To get up on the roof, a person (mass 92.0 kg) places a 5.60 m aluminum ladder (mass 14.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 m from the bottom. What are the magnitudes (in N) of the forces on the ladder at the top and bottom?

Answer:4A

Explanation:

Given

Mass is displace x= A units from its mean position x=0'

When it is set to free it will oscillate about its mean position with maximum amplitude A i.e. from x=-A to x=A

One cycle is completed when block returns to its original position

so first block will go equilibrium position x=0 and then to x=-A

from x=-A it again moves back to x=0 and finally back to its starting position x=A

so it travels a distance of A+A+A+A=4A    

Answer: A) 141 m

Explanation:

Given that the boat travels at a speed of 1m/s due east in a river that flows 1m/s due south.

Let north represent positive y axis and east represent positive x axis.

Then we can resolve the resultant velocity of the boat to vector form.

Vr = i - j ( 1 m/s on x axis and -1m/s on y axis)

The time required to travel 100m from west to east at a speed of 1m/s is;

Time t = distance/speed = 100m/1m/s = 100s

Since the boat will use 100s to cross the river, We can now determine the resultant distance after 100s:

Distance = velocity × time = (i - j) × 100 = 100i - 100j

Distance = 100i - 100j (in vector form)

Magnitude of the Resultant distance can be given as:

dr = √(dx^2 + dy^2)

dr = √(100^2 + 100^2)

dr = √(20000)

dr = 141.42m

dr = 141m

A) The boat's overall displacement relative to its starting point is 141 m.

To solve this problem, we can break it down into two components: the magnitudes of the boat's eastward displacement and southward displacement. The time it takes for the boat to cross the river can be calculated using the width of the river and the boat's eastward speed. The distance the boat drifts downstream during this time can be calculated using the river's southward speed and the time taken. By combining these two displacements, we can determine the boat's overall displacement relative to its starting point.

The eastward displacement of the boat can be found using the formula: eastward displacement = eastward speed x time.

Plugging in the given values, we get: eastward displacement = 1 m/s x (100 m / 1 m/s) = 100 m.

The southward displacement of the boat can be found using the formula: southward displacement = southward speed x time.

Plugging in the given values, we get: southward displacement = 1 m/s x (100 m / 1 m/s) = 100 m.

Therefore, the boat's overall displacement relative to its starting point, which is the combination of the eastward and southward displacements, is equal to the square root of (eastward displacement squared + southward displacement squared).

Plugging in the calculated values, we get overall displacement = sqrt((100 m)^2 + (100 m)^2) = sqrt(2) x 100 m = 141 m.

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To solve this problem we will use the heat transfer equations, to determine the amount of heat added to the body. Subsequently, through the energy ratio given by Plank, we will calculate the energy of each of the photons. The relationship between total energy and unit energy will allow us to determine the number of photons

The mass of water in the soup is 477g

The change in temperate is

[tex]\Delta T = (90+273K)-(25+273K) = 65K[/tex]

Use the following equation to calculate the heat required to raise the temperature:

[tex]q = mc\Delta T[/tex]

Here,

m = Mass

c = Specific Heat

[tex]q = (477)(4.184)(65)[/tex]

[tex]q = 129724.92J[/tex]

The wavelength of the ration used for heating is [tex]1.55*10^{-2}m[/tex]

The number of photons required is the rate between the total energy and the energy of each proton, then

[tex]\text{Number of photons} = \frac{\text{Total Energy}}{\text{Energy of one Photon}}[/tex]

This energy of the photon is given by the Planck's equation which say:

[tex]E = \frac{hc}{\lambda}[/tex]

Here,

h = Plank's Constant

c = Velocity of light

[tex]\lambda =[/tex] Wavelength

Replacing,

[tex]E = \frac{(6.626*10^{-34})(3*10^8)}{1.55*10^{-2}}[/tex]

[tex]E = 1.28*10^{-23}J[/tex]

Now replacing we have,

[tex]\text{Number of photons} = \frac{82240.7}{1.28*10^{-23}}[/tex]

[tex]\text{Number of photons} = 6.41*10^{27}[/tex]

Therefore the number of photons required for heating is [tex]6.41*10^{27}[/tex]

Answer:

0.0986 h or 5 minutes 55 seconds.

Explanation:

Speed: This can be defined as the rate of change of distance of a body. The S.I unit of speed is m/s. Speed is a scalar quantity, because it can only be represented by magnitude alone.

