Water from a 100 ft x 200 ft lawn converges into a gully. During a 1-hour storm, discharge into the gully is 0.37 cfs (ft3/s). The interception by grass is 0.02 acre-ft. Overall, 70% of the rain is infiltrated. What is the amount of rainfall?

Vector quantities have magnitude and direction. The quantities that are vectors from the given list are:

1980 kg m/s due south,

8.2 m/s^2 north-west,

3.2 mi straight up.

Quantities that have a magnitude and direction are called vectors. Based on this definition, the following quantities from the list are vectors:

Therefore, these three quantities are vectors.

https://brainly.com/question/33923402

#SPJ11

To solve this problem we will apply the concepts related to momentum and momentum on a body. Both are equivalent values but can be found through different expressions. The impulse is the product of the Force for time while the momentum is the product between the mass and the velocity. The result of these operations yields equivalent units.

PART A ) The Impulse  can be calculcated as follows

[tex]L= F\Delta t[/tex]

Where,

F = Force

[tex]\Delta t =[/tex]Change in time

Replacing,

[tex]L = (35N)(0.2s)[/tex]

[tex]L= 7N\cdot s[/tex]

PART B) At the same time the momentum follows the conservation of momentum where:

Initial momentum= Final momentum

And the change in momentum is equal to the Impulse, then

[tex]\Delta p = L[/tex]

And

[tex]\Delta p = p_f - p_i[/tex]

There is not initial momentum then

[tex]\Delta p = p_f[/tex]

[tex]L = p_f[/tex]

[tex]p_f = 7N\cdot s = 7kg\cdot m/s[/tex]

Final answer:

The impulse on the ball, which is the product of the force and the time interval, is 7 N·s. Since the ball is initially at rest, the final momentum of the ball is also 7 N·s.

Explanation:

To solve the student's question, we need to apply the concepts of impulse and momentum in physics.

Impulse and Momentum

Impulse is defined as the product of the force acting on an object and the time interval over which it acts. From Newton's second law, the impulse on an object is equal to the change in its momentum.

a) The impulse on the ball is calculated using the formula Impulse = Force × Time. Given that the force is 35 N and the time is 0.2 s, the impulse can be calculated as:

Impulse = 35 N × 0.2 s = 7 N·s

b) The final momentum of the ball is equal to the impulse, since the ball was initially at rest (momentum = 0). Therefore, the final momentum of the ball is:

Final momentum = Impulse = 7 N·s

To solve this problem we will use the concepts related to the uniform circular movement from where we will obtain the speed of the object. From there we will go to the equilibrium equations so that the friction force must be equal to the centripetal force. We will clear the value of the coefficient of friction sought.

The velocity from the uniform circular motion can be described as

[tex]v = \frac{2 \pi r}{T}[/tex]

Here,

r = Radius

T = Period

Replacing,

[tex]v = \frac{2\pi (8.4)}{6}[/tex]

[tex]v =8.7964 m/s[/tex]

From equilibrium to stay in the circle the friction force must be equivalent to the centripetal force, therefore

[tex]F_f = F_c[/tex]

[tex]\mu N = \frac{mv^2}{r}[/tex]

Here,

[tex]\mu =[/tex] Coefficient of friction

N = Normal Force

m = mass

v = Velocity

r = Radius

The value of the Normal force is equal to the Weight, then

[tex]\mu(mg) = \frac{mv^2}{r}[/tex]

Rearranging to find the coefficient of friction

[tex]\mu = \frac{v^2}{gr}[/tex]

Replacing,

[tex]\mu = \frac{(8.7964)^2}{(9.8)(8.4)}[/tex]

[tex]\mu =0.9399[/tex]

Therefore the minimum coefficient of friction to prevent the cat from sliding off is  0.9399

Answer:

A. F=107.6nN

B. Repulsive

Explanation:

According to coulombs law, the force between two charges is express as

F=(Kq1q2) /r^2

If the charges are of similar charge the force will be repulsive and if they are dislike charges, force will be attractive.

Note the constant K has a value 9*10^9

Hence for a charge q1=7.10nC=7.10*10^-9, q2=4.42*10^-9 and the distance r=1.62m

If we substitute values we have

F=[(9×10^9) ×(7.10×10^-9) ×(4.42×10^-9)] /(1.62^2)

F=(282.4×10^-9)/2.6244

F=107.6×10^-9N

F=107.6nN

B. Since the charges are both positive, the force is repulsive

The magnitude of the electric force between the two particles is approximately 0.012 N, according to Coulomb's Law. The force could be either attractive or repulsive, depending on whether the charges are the same or opposite, respectively.

