a person can throw a 300 gram ball at 25 m/s the maximum height the person can throw the ball on Earth?​

Final answer:

To determine how long it takes for the flywheel to come to a stop, calculate the angular acceleration using the torque and moment of inertia. Then, use the angular acceleration to find the time.

Explanation:

To determine how long it takes for the flywheel to come to a stop, we need to calculate the angular acceleration first. To do this, we can use the formula:

α = τ / I

Where α is the angular acceleration, τ is the torque, and I is the moment of inertia. The torque can be calculated using the formula:

τ = rF

Where r is the radius of the flywheel and F is the force exerted on it. Once we have the angular acceleration, we can use it to find the time it takes for the flywheel to come to a stop using the formula:

t = ω / α

Where t is the time, ω is the initial angular velocity, and α is the angular acceleration.

Answer:

The number of bright fringes per unit width on the screen is, [tex]x=\dfrac{\lambda D}{d}[/tex]      

Explanation:

If d is the separation between slits, D is the distance between the slit and the screen and [tex]\lambda[/tex] is the wavelength of the light. Let x is the  number of bright fringes per unit width on the screen is given by :

[tex]x=\dfrac{n\lambda D}{d}[/tex]

[tex]\lambda[/tex] is the wavelength

n is the order

If n = 1,

[tex]x=\dfrac{\lambda D}{d}[/tex]

So, the the number of bright fringes per unit width on the screen is [tex]\dfrac{\lambda D}{d}[/tex]. Hence, the correct option is (B).

Final answer:

The number of bright fringes per unit width in a Young's double-slit experiment is given by the reciprocal of the fringe spacing, which is d/(Dλ), corresponding to answer choice E.

Explanation:

In a Young's double-slit experiment, to find the number of bright fringes per unit width on the screen, we consider the separation between the slits (d), the distance from the slits to the screen (D), and the wavelength of light used (λ). The distance between adjacent bright fringes, or fringe spacing, is given by Δy = Dλ/d. From this relation, the number of bright fringes per unit width can be obtained by taking the reciprocal of the fringe spacing, which implies 1/Δy = d/(Dλ).

Therefore, the correct formula to calculate the number of bright fringes per unit width on the screen is the reciprocal of Δy, which is d/(Dλ), matching answer choice E.

Answer:

P + C = W if in equilibrium?

P + C - W = 0; P + C = W

Explanation:

Action and reaction are equal and opposite according to newton's third law of motion. a force is that which tends to change a body's state of rest or uniform motion in a straight line, recall

P is the upward force the person exerts on the crate,

C is the vertical contact force exerted on the crate by the floor,

W is the weight of the crate

let the sum of upward forces be on the left hand side of the equation and the sum of downward forces on the right hand side.

P+C=W

P+C-W=0 if they are in equilibrium

the forces are related by the above.

When a person is unsuccessfully attempting to lift a crate, the forces acting on the crate are balanced. The sum of the upward force (P) applied by the person and the contact force (C) from the floor equals the weight (W) of the crate, represented by the equation P + C = W.

In the situation described, the person is trying to lift a crate but is unable to do so. This means that the forces acting on the crate are in equilibrium, or balanced, as there is no resulting upward or downward motion of the crate. In this case, the force exerted by the person (P) and the vertical contact force from the floor (C) is opposing the weight of the crate (W).

In terms of magnitudes, the sum of P and C equals the weight W. This is based on the principle of equilibrium which states that an object at rest has equal and opposite forces acting on it. Therefore, P + C = W. This means the upward forces are equal to the downward force, hence the person can't lift the crate successfully.

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Answer:

A = 2 cm ,   λ = 8 cm

Explanation:

The amplitude of a wave is the maximum height it has, in this case the height is measured by the vertical ruler,

We are told the balance point is in the reading of 5 cm, that the maximum reading is 3 cm and the Minimum reading is 7 cm. Therefore, the distance from the ends of the ridge to the point of equilibrium is

          d = 7-5 = 2 cm

          d = 5-3 = 2 cm

          A = 2 cm

The wavelength is the minimum horizontal distance for which the wave is repeated, that is measured by the horizontal ruler.

The initial reading for 4 cm and the final reading for 8 cm, this distance corresponds to a crest of the wave, the complete wave is formed by two crests whereby the wavelength is twice this value

          Δx = 8-4 = 4 cm

          λ = 2 Δx

          λ = 8 cm

The amplitude of the wave is approximately [tex]2cm[/tex], measured from the equilibrium to the crest or trough. The wavelength of the wave is approximately [tex]8cm[/tex].

To determine the wavelength and amplitude of the given wave, we need to analyze the provided measurements.

