How many photons does a green (532nm) 1mW laser pointer emit? Can we still treat the light as a continuous electromagnetic wave in this case? (2 pts)

Explanation:

Make a table, listing the x and y coordinates of each square's center of gravity and its mass.  Multiply the coordinates by the mass, add the results for each x and y, then divide by the total mass.

[tex]\left\begin{array}{ccccc}x&y&m&xm&ym\\\frac{a}{2} &\frac{a}{2} &10&5a&5a\\\frac{3a}{2}&\frac{a}{2}&70&105a&35a\\\frac{a}{2}&\frac{3a}{2}&80&40a&120a\\\frac{3a}{2}&\frac{3a}{2}&50&75a&75a\\&\sum&210&225a&235a\\&&Avg&\frac{15a}{14}&\frac{47a}{42}\end{array}\right[/tex]

The x-coordinate of the center of gravity is 15/14 a.

The y-coordinate of the center of gravity is 47/42 a.

Answer:

0.833 m³/kg

-14.4 kJ

20 kJ/kg

Explanation:

[tex]\rho[/tex] = Density = 1.2 kg/m³

[tex]V[/tex] = Volume = 0.6 m³

t = Time taken = 1 hour

P = Power = 4 W

Mass is given by

[tex]m=\rho V\\\Rightarrow m=1.2\times 0.6\\\Rightarrow m=0.72\ kg[/tex]

As there is no change in kinetic and potential energy, the specific volume of the tank will be unaffected

[tex]V_{s}=\dfrac{0.6}{0.72}\\\Rightarrow V_s=0.833\ m^3/kg[/tex]

The specific volume at the final state is 0.833 m³/kg

Energy is given by

[tex]E=Pt\\\Rightarrow E=-4\times 1\times 3600\\\Rightarrow E=-14400\ J[/tex]

The energy transfer by work is -14.4 kJ

Change in specific internal energy is given by

[tex]E=-m\Delta u\\\Rightarrow \Delta u=-\dfrac{E}{m}\\\Rightarrow \Delta u=-\dfrac{-14.4}{0.72}\\\Rightarrow \Delta u=20\ kJ/kg[/tex]

The change in specific internal energy is 20 kJ/kg

Answers:

Answer:

7.23 m/s

Explanation:

From Newton's equation of motion,

v = u + at ...................... Equation 1.

Where v = final velocity, u = initial velocity, a = acceleration, t = time.

Also,

s = ut+ 1/2at²........................ Equation 2

Where s = distance.

Given: t = 8.8 s, s = 45.0 m.

Substitute into equation 2

Note: we find the value of a in terms of u

45 = u(8.8)+1/2a(8.8)²

45 = 8.8u+38.72a

38.72a = 45 -8.8u

38.72a = (45-8.8u)

a = (45-8.8u)/38.72

also, v = 3.00 m/s

Substituting into equation 1

3 = u + 8.8[(45-8.8u)/38.72)]

3 = u + (45-8.8u)/4.4

3×4.4 = 4.4u + 45 - 8.8u

13.2 - 45 = 4.4u - 8.8u

-31.8 = -4.4u

u = -31.8/-4.4

u = 7.23 m/s.

Hence the initial velocity = 7.23 m/s

Answer with Explanation:

We are given that

a.Length building=632 m

1 m=3.2808 feet

632 m=[tex]3.2808\times 632=2073.47 feet[/tex]

Length of building,l=2073.47 feet

Width of building,b=710 yards=[tex]710\times 3=2130 ft[/tex]

1 yard=3 feet

Height of building,h=112 ft

Volume of building=[tex]l\times b\times h[/tex]

Volume of building=[tex]2073.47\times 2130\times 112=494647003.2ft^3[/tex]

Volume of building=494647003.2  cubic feet

b. 1 cubic feet=0.028 cubic meter

Volume of building= [tex]494647003.2\times 0.028[/tex]cubic meters

Volume of building=13850116.0896 cubic meters

Answer:

a) The population of prairie dogs after nine months is 280.

b) P(t) = 30 + 30 · t - t²/4 for 0 ≤ t ≤ 60

Explanation:

Hi there!

We have the following information:

The initial population is P(0) = 30.

The rate of growth of the population is the following:

P´(t) = 30 - t/2 where

a) Let´s find the function of the population of prairie dogs P(t). For that, let´s integrate the P´(t) function between t = 0 and t and between P = 30 and P

P(t) = ∫P´(t)

P´(t) = dP/dt = 30 - t/2

Separating variables:

dP = (30 - t/2) dt

∫dP = ∫(30 - t/2) dt

P - 30 = 30 · t - t²/4

P(t) = 30 + 30 · t - t²/4

The population of prairie dogs at t = 9 months will be equal to P(9):

P(9) = 30 + 30(9) - (9)²/ 4

P(9) = 280 prairie dogs

The population of prairie dogs after nine months is 280.

b) P(t) = 30 + 30 · t - t²/4 (it was obtained in part a).

