnet.

Explanation:

The correct answer regarding magnetic field lines is B. Magnetic field lines form closed loops and never intersect. This observation can be explained through several key principles:

Therefore, options A, C, and D are not correct. The field lines start at the north pole and end at the south pole, not the other way around. Positive charges are affected by magnetic fields but not always along the magnetic field lines because the magnetic force is perpendicular to both the velocity of the charge and the magnetic field.

Answer:

9.2 A

Explanation:

V = IR

115 V = I (12.5 Ω)

I = 9.2 A

Answer:[tex]P_{bulb}=99.11\ kPa[/tex]

Explanation:

Given

When the bulb is squeezed then liquid level rises to height h i.e. pressure decreases inside the bulb which causes rises in the tube

for common elevation i.e. at liquid level pressure must be equal therefore

[tex]P_{bulb}+\rho gh=P_{atm}[/tex]

[tex]P_{bulb}=1.013\times 10^5-1490\times 9.8\times h[/tex]

for [tex]h=0.15\ m[/tex]

[tex]P_{bulb}=1.013\times 10^5-1490\times 9.8\times 0.15[/tex]

[tex]P_{bulb}=101.3\times 10^3-2.1903\times 10^3[/tex]

[tex]P_{bulb}=99.11\ kPa[/tex]

for [tex]h=0.1\ m[/tex]

[tex]P_{bulb}=101.3\times 10^3-1.4602\times 10^3[/tex]  

[tex]P_{bulb}=99.83\ kPa[/tex]

Answer:

867755.73 V/m

Explanation:

Detailed explanation and calculation is shown in the image below

Answer:

75.36 mph

Explanation:

The distance between the other car and the intersection is,

[tex]x=x_{0}+V t \\ x=\frac{1}{2}+V t[/tex]

The distance between the police car and the intersection is,

[tex]y=y_{0}+V t[/tex]

[tex]y=\frac{1}{2}-40 t[/tex]

(Negative sign indicates that he is moving towards the intersection)

Therefore the distance between them is given by,

[tex]z^{2}=x^{2}+y^{2}(\text { Using Phythogorous theorem })[/tex]

[tex]z^{2}=\left(\frac{1}{2}+V t\right)^{2}+\left(\frac{1}{2}-40 t\right)^{2} \ldots \ldots \ldots(1)[/tex]

The rate of change is,

[tex]2 z \frac{d z}{d t}=2\left(\frac{1}{2}+V t\right) V+2\left(\frac{1}{2}-40 t\right)(-40)[/tex]

[tex]2 z \frac{d z}{d t}=V+2 V^{2} t-40+3200 t \ldots \ldots \ldots[/tex]

Now finding [tex]z[/tex] when [tex]t=0,[/tex] from (1) we have

[tex]z^{2}=\left(\frac{1}{2}+V(0)\right)^{2}+\left(\frac{1}{2}-40(0)\right)^{2}[/tex]

[tex]z^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \\ z=\sqrt{\frac{1}{2}} \approx 0.7071[/tex]

The officer's radar gun indicates 25 mph pointed at the other car then, [tex]\frac{d z}{d t}=25[/tex] when [tex]t=0,[/tex] from

From (2) we get

[tex]2(0.7071)(25)=V+2 V^{2}(0)-40+3200(0)[/tex]

[tex]2(0.7071)(25)=V+2 V^{2}(0)-40[/tex]

[tex]35.36=V-40[/tex]

[tex]V=35.36+40=75.36[/tex]

Hence the speed of the car is [tex]75.36 mph[/tex]

To find the speed of the other car, we can use the concept of relative velocity. The police officer's radar gun indicates a speed of 25 mph, which is the relative velocity between the two cars. Since the police car is moving north at 40 mph, the other car must be moving east at a speed that results in a relative velocity of 25 mph.

To find the speed of the other car, we can use the concept of relative velocity. The police officer's radar gun indicates a speed of 25 mph, which is the relative velocity between the two cars. Since the police car is moving north at 40 mph, the other car must be moving east at a speed that results in a relative velocity of 25 mph.