The crest rises from the equilibrium position of [tex]5 cm `[/tex] to just before [tex]3 cm[/tex] , giving an amplitude measurement of approximately [tex]2 cm (5 cm - 3 cm)[/tex] .

Similarly, the trough falls from the equilibrium of [tex]5 cm[/tex] to just before [tex]7 cm[/tex], again giving a vertical displacement of approximately [tex]2 cm (7 cm - 5 cm)[/tex].

The horizontal distance from the start of the first crest to the start of the next crest should be measured. According to the provided data, the first crest aligns from [tex]0[/tex]  to just under [tex]4 cm[/tex] , and the next crest starts just under [tex]8 cm[/tex].

The wavelength is the horizontal distance covering one complete cycle, measuring just less than [tex]8 cm[/tex]. Therefore, the wavelength is approximately [tex]8 cm[/tex] .

Hence,

Answer:0.024 mm

Explanation:

Given

Fringe shifts by an amount to 30 dark bands i.e. [tex]m=30[/tex]

Wavelength [tex]\lambda =480 nm[/tex]

refractive index of mica [tex]n_2=1.6[/tex]

refractive index of air [tex]n_1=1[/tex]

Phase difference is given by

[tex]m=\frac{t\left [ n_2-n_1\right ]}{\lambda }[/tex]

[tex]30=\frac{t\left [ 1.6-1\right ]}{480\times 10^{-9}}[/tex]

[tex]t=\frac{30\times 480\times 10^{-9}}{1.6-1}[/tex]

[tex]t=0.024\ mm[/tex]

To determine the thickness of the mica in a Young's double-slit experiment, we can use the formula d = mλ / sinθ, where d is the thickness of the mica, λ is the wavelength of the light, θ is the angle of the shift in the fringe pattern, and m is the number of dark bands shifted. By substituting the given values into the formula and solving for d, we can calculate the thickness of the mica.

In a Young’s double-slit experiment, the shift in the center of the fringe pattern on the screen is caused by the introduction of a thin sheet of mica over one of the slits. This shift corresponds to 30 dark bands. To determine the thickness of the mica, we can use the formula for double-slit interference:

d sinθ = mλ,

where d is the distance between the slits, θ is the angle of the shift in the fringe pattern, m is the number of dark bands shifted, and λ is the wavelength of the light. Rearranging the formula, we have:

d = mλ / sinθ.

Given that the wavelength of the light is 480 nm and the index of refraction for the mica is 1.60, we can calculate the thickness of the mica by substituting the values into the formula and solving for d.

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Final answer:

The question pertains to the rotational kinetic energy of a rolling spherical object with non-uniform density and how it compares to the total kinetic energy using its moment of inertia.

Explanation:

The student is asking about the rotational kinetic energy of a spherical object with non-uniform density in comparison to its total kinetic energy. The given moment of inertia for the object is 0.59 M R2, where M is the mass, and R is the radius of the sphere. Using the formula for rotational kinetic energy, Krot = (1/2) I ω2, and the formula for translational kinetic energy, Ktrans = (1/2) m v2, we can find the energy components. However, because the spherical object rolls without slipping, there is a relationship between linear velocity (v) and angular velocity (ω) which is v = ωR. This allows us to compare the rotational kinetic energy to the total kinetic energy.

Answer:

a. Quadruped arm and opposite leg raise

Explanation:

Quadruped arm and opposite leg lift

- Kneel on the floor, lean forward and place your hands down.

- Keep your knees in line with your hips and hands directly under your shoulders.

- Simultaneously raise one arm and extend the opposite leg, so that they are in line with the spine.

- Go back to the starting position.

This method is usually used as an alternative to iso-abs exercise or also known as a bridge, which allows you to exercise the abdominal and spinal area at the same time.

It is also used together with other exercises for the treatment of hyperlordosis.

The suitable regression for a prone iso-abs exercise is the Quadruped arm and opposite leg raise.

The question asks for a regressed exercise alternative to the prone iso-abs exercise, which typically involves the individual maintaining a prone plank position, activating their abdominal core muscles without additional movement. When a client experiences difficulty with this exercise, an appropriate regression should reduce the demand on the core muscles while still enabling them to engage effectively.

The correct regression exercise from the options provided is a. Quadruped arm and opposite leg raise. This exercise involves the individual starting on their hands and knees (a quadruped position), then extending one arm and the opposite leg to create a straight line from fingertips to toes. This movement stabilises the core and has a reduced intensity compared to the prone iso-abs exercise.

The other options, b. Cable rotation, c. Rolling active resistance row, and d. Cable chop, are more dynamic exercises involving rotation and should not be confused with static core stabilisation exercises.