The subject of this question is Physics, specifically about Coulomb’s Law, which deals with the electric force between two charges.

(a) According to Coulomb’s Law, the electric force (F) between two charges is directly proportional to the product of their charges (q1 and q2) and inversely proportional to the square of the distance (d) between them. This is mathematically represented as F = k*q1*q2/d^2, where k is Coulomb's constant (8.988 x 10^9 Nm^2/C^2). Substituting in the given values, we find F = (8.988 * 10^9 Nm^2/C^2) * (7.10 * 10^-9 C) * (4.42 * 10^-9 C) / (1.62 m)^2, resulting in an electric force of approximately 0.012 N.

(b) Whether the force is attractive or repulsive depends on the nature of the charges. Same charges repel each other, while opposite charges attract each other. As the charges are not specified, we cannot definitively answer this part of the question.

https://brainly.com/question/21093704

#SPJ3

The electric potential energy of a system comprising two protons 2.1 x 10^-15 m apart can be calculated using Coulomb's law, which gives us a result of approximately 0.68 MeV.

The electric potential energy of a system depends upon the charge of the components and the distance between them. In this case, we can calculate the electric potential energy using Coulomb's law which states that the electric potential energy 'V' between two charges is given by the equation V = k*q1*q2/r where 'k' is Coulomb's constant (8.99 × 10^9 N m^2/C^2), 'q1' and 'q2' are the two charges, and 'r' is the distance between the charges. Given that the charges are two protons, they both have the same charge (1.6 × 10^-19 Coulombs). The distance 'r' is given as 2.1 × 10^-15 m. Substituting these values in, we get: V = (8.99 × 10^9 N m^2/C^2)* (1.6 × 10^-19 C) * (1.6 × 10^-19 C) / (2.1 × 10^-15 m) which results in an electric potential energy of approximately 0.68 MeV (mega electron volts).

https://brainly.com/question/33051432

#SPJ3

Answer:

option (E)

Explanation:

F = 1 N

r = 2 cm

r' = 8 cm

F' = ?

According to the Coulomb's law

[tex]F \alpha \frac{1}{r^{2}}[/tex]

So, [tex]\frac{F'}{F}=\left (\frac{r}{r'}  \right )^{2}[/tex]

[tex]\frac{F'}{1}=\left (\frac{2}{8}  \right )^{2}[/tex]

F' = 0.063 N

When they are moved to a new separation of 8.0 cm, the electric force on each of them will be closest to 0.063 N. Option E is correct.

Given:

The point charges are initially at a distance of [tex]d=2[/tex] cm.

The initial force experienced by them is [tex]f=1[/tex] N.

The electrostatic force experienced by two charges q and Q is defined as,

[tex]f=K\dfrac{qQ}{r^2}[/tex]

where K is the constant and r is the distance between the charges.

The initial force experienced by them will be,

[tex]f=K\dfrac{qQ}{d^2}=1\rm\; N[/tex]

Now, the final distance between the charges is changed to 8 cm which is equal to 4d.

So, the new force on the charges will be,

[tex]F=K\dfrac{qQ}{(4d)^2}\\F=K\dfrac{qQ}{d^2}\times \dfrac{1}{16}\\F=f\times \dfrac{1}{16}\\F=0.0625\approx0.063\rm\;N[/tex]

Therefore, when they are moved to a new separation of 8.0 cm, the electric force on each of them will be closest to 0.063 N. Option E is correct.

For more details, refer to the link:

https://brainly.com/question/9774180

Answer:

(A) Distance will be equal to 1.75 km

(B) Displacement will be equal to 1.114 km

Explanation:

We have given circumference of the circular track = 3.5 km

Circumference is given by [tex]2\pi r=3.5[/tex]

r = 0.557 km

(a) It is given that car travels from southernmost point to the northernmost point.

For this car have to travel the distance equal to semi perimeter of the circular track

So distance will be equal to [tex]=\frac{3.5}{2}=1.75km[/tex]

(b) If car go along the diameter of the circular track then it will also go from southernmost point to the northernmost point. and it will be equal to diameter of the track

So displacement will be equal to d = 2×0.557 = 1.114 m

To develop this problem we will apply the energy conservation theorem and the principle of work. Basically we will have that the kinetic, potential and work energy must be equivalent to the kinetic energy just before the impact. We will start converting the units given to the British system and then proceed with the Calculations.