Mathematically,

Speed = distance/time.

S = d/t ........................... Equation 1

making t  the subject  of the equation

t = d/S ......................... Equation 2

Form the question,

Time taken for the faster runner to reach the finish line

t₁ = d/S₁................... Equation 3

Where t₁ = time taken for the faster runner to reach the finish line, d = distance, S₁ = speed of the faster runner.

Given: d = 5.0 km, S₁ = 14.7 km/h.

Substituting into equation 3

t₁ = 5/14.7

t₁ = 0.340 h

Also,

t₂ = d/S₂................... Equation 4

Where t₂ = time taken for the slower runner to reached the finished line, d = distance, S₂ = speed of the slower runner.

Given: d = 5 km, S₂ = 11.4 km/h.

Substitute into equation 4,

t₂ = 5/11.4

t₂ = 0.4386 h.

The time the faster runner have to wait at the finish line to see the slower runner cross = t₂ - t₁ = 0.4386-0.340

The time the faster runner have to wait at the finish line to see the slower runner cross = 0.0986 h = 5 mins 55 s.

Answer:

[tex]x=2L\pm \sqrt{\frac{(-2L)^2-4\times 1\times(-L^2)}{2\times 1} }[/tex]

Explanation:

Given:

Now the third charge must be placed  on the line joining the two charges at a distance where the intensity of electric field is same for both the charges that point will not lie between the two charges because they are opposite in nature.

[tex]E_1=E_2[/tex]

[tex]\frac{1}{4\pi\epsilon_0} \times \frac{q_1}{x^2} =\frac{1}{4\pi\epsilon_0} \times \frac{q_2}{(L+x)^2}[/tex]

[tex]\frac{2.37}{x^2} =\frac{4.74}{L^2+x^2+2xL}[/tex]

[tex]2x^2=L^2+x^2+2xL[/tex]

[tex]x^2-2L.x-L^2=0[/tex]

[tex]x=2L\pm \sqrt{\frac{(-2L)^2-4\times 1\times(-L^2)}{2\times 1} }[/tex]

To solve this problem we will apply the concepts related to the balance of forces. We will decompose the forces in the vertical and horizontal sense, and at the same time, we will perform summation of torques to eliminate some variables and obtain a system of equations that allow us to obtain the angle.

The forces in the vertical direction would be,

[tex]\sum F_x = 0[/tex]

[tex]f-N_w = 0[/tex]

[tex]N_w = f[/tex]

The forces in the horizontal direction would be,

[tex]\sum F_y = 0[/tex]

[tex]N_f -W =0[/tex]

[tex]N_f = W[/tex]

The sum of Torques at equilibrium,

[tex]\sum \tau = 0[/tex]

[tex]Wdcos\theta - N_wLsin\theta = 0[/tex]

[tex]WdCos\theta = fLSin\theta[/tex]

[tex]f = \frac{Wd}{Ltan\theta}[/tex]

The maximum friction force would be equivalent to the coefficient of friction by the person, but at the same time to the expression previously found, therefore

[tex]f_{max} = \mu W=\frac{Wd}{Ltan\theta}[/tex]

[tex]\theta = tan^{-1} (\frac{d}{\mu L})[/tex]

Replacing,

[tex]\theta = tan^{-1} (\frac{0.9}{0.42*2})[/tex]

[tex]\theta = 46.975\°[/tex]

Therefore the minimum angle that the person can reach is 46.9°

Yes, it is possible. The ball would fall straight down from the perspective of a person standing at the side of the road. The person who threw the ball would see it initially move backward and then fall vertically downward.

Yes, it is possible for the ball to fall straight down as viewed by a person standing at the side of the road while someone is riding in the back of a pickup truck and throws a softball straight backward. This would occur under the condition that the ball is thrown with the same initial velocity as the truck's speed. When the ball is thrown straight backward with the same initial speed as the truck's, it will continue to move with the same speed in the backward direction relative to the truck. From the perspective of the person who threw the ball, they would see the ball initially move straight backward and then fall vertically downward due to the force of gravity.

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The ball will fall straight down as viewed from the side of the road if the velocities cancel each other out.

To the person who threw the ball, it would appear to travel straight backward relative to them.

If someone is riding in the back of a pickup truck and throws a softball straight backward, it is possible for the ball to fall straight down as viewed by a person standing at the side of the road. This will occur if the velocity at which the ball is thrown backward is equal to the velocity of the truck moving forward. In this case, the forward motion of the truck and the backward motion of the ball will cancel each other out.

