Once​ Kate's kite reaches a height of 52 ft ​(above her​ hands), it rises no higher but drifts due east in a wind blowing 6 ft divided by s. How fast is the string running through​ Kate's hands at the moment that she has released 105 ft of​ string?

Answer: a) Negative

Explanation: According to the question: The pKa of aspartic acid is 3.86.

For acidic amino acids,

pH > pKa 

When side chains are negatively charged, amino acid will be acidic

pH < pKa 

That is, when side chains are uncharged, amino acid will be neutral

For aspartic acid, when this polypeptide were in an aqueous solution with a pH of 7

pH (7.0)  > pKa (3.86)

Therefore the chains are negatively charged

Final answer:

In an aqueous solution at pH 7, the aspartic acid side chain would have a (a)negative charge because the pH of the environment is higher than the pKa of aspartic acid (3.86).

Explanation:

The question relates to the charge of the aspartic acid side chain in an aqueous solution at pH 7, given that the pKa of aspartic acid is 3.86. The pKa value represents the pH at which half of the aspartic acid side chains are deprotonated (negative charge) and half remain protonated (neutral). When the pH of the environment is higher than the pKa (as is the case here, pH 7 > pKa 3.86), the majority of the aspartic acid side chains will be deprotonated, thus carrying a negative charge. Consequently, in an aqueous solution at pH 7, the aspartic acid side chain would have (a)negative charge.

Answer:

Work done by the spring is negative

Explanation:

We can answer this question by thinking what is the force acting on the box.

In fact, the force acting on the box is the restoring force of the spring, which is given by Hooke's Law:

[tex]F=-kx[/tex]

where

k is the spring constant

x is the displacement of the box with respect to the equilibrium position of the spring

The negative sign in the equation indicates that the direction of the force is always opposite to the direction of the displacement: so, whether the spring is compressed or stretched, the force applied by the spring on the box is towards the equilibrium position.

The work done by the restoring force is also given by

[tex]W=Fx cos \theta[/tex]

where

F is the restoring force

x is the displacement

[tex]\theta[/tex] is the angle between the direction of the force and the displacement

Here we know that the force is always opposite to the displacement, so

[tex]\theta=180^{\circ} \rightarrow cos \theta =-1[/tex]

Which means that the work done by the spring is always negative, since the direction of the restoring force is always opposite to the direction of motion.

The work done by the spring on the box as it compresses is negative, due to the force applied by the spring being in the opposite direction of the box's displacement.

For the given problem, the work done by the spring on the box is negative. This is because the spring is exerting a force in the opposite direction to that of the box's motion, leading to a decrease in the kinetic energy of the box. According to the work-energy principle, if the force applied by the spring opposes the direction of motion, the work done is negative. When the spring compresses, it stores elastic potential energy, which is a conversion of the box's kinetic energy. The spring force acts to the left (assuming the right is the positive direction), while the displacement of the box due to that force is to the right, resulting in negative work done by the spring.

Answer:

Explanation:

a , b )

The problem is based on interference in thin films

formula for constructive interference

2μ t = ( 2n+ 1 ) λ / 2 , μ is refractive index of layer, t is thickness and λ is wavelength of light.

n is called the order of fringe . If we place n= 0 , 1 , 2  etc , the thickness also changes . So constructive interference is possible at more than one thickness .

Put the value of  λ = 116 nm . μ = 1.28 , t = 116 nm in the given equation

2 x 1.28 x 116 x 2 = ( 2n+ 1 ) λ

593.92 = ( 2n+ 1 ) λ

when n = 0

λ = 593.92 nm .

This falls in visible range .

c )

2μ t = ( 2n+ 1 ) λ / 2

Put λ = 593.92 nm , n = 1

2 x 1.28 t₁ = 3 x 593.92 / 2

t₁ = 348 nm .

Put n = 2

2 x 1.28 t₂ = 5 x 593.92 / 2

t₂ = 580 nm .

The speed of the eagle immediately after catching the pigeon is [tex]$19.0$[/tex] m/s.

To solve this problem, we need to apply the principle of conservation of momentum, which states that the total momentum of a closed system remains constant before and after an interaction.

Given information:

- Mass of the pigeon = [tex]$m_2$[/tex]

- Mass of the eagle = [tex]$2m_2$[/tex] (twice the mass of the pigeon)

- Initial velocity of the pigeon, [tex]$\vec{v}_{i,2} = 17.3 m/s[/tex] (northward)

- Initial velocity of the eagle, [tex]$\vec{v}_{i,1} = 33.7 m/s[/tex] at an angle [tex]$\theta = 39.9^\circ$[/tex] below the horizontal

We need to find the final velocity of the eagle after catching the pigeon, [tex]$\vec{v}_f$[/tex].

Step 1: Resolve the initial velocity vectors into components.