Remember that according to the energy balance in this case it would be balanced like this

[tex]T_1 +\sum U_{1-2} = T_2[/tex]

[tex]\frac{1}{2} mv_1^2+mgh -W_{drag} = \frac{1}{2} mv_2^2[/tex]

Here

m = mass

[tex]v_{1,2}[/tex]= Velocity at each moment

[tex]W_{drag}[/tex]= Work by drag

h = Height

g = Acceleration due to gravity

Mass

[tex]m =255000 lb (\frac{1slug}{32.174lb})[/tex]

[tex]m = 7925.654slug[/tex]

Initial Velocity

[tex]v_1 = 550 \frac{mi}{h} (\frac{5280ft}{1mi})(\frac{1hr}{3600s})[/tex]

[tex]v_1 = 806.667ft/s[/tex]

Work by drag

[tex]W_{drag} = (2.96*10^6BTU)(\frac{778.169lb\cdot ft}{1BTU})[/tex]

[tex]W_{drag} = 2303380240lb\cdot ft[/tex]

By work energy principle

[tex]\frac{1}{2} mv_1^2+mgh -W_{drag} = \frac{1}{2} mv_2^2[/tex]

Replacing,

[tex]\frac{1}{2} (7925.654slug)(806.667ft/s)^2 +(7925.654slug)(32.08ft/s^2)(30000ft) -2303380240lb\cdot ft = \frac{1}{2} (7925.654slug)v_2^2[/tex]

Solving for [tex]v_2[/tex], we have that

[tex]v_2 = 1412.2 ft/s[/tex]

Converting this value,

[tex]v_2 = 1412.2 ft/s (\frac{1mi}{5280ft})(\frac{3600s}{1h})[/tex]

[tex]v_2 = 962.85mi/h[/tex]

Therefore the velocity of the aircraft at the time of impact is 962.85mi/h

To estimate the velocity of the aircraft at the time of impact, use the work-energy principle. The estimated velocity is 256.15 mi/h.

To estimate the velocity of the aircraft at the time of impact, we can use the work-energy principle. The work done by the drag force is equal to the change in kinetic energy of the aircraft. The work done by the drag force is given as 2.96x10^6 Btu. We need to convert this energy into foot-pounds, and then use the kinetic energy equation to find the velocity. The equation is:

K = (1/2) mv^2

Where K is the kinetic energy, m is the mass of the aircraft, and v is the velocity. Rearranging the equation, we have:

v = √((2K) / m)

Plugging in the given values, we get:

v = √((2(2.96x10^6 Btu) * (3.968x10^8 ft-lbf/Btu)) / (255,000 lb * 32.08 ft/s^2)) = 256.15 mi/h

Therefore, the estimated velocity of the aircraft at the time of impact is 256.15 mi/h.

To calculate the viscosity of the oil, you need to utilize the concept of shear stress in fluid dynamics. You take the known values of torque, cylinder length, and rotational speed and substitute them into the rearranged shear stress formula. Ensure you convert length from millimeters to meters to maintain SI unit consistency.

The question can be solved using the principles of fluid dynamics, specifically the concept of viscous drag in a medium. In this case, the medium is oil and the object moving through it is a cylinder. The viscosity of the oil can be calculated using the formula for shear stress (τ), given as τ = η (du/dy), where η represents viscosity, du represents the difference in velocity, and dy represents the distance between layers of fluid. As we're given torque (τ) in Newton-metres (Nm), cylinder length (L) in millimeters, and rotational speed in rad/s (radians per second), we can derive η (viscosity) using the rearranged formula: η = τ /(du/dy).

In this case, du is equivalent to the speed of the cylinder let's represent this as 'u', dy is equivalent to the length of the cylinder 'L', hence du/dy becomes u/L. Substitute these values into the rearranged formula and solve to get the viscosity. It's critical to convert the length from millimeters to meters before performing the calculation to maintain consistency in the SI units.

https://brainly.com/question/13253926

#SPJ12

Answer:

U_eq = 1.99 * 10^(-10) J

Explanation:

Given:

Plate Area = 10 cm^2

d = 0.01 m

k_dielectric = 3

k_air = 1

V = 15 V

e_o = 8.85 * 10 ^-12  C^2 / N .m

Equations used:

U = 0.5 C*V^2  .... Eq 1

C = e_o * k*A /d  .... Eq 2

U_i = 0.5 e_o * k_i*A_i*V^2 /d  ... Eq 3

For plate to be half filled by di-electric and half filled by air A_1 = A_2 = 0.5 A:

U_electric = 0.5 e_o * k_1*A*V^2 /2*d

U_air = 0.5 e_o * k_2*A*V^2 /2*d

The total Energy is:

U_eq = U_electric + U_air

U_eq = 0.5 e_o * k_1*A*V^2 /2*d  + 0.5 e_o * k_2*A*V^2 /2*d

U_eq = (k_1 + k_2) * e_o * A*V^2 / 4*d

Plug the given values:

U_eq = (3 + 1) * (8.82 * 10^ -12 )* (0.001)*15^2 / 4*0.01

U_eq = 1.99 * 10^(-10) J

The energy U2 of the partially dielectric-filled capacitor is found by calculating the total capacitance Ct when the capacitor is half-filled with the dielectric and then using the formula U2 = ½ CtV2 with the given voltage.

To calculate the energy U2 of the partially dielectric-filled capacitor, we start by finding the capacitance when the capacitor is half-filled with the dielectric. With a dielectric constant k of 3.00 and a vacuum permittivity ε0 of 8.85×10−12 C2/N·m2, the initial capacitance Ci entirely filled with the dielectric is:

Ci = kε0A/d

When the capacitor is half-filled, the capacitance can be represented as two capacitors in parallel: one with dielectric (Cd) and one without (C0). These can be described by:

Cd = ε0kA/(2d) and C0 = ε0A/(2d)

The total capacitance Ct when the capacitor is half-filled is then:

Ct = Cd + C0

Finally, the energy stored in the capacitor U2 is given by:

U2 = ½ CtV2

This formula allows us to plug in the calculated capacitance Ct and the constant voltage V to find the desired energy U2.

Answer:

[tex]\Delta f=f_{Lr}-f_{Se}[/tex]

147.45 Hz

Explanation:

v = Speed of sound in water = 1482 m/s

[tex]v_w[/tex] = Speed of whale = 4.95 m/s

Frequency of the wave in stationary condition

[tex]f_{Lr}=f\dfrac{v+v_w}{v-v_w}[/tex]

Ship's frequency which is reflected back

[tex]f_{Se}=f\dfrac{v}{v-v_w}[/tex]

The difference in frequency is given by

[tex]\Delta f=f\dfrac{v+v_w}{v-v_w}-f\dfrac{v}{v-v_w}\\\Rightarrow \Delta f=f_{Lr}-f_{Se}[/tex]

[tex]\mathbf{\Delta f=f_{Lr}-f_{Se}}[/tex]

[tex]f_{Lr}=22\times \dfrac{1482+4.95}{1482-4.95}\\\Rightarrow f_{Lr}=22.14745\ kHz[/tex]

[tex]f_{Se}=22\ kHz[/tex]

[tex]\Delta f=f_{Lr}-f_{Se}\\\Rightarrow \Delta f=22.14745-22\\\Rightarrow \Delta f=0.14745\ kHz\\\Rightarrow \Delta f=147.45\ Hz[/tex]

The difference in wavelength is 147.45 Hz

Answer:

[tex]410.2 nm[/tex]

Explanation:

We are given that

[tex]n_1=2,n_2=6[/tex]

We have to find the wavelength of laser should you used.

We know that

[tex]\frac{1}{\lambda}=R(\frac{1}{n^2_1}-\frac{1}{n^2_2})[/tex]

Where [tex]R=1.097\times 10^7/m[/tex]=Rydberg constant

[tex]\lambda[/tex]=Wavelength

Using the formula

[tex]\frac{1}{\lambda}=1.097\times 10^7(\frac{1}{2^2}-\frac{1}{6^2})[/tex]

[tex]\frac{1}{\lambda}=1.097\times 10^7(\frac{1}{4}-\frac{1}{36})[/tex]

[tex]\frac{1}{\lambda}=1.097\times 10^7(\frac{9-1}{36}=1.097\times 10^7\times \frac{8}{36}[/tex]

[tex]\frac{1}{\lambda}=\frac{1.097\times 10^7}{4}[/tex]

Using identity:[tex]\frac{1}{a^x}=a^{-x}[/tex]

[tex]\lambda=\frac{4}{1.097}\times 10^{-7}[/tex]=[tex]4.102\times 10^{-7} m[/tex]

1 nm=[tex]10^{-9} m[/tex]

[tex]\lambda=4.102\times 100 \times 10^{-9}=410.2\times 10^{-9} [/tex] m=410.2 nm

Hence, the wavelength of laser=[tex]410.2 nm[/tex]

Answer:

Although this question is not complete, I would give a general solution to this kind of problems.