The population of the prairie dog community 9 months later would be 294 prairie dogs, and the population equation for the entire prairie dog community for t months is P(t) = 30t - t^2 / 4 + 30 for 0 ≤ t ≤ 60.

The equation given is P'(t) = 30 - t / 2. This is a differential equation, and to get P(t), we need to find the anti-derivative or integrate P'(t) with respect to time, t. The integral of P'(t) = ∫(30 - t / 2) dt = 30t - t^2/4.

Adding the initial condition, P(0) = 30, we find P(t) = 30t - t^2 / 4 + 30 for 0 ≤ t ≤ 60. Therefore, a. The population of the prairie dog community 9 months later would be P(9) = 30*9 - (9^2) / 4 + 30 = 294 prairie dogs. b. P(t) = 30t - t^2 / 4 + 30 for 0 ≤ t ≤ 60 is the population equation for the entire prairie dog community for t months.

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Answer:

(a) 2.33 J.

(b) 1.87 m/s.

Explanation:

(a)

[tex]F(x) = (2.5 - x^2)\^i[/tex]

The force as a function of position is given above. Since the force is a function of position, we can assume that we will use work-energy theorem.

[tex]W = \Delta K = K_2 - K_1\\\int\limits^2_0 {F(x)} \, dx = \frac{1}{2}mv_2 - 0\\\int\limits^2_0 {(2.5-x^2)} \, dx = (2.5x - \frac{x^3}{3})\left \{ {{x=2} \atop {x=0}} \right. = (5 - 8/3) = 7/3[/tex]

Therefore, the kinetic energy of the block is 2.33 J.

(b) In order to find the maximum speed in this interval, we need to investigate the acceleration of the block. Since acceleration is the derivative of velocity, velocity is at its maximum when acceleration is zero.

From Newton's Second Law:

[tex]F = ma\\a(x) = F(x)/m = 1.66 - 0.66x^2[/tex]

In order this to be zero:

[tex]1.66 - 0.66x^2 = 0\\x = 1.58~m[/tex]

The velocity of the block at x = 1.58 m can be found by work-energy theorem.

[tex]\int\limits^{1.58}_0 {(2.5-x^2)} \, dx = \frac{1}{2}(1.5)v^2\\ (2.5x-\frac{x^3}{3})\left \{ {{x=1.58} \atop {x=0}} \right. = 2.63 J\\2.63 = \frac{1}{2}(1.5)v^2\\v = 1.87~m/s[/tex]

Answer:

The focal length of the concave mirror is -15.5 cm

Explanation:

Given that,

Height of the object, h = 20 cm

Radius of curvature of the mirror, R = -31 cm (direction is opposite)

Object distance, u = -94 cm

We need to find the focal length of the mirror. The relation between the focal length and the radius of curvature of the mirror is as follows :

R = 2f

f is the focal length

[tex]f=\dfrac{R}{2}[/tex]

[tex]f=\dfrac{-31}{2}[/tex]

f = -15.5 cm

So, the focal length of the concave mirror is -15.5 cm. Hence, this is the required solution.

To solve this problem we will apply the concepts related to the conservation of momentum. For this purpose we have that the initial momentum must be equivalent to the final momentum of the system. Mathematically this can be expressed as

[tex]m_1v_1 = m_2v_2[/tex]

Here,

[tex]m_{1,2}[/tex] = Mass of each object

[tex]v_{1,2}[/tex]= Velocity of each object

Rearranging to find the speed of the heavier weight,

[tex]v_2 = \frac{m_1v_1}{m_2}[/tex]

[tex]v_2 = \frac{(2.3)(6)}{5.3}[/tex]

[tex]v_2 = 2.6m/s[/tex]

Therefore the speed of the heavier weight is 2.6m/s

Answer:

There are 5.98 electrons are missing from each ions.

Explanation:

Given that,

The electric force between two identical charges is, [tex]F=33\times 10^{-9}\ N[/tex]  

The distance between the charges, [tex]d=0.5\ nm=0.5\times 10^{-9}\ m[/tex]

The electric force between charges is given by :

[tex]F=\dfrac{kq^2}{d^2}[/tex]

[tex]q=\sqrt{\dfrac{Fd^2}{k}}[/tex]

[tex]q=\sqrt{\dfrac{33\times 10^{-9}\times (0.5\times 10^{-9})^2}{9\times 10^9}}[/tex]

[tex]q=9.57\times 10^{-19}\ C[/tex]

Let there are n number of electrons are missing from each ions. It is given by :

[tex]n=\dfrac{q}{e}[/tex]

[tex]n=\dfrac{9.57\times 10^{-19}}{1.6\times 10^{-19}}[/tex]

n = 5.98 electrons

So, there are 5.98 electrons are missing from each ions. Hence, this is the required solution.