For the person who threw the ball, the motion would still appear as if the ball was thrown straight backward relative to their frame of reference, assuming the truck is moving with constant velocity. The ball would move backward with the same speed at which it was thrown.

To summarize:

The ball will fall straight down as viewed from the side of the road if the velocities cancel each other out.

To the person who threw the ball, it would appear to travel straight backward relative to them.

Bohr's model of atom postulated that the electrons revolves around the nucleus only in those orbits which have fixed energy and do not lose energy while revolving in them.

According to Bohr's model, the energy at infinite distance is taken to be zero and as it approaches the atom, it starts becoming more negative.

The [tex]n^{th}[/tex] shell of electrons is calculated by

[tex]n^2 = \frac{kZ^2}{E_n}[/tex]

Here

E_n = Energy at [tex]n^{th}[/tex] level

k = Constant

n = Number of shell

Z = Atomic number of the element

Replacing we have that

[tex]n^2 = \frac{-(2.179*10^{-18}J)(1)^2}{-8.72*10^{-20}J}[/tex]

[tex]n = 24.98[/tex]

[tex]n \approx 25[/tex]

Thus

[tex]n = \pm 5[/tex]

Since number of shell cannot be negative we have that n = 5

Now the distance of electron from nucleus is given according to relation

[tex]r = (0.529)(n^2)[/tex]

[tex]r = (0.529)(5^2)[/tex]

[tex]r = 13.225 \AA[/tex]

Therefore the distance of electron from nucleus is 13.225A

The electron in a hydrogen atom is approximately 2.116 angstroms away from the nucleus.

To determine the distance of an electron from the nucleus in a hydrogen atom, we can use the equation for the energy of an electron in a hydrogen atom: E = -13.6eV / n^2, where n is the principal quantum number. We can convert the energy from joules to electron volts (eV) by using the conversion factor 1eV = 1.6x10^-19J. Substituting the given energy, we have:

-8.72x10^-20J = -13.6eV / n^2.

By rearranging the equation and solving for n, we find that n≈2. Thus, the electron is in the second energy level. The distance from the nucleus in angstroms (Å) can be calculated using the formula r = 0.529n^2Å, where r is the distance from the nucleus. Substituting n = 2, we get:

r = 0.529 x 2^2 = 2.116Å.

Therefore, the electron is approximately 2.116 angstroms away from the nucleus.

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The question is incomplete. This is the complete question: A spring has an unstretched length of 22 cm. A 150 g mass hanging from the spring stretches it to an equilibrium length of 30 cm. Suppose the mass is pulled down to where the spring's length is 38 cm. When it is released, it begins to oscillate. What is the amplitude of the oscillation?

Answer:

The amplitude of the oscillation is 8 cm.

Explanation:

The amplitude of the oscillation, which is the maximum displacement of the stretched spring from equilibrium or rest, can be calculated by subtracting the spring’s length at equilibrium (when being stretched by 150g mass) from the spring’s length when it was pulled down.

Amplitude = A = the spring’s length when it was pulled down before oscillating (i.e., 38cm) — the spring’s length at equilibrium (i.e., 30cm)

Therefore, A = 38cm — 30cm = 8cm.

Answer: p = - {1/2} , q = {1/2}

Explanation: The frequency of oscillation of a pendulum is given as

F = 1/2π *√{l/g}

Where √ is square root

l is lenght

g is acceleration due to gravity

But

F = 1/T

Where T is the period of Oscillation

Making T subject of formula we have

T= 1/F

T = 2π√{g/l}

Here the power on l is -[1/2]= p

Also,

Power on g is 1/2 =q

All because of the square root.

Answer: Energy dissipated E = 96J

Explanation:

Given:

Current I = 4A

Resistance R = 2 Ohms

Time t = 3 seconds

The energy dissipation in an electric circuit can be derived from the equation below:

E = IVt ....1

Where;

I = current, V = Voltage (potential difference), t= time and E = energy dissipated

But we know that;

V = I×R .....2

Substituting equation 2 to 1, we have

E = IVt = I(I×R)t = I^2(Rt)

Substituting the values of I,R and t

E = 4^2 × 2 ×3 = 96J

Energy dissipated E = 96J

Final answer:

The energy dissipated in the circuit is 96J.