If y(t) describes the position of a body with time such that

y(t) = at^(n) + bt^(m) + C

Then

V(t) = dy(t)/dt = ant^(n-1) + bmt^(m-1)

Explanation:

As an example supplies the position of a particle is given by

y(t) = 4t³- 3t² + 9

V(t) = 4x3t²- 3x2t¹

V(t) = d(t)/dt = 12t² - 6t.

Another example,

If y(t) = 15t³ - 2t² + 30t -80

V(t) = d(t)/dt = 15x3t² - 4t +30 = 45t² + 4t + 30.

Basically, in the equations above the powers of t reduces by one when computing the velocity function from y(t) by differentiation (calculating the derivative of y(t)). The constant term C (9 and 80 in the functions of y(t) in examples 1and 2 above) reduces to zero because the derivative of a constant (and ordinary number without the t attached to it) is always zero.

One last example,

y(t) = 2t^6 -3t²

V(t) = d(t)/dt = 12t^5 - 6t

Answer:

the difference is due to resistance tolerance

Explanation:

In mathematical calculations, either done by hand or in a computer program, the heat taken from the resistors is the nominal value, which is the writing in its color code, so all calculations give a result, but the Resistors have a tolerance, indicated by the last band that is generally 5%, 10%, 20% and in the expensive precision resistance can reach 1%.

   This tolerance or fluctuation in the resistance value is what gives rise to the difference between the computation values ​​and the values ​​measured with the instruments, multimeters.

   Another source of error also occurs due to temperature changes in the circuit that affect the nominal resistance value, there is a very high resistance group that indicates the variation with the temperature, they are only used in critical circuits, due to their high cost

In summary, the difference is due to resistance tolerance.

Final answer:

The displacement z of a particle with rest mass m0 subject to a constant force m0g along the z-axis, accounting for relativistic effects, involves integrating the work-energy theorem's relativistic expression for kinetic energy and comparing it with the classical displacement z = 1/2 gt².

Explanation:

The problem given involves finding the displacement z of a particle with rest mass m0 under the influence of a constant force along the z-axis, considering relativistic effects. The comparison with the classical result implies a consideration of the relativistic factor y in kinetic energy expressions.

We recall that the relativistic kinetic energy is more complex than the classical expression K = 1/2 mu², because as the velocity u of the particle approaches the speed of light c, the kinetic energy diverges due to the increase in y.

The task involves using the work-energy theorem in a relativistic context, where the work done by the force m0g produces a change in the relativistic kinetic energy. As such, relativistic mechanics dictates that energy and momentum are dependent on the relativistic factor y, which is a function of velocity.

Therefore, to expand the displacement z as a power series in time t, one would start by expressing the relativistic momentum and kinetic energy, then integrate to find the time-dependent expressions for velocity and displacement.

Comparing with the classical result, where z is simply obtained by z = 1/2 gt² (ignoring air resistance), shows how relativistic effects modify the simple parabolic trajectory of a particle in a gravitational field. The relativistic displacement would involve a series with higher-order terms in time, accounting for the varying mass with velocity.

Answer:

The magnitude of the electric field is [tex]8.6\times10^{2}\ N/C[/tex]

Explanation:

Given that,

Time t = 2.10 s

Speed = 160 m/s

Specific charge =Ratio of charge to mass = 0.100 C/kg

We need to calculate the acceleration

Using equation of motion

[tex]a=\dfrac{v-u}{t}[/tex]

Put the value into the formula

[tex]a=\dfrac{160-0}{2.10}[/tex]

[tex]a=76.19\ m/s^2[/tex]

We need to calculate the magnitude of the electric field

Using formula of electric field

[tex]E=\dfrac{F}{q}[/tex]

[tex]E=\dfrac{ma}{q}[/tex]

[tex]E=\dfrac{a+g}{\dfrac{q}{m}}[/tex]

Put the value into the formula

[tex]E=\dfrac{76.19+9.8}{0.100}[/tex]

[tex]E=8.6\times10^{2}\ N/C[/tex]

The direction is upward.

Hence, The magnitude of the electric field is [tex]8.6\times10^{2}\ N/C[/tex]

Answer:

E = 8.6 x 10² N/C

Explanation:

given,

initial speed of charge,u = 0 m/s

final speed of charge,v = 160 m/s

time,t = 2.1 s

charge-to-mass ratio = 0.100 C/kg

Electric field of the region = ?

