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Compare and contrast microscopic and macroscopic energy transfer. Give at least three comparisons for each. THX

It takes 30 s from launch to the top of the rocket's trajectory.

The total time of flight is calculated to be equal to 45 s.

Trajectory is the path that any object with mass in motion follows through space as function of time.

As we know, Vf = Vi + at

d = Vi × t + (1/2) × \(\mathrm{at^2}\)

Vf = final velocity  ; Vi = initial velocity  ; a = acceleration  ; t = time  d = distance

As, F = ma

So, F = mg

ma = mg

a = g

Vf = Vi + at

Vf = 0 + (\(\mathrm{9.81 m/s^2}\)) × 15 s

Vf = 147.15 m/s

So the velocity of the rocket at the instant the rocket motor shuts down is 147.15 m/s.

Vf = Vi + at

0 = 147.15 m/s + (\(\mathrm{-9.81 m/s^2}\)) × t

t = 15 s - (147.15 m/s)/(\(\mathrm{-9.81 m/s^2}\))

t = 30 s

So it takes 30 s from launch to the top of the rocket's trajectory.

total time of flight = 15 s (accelerating upwards) + 30 s (reaching maximum height and falling back to ground)

Total time of flight = 45 s

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Answer:

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

Explanation:

Given;

wheel rotates from rest with constant angular acceleration.

Initial angular speed v = 0

Time t = 2.50

Distance x = 8 rev

Applying equation of motion;

x = vt +0.5at^2 ........1

Since v = 0

x = 0.5at^2

making a the subject of formula;

a = x/0.5t^2 = 2x/t^2

a = angular acceleration

t = time taken

x = angular distance

Substituting the values;

a = 2(8)/2.5^2

a = 2.56 rev/s^2

velocity at t = 2.50

v1 = a×t = 2.56×2.50 = 6.4 rev/s

Through the next 5 second;

t2 = 5 seconds

a2 = 2.56 rev/s^2

v2 = 6.4 rev/s

From equation 1;

x = vt +0.5at^2

Substituting the values;

x2 = 6.4(5) + 0.5×2.56×5^2

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

Given, Voltage= 4V

           Charge=16C

q=cv

here q is directly proportional to v, but c is constant

so, c=q/v

     c=16/4

     c=4f

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Answer:

Explanation:

By Elaborating on the Question,

Q is directly proportional to V.

⇒ Q ∝ V

The constant of proportionality, in this case, is C

that is the capacitance.

Hence, Q = CV.

From the relation,

           ⇒ C = Q/V

           ⇒ C = 16/4

           ⇒ C = 4F

Hence, the Value of the Capacitor is 4F

The unit vector notation in the given direction is \(0.58 \hat x + 0.81 \hat y\).

A physical quantity that has both directions and magnitude is referred to as a vector quantity.

A lowercase letter with a "hat" circumflex, such as "û," is used to denote a vector with a magnitude equal to one. This type of vector is known as a unit vector.

In terms of subtracting a vector, it works the same as adding the vector's opposite.

Given that:

\(\vec A = 22 \hat x + (-14)\hat y,\\\vec B = 2.5\hat x + 13\hat y.\)

( two given vectors are written here.)

Therefore: the subtraction of these two vectors will be:

\(\vec A - \vec B = (= 22 \hat x + (-14)\hat y) - (2.5\hat x + 13\hat y)\\= 19.5\hat x - 27 \hat y\)

Unit vector in that direction of vectors  is =

\(= \frac{19.5\hat x - 27 \hat y}{\sqrt{19.5^2 + (-27)^2} } \\= 0.58 \hat x + 0.81 \hat y\)

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Answer:

un café y a seguir pensando porque no se la respuesta xd

Answer:

For a thermodynamic system to be in equilibrium, all intensive (temperature, pressure) and extensive thermodynamic properties (U, G, A, H, S, etc) must be constants. Hence, the total change in any of those properties (dℑ ) must be zero at equilibrium.

Explanation:

hope it helps :)

Answer: For a thermodynamic system to be in equilibrium, all intensive (temperature, pressure) and extensive thermodynamic properties (U, G, A, H, S, etc) must be constants. Hence, the total change in any of those properties (dℑ ) must be zero at equilibrium.

Explanation:

Answer:

Galileo thought that a ball, rolling or sliding down a hill without friction, would run up to the same height on an opposite hill.

Suppose that the opposite hill was horizontal. Would the ball's motion continue forever along the tangent, or forever parallel to the Earth's surface

Galileo's conclusion from this thought experiment was that no force is needed to keep an object moving with constant velocity.

Newton took this as his first law of motion.