Newton's Third Law states the forces between the car and the truck are equal and opposite. Thus, the force the car exerts on the truck is equal to the force the truck exerts on the car, making Choice A correct for all situations described, including when the car and truck are at cruising speed or if the truck puts on its brakes.

According to Newton's Third Law of Motion, every action has an equal and opposite reaction. This law explains the forces between two objects—the car and the truck—when one object exerts force on another. In this scenario, when the compact car pushes the large truck:

Therefore, for the situations described:

The first option is mathematically described as

[tex]x = x_0 +v_0 t +\frac{1}{2} at^2[/tex]

Here,

[tex]x_0 =[/tex]Initial position

[tex]v_0 =[/tex] Initial velocity

t = Time

a = Acceleration

As we can see in this equation, the position of a body is described taking into account its initial point with respect to the reference system, the initial velocity of this body and the acceleration when it is constant. All this depending on time.

The second option despises the initial position, so it does not allow the exact calculation of the position.

The third option does not consider the position, only the speed and acceleration with respect to time

The fourth option considers acceleration and distance but does not take into account the time of the object.

Therefore the correct answer is A.

Answer:

Walking into a brick wall at normal walking speed (1.4 m/s), you will come to a complete stop in a short time (0.1 s) and experience much more acceleration (14 m/s²) back the way you came, but because you are softer than the wall, the inelastic collision will probably cause you to bounce back off the wall, changing the actual experienced acceleration you feel.

Explanation:

Assuming normal human walking speeds and time it takes to come to rest.

Normal human walking speed = 5km/h = 1.4 m/s

Time it takes you to come to a complete stop = 0.1 seconds

acceleration = Δ Velocity/ Time

Δ Velocity = final velocity - initial velocity

Δ Velocity =  (1.4 - 0)m/s  = 1.4 m/s

acceleration = 1.4/0.1

acceleration = 14 m/s²

Walking into a brick wall at normal walking speed, you will experience much more acceleration back the way you came, but because you are softer than the wall, the inelastic collision will probably cause you to bounce back off the wall, changing the actual experienced acceleration you feel.

Acceleration of the subject.

The acceleration is the rate at which the body velocity changes with time. It can be referred to the body the speed and direction. The point of the object moves in the straight line as acceleration is both a magnitude and a direction. Estimated acceleration of the subject and yourself

thus the answer is 1.4 m/s and stops at 0.1s

Learn more about the acceleration.

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To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

[tex]\sum \tau = F*d[/tex]

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,

[tex]F*(27-6)= 6*600[/tex]

[tex]F = \frac{6*600}{21}[/tex]

[tex]F= 171.42 lb[/tex]

So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb

(a) The volume occupied by the water vapor is V = 1.08 m³.

(b) The amount of water vapor present is 277.78 gram-moles.

(c) The number of molecules is [tex]\rm\(1.67305 \times 10^{26}\)[/tex] molecules.

Given:

Specific volume (v) = 0.2160 m³/kg

Mass of water vapor (m) = 5 kg

One gram-mole of water vapor = 18 g

Avogadro's number [tex]\rm (\(N_A\))[/tex] = [tex]\rm \(6.023 \times 10^{23}\)[/tex]

(a) The volume occupied by the water vapor (V) can be calculated using the formula: [tex]\rm \[ V = m \times v \][/tex]

Substitute the values:

[tex]\rm \[ V = 5 \, \text{kg} \times 0.2160 \, \text{m³/kg} \]\\\rm V = 1.08 \, \text m\³[/tex]

(b) The amount of water vapour present in gram moles can be calculated using the formula:

[tex]\rm \[ \text{Amount of water vapor} = \frac{m}{\text{One gram-mole of water vapor}} \][/tex]

Substitute the value of one gram-mole of water vapor [tex](\(18 \, \text{g}\))[/tex]:

[tex]\[ \text{Amount of water vapor} = \frac{5 \, \text{kg} \times 1000}{18 \, \text{g}} \]\\\\\ \text{Amount of water vapor} = 277.78 \, \text{gram-moles} \][/tex]

(c) The number of molecules can be calculated using Avogadro's number:

[tex]\rm \[ \text{Number of molecules in 1 gram-mole} = N_A \\= 6.023 \times 10^{23} \, \text{molecules} \][/tex]

So, the number of molecules in 1 gram of water vapor is [tex]\rm \(\frac{N_A}{18}\)[/tex] molecules.