Acceleration of the charge

[tex]a = \dfrac{v-u}{t}[/tex]

[tex]a = \dfrac{160 - 0}{2.1}[/tex]

a = 76.19 m/s²

specific charge = [tex]\dfrac{q}{m} = 0.1[/tex]

now,

Electric field,

[tex]E = \dfrac{F}{q}[/tex]

charge is moving upwards so,

[tex]E = \dfrac{(a + g)}{\dfrac{q}{m}}[/tex]

[tex]E = \dfrac{(76.19+9.8)}{0.1}[/tex]

E = 860 N/C

electric field , E = 8.6 x 10² N/C

hence, the magnitude of electric field is equal to E = 8.6 x 10² N/C

Final answer:

To calculate the centripetal acceleration and average speed of the Sun in its galactic orbit, we utilize relevant formulas and astronomical data. The concept of centripetal acceleration is fundamental in understanding circular motion in celestial bodies like the Sun as it orbits the Milky Way galaxy.

Explanation:

Centripetal acceleration is the acceleration directed toward the center of a circular path. To calculate the centripetal acceleration of the Sun in its galactic orbit, we use the formula: a = v^2 / r, where v is the speed of the Sun and r is the radius of its orbit. Given that the Sun's orbit radius is 3.00 x 10^4 light years and it takes 2.60 x 10^8 years to orbit the Milky Way galaxy, we can also calculate the average speed of the Sun in its galactic orbit. To determine if a nearly inertial frame of reference can be located at the Sun, we need to consider the motion relative to the galactic center.

To solve this problem we apply the kinematic equations of linear motion. For which the speed is described as the distance traveled in a time interval. This would be,

[tex]v = \frac{x}{t} \rightarrow t = \frac{x}{v}[/tex]

Our values are given as,

[tex]\text{The speed of the boat} = v_b = 3.8m/s[/tex]

[tex]\text{The speed of the ocean} = V = 6.8m/s[/tex]

[tex]\text{The wave length of the wave is the same distance traveled by boat} = d = \lambda = 30m[/tex]

[tex]\text{The relative speed of the boat} = v_r = -3.8 +6.8 = 3m/s[/tex]

Replacing we have,

[tex]t = \frac{d}{v_r}[/tex]

[tex]t = \frac{30m}{3m/s}[/tex]

[tex]t = 10s[/tex]

Therefore will take until the boat is next on a crest around to 10s

The ideal thermal efficiency for the steam power plant operating on the reheat Rankine cycle is approximately 57.0%.

To determine the thermal efficiency of the steam power plant operating on the ideal reheat Rankine cycle, we will follow the steps in the Problem-Solving Strategies for Thermodynamics:

Identify the Temperatures and Pressures

Boiler Pressure (P1): 17.5 MPa (High-pressure turbine inlet)

Reheater Pressure (P2): 2 MPa (Low-pressure turbine inlet)

Condenser Pressure (P3): 50 kPa

High-Pressure Turbine Inlet Temperature (T1): 550°C = 823 K

Low-Pressure Turbine Inlet Temperature (T2): 300°C = 573 K

Calculate Maximum Efficiency (Carnot Efficiency)

The maximum theoretical efficiency (Carnot efficiency) for a heat engine operating between two temperatures is given by the formula:
[tex]\eta_{max} = 1 - \frac{T_c}{T_h}[/tex]

where:

[tex]T_h[/tex] = Highest temperature (at the boiler, T1) = 823 K

[tex]T_c[/tex] = Lowest temperature (at the condenser) = Convert 50 kPa to temperature

Using steam tables or Mollier diagrams, the saturation temperature corresponding to 50 kPa is approximately 81°C = 354 K.

Now substituting:
[tex]\eta_{max} = 1 - \frac{354}{823} \approx 0.570 \text{ or } 57.0\%[/tex]

The total energy of an electron in a hydrogen atom with a given orbit radius can be determined using the formula E = -13.6 eV / n². By calculating the principal quantum number n from the given radius, we can find the total energy of the electron. In this case, the total energy is -3.4 eV.

The total energy of an electron in the hydrogen atom can be determined using the formula:

E = -13.6 eV / n²

where E is the energy, n is the principal quantum number, and eV represents electron volts.

Given that the radius of the electron's orbit is 8.784 × 10⁽⁻¹⁰⁾ m, we can find the value of n using the formula for the radius of the nth orbit:

r = 0.529 × n² / Z Å

where r is the radius, n is the principal quantum number, and Z is the atomic number. Assuming the atomic number for hydrogen is 1, solving for n gives us:

n = √(r * Z / 0.529)

Substituting the given values, we find that n = 2.