To calculate the speed of the other car, we can use the Pythagorean theorem. The police car is moving in a straight line, so the relative velocity is the hypotenuse of a right triangle. The northward velocity of the police car is one side of the triangle, and the eastward velocity of the other car is the other side. Using the equation a² + b² = c², where a is the northward velocity, b is the eastward velocity, and c is the relative velocity, we can solve for b. Rearranging the equation, we get b = √(c² - a²). Plugging in the known values, we have b = √(25² - 40²) = √(625 - 1600) = √(-975)

The square root of a negative number is not a real number, so it is not possible to determine the speed of the other car using only the given information.

https://brainly.com/question/34099078

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Answer:

The initially known quantity, together with the wavelength, that is sufficient to calculate the stopping potential for electrons from the surface of a metal is called the WORK FUNCTION.

Explanation:

The stopping potential is defined as the potential that is required to stop electrons from being ejected from the surface of a metal when light with energy greater than the metal's work function/work potential is incident on the metal.

Given that light is known to be made up of photons, which carry energy in packets according to the frequencies of the light.

The photoelectric phenomenon explains that when light of a certain frequency that corresponds to an energy level that is higher than a metal's work function is incident on a metal, it will lead to electrons being ejected from the surface of the metal. The energy of the ejected electrons is then proportional to the difference between the energy level of the photons and the metal's work function.

Basically, it is the excess energy after overcoming the work function that rejects the electrons.

So, to prevent this excess energy from ejecting electrons from a metal's surface, an energy thay matches this excess must be in place to stop electrons from coming out. This energy/potential required to stop the ejection of electrons, is called the stopping potential.

The stopping potential is given as

eV₀ = hf - ϕ

The stopping potential (eV₀) them depends on the hf and the ϕ.

hf is the energy of the photons, where h is Planck's constant and f is the photons' frequency which is further given as

f = (c/λ)

c = speed of light (speed of the photons)

λ = wavelength of the photons.

The other quantity, ϕ, is the metal's work function; the amount of energy needed to be overcome by the photons before ejection of electrons is possible. It is the minimum energy that the light photoms must possess to even stand a chance of being able to eject electrons from a metal's surface.

So, the stopping potential is the difference between the energy of the photons (obtained using the photons' frequency, wavelength and/or speed) and the metal's work function.

Hope this Helps!!!!

Answer: A monochromatic light source is one that consists of many colors

Explanation:

Monochromatic comes from "mono - one, chromatic - color", so a  monochromatic wave has only one wavelength, this means that it has only one color (while it may have different shades or tones). So the correct option is the second one "A monochromatic light source is one that consists of many colors "

The other 3 statements are true.

The incorrect statement is that 'A monochromatic light source is one that consists of many colors', because a monochromatic light source indeed emits light of a single wavelength or color. The other claims regarding phase shift on reflection, and the location of highest intensity in single and double slit interference patterns are accurate.

The statement that is NOT true among the given options is: 'A monochromatic light source is one that consists of many colors'. In fact, a monochromatic light source emits light of a single wavelength or color.

Further addressing the related concepts: When a light wave encounters a boundary where it must reflect off a surface with a larger index of refraction, the phase of the wave indeed flips by pi. This is an important concept in understanding wave reflection and refraction.

Regarding the intensity of the central maximum in both single and double slit interference patterns, it is accurate to say that the intensity of the center maximum is the most significant. This consequence is a direct outcome of the interference pattern produced within these experiments. In the central maximum, most wave paths align constructively, resulting in the highest intensity.

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Answer:

Explanation:

The star is revolving the black hole like earth revolves around the sun .so time period of rotation  T is given by the following relation

T² = [tex]\frac{4\pi^2\times R^3}{GM }[/tex] , R is distance between black hole and star , M is mass of black hole

Given T = 4.8 hours

4.8² =  [tex]\frac{4\pi^2\times R^3}{GM }[/tex]

Using the same equation for earth sun system

24² =  [tex]\frac{4\pi^2\times (50R)^3}{GM_s }[/tex]  , Ms is mass of the sun and 50R is distance between the sun and the earth .

Dividing the equation

[tex](\frac{4.8}{24})^2[/tex] = [tex]\frac{M_s}{M}\times\frac{1^3}{50^3}[/tex]

[tex]\frac{M}{M_s}[/tex] = 2x 10⁻⁴

The mass of the black hole is [tex]2\times 10^{-4}[/tex] times the mass of our sun.

Given information:

An X-ray telescope detected a source called Cygnus X-3, whose intensity changed with a period of 4.8 hours (T).

The star is orbiting around the black hole.

The distance between the centers of the star and the black hole is one-fiftieth of the distance between the centers of the Earth and our Sun.