The number of molecules in [tex]\rm \(5 \, \text{kg}\)[/tex] of water vapor is:

[tex]\rm \[ \text{Number of molecules} = 5 \times 1000 \times \frac{N_A}{18} \, \text{molecules} \][/tex]

Now, calculate the number of molecules:

[tex]\rm \[ \text{Number of molecules} = 1.67305 \times 10^{26} \, \text{molecules} \][/tex]

Know more about Avogadro's number:

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Wilma's average speed over the entire trip is approximately 3.71 m/s, calculated by dividing the total distance by the total time. However, as she returned to her starting point, her total displacement is zero, and thus, her average velocity is 0 m/s.

The average speed is given by the total distance divided by the total time. Since Wilma I. Ball travels the same distance twice (A to B and back), we can say the total distance is 2d. Suppose the distance between A to B is 'd' m. The time taken for the first trip is d/4.51 s, and the time taken for the return trip is d/3.15 s. Therefore, the average speed is given by (2d) / ((d/4.51) +(d/3.15)). This simplifies to approximately 3.71 m/s.

However, average velocity is the total displacement divided by total time. Since she ends at her starting point, her total displacement is zero. Thus, her average velocity over her round trip is 0 m/s.

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Answer:

1.5 m

Explanation:

given,

wavelength of of the standing waves

λ₁ = 2 m

λ₂ = 1.5 m

to  find wavelength corresponding to higher mode.

we know,

     [tex]wavelength\ \alpha\ \dfrac{1}{n}[/tex]

where n is the mode number.

From the above expression we can say that wavelength is inversely proportional to mode.

Hence , the wavelength corresponding to higher mode is equal to 1.5 m

Answer:

1.962 rad / s², 1.6 s

Explanation:

Radius of the part = 1.0m / 2 = 0.5 m

angular speed = 30 rpm = 30 rpm × (2πrad / rev) × 1 minutes / 60 seconds = 3.142 rads⁻¹

μk = frictional force / normal ( mg )

normal is the force acting upward against the force of gravity

frictional force = - μk mg

since the body came to rest then

Fnet + Ff =  0

Fnet = - Ff

Fnet = ma

ma =  - μk mg

a =  - μkg where g = 9.81

a = - 0.1 × 9.81 = 0.981 m/s²

magnitude of angular acceleration = tangential acceleration / radius = 0.981 / 0.5 = 1.962 rad / s²

b) time for the train to come to rest = angular velocity  / angular angular acceleration = 3.142/ 1.962 = 1.6 s

The equation earlier derived answer this question

Fnet + Ff = 0 since the body came to a rest

Fnet = - Ff and Ff = - μk mg, Fnet = ma

ma = - μk mg

m cancel m on both side

a = - μkg since it magnitude

a = μkg

The coefficient of rolling friction multiplied by gravity gives the tangential acceleration in rolling motion due to the relationship between frictional force and the object's acceleration.

The reason why the coefficient of rolling friction multiplied by gravity gives the tangential acceleration is due to the relationship between the frictional force and the force responsible for the acceleration of the object. In the case of rolling motion, the frictional force opposes the rotation and contributes to the overall acceleration.

Answer:

3.170 miles

Explanation:

Conversion from yards to feet:

5580 yards = 5580 yard * 3 ft/yard = 16740 feet

Conversion from feet to miles:

16740 feet = 16740*(1/5280) mile/ft = 3.170 miles.

Therefore, that particular artillery piece has a range R = 3.170 miles

Answer:

The displacement is at x=1.59m and t=0.150s is [tex]-6.50\cdot10^{-2}\text{ m}.[/tex]

Out of the given points, the argument of the cosine is an integer multiple of [tex]\pi[/tex] for x=1.59m, 1.86m, 2.13m.

Explanation:

The displacement at x and t is given by y(x,t). We now need to find [tex]k[/tex] and [tex]\omega[/tex]. The speed of the wave is given by

[tex]c=\frac{\omega}{k}[/tex]

while the wavelength satisfies

[tex]k=\frac{2\pi}{\lambda}=\frac{2\pi}{0.540\text{ m}}=11.6\text{ m}^{-1}.[/tex]

Substituting this into the previous equation we find

[tex]\omega=ck=8.75\text{ m/s}\cdot 11.6\text{ m}^{-1}=102\text{ rad/s}.[/tex]

Now we have

[tex]y(1.59\text{ m},0.150\text{ s})=6.50\cdot10^{-2}\text{ m}\cos(11.6\cdot1.59-102\cdot0.150)=-6.50\cdot10^{-2}\text{ m}.[/tex]

Now we calculate [tex]kx-\omega t[/tex] at [tex]t=0.150\text{ s}[/tex] at each given x and check whether it is integer multiple of [tex]\pi[/tex].