Explanation:

Here ya go-

Answer:a vertical expanse of water stretching between the surface and the floor of a body of water

Explanation:

7.1 The reading on V₁ is 0 V.

7.2 The total resistance of the circuit is 6 Ω.

The reading on V₁ is zero because the two cells are connected in parallel. When cells are connected in parallel, the voltage across each cell is the same. In this case, the voltage across each cell is 3 V. Therefore, the reading on V₁ is 0 V.

The total resistance of the circuit is given by the following equation:

R_T = R₁ + R₂

where:

R_T is the total resistance of the circuit in ohms

R₁ is the resistance of the first resistor in ohms

R₂ is the resistance of the second resistor in ohms

In this case:

R_T = ?

R₁ = 2 Ω

R₂ = 4 Ω

Substituting these values into the equation:

R_T = 2 Ω + 4 Ω

R_T = 6 Ω

Therefore, the total resistance of the circuit is 6 Ω.

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The line current of the 3-phase 415 volt, 50Hz supply for a balanced load of 3000 Watts at 0.8 power factor connected in star is approximately 6.26 amps.

Lower current results in less torque, whereas higher current results in more torque. A motor with a delta connection will draw more line current and create more torque than a motor with a star connection. Delta connection is the response to this query.

To calculate the line current of a 3-phase system, we need to use the following formula:

Line current = Power / (√3 x Voltage x Power factor)

Where:

Power = 3000 Watts

Voltage = 415 volts

Power factor = 0.8 (assumed lagging since no indication is given)

Using the above formula, we get:

Line current = 3000 / (√3 x 415 x 0.8) = 6.26 amps (approx.)

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As the time elapses before the ball strikes the ground, the horizontal velocity of the ball remains the same.

The horizontal velocity of the ball remains constant because of the absence of any forces acting on it in the horizontal direction. The only force acting on the ball is gravity, which only affects the vertical direction of the ball's motion.

Is a fundamental concept in physics, as it explains why objects maintain a constant horizontal velocity when there are no forces acting upon them. This phenomenon can be observed in everyday life, such as when a ball is thrown in the air.

As the ball moves upwards, the only force acting upon it is the force of gravity, which causes the ball to slow down and eventually fall back to the ground. During this time, the ball's horizontal velocity remains constant.

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Answer:

a)  W = - 1.92 10⁻¹⁸ J, b)    U = 1.92 10⁻¹⁸ J, c)   v = 2.05 10⁶ m / s

Explanation:

a) Work is defined by

        W = F. x

 the electric force is

         F = q E

we substitute

          W = q E x

where the displacement is parallel to the electric field

all quantities must be in the SI system

         x = 3.20 cm = 0.0320 m

         W = - 1.6 10-19 375 0.0320  

         W = - 1.92 10⁻¹⁸ J

b) potential energy

          U = \(k \frac{q_1q_2}{r}\)

the electrical power is

          V = k q₁ / r

we can write this potential as a function of the electric field

           V = -E x

we substitute

           U = -q E x

           U = - (-1.6 10⁻¹⁹) 375 0.0320

           U = 1.92 10⁻¹⁸ J

Observe that the variation of the red potential is equal to the electrical work

c) let's use conservation of energy

starting point

          Em₀ = U = e E x

final point

          Em_f = K = ½ m v²

energy is conserved

          Em₀ = Em_f

          e E x = ½ m v²

          v² = 2 e E x / m

           v = \(\sqrt{ \frac{{2 \ 1.6 10^{-19} \ 375 \ 0.0320 } }{9.1 \ 10^{-31} } }\)

           v = \(\sqrt{4.2198 \ 10^{12}}\)

           v = 2.05 10⁶ m / s

f = μk.N (N = m.g)

f = 0.29 x 95 x 10

f = 275.5 N

Answer:

W = 47.04 N

Explanation:

It is given that,

A block of mass 4.8 kg is suspended by the pulley and it is assumed that it is frictionless and the mass of the strings to be negligible.

We need to find the reading of the spring scale if the system is in equilibrium.

At this position, the scale reading is equal to the weight of the block such that,

W = mg, where g is the acceleration due to gravity

W = 4.8 kg × 9.8 m/s²

W = 47.04 N

So, the reading of the spring scale is 47.04 N.

Answer:

A and B is positive charge

C_negative

Explanation:

because when an ebonite is rubbed with fur produce negative charge due to law of electrostatic like charge repel and unlike attract

Answer:

I think the answer is 19.8 potential energy

Explanation:

NONE.