For the pigeon:

[tex]$\vec{v}_{i,2} = (0, 17.3) m/s[/tex]

For the eagle:

[tex]$\vec{v}_{i,1} = (33.7 \cos 39.9^\circ, -33.7 \sin 39.9^\circ) m/s[/tex]

[tex]$\vec{v}_{i,1} = (22.8, -25.4) m/s[/tex]

Step 2: Calculate the initial total momentum of the system.

Initial total momentum, [tex]$\vec{p}_i = m_2 \vec{v}_{i,2} + (2m_2) \vec{v}_{i,1}$[/tex]

[tex]$\vec{p}_i = m_2 (0, 17.3) + 2m_2 (22.8, -25.4)$[/tex]

[tex]$\vec{p}_i = m_2 (45.6, -33.8) kg⋅m/s[/tex]

Step 3: Calculate the final total momentum of the system.

Since momentum is conserved, the final total momentum is equal to the initial total momentum.

[tex]$\vec{p}_f = \vec{p}_i = m_2 (45.6, -33.8) kg⋅m/s[/tex]

Step 4: Calculate the final velocity of the combined system (eagle + pigeon).

Let [tex]$\vec{v}_f = (v_{f,x}, v_{f,y})$[/tex] be the final velocity of the combined system.

Total final momentum = [tex]$(m_2 + 2m_2) \vec{v}_f = 3m_2 \vec{v}_f$[/tex]

[tex]3m_2 \vec{v}_f = m_2 (45.6, -33.8)$$\vec{v}_f = \left(\frac{45.6}{3}, -\frac{33.8}{3}\right)$ m/s\\$\vec{v}_f = (15.2, -11.3)$ m/s[/tex]

Step 5: Calculate the speed of the eagle after catching the pigeon.

Speed = [tex]$\sqrt{v_{f,x}^2 + v_{f,y}^2}$[/tex]

Speed = [tex]$\sqrt{15.2^2 + (-11.3)^2} m/s[/tex]

Speed = [tex]$19.0 m/s[/tex]

Answer:

a

The velocity is  [tex]v =17.98 \ m/s[/tex]

b

The diameter is  [tex]d = 0.00184m[/tex]

Explanation:

The diagram of the set up is shown on the first uploaded image  

From the question we are told that

    The height of the water tank is [tex]h = 20.0 \ m[/tex]

    The position of the hole [tex]p_h = 16.5m[/tex] below  water level

     The  rate  of water flow [tex]\r V = 2.90 *10^{-3} m^3 /min = \frac{2.90 *10^{-3}}{60} = 0.048*10^{-3} m^3/s[/tex]

  According to Bernoulli's theorem position of the hole

              [tex]\frac{P_o + h \rho g}{\rho} + \frac{1}{2} u^2 = \frac{P_o}{\rho } + \frac{1}{2} v^2[/tex]

Where  u is  the initial speed the water through the hole = 0 m/s

              [tex]P_o[/tex] is the atmospheric pressure

            [tex]\frac{P_o }{\rho} + \frac{ h \rho g}{\rho} + 0 = \frac{P_o}{\rho } + \frac{1}{2} v^2[/tex]

                   [tex]v = \sqrt{2gh}[/tex]

Substituting value

           [tex]v = \sqrt{2 * 9.8 * 16.5 }[/tex]

              [tex]v =17.98 \ m/s[/tex]

The Volumetric flow rate is mathematically represented as

          [tex]\r V = A * v[/tex]

     Making A the subject

              [tex]A = \frac{\r V}{v}[/tex]

 substituting value  

             [tex]A = \frac{0.048 *10^{-3}}{17.98}[/tex]

                 [tex]= 2.66*10^{-6}m^2[/tex]

Area is mathematically represented as

        [tex]A = \frac{\pi d^2}{4}[/tex]

  making d the subject

         [tex]d = \sqrt{\frac{4*A}{\pi} }[/tex]

  Substituting values

        [tex]d = \sqrt{\frac{4 * 2.67 *10^{-6}}{3.142} }[/tex]

          [tex]d = 0.00184m[/tex]

Answer:

Explanation:

Given the following information,

Resistor of resistance R = 13.6Ω

Capacitor of capacitance C = 11.9-μF

C = 11.9 × 10^ -6 F

Inductor of inductance L = 19.1-mH

L = 19.1 ×10^-3 H

All this connected in series to a generator that generates Vrms= 117V

Vo = Vrms√2 = 117√2

Vo = 165.463V

a. Frequency for maximum current?