Answer:

option A

Explanation:

given,

frequency is increased by 20%

we know,

[tex]\dfrac{x_n}{L}d = (n-\dfrac{1}{2})\times \lambda[/tex]...........(1)

where

x_n is the perpendicular distance between the point the interference pattern is obtained,

L is the distance between the center of the two point sources

and λ is  the wavelength of light.

If the frequency is increased by 20%, then the number of nodal lines is increased by 20%.

From equation (1),we observe that the frequency is directly proportional to the number of nodal lines.

Hence, the correct answer is option A

Answer:

d=1.49×1011m

Explanation:

Velocity is defined as the rate of travel, and can be found using the distance formula.

velocity=distancetime

Rearranging this formula we can solve for distance given velocity and time of travel.

d=vt

We are given velocity and time, and so can solve for distance, but if we plug in the values given;

d=(3.00×108m/s)(8.3minutes)

We can see that the units do not match up. Since seconds are the SI unit for time, we will need to convert 8.3 minutes to seconds.

t=(8.3minutes)(60seconds/minute)=(498s)

Now our units work out and we can solve for distance.

= 15.85

(8.3 min)/(1 AU) = (T)/(1.52 AU)

(8.3 min)x(1.52 AU) = (T x 1 AU)

T = (8.3 min x 1.52 AU) / (1 AU)

T = 12.62 minutes

The ratio of the low temperature to the high temperature in the Carnot engine, given that the real engine's efficiency is 55% of a Carnot engine's efficiency, is calculated to be 0.3535.

The question involves thermodynamics and specifically deals with the operation and efficiency of a Carnot engine.

The given heat engine absorbs 450 J of heat from the high-temperature reservoir and expels 290 J to the low-temperature reservoir. The efficiency (efficiency) of this engine is given as 55% of a Carnot engine's efficiency. Using the first law of thermodynamics, wecan calculate the work done (W) by the engine:

W = Qh - Qc = 450 J - 290 J = 160 J.

The efficiency of the engine is the ratio of the work done to the heat absorbed:

efficiency = W / Qh = 160 J / 450 J = 0.3556, or 35.56%.

Now, the efficiency of a Carnot engine is defined as:

efficiencyCarnot = 1 - (Tc / Th).

The problem states that the engine's efficiency is 55% of a Carnot engine's efficiency, which means:

0.3556 = 0.55 * efficiencyCarnot

From this equation, we can solve for efficiencyCarnot and then use it to calculate the ratio of the low temperature to the high temperature in the Carnot engine:

efficiencyCarnot = 0.3556 / 0.55

efficiencyCarnot = 0.6465, or 64.65%

Thus:

0.6465 = 1 - (Tc / Th)

Tc / Th = 1 - 0.6465 = 0.3535.

Therefore, the ratio of the low temperature to the high temperature in the Carnot engine is 0.3535.

The ratio of the low temperature to the high temperature in the Carnot engine is  [tex]\( \frac{1}{3} \)[/tex].

The correct ratio of the low temperature to the high temperature in the Carnot engine is given by:

[tex]\[ \frac{T_{\text{low}}}{T_{\text{high}}} = 1 - \frac{W_{\text{actual}}}{Q_{\text{in}}} \cdot \frac{1}{\eta_{\text{Carnot}}} \][/tex]

where [tex]\( W_{\text{actual}} \)[/tex] is the work done by the actual engine, [tex]\( Q_{\text{in}} \)[/tex] is the heat absorbed from the high-temperature reservoir, and [tex]\( \eta_{\text{Carnot}} \)[/tex]  is the efficiency of the Carnot engine.

First, we calculate the actual efficiency of the given heat engine using the provided values:

[tex]\[ W_{\text{actual}} = Q_{\text{in}} - Q_{\text{out}} \][/tex]

[tex]\[ W_{\text{actual}} = 450 \, \text{J} - 290 \, \text{J} \][/tex]

[tex]\[ W_{\text{actual}} = 160 \, \text{J} \][/tex]

The actual efficiency [tex]\( \eta_{\text{actual}} \)[/tex] is then:

[tex]\[ \eta_{\text{actual}} = \frac{W_{\text{actual}}}{Q_{\text{in}}} \][/tex]

[tex]\[ \eta_{\text{actual}} = \frac{160 \, \text{J}}{450 \, \text{J}} \][/tex]

[tex]\[ \eta_{\text{actual}} = \frac{16}{45} \][/tex]

Given that the efficiency of the actual engine is 55% of the efficiency of a Carnot engine operating between the same two temperatures, we can write:

[tex]\[ \eta_{\text{actual}} = 0.55 \cdot \eta_{\text{Carnot}} \][/tex]

The efficiency of a Carnot engine is given by:

[tex]\[ \eta_{\text{Carnot}} = 1 - \frac{T_{\text{low}}}{T_{\text{high}}} \][/tex]

Combining the two equations, we get:

[tex]\[ \frac{16}{45} = 0.55 \left( 1 - \frac{T_{\text{low}}}{T_{\text{high}}} \right) \][/tex]

Solving for [tex]\( \frac{T_{\text{low}}}{T_{\text{high}}} \)[/tex]:

[tex]\[ \frac{16}{45} = 0.55 - 0.55 \cdot \frac{T_{\text{low}}}{T_{\text{high}}} \][/tex]

[tex]\[ 0.55 \cdot \frac{T_{\text{low}}}{T_{\text{high}}} = 0.55 - \frac{16}{45} \][/tex]

[tex]\[ \frac{T_{\text{low}}}{T_{\text{high}}} = \frac{0.55 - \frac{16}{45}}{0.55} \][/tex]

[tex]\[ \frac{T_{\text{low}}}{T_{\text{high}}} = \frac{0.55 \cdot \frac{45}{45} - \frac{16}{45}}{0.55} \][/tex]

[tex]\[ \frac{T_{\text{low}}}{T_{\text{high}}} = \frac{\frac{24.75}{45} - \frac{16}{45}}{0.55} \][/tex]

[tex]\[ \frac{T_{\text{low}}}{T_{\text{high}}} = \frac{\frac{8.75}{45}}{0.55} \][/tex]

[tex]\[ \frac{T_{\text{low}}}{T_{\text{high}}} = \frac{8.75}{45 \cdot 0.55} \][/tex]

[tex]\[ \frac{T_{\text{low}}}{T_{\text{high}}} = \frac{8.75}{24.75} \][/tex]

[tex]\[ \frac{T_{\text{low}}}{T_{\text{high}}} = \frac{1}{3} \][/tex]

Answer:

D. Graphing the force as a function of distance and calculating the area under the curve.

Explanation:

Answer:

The correct option is (D). "Graphing the force as a function of distance and calculating the area under the curve"

Explanation:

1) As we know from the definition of Work done which is Work = Force × Displacement. Time doesn't have any use in the mentioned equation and thus the stop watch is of no use to us.

2) Using the meter stick, we can measure the distance for which the block is pulled horizontally i.e "Displacement"

3) Using the spring scale, we can calculate the "Force" applied on the block of wood to move it horizontally.

4) For example, lets say that for a constant force of 4 Newtons, the wooden block is pulled horizontally 4 meters. Plotting Force vs Displacement on a graph would yield a horizontal line as shown in the attachment. Area under the F vs D graph will give us the total work done.

Answer:

The degradation of energy is the loss of useful energy: energy is conserved in changes, but tends to be transformed into thermal energy, which is a less usable form of energy.

Explanation:

Some forms of energy can be transformed into others. In these transformations, energy degrades, loses quality. In any transformation, part of the energy is converted into heat or heat energy.

Any type of energy can be completely transformed into heat; but, this cannot be completely transformed into another type of energy. It is said, then, that heat is a degraded form of energy. Are examples:

- The electrical energy, when going through a resistance.

- Chemical energy, in the combustion of some substances.

- Mechanical energy, by shock or friction.

Therefore, the Yield is defined as the ratio (in% percent) between the useful energy obtained and the energy contributed in a transformation.

R = ( useful energy  / total energy) * 100

where R is the yield

Final answer:

Energy degradation during the generation of electrical power reflects the decrease in usability of energy as it is transformed, with much of it becoming waste heat. This affects the efficiency of power systems, indicating that more energy must be input to produce desired effects.

Explanation:

In the context of generating electrical power, energy degradation refers to the decrease in the quality or usefulness of energy as it undergoes various transformations. Although energy is conserved according to the first law of thermodynamics, during transformations, some energy is inevitably converted to forms that are less useful for doing work, primarily heat energy.

The phenomenon of energy degradation is a fundamental concept in thermodynamics and has significant implications for the efficiency of power generation systems. It points to the reality that while energy is not destroyed, its capacity to do work diminishes, requiring more energy inputs to achieve the desired outputs in practical applications.

Answer:

d) kinetic energy transformed to mechanical energy

Explanation:

Wind energy comes from its movement, so kinetic energy

         Em = K = ½ m v²2

This energy spins the mill aspadle that this movement of the rotor within a magnetic field creates electricity in accordance with Faraday's law.

Consequently, from the above we should make a graph of the wind speed (kinetic energy) according to the electricity produced

The correct answer is d

kinetic energy transformed to mechanical energy this best completes the chart. Hence option D is correct.