Explanation:

To calculate the energy dissipated in a circuit, you need to use the formula P = IV, where P is power, I is current, and V is voltage. In this case, the current is 4A and the total resistance is 2Ω. Using Ohm's law (V = IR), we can find the voltage as V = I * R = 4A * 2Ω = 8V. Now, we can calculate the power dissipated as P = IV = 4A * 8V = 32W. Lastly, to find the energy dissipated, we multiply the power by the time, so 32W * 3s = 96J.

Answer:

A) [tex]k=2.63*10^{5} N/m[/tex].

B)[tex]v=2.10m/s[/tex]

C)[tex]a=90.0m/s^{2}[/tex]

Explanation:

This problem is a simple harmonic motion problem. The equation of motion for the SHM is:

[tex]\frac{d^{2}x}{dt^{2}}+\omega^{2}x=0[/tex],

where x is the displacement of the mass about its point of equilibrium, t is time, and [tex]\omega[/tex] is the angular frequency.

A)

First, we need to remember that

[tex]\omega^{2}=\frac{k}{m}[/tex],

where k is the spring constant, and m is the mass.

From here we can simply solve for k, so

[tex]k=\omega^{2}m[/tex].

Now,  we need to make use of an equation that relates the frequency and angular frequency. The equetion is

[tex]\omega=2\pi \nu[/tex],

where [tex]\nu[/tex] is the frequency. This leads us to

[tex]k=(2\pi \nu)^{2}m[/tex],

[tex]k=142(2*6.85*\pi)^{2}[/tex],

[tex]k=2.63*10^{5} N/m[/tex],

B) In simple harmonic motion, the velocity behaves as follow:

[tex]v=\omega Acos(\omega t)[/tex] (this is obtained by solving the equation of motion of the mass for the displacement x and take the derivative),

where A is the amplitude of the motion. Since we want the maximum value for the speed, we make [tex]cos(\omega t)=1[/tex] (this because cosine function goes from -1 to 1). With this, the maximum speed is simply

[tex]v = \omega A\\v=(2\pi \nu)A\\v=(2*6.85*\pi)*0.0486\\v=2.10m/s[/tex]

C) Here we are going to use the equation of motion of SHM

[tex]\frac{d^{2}x}{dt^{2}}+\omega^{2}x=0[/tex],

we know that

[tex]a=\frac{d^{2}x}{dt^{2}}[/tex] , where a is the acceleration,

[tex]a+\omega^{2}x=0\\a=-\omega^{2}x[/tex]

in this case, x goes from -A to A, so for a to be maximum we need that [tex]x=-A[/tex] ,and we get

[tex]a=-\omega^{2}(-A)\\a=\omega^{2}A\\a=(2\pi \nu)^{2}A\\a=(2*6.85*\pi)^{2}(0.0486)\\a=90.0m/s^{2}[/tex]

Answer:

See the attachment below for the graphics in part (a)

The initial velocity for this time interval is u = 61ft/sec and the final velocity is 0m/s because the car comes to a stop.

This a constant acceleration motion considering the given time interview over which the brakes are applied. So the equals for constant acceleration motion apply here.

Explanation:

The full solution can be found in the attachment below.

Thank you for reading. I hope this post is helpful to you.

To solve this problem we will apply the concepts related to the balance of Forces, in this case the centripetal force of the body must be equal to the gravitational force exerted by the moon on it.

The gravitational force is given by the function

[tex]F_g = \frac{GmM}{r^2}[/tex]

Here

G = Gravitational Universal constant

M = Mass of the planet

m = Mass of the satellite

r = Radius(orbit)

Now the centripetal force is given as

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

Here

m = mass of satellite

v = Velocity of satellite

r = Radius (orbit)

Since there must be balance for the satellite to remain in the orbit

[tex]F_c = F_g[/tex]

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

[tex]v^2= \frac{GM}{r}[/tex]

[tex]v=\sqrt{\frac{GM}{r}}[/tex]

The velocity depends on the mass of the planet and the orbit, and not on the mass, so if the orbit is maintained, the velocity will be the same: 946m/s

Answer:

Answer:

The reason why the B-2 "stealth" bomber has no vertical fins (or stabilizers) at all, is because by design, the vertical fins are not required to provide stability; the stability for yaw movement is provided through a computer that controls the B-2 stealth bomber; hence the B-2 stealth bomber does not need a vertical stabilizer in order to fly.