Maximum current occurs at resonance

I.e Xc = XL

At maximum current, the frequency is given as

f = 1/(2π√LC)

Then,

f = 1/(2π√(19.1×10^-3 × 11.9×10^-6)

f = 1/(2π√(2.2729×10^-7))

f = 1/(2π × 4.77 ×10^-4)

f = 333.83Hz

Then, the frequency is 333.83Hz.

b. Since we know the frequency,

Then, we need to find the capacitive and inductive reactance

Capacitive reactance

Xc = 1/2πfC

Xc = 1/(2π × 338.83 × 11.9×10^-6)

Xc = 1/ 0.024961

Xc = 40.1Ω

Also, Inductive reactance

XL = 2πfL

XL = 2π × 333.83 × 19.1×10^-3

XL = 40.1Ω

As expected, Xc=XL, resonance

Then, the impedance in AC circuit is given as

Z = √ (R² + (Xc—XL)²)

Z = √ 13.6² + (40.1-40.1)²)

Z = √13.6²

Z = 13.6 ohms

Then, using ohms las

V = IZ

Then, I = Vo/Z

Io = 165.46/13.6

Io = 12.17Amps

The current is 12.17 A

Answer:

a) Current is maximum at frequency, f₀ = 333.83 Hz

b) Maximum current = 12.17 A

Explanation:

Inductance, L = 19.1 mH = 19.1 * 10⁻³ H

Capacitance, C = 11.9 μF =11.9 * 10⁻⁶ F

a) Current is maximum at resonant frequency, f₀

[tex]f_{0} = \frac{1}{2\pi\sqrt{LC} }[/tex]

[tex]f_{0} = \frac{1}{2\pi\sqrt{11.9 * 10^{-6}* 19.1 * 10^{-3} } }[/tex]

[tex]f_{0} = 333.83 Hz[/tex]

b) Maximum value of the RMS current

[tex]V_{RMS} = 117 V\\V_{max} = \sqrt{2} V_{RMS}\\V_{max} = \sqrt{2} * 117\\V_{max} = 165.46 V[/tex]

[tex]I_{max} = \frac{V_{max} }{R} \\I_{max} = \frac{165.46}{13.6} \\I_{max} = 12.17 A[/tex]

Answer:

[tex]8.0\mu C[/tex]

Explanation:

We are given that

[tex]f=1.6 Hz[/tex]

[tex]q=3.0\mu C=3.0\times 10^{-6} C[/tex]

[tex]1\mu C=10^{-6} C[/tex]

Current,I=[tex]75\mu A=75\times 10^{-6} A[/tex]

[tex]1\mu A=10^{-6} A[/tex]

We have to find the maximum charge of the capacitor.

Charge on the capacitor,[tex]q=q_0cos\omega t[/tex]

[tex]\omega=2\pi f=2\pi\times 1.6=3.2\pi rad/s[/tex]

[tex]3\times 10^{-6}=q_0cos3.2\pi t[/tex]....(1)

[tex]I=\frac{dq}{dt}=-q_0\omega sin\omega t[/tex]

[tex]75\times 10^{-6}=-q_0(3.2\pi)sin3.2\pi t[/tex]....(2)

Equation (2) divided by equation (1)

[tex]-3.2\pi tan3.2\pi t=\frac{75\times 10^{-6}}{3\times 10^{-6}}=25[/tex]

[tex]tan3.2\pi t=-\frac{25}{3.2\pi}=-2.488[/tex]

[tex]3.2\pi t=tan^{-1}(-2.488)=-1.188rad[/tex]

[tex]q_0=\frac{q}{cos\omega t}=\frac{3\times 10^{-6}}{cos(-1.188)}=8.0\times 10^{-6}=8\mu C[/tex]

Hence, the maximum charge of the capacitor=[tex]8.0\mu C[/tex]

Given:

Now,

The mass of the steel hemisphere will be:

→ [tex]m_h = \rho_s\times \frac{2}{3} \pi r^2[/tex]

By putting the values, we get

         [tex]= 7.8\times 10^3\times \frac{2}{3} \pi (0.16)^2[/tex]

         [tex]= 66.9 \ kg[/tex]

and,

The mass of aluminum cylinder will be:

→ [tex]m_c = \rho_a \times \pi r^2 h[/tex]

        [tex]= 2.7\times 10^3\times \pi (0.08)^2 (0.18)[/tex]

        [tex]= 9.7 \ kg[/tex]

Now,

The mass center of steel hemisphere will be:

→ [tex]z_1 = r - \frac{3r}{8}[/tex]

       [tex]= 0.16-(3\times \frac{0.16}{8} )[/tex]

       [tex]= 0.1 \ m[/tex]

The mass center of aluminum hemisphere will be:

→ [tex]z_2 = r+\frac{h}{2}[/tex]

       [tex]= 0.16+\frac{0.18}{2}[/tex]

       [tex]= 0.25 \ m[/tex]

hence,

The center of mass will be:

→ [tex]Z = \frac{m_h z_1 +z_2 m_c}{m_s+m_c}[/tex]

By putting the values, we get

      [tex]= \frac{66.9\times 0.1+9.76\times 0.25}{66.9+9.76}[/tex]

      [tex]= 0.12 \ m[/tex]

Thus the above answer is right.

Learn more about mass center here:

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To find the location (z) of the mass center of a steel hemisphere and an aluminum cylinder assembly, we first calculate the mass of each part using the given densities and geometric properties. We then use a formula to calculate 'z' based on the mass and z-coordinate of the center of mass of each part. The center of mass assumes the symmetric mass distributions.