Kinetic energy is the energy associated with a motion of a body. When a body is in motion having mass m then it has kinetic energy. Kinetic energy is denoted by K or T. it is expressed in joules. Kinetic energy is given by,

K = 1/2 mv²

Hence an object having zero mass or zero velocity have zero kinetic energy.

In line with Faraday's law, when the mill asp rotates under the influence of this energy, electricity is generated.

In light of the foregoing, we should create a graph showing the wind speed (kinetic energy) in relation to the power generated.

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Answer:E)Valley Fever;meningitis

Explanation:Valley fever is a a fungal lung infection while the meningitis is a viral or bacterial disease that affect the meninges.

Final answer:

The speed of sound in the metal is 444.44 m/s.

Explanation:

To determine the speed of sound in the metal, we can use the information given and the speed of sound in air. The pulse traveling through the metal arrives 9ms earlier than the pulse traveling through air. From this, we can calculate the time difference it takes for the pulses to travel through the length of the metal bar. The speed of sound in air is given as 343 m/s. We can use the formula: speed = distance/time. Rearranging the formula, we have: time = distance/speed. As the distance is given as 4.00 m and the time difference is given as 9 ms (0.009 s), we can calculate the speed of sound in the metal by dividing the distance by the time difference: speed = 4.00 m / 0.009 s = 444.44 m/s.

The second player is 91 meters far from the first player.

Why?

First, let be the +y the North, so, to solve the problem we can use the following formula:

[tex]y=yo+v_o*t+\frac{1}{2}*a*t[/tex]

Now, subsituting the given information, we have(assuming that the speed is constant):

\\\\y-yo=13\frac{m}{s}*7s+\frac{1}{2}*0*7s\\\\y-yo=91m\\\\distance=91m[/tex]

Hence, we have that the second player is 91m far from the first player.

Have a nice day!

Answer:C. 10N

Explanation:

The applied force needed to maintain a constant velocity is 10N. This is because the moving force acting on the body will always be equal to the frictional force if the velocity of the body is constant.

But note that this is not the case if the body ia accelerating. If the body is accelerating then the frictional force will not be able to overcome the moving force acting on the body and such the moving force will be greater than the frictional force in that regards.

The applied force needed to maintain a constant velocity is 10 N.

The given parameters;

The net horizontal force on the object is calculated by applying Newton's second law of motion as shown below;

∑F = ma

[tex]F- F_k = ma[/tex]

where;

At constant velocity, the acceleration of the object is zero.

[tex]F-F_k = m(0)\\\\F-F_k = 0\\\\F= F_k\\\\F = 10 \ N[/tex]

Thus, the applied force needed to maintain a constant velocity is 10 N.

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Answer:

1000 mph

Explanation:

P = Perimeter = 24000 mi

r = Radius of the equator

t = Time taken to complete one rotation = 24 h

Perimeter of a circle is given by

[tex]P=2\pi r\\\Rightarrow r=\dfrac{P}{2\pi}\\\Rightarrow r=\dfrac{24000}{2\pi}\ mi[/tex]

Angular speed is given by

[tex]\omega=\dfrac{2\pi}{t}\\\Rightarrow \omega=\dfrac{2\pi}{24}[/tex]

Velocity if given by

[tex]v=r\omega\\\Rightarrow v=\dfrac{24000}{2\pi}\times \dfrac{2\pi}{24}\\\Rightarrow v=1000\ mph[/tex]

The person would be going at a speed of 1000 mph

Answer:

354 cm³

Explanation:

Step 1: calculate the density of PEG 400

[tex]Specific gravity({\gamma}) = \frac{density of liquid (\rho _{l})}{density of water (\rho_{w})}[/tex]

[tex]1.13=\frac{\rho _{l}}{1(\frac{g}{m^3})}[/tex]

[tex]{\rho _{l}[/tex] = 1.13X1 = 1.13 [tex]\frac{g}{cm^3}[/tex]

Step 2:

Molecular weight of PEG 400 = 400g/mol

Step 3: calculate the volume of PEG 400 to be measured

[tex]Volume _{PEG 400} =\frac{mass of _{PEG 400}}{density of _{PEG 400}}[/tex]

[tex]Volume _{PEG 400}=\frac{400}{1.13}(cm^3)[/tex]

                                       = 353.98 cm³

                                       ≅354 cm³

353.4 cm³ volume should be measured for the preparation.

Given:

Specific gravity=1.13

To find:

Volume=?

Specific gravity is the ratio of the density (mass of a unit volume) of a substance to the density of a given reference material, often a liquid.

On substituting the values, we will get:

[tex]\text{Density of liquid}=1.13*1=1.13 gcm^{-3}[/tex]

Molecular weight of PEG 400 =400g/mol

Now, we have the values of density and mass thus volume can be calculated easily:

The density of a substance is its mass per unit volume.