[tex]11.6\text{ m}^{-1}\cdot 0.795\text{ m}-102\text{ rad/s}\cdot0.150\text{ s}=-6.08=-1.93\pi\text{ not an integer multiple of }\pi;[/tex]

[tex]11.6\text{ m}^{-1}\cdot 1.59\text{ m}-102\text{ rad/s}\cdot0.150\text{ s}=3.14=\pi\text{ it is an integer multiple of }\pi;[/tex]

[tex]11.6\text{ m}^{-1}\cdot 1.73\text{ m}-102\text{ rad/s}\cdot0.150\text{ s}=4.77=1.52\pi\text{ not an integer multiple of }\pi;[/tex]

[tex]11.6\text{ m}^{-1}\cdot 1.86\text{ m}-102\text{ rad/s}\cdot0.150\text{ s}=6.28=2\pi\text{ it is an integer multiple of }\pi;[/tex]

[tex]11.6\text{ m}^{-1}\cdot 2.00\text{ m}-102\text{ rad/s}\cdot0.150\text{ s}=7.90=2.51\pi\text{ not an integer multiple of }\pi;[/tex]

[tex]11.6\text{ m}^{-1}\cdot 2.13\text{ m}-102\text{ rad/s}\cdot0.150\text{ s}=9.41=3\pi\text{ it is an integer multiple of }\pi;[/tex]

Final answer:

The wave function y(x, t) = Acos(kx−ωt) can be used to find x positions of maximum displacement for a transverse wave on a string.

Explanation:

The wave function y(x, t) = Acos(kx−ωt) represents a transverse wave on a string. Given the values of k = 11.6 rad/m and ω = 102 rad/s, we can find the x positions that correspond to points of maximum displacement by setting the argument of the cosine function to be an integer multiple of π.

Using the wave equation, the x positions that correspond to points of maximum displacement are:

0.795 m

1.59 m

2.00 m

2.13 m

To solve this problem we will apply the first law of thermodynamics which details the relationship of energy conservation and the states that the system's energy has. Energy can be transformed but cannot be created or destroyed.

Accordingly, the rate of work done in one cycle and the heat transferred can be expressed under the function,

[tex]\dot{U} = \dot{Q}-\dot{W}[/tex]

Substitute 1W for [tex]\dot{Q}[/tex] and 1.5 W for [tex]\dot{W}[/tex]

[tex]\dot{U} = 1-1(1.5)[/tex]

[tex]\dot{U} = 2.5W[/tex]

Now calculcate the rate of specific internal energy increase,

[tex]\dot{u} = \frac{\dot{U}}{m}[/tex]

[tex]\dot{u} = \frac{2.5}{1.5}[/tex]

[tex]\do{u} = 1.6667W/kg[/tex]

The rate of specific internal energy increase is 1.6667W/kg

To answer the student's question, combine the total power input from heat and stirring, then calculate the temperature increase of the oil using its specific heat capacity with the energy conservation principle.

The student is asking about the rate of specific energy increase when heat is transferred to motor oil while work is being done on it by stirring. In thermodynamics, this relates to the conversion of work into thermal energy. The first step is to calculate the total energy input per second (power), which is the sum of heat transfer and the work done by stirring. The next step is to use the specific heat capacity to determine the rate of temperature increase for the oil.

Given the specific heat capacity of water and a weight lifting scenario, the student needs to apply the energy conservation principle, converting the work done into the thermal energy, which will result in raising the temperature of the water. To find this temperature increase, one should use the formula Q = mcΔT, where Q is the energy transferred as heat, m is the mass of the water, c is the specific heat capacity, and ΔT is the change in temperature.

The speed of third fragment of the given gadget is 8.62 m/s.

The given parameters:

The mass of the third fragment is calculated as follows;

[tex]m_3 = 21.85-(6.42 + 8.26)\\\\m_3 = 7.17 \ kg[/tex]

Apply the principle of conservation of linear momentum to determine the speed of the third fragment as follows;

[tex]m_3v_3 = m_1v_1 cos(\frac{\theta}{2} ) \ + \ m_2v_2 cos(\frac{\theta}{2} )\\\\7.17v_3 = 6.42\times 6.8 \times cos(\frac{64}{2} ) \ + 8.26 \times 6.8 \times cos(\frac{64}{2} )\\\\7.17 v_3 = 61.82 \\\\v_3 = \frac{61.82}{7.17} \\\\v_3 = 8.62 \ m/s[/tex]

Thus, the speed of third fragment of the given gadget is 8.62 m/s.

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Using conservation of momentum and resolving in x and y directions, we find the speed of the third fragment to be approximately 8.69 m/s. Initial conditions ensure total momentum is zero.

To solve this, we use the conservation of momentum. The total initial momentum of the system is zero since the gadget was initially at rest. For the fragments, the equation can be set up as follows:

m₁ * v₁ + m₂ * v₂ + m₃ * v₃ = 0

First, we resolve this in the x and y directions.