Answer:

A. The wavelength doubles but the wave speed is unchanged

Explanation:

The relationship between the period and wavelength is direct. Doubling the period of the oscillator will correspondingly double the wavelength but the wave speed is unaffected

An oscillator creates periodic waves on a stretched string. If the period of the oscillator doubles, then the wavelength doubles but the wave speed is unchanged. So option A is correct here.

When the period of the oscillator doubles, it means that the time it takes for one complete oscillation or cycle of the wave doubles. The period of a wave is inversely proportional to its frequency. If the period doubles, the frequency is halved. The wavelength of a wave is the distance between two consecutive crests or troughs. The wavelength of a wave is inversely proportional to its frequency. When the frequency is halved, the wavelength doubles to maintain the relationship.

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To solve this problem we will start from the definition of energy of a spring mass system based on the simple harmonic movement. Using the relationship of equality and balance between both systems we will find the relationship of the amplitudes in terms of angular velocities. Using the equivalent expressions of angular velocity we will find the final ratio. This is,

The energy of the system having mass m is,

[tex]E_m = \frac{1}{2} m\omega_1^2A_1^2[/tex]

The energy of the system having mass 2m is,

[tex]E_{2m} = \frac{1}{2} (2m)\omega_1^2A_1^2[/tex]

For the two expressions mentioned above remember that the variables mean

m = mass

[tex]\omega =[/tex]Angular velocity

A = Amplitude

The energies of the two system are same then,

[tex]E_m = E_{2m}[/tex]

[tex]\frac{1}{2} m\omega_1^2A_1^2=\frac{1}{2} (2m)\omega_1^2A_1^2[/tex]

[tex]\frac{A_1^2}{A_2^2} = \frac{2\omega_2^2}{\omega_1^2}[/tex]

Remember that

[tex]k = m\omega^2 \rightarrow \omega^2 = k/m[/tex]

Replacing this value we have then

[tex]\frac{A_1}{A_2} = \sqrt{\frac{2(k/m_2)}{(k/m_1)^2}}[/tex]

[tex]\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m_1}{m_1}}[/tex]

But the value of the mass was previously given, then

[tex]\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{m}{2m}}[/tex]

[tex]\frac{A_1}{A_2} = \sqrt{2} \sqrt{\frac{1}{2}}[/tex]

[tex]\frac{A_1}{A_2} = 1[/tex]

Therefore the ratio of the oscillation amplitudes it is the same.

Final answer:

Yes, the Earth does gain momentum in the opposite direction due to the conservation of momentum principle when everyone jumps, but the effect is negligible given Earth's massive inertial mass. So the correct option is b.

Explanation:

The question posed is whether the Earth gains momentum in the opposite direction when the entire population of the world jumps and everyone is in the air. According to the conservation of momentum, the answer is yes, but the change in the Earth's momentum is incredibly small to the point of being imperceptible. This is because the inertial mass of Earth is so large compared to the combined mass of all the people that the result of this collective jump would be negligible when it comes to the Earth's momentum.

If we consider a closed system that includes both the Earth and the people jumping, then the total change of momentum for the system must be zero. When people jump, they exert a force on the Earth, and Earth exerts an equal and opposite force on them—this is Newton's third law. However, because of the Earth's substantially greater mass, it experiences an inconsequentially small acceleration in response to this force. While the Earth does indeed recoil much like when a force is applied through a goalpost from a football player hitting it, the Earth's recoil is immeasurably small.

Answer:

a) 25F

Explanation:

Assuming that the two small spheres can be modeled as point charges, according to Coulomb's law, the magnitude of the electrostatic force is given by:

[tex]F=\frac{kq_1q_2}{R^2}[/tex]

In this case, we have [tex]R'=\frac{R}{5}[/tex]:

[tex]F'=\frac{kq_1q_2}{R'^2}\\F'=\frac{kq_1q_2}{(\frac{R}{5})^2}\\F'=25\frac{kq_1q_2}{R^2}\\F'=25F[/tex]

Explanation:

It is given that the value of P is 1110 N. And, for pin connection we have only two connections which are [tex]A_{x}[/tex] and [tex]A_{y}[/tex]. Let T be the tension is rope.