[tex]\text{Volume}=\frac{\text{mass}}{\text{density}}=\frac{400}{1.13} \\\\\text{Volume}=353.4 cm^{3}[/tex]

353.4 cm³ volume should be measured for the preparation.

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Explanation:

Plate tectonics is the idea that the earth's outer solid crust (the lithosphere) is fragmented into a few dozen "plates" that pass relative to each other across the earth's surface, like ice slabs on a lake. 

A scientist named Alfred Wegener in 1915 proposed that the continents rammed through ocean basin crust, which would clarify why the shapes of many coastlines (such as South America and Africa) seem to match together like a puzzle.

The rate of change of the volume of the cone at the specified instant is approximately [tex]\(6.03 \times 10^{9} \pi\)[/tex] cubic centimeters per second.

To find the rate of change of the volume of the cone, you can use the formula for the volume of a cone:

[tex]\[V = \frac{1}{3}\pi r^2 h\][/tex]

Where:

- V is the volume of the cone.

- r is the radius of the cone.

- h is the height of the cone.

You are given that the radius is increasing at a rate of 333 cm/s (dr/dt = 333 cm/s) and the height is decreasing at a rate of 444 cm/s (dh/dt = -444 cm/s). At a certain instant, the radius is 888 cm (r = 888 cm) and the height is 101010 cm (h = 101010 cm).

To find the rate of change of the volume (dV/dt) at that instant, you can use the product rule for differentiation:

[tex]\[ \frac{dV}{dt} = \frac{1}{3} \pi \left(2rh\frac{dr}{dt} + r^2\frac{dh}{dt}\right) \][/tex]

Now, plug in the values:

[tex]\[ \frac{dV}{dt} = \frac{1}{3} \pi \left(2(888 cm)(101010 cm)(333 cm/s) + (888 cm)^2(-444 cm/s)\right) \][/tex]

Calculate:

[tex]\[ \frac{dV}{dt} = \frac{1}{3} \pi \left(5910088880 cm^3/s + (-157593984 cm^3/s)\right) \][/tex]

Now, simplify:

[tex]\[ \frac{dV}{dt} = \frac{1}{3} \pi \cdot 5752494896 cm^3/s \][/tex]

[tex]\[ \frac{dV}{dt} \approx 6.03 \times 10^{9} \pi \; cm^3/s \][/tex]

So, the rate of change of the volume of the cone at that instant is approximately [tex]\(6.03 \times 10^{9} \pi\)[/tex] cubic centimeters per second.

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Answer:

Explanation:

Given

Length of segment is [tex]L=1 unit[/tex]

inclination of segment [tex]\theta =30^{\circ}[/tex]

To calculate the coordinates of segment we need to resolve its component in x and y co-ordinates

such that in triangle OPQ

[tex]\sin \theta =\frac{y}{L}[/tex]

[tex]y=L\sin 30[/tex]

[tex]y=1\times \frac{1}{2}=0.5 unit[/tex]

[tex]\cos \theta =\frac{x}{L}[/tex]

[tex]x=L\cos \theta [/tex]

[tex]x=1\times \cos 30[/tex]

[tex]x=1\times \frac{\sqrt{3}}{2}[/tex]

[tex]x=\frac{\sqrt{3}}{2} units[/tex]      

Answer:

B. increasing the temperature of the atmosphere

Explanation:

Generally, atmospheric loss can be defined as the loss of the gases in the atmosphere to outer space. This process usually occur through either thermal escape or non-thermal escape. Atmospheric loss of gases to outer space by thermal escape occurs when the molecular velocity due to thermal energy is considerably high. One of the factors that can lead to increase in thermal energy and ultimately increase in atmospheric loss is increase in the temperature of the atmosphere. Therefore, the correct answer is option B.

The work done by air resistance on a skydiver with a constant velocity is -2.93 × 10⁵ J, as the air resistance force is equal but opposite to the gravitational force, and work is calculated as force times distance in the direction of the force.

The work done by a nonconservative force such as air resistance can be calculated as the product of the force and the distance over which it acts, in the direction of the force. Since the skydiver is falling with a constant velocity, the force of air resistance must be equal and opposite to the gravitational force acting on the skydiver, resulting in no net force and therefore no acceleration. The work done by air resistance is thus equal to the gravitational force times the distance fallen, with a negative sign because the force of air resistance acts in the opposite direction to the displacement.