Let R be the center to center distance between the black hole and the sun.

So, the time period T of the sun around the black hole will be,

[tex]T^2=\dfrac{2\pi^2R^3}{GM}\\4.8^2=\dfrac{2\pi^2R^3}{GM}[/tex]

where M is the mass of the black hole.

The distance between the earth and our sun will be 50R.

So, the time period of the earth will be,

[tex]T^2=\dfrac{2\pi^2R^3}{GM}\\24^2=\dfrac{2\pi^2(50R)^3}{GM_s}[/tex]

where [tex]M_s[/tex] is the mass of our sun.

Now, compare the above two relations to get the mass of black hole in terms of mass of our sun as,

[tex]4.8^2=\dfrac{2\pi^2R^3}{GM}\\24^2=\dfrac{2\pi^2(50R)^3}{GM_s}\\\dfrac{24^2}{4.8^2}=\dfrac{50^3}{1}\times \dfrac{M}{M_s}\\\dfrac{M}{M_s}=0.0002\\\dfrac{M}{M_s}=2\times 10^{-4}[/tex]

Therefore, the mass of black hole is [tex]2\times 10^{-4}[/tex] times the mass of our sun.

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Answer:

Explanation:

a ) angular frequency ω = [tex]\sqrt{\frac{k}{m} }[/tex]

k is spring constant and m is mass attached

ω = [tex]\sqrt{\frac{20}{1.5} }[/tex]

= 3.6515 rad / s

frequency of oscillation n = 3.6515 / (2 x 3.14)

= .5814 s⁻¹

x = .1 mcos(ωt)

= .1 mcos(3.6515t)

b ) maximum speed = ωA , A is amplitude

= 3.6515 x .1

= .36515 m /s

36.515 cm /s

maximum acceleration = ω²A

= 3.6515² x .1

= 1.333 m / s²

c ) Kinetic energy at displacement x

= 1/2 m ω²( A²-x²)

potential energy =1/2 m ω²x²

so 1/2 m ω²( A²-x²) = 1/2 m ω²x²

A²-x² = x²

2x² = A²

x = A / √2

The frequency of the oscillation is 0.58 Hz and the displacement x varies as a cosine function. The maximum speed is 0.366 m/s and maximum acceleration is 0.427 m/s². The kinetic and potential energies are equal when the mass is 0.0707m away from the equilibrium point.

(a) The frequency of the oscillations is given by the formula f = 1/2π √(k/m), where k is the spring constant and m is the mass. Substituting the given values we get f = 1/2π √(20.0 N/m / 1.5 kg) = 0.58 Hz. The displacement as a function of time is given by the equation x(t) = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase angle. Given the mass was released from rest, the phase angle is 0 and the equation becomes x(t) = (0.10 m) cos(2πft).

(b) The maximum speed is given by the formula vmax = Aω, substituting the values, vmax = (0.10 m)(2πf) = 0.366 m/s. The maximum acceleration is given by the formula amax = Aω² which is amax = (0.10 m)(2πf)² = 0.427 m/s².

(c) The kinetic energy and potential energy of the system are the same when the displacement is equal to the amplitude divided by √2: x = A/√2, that is x = 0.10m /√2 = 0.0707 m from the equilibrium point.

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Answer:

This is because of the inertia of the object.

Explanation:

This is because of the inertia of the object.

When you push a static object you must overcome the static friction force of the object. Once you have overcome the static friction the inertia law demands that the object tend to conserve its motion. That is the reason why you need less force when the object is already in motion. In other words, the inertia gained by the object with the initial force, "helps" you with the work of moving the object. This is also the reason why the kinetic friction of an object in motion over a surface is lower than the static friction.

Answer:oxygen

Explanation:algae are plants that lives in aquatic habitat,with a few of them occurring in land. Algae have chlorophyll and as a result are autotrophic in nutrition. Algae uses carbon dioxide as a raw material for photosynthesis which is the process where they produce food. They also give off oxygen which results from the splitting of water by light.

This oxygen given off is used by organisms for cellular respiration.the mitochondria is the organelle responsible it's utilization in respiration and carbon dioxide is given off.oxygen serves as an electron acceptor in the energy producing process in the mitochondria. It is an important gas for aerobic respiration.

Final answer:

Carbon dioxide is a metabolic waste of photosynthesis in algae that can be recycled back into the ecosystem through cellular respiration.