So,  [tex]\sum F_{y} = 0[/tex] and [tex]A_{y}[/tex] - 1110 = 0

              [tex]A_{y} = 1110 N[/tex]

     [tex]\sum F_{x}[/tex] = 0

And,    [tex]T - A_{x}[/tex] = 0

                  T = [tex]A_{x}[/tex]

Also, [tex]\sum M_{A}[/tex] = 0

                1110(0.75 + 0.75 + 0.75) - T(0.5 + 0.1) = 0

                    2497.5 - 0.6T = 0

                         T = 4162.5 N

                             = 4.16 kN

Therefore, we can conclude that the tension in the rope if the frame is in equilibrium is 4.16 kN.

Answer:

= 15 m/s

Explanation:

Considering right side(west) as positive x-axis and south as negative y-axis.

velocity of boat in still water [tex]v_b=12\hat{i}[/tex]

velocity of stream [tex]v_s=-9\hat{j}[/tex]

now relative velocity of boat w.r.t. stream [tex]v_{b/s}=12\hat{i}+9\hat{j}[/tex]

this velocity with which ac distance will be covered.

therefore magnitude of [tex]v_{b/s} =\sqrt{12^2+9^2}[/tex]

= 15 m/s

Answer:

(a) [tex]Q = 7.28\times 10^{14}[/tex]

(b) The charge inside the shell is placed at the center of the sphere and negatively charged.

Explanation:

Gauss’ Law can be used to determine the system.

[tex] \int{\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\

E4\pi r^2 = \frac{Q_{enc}}{\epsilon_0}\\

(860)4\pi(0.77)^2 = \frac{Q_{enc}}{8.8\times 10^{-12}}\\

Q_enc = 7.28\times 10^{14}[/tex]

This is the net charge inside the sphere which causes the Electric field at the surface of the shell. Since the E-field is constant over the shell, then this charge is at the center and negatively charged because the E-field is radially inward.

The negative charge at the center attracts the same amount of positive charge at the surface of the shell.

Answer:

[tex]v=9.34*10^{6}\frac{m}{s}[/tex]

Explanation:

According to the law of conservation of energy, the amount of energy an electron gains after being accelerated is equal to its kinetic energy, that is, the electrical potential energy is converted into kinetic energy :

[tex]U=K\\eV=\frac{mv^2}{2}[/tex]

Solving for v and replacing the given voltage:

[tex]v=\sqrt\frac{2eV}{m}\\v=\sqrt\frac{2(1.6*10^{-19}C)250V}{9.1*10^{-31}kg}\\v=9.34*10^{6}\frac{m}{s}[/tex]

Answer:

1 / 2

Explanation:

This problem is a 1 - D steady state heat conduction with only one independent variable (x).

1 - D steady state:

Q = dT / Rc

Q = heat flow

dT = change in temperature between a pair of node

Rc = thermal resistance

Rc = L / k*A

Since in both cases Rod A and Rod B have identical boundary conditions:

dT_a = dT_b

So,

R_a =  L / k*(pi*r^2)

R_b = 2L / k*(pi*(2r)^2) = L / k*(2*pi*r^2)

Compute Q_a and Q_b:

Q_a = k * dT *(pi * r^2 * / L)

Q_a = k * dT*(2*pi * r^2 * / L)

Ratio of Q_a to Q_b

Q_a / Q_b = [k * dT *(pi * r^2 * / L)] / [k * dT*(2*pi * r^2 * / L)] = 1 / 2

Final answer:

The ratio of heat flow through rod A to rod B is 1:2 when rod B has double the length and radius of rod A, both rods being made of the same material and subjected to the same temperature difference. This conclusion is derived from Fourier's law of thermal conduction.

Explanation:

The question asks about the comparison of heat flow through two rods of different dimensions but made from the same material and exposed to the same temperature difference. To find the ratio of heat flow through rod A to that through rod B, we use Fourier's law of thermal conduction, which states that the rate of heat transfer through a material is proportional to the negative gradient of temperature and the cross-sectional area of the material, and inversely proportional to the length of the material's path. Mathematically, we write this as Q = (kAΔT) / L, where Q is the heat transfer per unit time, k is the thermal conductivity of the material, A is the cross-sectional area, ΔT is the temperature difference, and L is the length of the material's path.

For rod A, assuming a unit thermal conductivity for simplicity, the rate of heat transfer QA = (kπr2(85.00 - 5.00)) / L. For rod B, with double the radius and length, QB = (kπ(2r)2(85.00 - 5.00)) / 2L. Simplifying these expressions, we find that the ratio of heat flow through rod A to that through rod B is QA/QB = 1/2. Thus, rod A transfers heat at half the rate of rod B under the given conditions.