First, we need to calculate the gravitational force acting on the skydiver:

F_gravity = mass × acceleration due to gravity

F_gravity = 92.0 kg × 9.8 m/s²

F_gravity = 901.6 N

Now, we can calculate the work done by air resistance:

Work = force × distance × cos(ϸ)

Since the angle (ϸ) between the force of air resistance and the direction of displacement is 180 degrees (opposite directions), cos(ϸ) is -1. Therefore:

Work = -901.6 N × 325 m × -1

Work = -2.93 × 10⁵ J

The correct answer is D) -2.93 times 10⁵ J, which means the work done by air resistance is negative because it acts in the direction opposite to the displacement.

Answer:

188.7 m

Explanation:

height of bridge above water (h) = 393 m

mass of bungee jumper (m) = 150 kg

length of cord (L) = 78 m

acceleration due to gravity (g) = 9.8 m/s  

initial energy = mgh = 150 x 9.8 x 393 = 577,710 J

since the jumper barely touches the water, the maximum extension of the cord (x) = 393 - 78 = 315 m

from the conservation of energy mgh = [tex](\frac{1}{2})kx^{2}[/tex]

therefore

577,710 = [tex](\frac{1}{2})kx315^{2}[/tex]

k = 11.64 N/m

from Hooke's law, force (f) = kx' ⇒ mg = kx'

where x' is the extension of the cord when it comes to rest

150 x 9.8 = 11.64 × x'

x' = 126.3 m

the final height at which the cord comes to a rest = height of the bridge - length of the cord - extension of the cord when it comes to rest

the final height at which the cord comes to a rest = 393  - 78 - 126.3 = 188.7 m

Answer:

2. The hydrogen atom has quantized energy levels.

Explanation:

The Bohr model of the atom states that the structure of the atom is quantized, that is, that electrons can only orbit the nucleus in specific orbits with a fixed radius. Therefore, the electron cannot be in energy levels that do not correspond to these quantized levels.

Answer:

the point between Earth and the Moon where the gravitational pulls of Earth and Moon are equal is E)3.45 × 10⁸ m

Explanation:

The force that the Earth exerts on a mass m is

F_e = (G M_e m) / R_e²

where

The force that the Moon exerts on a mass m is

F_m = (G M_m m) / R_m²

where

Therefore, the point where the gravitational pulls of Earth and Moon are equal is:

F_e = F_m

R_e + R_m = R = 3.84×10⁸ m

Thus,

(G M_e m) / R_e² = (G M_m m) / R_m²

M_e / R_e² = M_m / (R - R_e²)

(R - R_e²) / R_e² = M_m / M_e

(R - R_e) / R_e = (M_m / M_e)^1/2

R_e(R/R_e -1) / R_e = (M_m / M_e)^1/2

R/ R_e = (M_m / M_e)^1/2 + 1

R_e = R / [(M_m / M_e)^1/2 + 1]

R_e = (3.84×10⁸ m) / [(7.35 x 10²² kg / 5.97 x 10²⁴ kg )^1/2 + 1]

R_e = 3.45 × 10⁸ m

Therefore, the  point between Earth and the Moon where the gravitational pulls of Earth and Moon are equal is 3.45 × 10⁸ m.

Answer:

a) I = 0.363 kg*[tex]m^{2}[/tex]

b) [tex]I_{T}[/tex] = 0.82385 kg*[tex]m^{2}[/tex]

Explanation:

a) If we approximate the skater how a cylinder his moment of inertia is:

I = [tex]\frac{mr^{2} }{2}[/tex]

I = [tex]\frac{(60)(0.110)^{2} }{2}[/tex]

I = 0.363 kg*[tex]m^{2}[/tex]

b)  If the skater has his arms extended then:

[tex]I_{T} = I_{B} + I_{A}[/tex]

       where   [tex]I_{B}[/tex]: Body’s moment of inertia

                     [tex]I_{A}[/tex]: Moment of inertia of the arms

[tex]I_{B}[/tex] = [tex]\frac{mr^{2} }{2}[/tex]

[tex]I_{B}[/tex] = [tex]\frac{(52.5)(0.110)^{2} }{2}[/tex] = 0.3176 kg*[tex]m^{2}[/tex]

[tex]I_{A}[/tex] = 2 * [tex]\frac{mL^{2} }{12}[/tex]

[tex]I_{A}[/tex] = 2 * [tex]\frac{(3.75)(0.9)^{2} }{12}[/tex] = 0.50625 kg*[tex]m^{2}[/tex]

[tex]I_{T}[/tex]  = 0.3176 + 0.50625 = 0.82385 kg*[tex]m^{2}[/tex]

The moment of inertia for a skater can be calculated using the mass and radius in the formula for the moment of inertia of a cylinder. If the skater extends their arms, the moment of inertia increases, and this is calculated by adding the moment of inertia of the skater to the moments of inertia of the arms.