To determine the location z of the mass center of the assembly, we first need to calculate the mass of each part of the assembly using their given densities and geometrical properties.

The mass of the steel hemisphere (mSt) is calculated as density times the volume. The volume of the hemisphere can be calculated using the formula 2/3*Pi*r^3, where r is the radius of the hemisphere.

The mass of the aluminum cylinder (mAl) is calculated the same way, as the product of the cylinder's density and volume. The volume of a cylinder is given by the formula Pi*r^2*h, where r is the radius and h is the height of the cylinder, which is given as 180 mm or 0.12 m.

Once we have these masses, we then calculate the mass center z of the assembly, which is given by the formula z = (mSt * zSt + mAl * zAl) / (mSt + mAl), where zSt and zAl are the z-coordinates of the center of the mass of the steel hemisphere and the aluminum cylinder respectively.

Since the mass of each body acts as if it were at the center of the body, we can assume that the z-coordinate of the hemisphere (zSt) is at its geometric center, whereas for the cylinder (zAl), it would be at h/2 from the bottom end.

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The final temperature of the refrigerant in the piston-cylinder device can be determined via lookup and interpolation from standard tables, based on the final pressure of 0.4 MPa. The work done by the refrigerant as it expands and drops in pressure can be calculated via the first law of thermodynamics, specifically accounting for changes in internal energy as work done by the system.

This is an example of a reversible adiabatic expansion process involving a refrigerant (refrigerant-134a) in a piston-cylinder device. The process follows the behavior of ideal gases, and so we can make use of the first law of thermodynamics and other gas laws to determine the final temperature and the work done by the refrigerant.

To find the final temperature, we incorporate the specifics of the refrigerant-134a, which is defined by specific tables found in thermodynamics textbooks, providing details of temperature at various pressures. Interpolation between table values may be necessary to find the exact temperature at 0.4 MPa. Generally, the temperature will fall as the refrigerant expands and the pressure lowers, as defined by the ideal gas law (pV=nRT).

For the work done by the refrigerant, we'd need to apply the first law of thermodynamics, which in this case can be simplified as ΔU = W because there's no heat transfer in an adiabatic process (dQ=0). Changes in the internal energy of the gas would translate directly into work done on the piston. We calculate work using the formula -P(ΔV), where the negative sign indicates work being done by the system.

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(a) The final temperature in the cylinder is approximately 18.0°C.

(b) The work done by the refrigerant during the expansion is approximately 7.72 kJ.

To solve this problem, we need to determine two things: the final temperature of the refrigerant-134a vapor and the work done by the refrigerant during the expansion.

Given that the process is reversible and the system is insulated, this suggests that the expansion process is isothermal for an ideal gas, but since refrigerant-134a is not an ideal gas, we will use the refrigerant tables to find the properties.

Given Data:

Step-by-Step Solution:

(a) Determine the Final Temperature in the Cylinder:

1. Find the Initial and Final States Using Refrigerant-134a Tables:

Look up the refrigerant-134a property tables for [tex]\( P = 0.8 \)[/tex] MPa.

At [tex]\( P = 0.8 \)[/tex] MPa, the refrigerant-134a is in the saturated vapor region. The saturated temperature and properties at this pressure are:

Similarly, look up the refrigerant-134a property tables for [tex]\( P = 0.4 \)[/tex] MPa.

At [tex]\( P = 0.4 \)[/tex] MPa, the refrigerant-134a is also in the saturated vapor region. The properties at this pressure are:

The final temperature of the refrigerant-134a after the expansion, which is at 0.4 MPa, is approximately [tex]\( 18.0^\circ \text{C} \)[/tex].

(b) Determine the Work Done by the Refrigerant:

1. Calculate the Initial and Final Mass:

Using the specific volume [tex]\( v_g \)[/tex] to find the mass:

[tex]\[ m = \frac{V_1}{v_g (P_1)} \][/tex]

For [tex]\( P_1 = 0.8 \)[/tex] MPa:

[tex]\[ m = \frac{0.05 \text{ m}^3}{0.0769 \text{ m}^3/\text{kg}} \approx 0.650 \text{ kg} \][/tex]

For [tex]\( P_2 = 0.4 \)[/tex] MPa, verify if the mass is the same (which it is for an ideal gas in a closed system):

[tex]\[ V_2 = m \times v_g (P_2) = 0.650 \text{ kg} \times 0.0917 \text{ m}^3/\text{kg} \approx 0.0597 \text{ m}^3 \][/tex]

The new volume [tex]\( V_2 \)[/tex] can be verified as:

[tex]\[ V_2 = \text{initial volume} \times \frac{v_g (P_2)}{v_g (P_1)} \][/tex]

Which confirms our calculation.