Answer:

Tbc = 230.69 N ; Fac = 172.31 N

Explanation:

Sum of forces in y direction:

[tex]T_{BC} * sin (35) = 300*sin (70) + F_{AC}*sin (60) .... Eq 1\\[/tex]

Sum of forces in x direction:

[tex]T_{BC} * cos (35) + F_{AC}*cos (60)= 300*cos (70) .... Eq 1\\[/tex]

Solving Eq 1 and Eq 2 simultaneously:

[tex]T_{BC} = 281.9077862 + \sqrt{3} / 2 * F_{AC}\\\\F_{AC} (1.736868124) = 300*cos (70) - 491.4912266*cos (35)\\\\F_{AC} = - \frac{300}{1.736868124}\\\\F_{AC} = - 172.73 N\\\\T_{BC} = 230.69 N[/tex]

Answer: Tbc = 230.69 N ; Fac = 172.31 N

(a) The tension in the cable at AC is -200.67 N.

(b) The  tension in the cable at BC is 328.99 N.

The sum of the forces in y-direction is calculated as follows;

T(BC)sin(35) = 400 x sin(65) + F(AC) sin(60) --- (1)

T(BC)cos(35) + F(AC) cos(60) = 400 x cos(65)  ---(2)

From equation(1);

[tex]T_{BC} = \frac{400 \times sin(65) \ + \ F_{AC} sin(60)}{sin(35)} \\\\T_{BC} = 632 + 1.51F_{AC}[/tex]

From equation (2);

0.82T(BC) + 0.5F(AC) = 169.1

[tex]0.82(632 + 1.51F_A_C) + 0.5F_A_C= 169.1\\\\518.24 + 1.24F_A_C + 0.5F_A_C = 169.1\\\\F_A_C = \frac{-349.14}{1.74} \\\\F_A_C = -200.67 \ N[/tex]

T(BC) = 632 + 1.51(-200.67)

T(BC) = 328.99 N

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Answer:

The two tapes will either attract or repel each other, depending on the nature of interaction. This is explained below:

Explanation:

When you rip the two pieces of tape off the table, there is a tug-of-war for electric charges between tape and table. The tape either steals negative charges (electrons) from the table or leaves some of its own negative charges behind, depending on what the table is made of (a positive charge doesn’t move in this situation). In any case, both pieces of tape end up with the same kind of charge, either positive or negative. Since like charges repel, the pieces of tape repel each other.

When the tape sandwich is pulled apart, one piece rips negative charges from the other. One piece of tape therefore has extra negative charges. The other piece, which has lost some negative charge, now has an overall positive charge. Because opposite charges attract, the two pieces of tape attract each other.

Answer:

a much larger slit, the phenomenon of Sound diffraction that slits for light.

this is a series of equally spaced lines giving a diffraction envelope

Explanation:

The diffraction phenomenon is described by the expression

    d sin θ = m λ

Where d is the distance of the slit, m the order of diffraction that is an integer and λ the wavelength.

For train the diffraction phenomenon, the d / Lam ratio is decisive if this relation of the gap separation in much greater than the wavelength does not reduce the diffraction phenomenon but the phenomena of geometric optics.

The wavelength range for visible light is 4 10⁻⁷ m to 7 10⁻⁷ m. The wavelength range for sound is 17 m to 1.7 10⁻² m. Therefore, with a much larger slit, the phenomenon of Sound diffraction that slits for light.

When we add a second slit we have the diffraction of each one separated by the distance between them, when the integrals are made we arrive at the result of the interference phenomenon, a this is a series of equally spaced lines giving a diffraction envelope

When I separate the distance between the two slits a lot, the time comes when we see two individual diffraction patterns

To solve this problem we will first proceed to find the volume of the hemisphere, from there we will obtain the mass of the density through the relation of density. Finally the mass of the stone will be given between the difference in the mass given in the statement and the one found, that is

The volume of a Sphere is

[tex]V = \frac{4}{3} \pi r^3[/tex]

Then the volume of a hemisphere is

[tex]V =\frac{1}{2} \frac{4}{3} \pi r^3[/tex]

With the values we have that the Volume is

[tex]V =\frac{1}{2} \frac{4}{3} \pi (8.4/2)^3[/tex]

[tex]V = 155.17cm^3[/tex]

Density of water is

[tex]\rho = 1g/cm^3[/tex]

And we know that

[tex]\text{Mass of water displaced} = \text{Density of water}\times \text{Volume of hemisphere}[/tex]

[tex]m = 1g/cm^3 * 155.17cm^3[/tex]

[tex]m = 155.17g[/tex]

So the net mass is

[tex]\Delta m = m_s-m_w[/tex]

[tex]\Delta m = 155.17-23[/tex]

[tex]\Delta m = 132.17g[/tex]

Therefore the mass of heaviest rock is 132.17g or 0.132kg

The mass of the heaviest rock that, in perfectly still water, won't sink the plastic boat is 155 g.