The moment of inertia of a body is a measure of its resistance to rotational motion. It depends on the mass and how that mass is distributed relative to the axis of rotation. In the case of the skater (approximated as a cylinder), you calculate this using the formula for the moment of inertia of a cylinder, which is I=0.5MR², where M is the mass and R is the radius.

(a) For the skater without extended arms, substitute the given mass (60 kg) and radius (0.11 m) into the formula to get: I = 0.5 * 60 kg * (0.11 m)² = 0.363 kg-m².

(b) For the skater with extended arms, first, calculate the skater's body moment of inertia as before, but now with 52.5 kg mass. Secondly, calculate the moment of inertia of each arm (approximated as a rod rotating about one end) with the formula I=1/3mL², where m is the mass of an arm and L is the length. Add those values to get the total moment of inertia.

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Answer

given,

mass of block = 0.21 Kg

speed = 1.70 m/s

spring constant = k = 4.50 N/m

using conservation of energy

a)           K.E  =  P.E

 [tex]\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2[/tex]

 [tex]\dfrac{1}{2}\times 0.21 \times 1.7^2 = \dfrac{1}{2}\times 4.5 \times x^2[/tex]

 [tex]0.1348= x^2[/tex]

        x = 0.367 m

b) if the track is not friction less the maximum compression will be same as the compression in the part a.

In a frictionless track, the maximum compression of the spring is 0.494 m, calculated based on the energy conservation principle. If there's friction on the track, the maximum compression will be less, because part of the block's kinetic energy is used to overcome friction, leaving less to compress the spring.

This physics problem is rooted in concepts of energy conservation, specifically kinetic energy and potential energy. In part (a), when the block collides with the spring, the kinetic energy of the block is converted into potential energy stored in the compressed spring. The formula we need here is the conservative energy formula which states that: kinetic energy + potential energy = constant. Hence the kinetic energy before collision equals the potential energy at maximum compression.

So, 1/2 * mass * velocity^2 = 1/2 * k * compression^2. Substituting given values: 1/2 * 0.210 kg * (1.70 m/s)^2 = 1/2 * 4.50 N/m * x^2. Solving this equation yields x = 0.494 m, which is the maximum compression of the spring.

In part (b), if there is friction on the track, then some of the block's kinetic energy is used to overcome friction, which means less energy is available to compress the spring. Therefore, the friction on the track would result in a less than value for the maximum compression of the spring compared to the frictionless situation in part (a).

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Answer:

option D

Explanation:

given,

uniform length of cylinder = 1 m

diameter of the cylinder = 10 cm = 0.1 m

Eels have been recorded to spin = 14 rev/s

camera records at = 120 frames per second

time = [tex]\dfrac{1}{120}\ s/frame[/tex]

angle at which eel rotate = ?

ω = 14 rev/s

ω = 14 x 2 π rad/s

ω = 28 π rad/s

angle at which eel rotate

 θ = ω t

θ = [tex]28\pi\times \dfrac{1}{120}[/tex]

θ = 0.733 rad

θ =[tex]0.733 \times \dfrac{180^0}{2\pi}[/tex]

θ =[tex]42^0[/tex]

Hence, the correct answer is option D

The angle of rotation is the angle at which the object is rotate about the fixed point.

The angle at which the eel rotates from one frame to next frame is 42 degree.

Given that, we can treat the eel as a uniform cylinder.

So, uniform length of cylinder = 1 [tex]\rm m[/tex]  and diameter of the cylinder = 10 [tex]\rm cm[/tex] = 0.1 [tex]\rm m[/tex]. Eels have been recorded to spin at up to 14 rev/s when feeding in this way. The camera records at 120 frames per second.

Time taken by the camera to record one frame is [tex]t\;=\; \dfrac {1} {120} \;\rm s/\rm frame[/tex]

Revolution done by eel in radian per second is 14.

Angular Velocity is  [tex]\omega\;=\; 14\;\times\;2\pi\;\times\;0.1\;\rm rad/s[/tex].

The angle of rotation can be calculated by the formula given below.

Angle of rotation   [tex]\theta=\omega\;\times\;t[/tex].

Substituting the values in the above formula, the angle of rotation is,

[tex]\theta\;=\;14\;\times\;2\;\times\;3.14\;\times\;0.1\;\times\;\dfrac {1}{120}\;\rm rad[/tex]

 [tex]\theta=0.0733\;\rm rad[/tex]

[tex]\theta=0.0733\;\times\;\dfrac {360^\circ}{2\pi}[/tex]

[tex]\theta=41.5^\circ[/tex] or [tex]42^\circ[/tex].

The angle at which the eel rotates from one frame to next frame is 42 degree.

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