2. Calculate the Work Done During Expansion:

For a reversible expansion in an insulated system (isoenthalpic process), the work done \( W \) can be approximated using:

[tex]\[ W = \int_{V_1}^{V_2} P \, dV \][/tex]

The work done for an ideal gas (but approximate here for refrigerants) can be simplified using:

[tex]\[ W = \text{P}_1 \times V_1 \ln \left( \frac{V_2}{V_1} \right) \][/tex]

For refrigerants, use specific volume:

[tex]\[ W = \text{P}_1 \times V_1 \left[ \frac{v_g (P_2)}{v_g (P_1)} - 1 \right] \][/tex]

Substituting:

[tex]\[ W = 0.8 \text{ MPa} \times 0.05 \text{ m}^3 \left( \frac{0.0917}{0.0769} - 1 \right) \][/tex]

[tex]\[ W = 0.8 \times 10^3 \text{ Pa} \times 0.05 \text{ m}^3 \left( 1.193 - 1 \right) \][/tex]

[tex]\[ W = 40 \text{ kJ} \times 0.193 \approx 7.72 \text{ kJ} \][/tex]

Explanation:

A paper circuit is a functioning electronic circuit built on a paper surface. Projects involving paper circuits are unique because of the use of traditional art techniques and some unique materials to create a circuit that combines aesthetics and functionality.

Paper circuit are sometimes carefully designed and then transfered on the paper, and in some cases, they are designed directly on the paper (usually by experts that already know what they're doing).

To make designing circuit on paper easier, there are three commonly used materials for the circuitry and they are, conductive paints, conductive tapes and conductive inks.

Conductive tapes are made from metal strip (usually copper) that are taped to the paper. They are good to work with since they allow components to be soldered on them creating a stronger and more reliable joint.

Conductive paints are special paints that can be used to outline circuit path and also serve as the circuit. The only problem with conducting paints is that it can be messy, and needs time to dry.

Conductive inks are easier to use as they need no drying time. They are far less messy and allows the drawing of elaborate and more intricate circuit on the paper.

There are also components that have been modified (like led lights etc) and available to make paper circuit design easier.

Answer:

Explanation:

Given that,

Three children of masses and their position on the merry go round

M1 = 22kg

M2 = 28kg

M3 = 33kg

They are all initially riding at the edge of the merry go round

Then, R1 = R2 = R3 = R = 1.7m

Mass of Merry go round is

M =105kg

Radius of Merry go round.

R = 1.7m

Angular velocity of Merry go round

ωi = 22 rpm

If M2 = 28 is moves to center of the merry go round then R2 = 0, what is the new angular velocity ωf

Using conservation of angular momentum

Initial angular momentum when all the children are at the edge of the merry go round is equal to the final angular momentum when the second child moves to the center of the merry go round

Then,

L(initial) = L(final)

Ii•ωi = If•ωf

So we need to find the initial and final moment of inertia

NOTE: merry go round is treated as a solid disk then I= ½MR²

I(initial)=½MR²+M1•R²+M2•R²+M3•R²

I(initial) = ½MR² + R²(M1 + M2 + M3)

I(initial) = ½ × 105 × 1.7² + 1.7²(22 + 28 + 33)

I(initial) = 151.725 + 1.7²(83)

I(initial) = 391.595 kgm²

Final moment of inertial when R2 =0

I(final)=½MR²+M1•R²+M2•R2²+M3•R²

Since R2 = 0

I(final) = ½MR²+ M1•R² + M3•R²

I(final) = ½MR² + (M1 + M3)• R²

I(final)=½ × 105 × 1.7² + ( 22 +33)•1.7²

I(final) = 151.725 + 158.95

I(final) = 310.675 kgm²

Now, applying the conservation of angular momentum

L(initial) = L(final)

Ii•ωi = If•ωf

391.595 × 22 = 310.675 × ωf

Then,

ωf = 391.595 × 22 / 310.675

ωf = 27.73 rpm

So, the final angular momentum is 27.73 revolution per minute

Answer:

B. Forward

Explanation:

The net force in the system is initially balanced and the system is in equilibrium.

1) As John ducks, two things happens. First, there is a sudden impulse from his end which will change the momentum of the system from zero to an amount Mv proportional to the impulse Ft, which will be directed away from his direction because,

2) the process of ducking pushes his center of mass forward, which also moves the center of mass of the whole system away from him. This two effect will cause the skateboard to move forward.

Answer:

Check the explanation

Explanation:

So far there’s an increases in the speed of the kangaroo, then the tendons will stretch more thereby enabling them to store more energy. For this reason, they will have a additional time in the air propelled by greater spring energy. In contact with the ground, it will turn out to be like a spring in simple harmonic motion. There will be increases in their agility rate in hopping the amplitude of the oscillation, but that does not in any way affect the time, or period in contact with the ground.

Kangaroos convert kinetic energy to elastic potential energy during landing which is immediately converted back to kinetic energy for the next leap. This energy transformation cycle, aided by cushioning their landing by bending their hind legs, allows kangaroos to spend more time airborne without affecting the ground contact time, irrespective of their speed.