Answer:

it's b

Explanation:

no shade, direct sunlight

The height that will illustrate the distance will be d = 6.36m

Based on the information given, the length of the vine will be:

L = ✓(16² + 27.7)²

L = 32m

The velocity of Jane when she reaches position B will be:

V = ✓2gh

V = ✓(2 × 9.8 × 4.3)

V = 9.18m/s

We will apply the conversation of momentum. This will be:

50 × 9.18 = (50 + 80)V1

V1 = 3.53m/s

Therefore, the height that will illustrate the distance will be:

31.36² + d² = 32²

d² = 32² - 31.36²

d = 6.36m

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There are six half lives that have passed within 34,200 years. When we assume that we have some of the products present initially, then we overestimate the age of the sample.

The half life is the time taken to obtain only half of the number of atoms that were originally present in the radioactive material. We know that living things do contain the carbon- 14 isotope together with the carbon - 12 isotope in the substance. When the organism dies, the carbon - 14 which is radioactive begins to decay and its half life could be used to estimate the age of the sample as we know.

Given that after each half life, we have only half of the original amount that remains. We would have 1/64 of the original amount left after six half lives and this means that 34,200 years have passed.

If we  the daughter product of carbon-14 is present in the sample when it forms (even before any radioactive decay happens), then we would over estimate the  age of the sample.

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(a) The induced current in the circuit is clockwise.

b. To find the power (P) using P = Fd/t or P = Fv (since d/t = v). Here, F = ILB (from the Lorentz force), so P = (ILB)v.

(a) The induced current in the circuit is clockwise.

This can be determined using the right-hand rule.

As the metal bar moves to the left through the magnetic field directed out of the plane, the generated force on the electrons (Lorentz force) will push them toward the top rail, creating a clockwise current.

(b) To find the rate at which the applied force is doing work on the bar, first calculate the induced EMF (ε) using Faraday's law:

induced EMF (ε) using Faraday's law:

ε = BLv

= (0.65 T) * (0.36 m) * (5.9 m/s)

= 1.389 Tm²/s

= 1.389 V (since 1 Tm²/s = 1 V)

induced current (I) using Ohm's law:

I = ε/R

= 1.389 V / 45 Ω

= 0.03086 A

force (F) from the Lorentz force law, where F = ILB:

F = ILB

= (0.03086 A) * (0.36 m) * (0.65 T)

= 0.00723 N

Finally, we find the power (P) using P = Fv:

P = Fv

= (0.00723 N) * (5.9 m/s)

= 0.04266 W

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The earthquake would occur 13 minutes before 10:05 a.m. which will be at 9.52 am.

The p-waves travel with a constant velocity of 7 km/s

The time can be calculated by using the formula

t = d / v

where

T1 =  10:05 a.m

d is the distance they take to travel from the epicenter

v is the speed of the p-waves

On average, the speed of p-waves is

v = 7 km/s

d = 5600 km (given)

Substituting the values in the formula;

t = d / v

t = 5600 ÷ 7

t = 800 seconds

Converting into minutes,

t = 800 ÷ 60

t = 13.3

≈ 13 mins

T1 -  13 mins = T2

10:05 - 13 mins = 9.52 am

It means the earthquake occurred prior 13 minutes, that is at 9.52 am.

Therefore, the earthquake occurred at 9.52 am.

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1. The force between the two charges is 1.78 × 10⁻⁵ pounds, with opposite signs indicating attraction between the charges.

2. The resistance of 10 gauge aluminum wire over a 40 ft distance would be 0.506 ohms.

3. The cost of electricity from the automobile battery is 38.6 cents per kWh.

4. The current that would be carried through the body is 0.8 mA if dry.

1. The force between two point charges can be calculated using Coulomb's law, which states that the force is proportional to the product of the charges and inversely proportional to the square of the distance between them.

Using the values given, the force can be calculated as F = (k * q1 * q2) / r², where k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between them. Plugging in the values, the force can be calculated as 1.78 × 10⁻⁵ pounds, with opposite signs indicating attraction between the charges.

2. The resistance of a wire is determined by its length, cross-sectional area, and resistivity. The resistivity of aluminum is higher than that of copper, so a larger cross-sectional area is required to achieve the same resistance. Using the gauge size conversion chart, 10 gauge aluminum wire has a cross-sectional area of 5.26 mm², which is approximately 83% of the cross-sectional area of 12 gauge copper wire.

Thus, the resistance of 10 gauge aluminum wire over a 40 ft distance can be calculated as R = (rho * L) / A, where rho is the resistivity of aluminum, L is the length, and A is the cross-sectional area. Plugging in the values, the resistance can be calculated as 0.506 ohms.