Explanation:

Among the metabolic wastes produced by algae during photosynthesis, carbon dioxide (CO₂) is a byproduct that can be recycled and used in cellular respiration. Photosynthesis is the process in which algae absorb light energy to create carbohydrates within chloroplasts, releasing oxygen as a byproduct. Alternatively, cellular respiration involves using oxygen to break down carbohydrates, primarily in the cytoplasm and mitochondria, releasing ATP and carbon dioxide. These two processes are synchronously linked in a biological cycle that allows organisms to harness energy from the sun, guide by electron transport chains that drive cellular reactions.

Answer:

Maximum Power = 144.3 D

The associated focal length of the lens = [tex]6.92*10^{-3} m[/tex]

Explanation:

According to the Lens maker's Formula:

[tex]\frac{1}{f} = (n-1) (\frac{1}{R_1}-\frac{1}{R_2} )[/tex]

where;

[tex]n_1[/tex] = the refractive index of the medium

[tex]R_1[/tex] and [tex]R_2[/tex] = radius of curvature on each surface

For a convex lens, The radius of curvature in the front surface will be positive and that of the second surface will be negative . Therefore;

[tex]\frac{1}{f} = (n-1) (\frac{1}{R_1}-\frac{1}{-R_2} ) \\ \\ \frac{1}{f} = (n-1) (\frac{1}{R_1}+\frac{1}{R_2} )[/tex]

At maximum power

[tex]\frac{1}{f} = (1.430-1) (\frac{1}{6.50 \ mm}-\frac{1}{5.50 \ mm} )[/tex]

= [tex]0.144 \ mm^{-1}[/tex]

This Implies

[tex]f = 6.92 mm\\f = 6.92*10^{-3} \ m[/tex]

Therefore; the power is given by the formula:

[tex]P_{max} = \frac{1}{f}[/tex]

[tex]P_{max}= \frac{1}{6.92*10^{-3}}[/tex]

= 144.3 D

The maximum power is 144.3 D and the associated focal length of the lens is 6.92 [tex]\rm \times 10^{-3}[/tex] m and this can be determined by using the lens maker's formula.

Given :

The lens maker's formula can be used in order to determine the maximum power and associated focal length of the lens.

The lens maker's formula is given below:

[tex]\rm \dfrac{1}{f}=(n - 1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)[/tex]

where f is the focal length of the lens, n is the refractive index, and [tex]\rm R_1[/tex] and [tex]\rm R_2[/tex] are the radius of curvature on each surface.

The radius of curvature of the first surface is positive and the radius of curvature of the second surface is negative. So, the lens maker's formula becomes:

[tex]\rm \dfrac{1}{f}=(n - 1)\left(\dfrac{1}{R_1}+\dfrac{1}{R_2}\right)[/tex]

Now, substitute the values of the known terms in the above formula.

[tex]\rm \dfrac{1}{f}=(1.430 - 1)\left(\dfrac{1}{6.50}+\dfrac{1}{5.50}\right)[/tex]

Now, simplify the above equation in order to determine the value of 'f'.

[tex]\rm \dfrac{1}{f}=0.144\;mm^{-1}[/tex]

f = 6.92 mm = 6.92 [tex]\rm \times 10^{-3}[/tex] m

Now, the maximum power is given by the formula:

[tex]\rm P_{max} =\dfrac{1}{f}[/tex]

[tex]\rm P_{max} = 144.3 \;D[/tex]

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Answer:

 h ’= k  (h +L/2)²/ (2 M g )

Explanation:

For this exercise, one of the best methods to solve it is with energy conservation.

Starting point. Lower, spring with maximum compression

              Em₀ = Ke = ½ k x²

Final point. Higher after bounce

              = U = M g h ’

             Em₀ = Em_{f}

             ½ k x² = M g h’

the compressed distance is

          x = h+ L / 2

where h is the distance that compresses the spring by the height where it comes from and L/2 the additional compression

        h ’= ½ k x² / M g

we calculate

       h ’= k  (h +L/2)²/ (2 M g )

       h' = (k/2Mg) h2 (1 + L/2h)2

Answer:

Explanation:

a ) Let angle of incidence and angle of refraction be i and r respectively .

submarine is 300 m away from the shore . The point where laser strikes the surface of sea is 90 m horizontally away .