The phenomenon in question about kangaroo jumping is rooted in physics and biomechanics. As kangaroos gain speed, more kinetic energy is stored and thus converted into potential energy when they're in the air. This essentially means the kangaroo uses most of their energy while airborne rather than contacting the ground. Hence, ground contact time stays the same no matter what the kangaroo's speed is.

Firstly, the kangaroo's tendons work like a spring; as it lands, much of its kinetic energy is converted to elastic potential energy in the tendons. This act transforms the downward motion (gravitational potential energy) into an upward motion (kinetic energy) through elastic potential energy, which is stored as the tendons stretch. This is similar to the motion observed in a bungee jumper or an elastic band when stretched.

Additionally, the kangaroo's landing is cushioned by bending its hind legs, reducing the impact force over the contact time. With each hop, the kangaroo's kinetic energy increases when it leaves the ground while potential energy increases during 'flight' due to gravity. This establishes a cycle between kinetic and potential energy, with energy oscillating back and forth as the kangaroo continues to hop. Such energy efficiency is what enables the kangaroo to maintain the same ground contact time irrespective of its speed.

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Answer:

The message (signal) coming from the earth was sent first.

Explanation:

The radio messages reach Mars simultaneously, because the distance from Earth to Mars and that of from Mars to Asteroid is same. But the passenger in the spaceship is moving relative to Mars towards Earth and hence, the message from Earth reaches first.

According to passenger frame, the message (signal) coming from the Earth was sent first compared to message coming from Asteroid.

Answer:

A) 100

B) 10

C) 0.01

Explanation:

Let

Fs = force on slave cylinder

Fm = force on master cylinder

Given that the designed to exert a force 100 times as large as the one put into it. That is

Fs = 100 Fm

Using Pascal's law

Fs/As = Fm/Am

100Fm/As = Fm/Am

100/As = 1/Am

Cross multiply

As= 100Am

As/Am = 100

b) What must be the ratio of their diameters

Using their areas ratio, area of a cylinder is  πr^2

A = πr^2 =  π(D/2)^2

 π(Ds/2)^2 ÷  π(Dm/2)^2 = 100

(Ds/2)^2 ÷ (Dm/2)^2 = 100

(Ds/Dm)^2 = 100

Ds/Dm = 10

(c) By what factor is the distance through which the output force moves reduced relative to the distance through which the input force moves? Assume no losses to friction

Let us consider the workdone at the input and at the output.

Work done = force × distance

Fs × Hs = Fm × Hm

100Fm × Hs = Fm × Hm

100Hs = Hm

100Hs/Hm = 1

Hs/Hm = 1/100

Hs/Hm = 0.01

Given Information:

Energy of laser pulses = E = 100 mJ = 100×10⁻³ Joules

Time = t = 1 ns = 1×10⁻⁹ seconds

Required Information:

Instantaneous power = P = ?

Answer:

[tex]Instantaneous \: Power = 100 \: Mwatt[/tex]

Explanation:

The instantaneous power is the power dissipated at any instant of time whereas the average power is the power dissipated over a given time interval.  

The instantaneous laser power during each pulse is given by

[tex]Instantaneous \: Power = \frac{E}{t}[/tex]

Where E is the energy of the laser pulses and t is the time that each pulse lasts.

[tex]Instantaneous \: Power = \frac{100\times10^{-3}}{1\times10^{-9}}[/tex]

[tex]Instantaneous \: Power = 1\times10^{8} \: watt[/tex]

or

[tex]Instantaneous \: Power = 100 \: Mwatt[/tex]

Therefore, the instantaneous power of each pulse is 100 Mwatt.

Answer:

Explanation:

For any point stationary in front of ambulance , the ambulance is approaching the point . In this way the case is similar to source of sound approaching the observer. The doppler's effect can be applied to know the apparent frequency.

velocity of sound V = 343 m /s

speed of source Vs = 48.4

observer is stationary.

apparent frequency = real frequency x V / ( V - Vs )

= 620 x 343 / ( 343 - 48.4 )

= 722 Hz approx.

Answer:

a) True. The image of the mite is virtual

e) True. The image must be within the focal length of the eyepiece len

Explanation:

Let's review the general characteristics of compound microscopes

Formed by two converging lenses

Magnification is

       M = -L/fo   0.25/fe

Where fo is the focal length of the objective lens and fe is the focal length of the ocular lens, L is the tube length

Let's review the claims

a) True. The image of the mite is virtual

b) False. The effect is the opposite of the magnification equation

c) False. The objective lens forms a real image

d) False. As the seal distance increases the magnification decreases

e) True. The image must be within the focal length of the eyepiece len

A compound microscope uses an objective lens to create a real, inverted image larger than the object, and an eyepiece (ocular) to magnify this image further into a virtual image. The correct statements include that the eyepiece forms a virtual image, and the image from the objective lens should be within the eyepiece's focal length for further magnification.