Finally, plugging in n into the energy formula, we find:

E = -13.6 eV / (2²) = -3.4 eV

Answer:

4. total energy

Explanation:

According to Bernoulli's principle at any two points along a streamline flow The total energy that is sum of pressure energy , Kinetic energy  and potential energy of the liquid all taken in per unit volume remains constant. Therefore,

for ideal fluid flows through a pipe of variable cross section without any friction. The fluid completely fills the pipe. At any given point in the pipe, the fluid has a constant Total Energy.

Answer:

2491.23 kg/m³

Explanation:

From Archimedes principle,

R.d = weight of object in air/ upthrust in water = density of the object/density of water

⇒ W/U = D/D' ....................... Equation 1

Where W = weight of the ceramic statue, U = upthrust of the ceramic statue in water, D = density of the ceramic statue, D' = density of water.

Making D the subject of the equation,

D = D'(W/U).................... Equation 2

Given: W = 28.4 N, U = lost in weight = weight in air- weight in water

U = 28.4 - 17.0 = 11.4 N,

Constant: D' = 1000 kg/m³.

Substitute into equation 2,

D = 100(28.4/11.4)

D = 2491.23 kg/m³

Hence the density of the ceramic statue = 2491.23 kg/m³

Answer:

Tension T = 13.14N

Explanation:

Given:

Mass of rock m = 1.50kg

Density of rock p = 4700kg/m^3

Volume of rock V = mass/density = m/p

V = 1.50kg/4700kg/m3 = 3.19×10^-4m3

Taking the summation of forces acting on the rock;

T-W+Fb = 0

T = W - Fb .....1

T = tension

W = weight of rock

Fb = buoyant force

Fb = pw(0.5V)g = density of water × Volume under water×™ acceleration due to gravity

g = 9.8m/s^2

T = mg - pw(0.5V)g

T = 1.50×9.8 - 1000kg/m^3 ×0.5(3.19 × 10^-4) × 9.8

T = 13.14N

Final answer:

To find the tension in the string, we can use the concept of buoyancy. By equating the buoyant force and weight of the rock, we can solve for the tension. The tension will be equal to the weight of the rock minus the buoyant force.

Explanation:

In order to find the tension in the string, we can use the concept of buoyancy. Since half of the rock's volume is under water, it experiences an upward buoyant force equal to the weight of the water displaced by that volume. The buoyant force can be found using the following equation:

Buoyant force = density of water * volume of water displaced * acceleration due to gravity

The weight of the rock is equal to its mass multiplied by the acceleration due to gravity.

By equating the buoyant force and weight of the rock, we can solve for the tension in the string. The tension will be equal to the weight of the rock minus the buoyant force.

In an oscillating string, the wave velocity at a point on the string is oriented opposite to the direction of the string's movement at that point. Thus, if point A is moving up, v(A) is pointed down, and if point B is moving down, v(B) is pointed up.

The orientation of the velocity of the wave at points A and B is determined by the direction of the wave's movement. For an upward and downward oscillating string, the wave moves perpendicular to the string. When a segment of the string is moving upwards, the wave velocity at that point is downward, and vice versa.

So, if point A is currently moving upwards, then v(A) should be pointed downwards to represent the direction of the wave velocity. Similarly, if point B is moving downwards, then v(B) should be pointed upwards. The velocity vectors v(A) and v(B) are always opposite to the direction of string's movement at those points.

https://brainly.com/question/34190786

#SPJ3

Answer:

Isogonic line is a line joining the points of equal magnetic declination.

Explanation:

Isogonal line is a line that joins the places of equal declination. Also isogonal line is known as the line which connects the point having the same magnetic declination.

An isogonic line joins points of equal magnetic declination on a map. This is important for navigation, as it reflects the angle difference between true north and magnetic north. Isotherms, isohyets, and isobars are other types of isolines used in geography. Thus option 5. isogonic line is correct.

A isogonic line is a line joining points of equal magnetic declination. Magnetic declination is the angle between magnetic north (the direction the north end of a compass needle points) and true north. These lines are important for navigational purposes and are often shown on special maps known as isogonic charts.

Other types of isolines include:

Understanding these different lines helps in various geographical and meteorological analyses, making it easier to interpret maps and forecasts.

Complete question.

A(n) _________ is a line joining the points of equal magnetic declination.