In the x-direction:

6.42 * 6.8 * cos(0°) + 8.26 * 3.54 * cos(64°) + 7.17 * v₃x = 0

⇒ 6.42 * 6.8 * 1 + 8.26 * 3.54 * 0.438 + 7.17 * v₃x = 0

⇒ 43.656 + 12.807 + 7.17 * v₃x =0

⇒ 7.17 * v₃x  = - 56.463

v₃x = -7.88 m/s

In the y-direction:

6.42 * 6.8 * sin(0°) + 8.26 * 3.54 * sin(64°) + 7.17 * v₃y = 0

⇒ 6.42 * 6.8 * 0 + 8.26 * 3.54 * 0.899 + 7.17 * v₃y = 0

⇒ 26.281 + 7.17 * v₃y =0

⇒ 7.17 * v₃y  = - 26.281

v₃y = -3.67 m/s

Then, using Pythagorean theorem to find v3:

v₃ = √(v₃x² + v₃y²)

= √((-7.88)² + (-3.67)²)

≈ √(62.1+ 13.5)

≈ √75.6

≈ 8.69 m/s

Thus, the speed of the third fragment is approximately 8.69 m/s.

Answer:

[tex]P_2=4091\ KPa[/tex]

Explanation:

Given that

T₁ = 290 K

P₁ = 100 KPa

Power P =5.5 KW

mass flow rate

[tex]\dot{m}= 0.01\ kg/s[/tex]

Lets take the exit temperature = T₂

We know that

[tex]P=\dot{m}\ C_p (T_2-T_1)[/tex]

[tex]5.5=0.01\times 1.005(T_2-290})\\T_2=\dfrac{5.5}{0.01\times 1.005}+290\ K\\\\T_2=837.26\ K[/tex]

If we assume that process inside the compressor is adiabatic then we can say that

[tex]\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{0.285}[/tex]

[tex]\dfrac{837.26}{290}=\left(\dfrac{P_2}{100}\right)^{0.285}\\2.88=\left(\dfrac{P_2}{100}\right)^{0.285}\\[/tex]

[tex]2.88^{\frac{1}{0.285}}=\dfrac{P_2}{100}[/tex]

[tex]P_2=40.91\times 100 \ KPa[/tex]

[tex]P_2=4091\ KPa[/tex]

That is why the exit pressure will be 4091 KPa.

Answer:

Trial and error.

Explanation:

This approach of Thomas Edison where in testing of thousand of different material for light bulb filament are used to find the most suitable material is called trial and error approach of problem solving.

Trial and error pursue a method, seeing if it performs, and not trying a new mechanism. This mechanism will be repeated until a solution or success is attained. Assume moving a large object like a couch into your house for example.

Answer: B. The speed at the bottom is the same for both hills.

Explanation A sled which can also be called a sledge is a vehicle built with a smooth body on the side touching the ground underside which makes it easy for it to be able to slide towards the ground when on a slant hill. The velocity of the sled after it has slid down the Hill and the velocity at the bottom of the Hill will be the same because both point have the same level of steepness. Velocity of a material is directly proportional to the mass of a body and inversely proportional to the time traveled.

The speed of the sled at the bottom of both hills will be the same because the total energy, determined by the height of the hill and not its steepness, remains constant.

The best answer to this question is B. The speed at the bottom is the same for both hills. This is because the total energy of the sled (kinetic plus gravitational potential energy) must remain constant if we assume no friction or air resistance. The height of the hill, not its steepness, determines the total energy. The sled will convert all its potential energy (which depends on the height) into kinetic energy (which affects the speed) as it travels downhill. Hence, if the two hills are of the same height, the sled will have the same speed at the bottom of both hills.

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When an object moves with constant velocity, its acceleration is zero. So, the hay bale's velocity is 1 ft/s, and its acceleration is 0 ft/s2. These values persist irrespective of the horse's distance from the barn.

In the case of the farmer's horse and the hay bale, the horse is moving with a constant velocity of 1 ft/s. When an object moves with constant velocity, its acceleration is zero. This principle stems from the fundamental definition of acceleration as the rate of change of velocity over time. Since the horse's velocity is not changing, acceleration is zero.

Moving on to the velocity of the hay bale: if we suppose that the horse's movement directly influences the lifting of the hay bale, the bale's velocity would also be 1 ft/s. Provided that the system of lifting the bale is suitable for the task, it does not matter how far the horse is away from the barn; the velocity and acceleration values persist. Therefore, regardless of whether the horse is 10 ft away or 100 ft away, the velocity of the hay bale remains 1 ft/s and acceleration 0 ft/s2.

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The hay bale being lifted by the farmer's horse, which is walking at a constant velocity of 1 ft/s, would have the sameconstant velocity with an acceleration of 0 m/s², because there is no change in velocity. Velocity vectors for the horse's movement would remain consistent, indicating the absence of acceleration.