To understand how a compound microscope works, we analyze the function of its two lenses: the objective lens and the eyepiece. Initially, an object just beyond the objective lens's focal length (fobj) creates a real, inverted image that is larger than the object. This initial image acts as the object for the eyepiece lens, or ocular, which is placed within its own focal length (feye) to further magnify the image, effectively functioning as a magnifying glass. The resulting image is a virtual, larger, and further magnified version of the initial image, making it more comfortable for viewing as the eye is relaxed when looking at distant objects.

If we consider the statements provided, the correct statements are as follows:

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Answer:

One way to categorize waves is on the basis of the direction of movement of the individual particles of the medium relative to the direction that the waves travel. Categorizing waves on this basis leads to three notable categories: transverse waves, longitudinal waves, and surface waves.

I've read about it but never seen one. The way I understand it, it's a coordinated physical motion executed by a great number of people, as in a large crowd at a sporting event, timed so that it appears to propagate from one end of the crowd to the opposite end.

I would classify it as a cooperative community activity, involving liberty, equality, and fraternity, executed for the common good.

Answer:

A) angular velocity; ω = 1.598 rad/s

B) linear velocity;V = 8.31 m/s

C) Tangential Acceleration;a_t = 1.768 m/s²

D) Radial Acceleration;a_r = 13.28 m/s²

Explanation:

We are given that;

Radius; r = 5.2m

Time;t = 4.7 sec

θ = 0.170t²

Thus, angular acceleration would be the second derivative of θ which is d²θ/dt²

Thus,α = d²θ/dt² = 0.34 rad/s²

A) Formula for angular velocity is;

ω = αt

Where α is angular acceleration and t is time.

Thus;ω = 0.34 x 4.7

ω = 1.598 rad/s

b) formula for linear velocity is given by; V = ωr

We have ω = 1.598 rad/s and r = 5.2m

Thus; V = 1.598 x 5.2

V = 8.31 m/s

c) formula for tangential acceleration is;

a_t = αr

a_t = 0.34 x 5.2

a_t = 1.768 m/s²

d) formula for radial acceleration is;

a_r = rω²

a_r = 5.2 x 1.598²

a_r = 13.28 m/s²

To find the magnitudes of the astronaut's angular velocity, linear velocity, tangential acceleration, and radial acceleration, differentiate the angular position equation, multiply the radius and the angular velocity to find the linear velocity, and use the rate of change of linear velocity to find the tangential acceleration. The radial acceleration is the product of the angular velocity and the linear velocity.

The angular velocity can be found by differentiating the angular position equation with respect to time. The linear velocity is equal to the product of the radius and the angular velocity. The tangential acceleration is equal to the rate of change of linear velocity, while the radial acceleration is equal to the product of the angular velocity and the linear velocity.

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Answer:

Explanation:

We shall apply Lenz's law to solve the problem . This law states that direction of induced current is such that it opposes the change that creates it. Since current increases in the coil it creates increasing magnetic field in the other coil . So the current will be induced in it  so that it opposes this increase . It can be done only  if current in opposite direction is induced in it . Hence in the first case,  current will be induced in opposite direction .

In this case, current is decreasing in the primary coil and current will be induced in the secondary coil. Decreasing current will create decreasing magnetic field . So induced current will try to increase it . In can be done if current in the same direction is induced in the secondary coil.

Hence in the second case , current will be induced in the same direction .

Final answer:

The tension in the cable connecting the drum to the bucket is equal to the mass of the bucket times the difference between the acceleration due to gravity and the bucket's downward acceleration.

Explanation:

To determine the tension in the cable between the drum and the bucket, we can apply Newton's second law to the hanging mass. The net force acting on the bucket is the difference between the weight of the bucket and the tension in the rope:

[tex]F_{net}[/tex] = mg - T

Since we know the bucket has a downward acceleration a, we can write Newton's second law as:

ma = mg - T

Where m is the mass of the bucket, g is the acceleration due to gravity, and T is the tension in the rope. We can rearrange the equation to solve for T:

T = mg - ma

T = m(g - a)

So the tension in the cable is equal to the mass of the bucket times the difference between the acceleration due to gravity and the downward acceleration of the bucket.

Answer:

You have been hired as part of a research team consisting of ... Torque: After watching a news story about a fire in a high rise apartment building, you and your friend decide to design an emergency escape device from the top ... To avoid engine failure, your friend suggests a gravitational powered elevator.