Tan r = 90 / 120

= 3 / 4

.75

r = 37 degree

c ) sini / sin37 = 1.333

sini = .8

i = 53 degree

Tan 53 = 210 / h , h is height of the building .

h = 210 / tan 53

= 158 m

Answer:[tex]\omega _f=1.185\ rad/s[/tex]

Explanation:

Given

mass of objects [tex]m=5\ kg[/tex]

Initially mass is at [tex]r=0.9\ m[/tex]

Initial angular speed [tex]\omega_i=0.66\ rad/s[/tex]

Moment of inertia of student and  stool is [tex]I_s=8\ kg-m^2[/tex]

Finally masses are at a distance of [tex]r_f=0.31\ m[/tex] from axis

[tex]I_i=I_p+I_m[/tex]

[tex]I_i=8+2\times 5\times (0.9)^2[/tex]

[tex]I_i=16.1\ kg-m^2[/tex]

Final moment of inertia of the system

[tex]I_f=I_s+I_m[/tex]

[tex]I_f=8+2\times 5\times (0.31)^2[/tex]

[tex]I_f=8+0.961=8.961\ kg-m^2[/tex]

As there is no external torque therefore moment of inertia is conserved

[tex]I_i\omega _i=I_f\omega _f[/tex]

[tex]\omega _f=\frac{16.1}{8.96}\times 0.66[/tex]

[tex]\omega _f=1.796\times 0.66[/tex]

[tex]\omega _f=1.185\ rad/s[/tex]

Final answer:

When the student pulls the objects closer, his angular speed increases to conserve angular momentum.

Explanation:

Angular momentum is conserved in this scenario. Initially, the student with the objects has a certain angular speed and moment of inertia. When he pulls the objects closer, his moment of inertia decreases, resulting in an increase in angular speed to conserve angular momentum.

Answer:

Explanation:

a )

Amplitude A = 14 mm , angular frequency ω = 2π / T

= 2π / .5

ω = 4π rad /s

φ₀ = initial phase

Putting the given values in the equation

14 = 25 cos(ωt + φ₀ )

14/25 = cosφ₀

φ₀ = 56 degree

x(t) = 25cos(4πt + 56° )

b )

maximum velocity = ω A

=  4π  x 25

100 x 3.14 mm /s

= 314 mm /s

At x = 0 ( equilibrium position or middle point , this velocity is achieved. )

maximun acceleration = ω² A

= 16π² x A

= 16 x 3.14² x 25

= 3943.84 mm / s²

3.9 m / s²

It occurs at x = A or at extreme position.

Answer:

Explanation:

Flux through the coil = nBA , n is no of turns , B is magnetic flux and A a is area of the coli

= 200 x 5.6 x 10⁻⁵ x 11.8 x 10⁻⁴

=  13216 x 10⁻⁹ weber .

b ) When the coil becomes parallel to magnetic field  , flux through it will become zero.

c ) e m f induced = change in flux / time

= 13216 x 10⁻⁹ / 4.9 x 10⁻²

= 2697.14 x 10⁻⁷ V

= 269.7 x10⁻⁶

269.7 μV.

Answer:

The power of the pump is 337.5 W.

Explanation:

Power is the rate of transfer of energy or the rate at which work is done. It is measured in watts (W).

                       Power = [tex]\frac{Energy}{Time}[/tex]

But the energy here is the kinetic energy (K.E), which is the energy possessed by a moving body or object.

           K.E = [tex]\frac{1}{2}[/tex] m[tex]v^{2}[/tex]

where m is the mass = 90 kg and v is the velocity = 15 m/s.

    K.E =  [tex]\frac{1}{2}[/tex] × 90 × [tex]15^{2}[/tex]

         = 10125 Joules

  K.E = 10125 J = 10.125 KJ

So that the power of the pump when t = 30 s is;

              Power = [tex]\frac{10125}{30}[/tex]

                         = 337.5 W

The power of the pump is 337.5 W.