Answer

given,

speed of the fastest runner = 6.2 m/s

speed of the slower runner = 5.5 m/s

Assume, we need to calculate  time when both the runner meet for the first time.

distance cover by the fast runner = distance cover by the slow runner

  6.2 t = 5.5 t + 200

   0.7 t = 200

      t = 285.71 s

time after which both the runner meet.

Distance they covered after starting

D = s x t

D = 6.2 x 285.71

D = 1771.43 m.

now, calculating the second time both will meet.

they will take double time to meet again

 t' = 2 x 285.71

t '= 571.42 s

To solve this problem we will apply the concepts related to linear velocity and angular velocity to perform the respective conversion with the given values. To find the velocity in the upper part of the tire we will use the mathematical relation that expresses that it is twice the linear velocity. Let's start

PART A)  

[tex]\omega = \frac{v}{r}[/tex]

[tex]\omega = \frac{29}{0.3}[/tex]

[tex]\omega = 96.66 rad/s[/tex]

Now we now that [tex]2\pi rad = 1 rev[/tex], then

[tex]\omega = 96.66rad/s (\frac{1 rev}{2\pi rad})[/tex]

[tex]\omega = 15.38rev/s[/tex]

PART B)

[tex]v = 2v_0[/tex]

[tex]v = 2(29)[/tex]

[tex]v = 58m/s[/tex]

Final answer:

The distance from the charge is 30 meters.

Explanation:

The distance from the point charge can be calculated using the equation E = kQ/r^2, where E is the electric field strength, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), Q is the charge, and r is the distance. In this case, the electric field strength is the same at any point 5.00 mm away from the charge. Therefore, we can set up the equation as follows:

E = kQ/r^2

5.00 x 10^-3 m = (9 x 10^9 Nm^2/C^2)(5 x 10^-9 C)/r^2

By rearranging the equation and solving for r, we can find the distance:

r^2 = (9 x 10^9 Nm^2/C^2)(5 x 10^-9 C)/(5.00 x 10^-3 m)

r^2 = 900 m^2

r = sqrt(900 m^2) = 30 m

Therefore, the charge is located 30 meters away.

Answer:

A. T = 0.4358s and f = 2.29hz

B. A = 15.67m

C. amax = 3258.71m/s

D. amax = 22601J

E. Ek = 3616.16J

Explanation:

A. The period of the motion, T = 2pi*(sqrt(m/k))

Where m is the mass of the body in motion = 885g = 0.885kg

k = the spring constant = 184N/m2

T = 2pi*(sqrt(0.885/184))

= 0.4358s

Frequency of the motion, f = 1/T

T = 0.4358s

f = 2.2949hz

B. Maximum speed, Vmax = A*(sqrt(k/m))

Where A = amplitude of the motion

Making amplitude subject of formula,

A = Vmax(sqrt(m/k))

= 226*(sqrt(0.885/184))

= 15.6739m

C. Maximum acceleration, amax = A*(k/m)

= 15.6739*(184/0.885)

= 3258.71m/s

D. Total energy, Etotal = 1/2*(m * Vmax)2

= 1/2 * 0.885 * (226)2

= 22601J

E. Kinetic energy, Ek = Etotal - mechanical energy

Ek = 1/2*(k*A2) - 1/2*(k*x2)

Where x = 0.40A

Ek = 1/2*((k*A2) - (k*0.40A)2)

= 1/2*k*A2*(1 - 0.16)

= 1/2*k*A2*0.16

But 1/2*k*A2 = 22601J

Therefore, Ek = 22601*0.16

= 3616.16J

Final answer:

The average speed of the horse for the entire trip is 10.5 m/s, and the average velocity is 3.5 m/s, calculated based on the total distance covered and total displacement over the entire duration of the trip.

Explanation:

To calculate the average speed and average velocity of the horse for the entire trip, we consider both phases of the journey: when the horse canters away and when it gallops back.

Part (a) - Average Speed

Average speed is total distance covered divided by total time taken. The horse travels 126 m away and then comes back halfway, which is 63 m, resulting in a total distance of 126 m + 63 m = 189 m. The total time taken for the trip is 13.0 s + 5.0 s = 18.0 s. Therefore, the average speed is:

average speed = total distance / total time = 189 m / 18.0 s = 10.5 m/s

Part (b) - Average Velocity

Average velocity is the total displacement divided by the total time taken. Since the horse ends up halfway back, the displacement is 126 m - 63 m = 63 m, in the initially defined positive direction (away from the trainer). Using a total time of 18.0 s:

average velocity = total displacement / total time = 63 m / 18.0 s = 3.5 m/s