Explanation:

Answer:

The correct answer is 5.77 cm

Explanation:

Solution

From the question given, let us recall the following,

A sunny day = 60°F and

The sun  above the horizon is = 60°

The surface of a shallow pond is = 20 cm deep

The stick above sand = 10 cm

The next step is to find the length of the shadow

Now,

let CD = stick

BC= Length of shadow

Thus,

CD/BC = Tan 60°

BC = CD/Tan 60°

= 10/√3 = 5.77 cm

Therefore, the length of the shadow is 5.77 cm

Answer:

The speed of sound theoretically is 267.148 m/s which is clearly less than slope calculated speed 365.0 m/s

Explanation:

check the attached picture

Answer:

Check the explanation

Explanation:

When we have an object in periodic motion, the amplitude will be the maximum displacement from equilibrium. Take for example, when there’s a back and forth movement of a pendulum through its equilibrium point (straight down), then swings to a highest distance away from the center. This distance will be represented as the amplitude, A. The full range of the pendulum has a magnitude of 2A.

position = amplitude x sine function(angular frequency x time + phase difference)

x = A sin(ωt + ϕ)

x = displacement (m)

A = amplitude (m)

ω = angular frequency (radians/s)

t = time (s)

ϕ = phase shift (radians)

Kindly check the attached image below to see the step by step explanation to the question above.

Answer: velocity has a direction, while speed is the distance travelled by an object.

Explanation:

Answer:

Velocity includes Displacement instead of distance.

Explanation:

Velocity's formula is change in position divided(displacement) by change in time(Delta x/Delta t). Whereas speed is distance over change in time(d/Delta t).

Answer:

Using equations

R will be 1.25 Kilo ohm

To get the value of resistance, R, we need to use Ohm's law which is R = V/I', where I' is the current generated when the coil balances with the added mass. The exact value of R depends on parameters related to the coil and the magnetic field.

The question is asking for the value of the resistance R in a given circuit, when the circuit produces a current which generates a balance with an added mass through a coil in a magnetic field. To find the value of R, we need to use Ohm's law, which expresses the relationship between voltage (V), current (I), and resistance (R) as V=IR.

Here, the electromotive force or emf can be treated as the voltage in the circuit. In this case, emf = 20.0 V. When this coil creates a balance with an added 51.0 g mass, an equivalent current will be generated, let's denote it by I'

Using Ohm's law, we can solve for R by rearranging the formula to R = V/I'. The exact value of I' and thus R will, however, depend on additional factors involving the parameters of the coil and magnetic field which are not provided in the question. The law used here, Lenz's law, is a manifestation of the conservation of energy, dictating that the current induced in a circuit due to a change in the magnetic field will create a magnetic field that opposes the initial changing magnetic field.

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Answer:

a The kinetic energy is  [tex]KE = 0.0543 J[/tex]

b The height of the center of mass above that position is  [tex]h = 1.372 \ m[/tex]    

Explanation:

From the question we are told that

  The length of the rod is  [tex]L = 1.4m[/tex]

   The mass of the rod [tex]m = 140 = \frac{140}{1000} = 0.140 \ kg[/tex]  

   The angular speed at the lowest point is [tex]w = 1.09 \ rad/s[/tex]

Generally moment of inertia of the rod about an axis that passes through its one end is

                   [tex]I = \frac{mL^2}{3}[/tex]  

Substituting values

               [tex]I = \frac{(0.140) (1.4)^2}{3}[/tex]

               [tex]I = 0.0915 \ kg \cdot m^2[/tex]

Generally the  kinetic energy rod is mathematically represented as

             [tex]KE = \frac{1}{2} Iw^2[/tex]

                    [tex]KE = \frac{1}{2} (0.0915) (1.09)^2[/tex]

                           [tex]KE = 0.0543 J[/tex]

From the law of conservation of energy

The kinetic energy of the rod during motion =  The potential energy of the rod at the highest point

   Therefore

                   [tex]KE = PE = mgh[/tex]

                        [tex]0.0543 = mgh[/tex]

                             [tex]h = \frac{0.0543}{9.8 * 0.140}[/tex]

                                [tex]h = 1.372 \ m[/tex]    

Answer:

a) Kr = 0.0543 J

b) Δy = 0.0396 m

Explanation:

a) Given

L = 1.4 m

m = 140 g = 0.14 kg

ω = 1.09 rad/s

Kr = ?

We have to get the rotational inertia as follows

I = Icm + m*d²

⇒ I = (m*L²/12) + (m*(L/2)²)

⇒ I = (0.14 kg*(1.4 m)²/12) + (0.14 kg*(1.4 m/2)²)

⇒ I = 0.09146 kg*m²

Then, we apply the formula

Kr = 0.5*I*ω²

⇒ Kr = 0.5*(0.09146 kg*m²)*(1.09 rad/s)²

⇒ Kr = 0.0543 J

b) We apply the following principle

Ei = Ef

Where the initial point is the lowest position and the final point is at the maximum height that its center of mass can achieve, then we have

Ki + Ui = Kf + Uf

we know that ωf = 0 ⇒ Kf = 0

⇒ Ki + Ui = Uf

⇒ Uf - Ui = Ki

⇒ m*g*yf - m*g*yi = Ki

⇒ m*g*(yf - yi) = Ki

⇒ m*g*Δy = Ki

⇒ Δy = Ki/(m*g)

where

Ki = Kr = 0.0543 J

g = 9.81 m/s²

⇒ Δy = (0.0543 J)/(0.14 kg*9.81 m/s²)

⇒ Δy = 0.0396 m