The question involves determining the velocity and acceleration of a hay bale being lifted into a barn loft by a horse walking at a constant velocity. When dealing with such problems in physics, we typically look for changes in velocity over time to find acceleration. In this scenario, since the horse is moving with a constant velocity of 1 ft/s and no information is provided about forces acting on the hay bale or if there is any change in the speed of the horse, we can infer that the hay bale being lifted is moving at the same constant velocity of 1 ft/s, with an acceleration of 0 m/s² (or ft/s²) since there is no change in speed. This is because acceleration is defined as the rate of change of velocity over time, and a constant velocity means no change in velocity.

To conceptualize this, imagine sketching out a series of velocity vectors for the horse's path from one point to the next, indicating consistent motion with no alterations above or below the horizontal axis, representing that there is no acceleration occurring. An example in the given information refers to a racehorse accelerating from rest to 15.0 m/s, indicating that its acceleration can be calculated by dividing the change in velocity by the change in time, giving an average acceleration. But for the farmer's horse walking forward at a constant pace, the situation is quite different since there is no increase in speed.

To reiterate, if the horse maintains a constant velocity, the hay bale will have the same velocity (1 ft/s) and zero acceleration, provided that the system remains unaltered. Calculating acceleration in such cases is straightforward when there's a change in velocity; for constant velocity, the acceleration is simply zero.

Answer:

A) v = 1.885 m/s

B) v = 0.39 m/s

C) E = 0.03 J

D) [tex]x(t) = (0.15m)\cos(2\pi (2.0Hz)t)[/tex]

Explanation:

Part A

We will use the conservation of energy to find the speed at equilibrium.

[tex]K_{eq} + U_{eq} = K_A + U_A\\

\frac{1}{2}mv^2 + 0 = 0 + \frac{1}{2}kA^2\\

v = \sqrt{\frac{k}{m}}A[/tex]

where [tex]\omega = \sqrt{k/m}[/tex] and [tex]\omega = 2\pi f[/tex]

Therefore,

[tex] v = 2\pi f A = 2(3.14)(2)(0.15) = 1.885~m/s[/tex]

Part B

The conservation of energy will be used again.

[tex]K_1 + U_1 = K_2 + U_2\\

\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}kA^2\\

mv^2 + kx^2 = kA^2\\

(0.23)v^2 + k(0.10)^2 = k(0.15)^2\\

v^2 = \frac{k(0.15)^2-(0.10)^2}{0.23}\\

v = \sqrt{0.054k}[/tex]

where [tex]k = \omega^2 m = (2\pi f)^2 m = 2(3.14)(2)(0.23) = 2.89[/tex]

Therefore, v = 0.39 m/s.

Part C

Total energy of the system is equal to the potential energy at amplitude.

[tex]E = \frac{1}{2}kA^2 = \frac{1}{2}(2.89)(0.15)^2 = 0.03~J[/tex]

Part D

The general equation of motion in simple harmonic motion is

[tex]x(t) = A\cos(\omega t + \phi)\\

x(t) = (0.15m)\cos(2\pi (2.0Hz)t + \phi)[/tex]

where [tex]\phi[/tex] is the phase angle to be determined by the initial conditions. In this case, the initial condition is that at t = 0, x is maximum. Therefore,

[tex]x(t) = (0.15m)\cos(2\pi (2.0Hz)t)[/tex]

The speed of the mass when passing the equilibrium point is approximately 1.88 m/s, and when it is 0.10 m from equilibrium, the speed is approximately 1.26 m/s. The equation describing the motion of the mass when x was a maximum at t=0 is x(t) = (0.15 m)cos[2π(2.0 Hz)t].

Part A: To determine the speed when the mass passes the equilibrium point, we can use the formula for the velocity of an object in simple harmonic motion, which is v = ωA, where ω is the angular frequency and A is the amplitude. In this case, the angular frequency is given by ω = 2πf, where f is the frequency. So, ω = 2π(2.0 Hz) = 4π rad/s. The amplitude is 0.15 m. Substituting these values into the formula, we get v = (4π rad/s)(0.15 m) ≈ 1.88 m/s.

Part B: To determine the speed when the mass is 0.10 m from equilibrium, we can again use the formula for velocity. Using the same angular frequency ω, we find that v = (4π rad/s)(0.10 m) ≈ 1.26 m/s.

Part C: The total energy of the system can be found using the formula for the total energy of an object in simple harmonic motion, which is E = (1/2)kA², where k is the spring constant and A is the amplitude. In this case, k is not given, so we cannot determine the total energy.

Part D: The equation describing the motion of the mass, assuming that at t=0, x was a maximum, is given by x(t) = (0.15 m)cos[2π(2.0 Hz)t].

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Final answer:

Type m · a to correctly enter the multiplication of m and a, ensuring clarity in expressions, especially when dealing with scientific notation which simplifies multiplication and division by manipulating coefficients and powers of ten separately.