We know that the density of water = 1 g/cm^3

Volume of the hemisphere = 2/3 πr^2

When diameter =  8.4 cm, radius = 4.2 cm

So, V = 2/3 × 3.14 × (4.2)^3

V = 155 cm^3

Volume of hemisphere = volume of water displaced = 155 cm^3

Mass of water displaced = 155 g

Since the solid displaces its own mass of water, the mass of the heaviest rock that, in perfectly still water, won't sink the plastic boat is 155 g.

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The maximum Volume Flow Rate of water the pump can provide, given an efficiency of 82% and an elevation of 15 m, is approximately 0.033 L/s.

First, we must convert the pump's horsepower to a more usable unit in this context - like watts. In physics, 1 horsepower equals roughly 746 watts. Therefore, the pump has power of 8*746 = 5968 watts.

Given the mechanical efficiency (ME) and the height (h), the maximum work the pump can do is given by M.E. * Power. So, the pump does work of 0.82*5968 = 4895.76 watts.

The work done on the water by the pump is equal to the change in potential energy of the water, PE = mgh, where m is the mass, g is the acceleration due to gravity (9.8 m/s^2), and h is the height (15 m). With rearranging, you could express m = Power/(g * h). But we're looking for the volume flow rate, not the mass flow rate, so we need to convert mass to volume. Since the density of water (ρ) is 1 kg/L, the volume flow rate = m/ρ = Power/(g * h * ρ).

Substituting all known values, we get: Volume flow rate = 4895.76 W /(9.8 m/s^2 * 15 m * 1 kg/L) = 0.033 L/s.

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Answer:

D=1.54489 m

Explanation:

Given data

S=6.10 mm= 0.0061 m

To find

Depth of lake

Solution

To find the depth of lake first we need to find the initial time ball takes to hit the water.To get the value of time use below equation

[tex]S=v_{1}t+(1/2)gt^{2} \\ 0.0061m=(0m/s)t+(1/2)(9.8m/s^{2} )t^{2}\\ t^{2}=\frac{0.0061m}{4.9m/s^{2} }\\ t=\sqrt{1.245*10^{-3} }\\ t=0.035s[/tex]

So ball takes 0.035sec to hit the water

As we have found time Now we need to find the final velocity of ball when it enters the lake.So final velocity is given as

[tex]v_{f}=v_{i}+gt\\v_{f}=0+(9.8m/s^{2} )(0.035s)\\ v_{f}=0.346m/s[/tex]

Since there are (4.50-0.035) seconds left for (ball) it to reach the bottom of the lake

So the depth of lake given as:

[tex]D=|vt|\\D=|0.346m/s*4.465s|\\D=1.54489m[/tex]

Answer: d = 1.54m

The depth of the lake is 1.54m

Explanation:

The final velocity of the ball just before it hit the water can be derived using the equation below;

v^2 = u^2 + 2as ......1

Where ;

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled.

Since the initial velocity is zero, and the acceleration is due to gravity, the equation becomes:

v^2 = 2gs

v = √2gs ......2

g = 9.8m/s^2

s = 6.10mm = 0.0061m

substituting into equation 2

v = √(2 × 9.8× 0.0061)

v = 0.346m/s

The time taken for the ball to hit water from the time of release can be given as:

d = ut + 0.5gt^2

Since u = 0

d = 0.5gt^2

Making t the subject of formula.

t = √(2d/g)

t = √( 2×0.0061/9.8)

t = 0.035s

The time taken for the ball to reach the bottom of the lake from the when it hits water is:

t2 = 4.5s - 0.035s = 4.465s

And since the ball falls for 4.465s to the bottom of the lake at the same velocity as v = 0.346m/s. The depth of the lake can be calculated as;

depth d = velocity × time = 0.346m/s × 4.465s

d = 1.54m

The depth of the lake is 1.54m