Answer:

Explanation:

Given

mass of rod is M

Distance between rails is L

If the direction of magnetic field is directed upwards

If rod is moving right then,

Area enclosed is increasing therefore a Clockwise current will induce

and if rod is moving left then enclosed area is decreasing therefore an anticlockwise current will setup in the circuit

(b)Induced EMF will be

[tex]E=BvL[/tex]

where v=velocity of rod

and current is [tex]I=\frac{V}R}[/tex]

[tex]I=\frac{BvL}{R}[/tex]

(c)Force necessary to move rod with constant velocity v is

[tex]F_{net}=F-BIL=0[/tex]

[tex]F=BIL[/tex]

[tex]F=B\times \frac{BvL}{R}\times L[/tex]

[tex]F=\frac{B^2L^2v}{R}[/tex]

Answer:

Option C ⇒ D=2λr is the correct answer, since it has the largest aperture diameter.

Explanation:

Regarding Rayleigh's criterion, the angular resolution is given as below:

θ = 1.22λ/D

From this expression, it is observed that the larger the aperture size, the smaller will be the value of the angular resolution, and the better the device will be.

This signifies that at very high angular differences, the precision for distinguishing two points is high.

Therefore, option C ⇒ D=2λr is the correct answer, since it has the largest aperture diameter.

Answer:[tex]c=3.99\approx 4\ kJ/kg-K[/tex]

Explanation:

Given

Current [tex]I=300\ A[/tex]

Voltage [tex]V=270\ V[/tex]

Mass flow rate [tex]m=0.307\ kg/s[/tex]

Inlet temperature is [tex]T_i=15^{\circ}C\approx 288\ K[/tex]

[tex]T_{out}\leq 81^{\circ}C[/tex]

Here heat of Electromagnet is absorbed by liquid

Heat rate of Electromagnet [tex]\dot{Q}=VI[/tex]

[tex]\dot{Q}=270\times 300=81\ kJ/s[/tex]

Heat absorbed by liquid [tex]\dot{Q}=mc(\Delta T)[/tex]

[tex]\dot{Q}=0.307\times c\times (81-15)[/tex]

[tex]81\times 10^3=0.307\times c\times (81-15)[/tex]

[tex]c=3.99\ kJ/kg-K[/tex]

Answer:

Explanation:

Total time of transfer oil fuel = 2.45 x 9 = 22.05 minutes

= 22.05 x 60

= 1323 s

Total volume of ful transferred = 687 gallon

= .0037854 x 687

= 2.6 m³

radius of pipe = .5 x 1.45 inch

= .5 x 1.45 x 2.54 x 10⁻² m

r = .018415 m

cross sectional area

= π r²

a = 3.14 x .018415²

= 10.648 x 10⁻⁴ m²

If v be the velocity

volume of fuel coming out  a x v x t  , a is cross sectional area , v is velocity and t is time .

10.648 x 10⁻⁴ x v x 1323  = 2.6

v = 1.845 m / s

The velocity of the fuel through the refueling nozzle was calculated to be approximately 1.845 meters per second, using the area of the nozzle and the volume of fuel transferred over the total duration of the refuelings.

To calculate the velocity of the fuel through the nozzle during the in-flight refueling mentioned, we will need to use the formula for the volume flow rate, which is Q = A × v, where Q is the volume flow rate, A is the cross-sectional area of the nozzle, and v is the velocity of the fuel.

The cross-sectional area A of the nozzle can be found using the formula for the area of a circle, A = πr^2, where r is the radius of the nozzle. Given the diameter of the nozzle is 1.45 inches, the radius r is half of that, which is 0.725 inches. However, we should convert this to meters to use SI units: 0.725 inches × 0.0254 meters/inch = 0.018415 meters. So, A = π × (0.018415 meters)^2 ≈ 1.066 × 10^{-3} m^2.

To find the total volume of fuel transferred, we need to convert 687 gallons to cubic meters. There are 3.78541 liters in a gallon and 1,000 liters in a cubic meter, so the volume V in cubic meters is 687 gallons × 3.78541 liters/gallon × 1 m^3/1000 liters = 2.600 cubic meters.

Given that each refueling took 2.45 minutes and there were 9 refuelings, the total time t taken for the fuel transfer in seconds is 9 transfers × 2.45 minutes/transfer × 60 seconds/minute = 1321.5 seconds.

Now we can calculate the volume flow rate Q = V / t, which equals 2.600 m^3 / 1321.5 s = 1.967 × 10^{-3} m^3/s.

Finally, we can solve for the velocity v using Q = A × v, so v = Q / A which is (1.967 × 10^{-3} m^3/s) / (1.066 × 10^{-3} m^2) = 1.845 m/s.

Thus, the velocity of the fuel through the nozzle was approximately 1.845 meters per second.

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