Explanation:

When entering the expression m times a, you should use proper notation to differentiate between multiplication and adjacent variables or numbers. Instead of using 'x' as the multiplication symbol, as you might on paper, use the multiplication dot, represented by typing m · a with your keyboard. This avoids confusion, especially in contexts like scientific notation, where clarity is crucial. For example, 1·105 is not the same as 1 followed by 105. So, it is important to insert the multiplication dot to clarify that you are multiplying 1 by 10 raised to the 5th power.

In scientific notation, multiplication and division are simplified by separately manipulating the coefficients and the powers of ten. For instance, to multiply two numbers in scientific notation, such as (3 × 105) × (2 × 10°), you would multiply 3 by 2 to get 6, and then add the exponents (5 + 0) to get 5, resulting in an answer of 6 · 105. This simplifies calculations incredibly, especially with large numbers.

Answer:

Electric field, E = 45833.33 N/C

Explanation:

Given that,

Separation between the plates, d = 4.8 mm = 0.0048 m

The potential difference between the plates, V = 220 volts

We need to find the electric field between two parallel plates. The relation between the electric field and electric potential is given by :

[tex]V=E\times d[/tex]

[tex]E=\dfrac{V}{d}[/tex]

[tex]E=\dfrac{220}{0.0048}[/tex]

E = 45833.33 N/C

So, the electric field between two parallel plates is 45833.33 N/C. Hence, this is the required solution.

The electric field between the plates is 45833.33 N/C.

Given that the separation d between the plates is 4.8 mm and the potential difference V between the plates is 220 V.

The electric field E between the plates can be calculated by the formula given below.

[tex]E =\dfrac {V}{d}[/tex]

Substituting the values in the above equation, we get

[tex]E = \dfrac {220}{0.0048}[/tex]

[tex]E = 45833.33 \;\rm N/C[/tex]

Hence we can conclude that the electric field between the plates is 45833.33 N/C.

To know more about the electric field, follow the link given below.

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Explanation:

Earth or any planet are actually born from huge clouds of gas and dust. Their stellar mass are fairly distributed at a radius from the axis of rotation. Gravitational force cause the cloud to come together. Now the whole gathered in smaller area. Now, individual particles come close to the roational axis. Thus, decreasing the moment of inertia of the planet.

As

I=mr^2

reducing r reduces I. However, the angular moment of the system remains always conserved. So, to conserve the angular momentum the angular velocity of the planet increases and so did the  otational kinetic energy

The additional rotational kinetic energy of the Earth compared to the initial cloud comes from the gravitational potential energy that was present in the cloud before it collapsed. The conservation of angular momentum dictates that as the cloud shrinks, its rotation speed must increase, leading to an increase in rotational kinetic energy.

The Earth has more rotational kinetic energy now than did the cloud of gas and dust from which it formed because of the conservation of angular momentum. As the cloud collapsed under its own gravity, the rotation rate increased to conserve angular momentum, which is the product of moment of inertia and rotational velocity. Since the moment of inertia decreases as the mass moves closer to the axis of rotation (due to the shrinking size of the collapsing cloud), the rotational velocity must increase to keep the angular momentum constant. This results in an increase in rotational kinetic energy, which is given by the equation [tex]\( KE_{rot} = \frac{1}{2} I \omega^2 \)[/tex], where [tex]\( I \)[/tex] is the moment of inertia and [tex]\( \omega \)[/tex] is the angular velocity.

The gravitational potential energy of the cloud is converted into kinetic energy as the cloud collapses. The initial potential energy is high because the particles in the cloud are far from the center of mass. As the cloud contracts, this potential energy is transformed into kinetic energy, both linear and rotational, due to the conservation of energy. The increase in rotational kinetic energy is a direct consequence of this conversion and the conservation of angular momentum.

In summary, the additional rotational kinetic energy of the Earth compared to the initial cloud comes from the gravitational potential energy that was present in the cloud before it collapsed. The conservation of angular momentum dictates that as the cloud shrinks, its rotation speed must increase, leading to an increase in rotational kinetic energy.

The complete question is:

The Earth has more rotational kinetic energy now than did the cloud of gas and dust from which it formed. Where did this energy come from?

Answer:

(a) A = m/s^3, B = m/s.

(b) dx/dt = m/s.

Explanation:

(a)

[tex]x = At^3 + Bt\\m = As^3 + Bs\\m = (\frac{m}{s^3})s^3 + (\frac{m}{s})s[/tex]

Therefore, the dimension of A is m/s^3, and of B is m/s in order to satisfy the above equation.

(b) [tex]\frac{dx}{dt} = 3At^2 + B = 3(\frac{m}{s^3})s^2 + \frac{m}{s} = m/s[/tex]

This makes sense, because the position function has a unit of 'm'. The derivative of the position function is velocity, and its unit is m/s.