3. To calculate the cost of electricity per kWh, the total cost and the total amount of energy supplied must be known. Since the battery supplies 12 V and 51 A for one hour, the total energy supplied can be calculated as E = V * I * t, where V is the voltage, I is the current, and t is the time.

Plugging in the values, the total energy supplied can be calculated as 612 watt-hours (Wh). Since one kWh is equal to 1000 Wh, the total energy supplied can be converted to 0.612 kWh. Dividing the total cost by the total energy supplied gives the cost per kWh, which is 38.6 cents.

4. The current through the body can be calculated using Ohm's law, which states that current is equal to voltage divided by resistance. Using the values given, the resistance can be either 100,000 ohms or 300 ohms depending on whether the skin is dry or wet.

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The largest angular magnification is 0.9 of the refracting telescope has an objective lens of focal length 15.4 in and eyepieces of focal lengths 18 mm, 24 mm, 36 mm, and 85 mm.

The formula for angular magnification is M = F_o/F_e, where F_o is the focal length of the objective lens and F_e is the focal length of the eyepiece.

The largest angular magnification is achieved with the shortest focal length eyepiece, which in this case is 18mm.

Plugging the values into the formula, we get M = 15.4/18 = 0.86.

Expressing the answer to two significant figures, we get M = 0.9.

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The complete question is written here

a refracting telescope has an objective lens of focal length 15.4 in and eyepieces of focal lengths 18 mm, 24 mm, 36 mm, and 85 mm.What is the largest angular magnification you can achieve with this instrument?Express your answer to two significant figures.

The total distance is 105 kilometers and the total displacement is 15 kilometers east. Option C

To calculate the total distance, we add the distances traveled in each leg of the journey: 60 kilometers (from A to B) + 45 kilometers (from B back to A) = 105 kilometers.

However, displacement refers to the change in position of an object in a straight line from its starting point to its ending point. In this case, since the boat starts and ends at the same point (A), the total displacement is zero.

Hence The total distance is 105 kilometers and the total displacement is 15 kilometers east.

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The net frictional force acting on the object on a rough horizontal surface is 5 N.

This is because the maximum static frictional force is 3 N and the kinetic frictional force is 2 N. Therefore, the total of the two forces is 5 N.

The frictional force is always a resistive force between two surfaces in contact that opposes their relative motion. It is a non-conservative force, meaning it does not obey the laws of conservation of energy. The magnitude of the frictional force is dependent on a number of factors, including the nature of the two surfaces in contact, the normal force between them, the speed of the relative motion, and the coefficient of friction between them.

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Answer:

To solve this problem, you'll need to break the initial velocity of the seed into its horizontal and vertical components, then use the equations of motion to find the time of flight and horizontal distance.

The initial velocity (v) of the seed is 2.65 m/s. The angle it's launched at (θ) is 30.0 degrees below the horizontal. The height (h) it's launched from is 0.460 m.

First, calculate the horizontal (v_x) and vertical (v_y) components of the velocity. Because the seed is launched downward, the vertical component will be negative:

v_x = v * cos(θ) = 2.65 m/s * cos(30.0) = 2.29 m/s

v_y = v * sin(θ) = -2.65 m/s * sin(30.0) = -1.325 m/s

Next, use the equation of motion to find the time it takes for the seed to hit the ground:

h = v_y * t + 0.5 * g * t^2

Where g is the acceleration due to gravity, which is approximately 9.8 m/s². Solving the equation for t gives:

t = (-v_y - sqrt((v_y)^2 - 4 * 0.5 * g * (-h))) / (2 * 0.5 * g)

Plugging in the values:

t = (1.325 + sqrt((-1.325)^2 - 4 * 0.5 * 9.8 * (-0.460))) / (2 * 0.5 * 9.8)

t = 0.182 seconds

Finally, use the horizontal velocity and time of flight to find the horizontal distance the seed covers:

x = v_x * t = 2.29 m/s * 0.182 s = 0.417 m

So, the seed lands after approximately 0.182 seconds and travels approximately 0.417 meters horizontally.

Answer:

If I say the correct answer is B.280

Answer:

The answer should be 3-At sunset blue light is scattered by air molecules at higher altitudes. The longer red and yellow wavelengths must travel further through the atmosphere when the Sun is near the horizon. At some point the reds and yellows finally scatter in the lower atmosphere causing the reds and yellows.

Answer:

3. At sunset blue light is scattered by air molecules at higher altitudes. The longer red and yellow wavelengths must travel further through the atmosphere when the Sun is near the horizon. At some point the reds and yellows finally scatter in the lower atmosphere causing the reds and yellows.