What determines whether the equilibrium temperature of a mixture of two amounts of water will be closer to the initially cooler or warmer water?

Answer:

225.56V

Explanation:

Q=cv

6.93 µF capacitor was charged to 94 V, therefore Q₁ =6.93 *10^-6 x 94 =651.42µC

1.76 µF capacitor charged to 81 V, therefore Q₂ =1.76 *10^-6 x 81 =142.56µC

When the capacitor are brought together, the charges on them move to maintain equilibrum, so therefore, each capacitor has 0.5(142.56µC + 651.42µC )= 396.99µC

The final potential difference across the 1.76 µF capacitor =  Q/c = 396.99µC/1.76 µF = 225.56V

To find the final potential difference across the 1.76 µF capacitor, use charge conservation by equating the total charge before and after connection. After calculating the charges on each capacitor, rearrange the equation to solve for the final potential difference across the 1.76 µF capacitor.

In order to find the final potential difference across the 1.76 µF capacitor, we can use the concept of charge conservation.

First, let's calculate the total charge on the capacitors before they are connected. The charge on a capacitor is given by Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference across the capacitor.

For the 6.93 µF capacitor, the charge is Q1 = (6.93 µF)(94 V) = 650.42 µC. For the 1.76 µF capacitor, the charge is Q2 = (1.76 µF)(81 V) = 142.56 µC.

After connecting the capacitors, the total charge remains the same. So, we have Q1 + Q2 = Qf, where Qf is the final charge on the capacitors. Rearranging the equation, we can find the final potential difference across the 1.76 µF capacitor:

Vf = (Qf - Q1) / C2 = (142.56 µC - 650.42 µC) / (1.76 µF).

Calculating this expression will give us the final potential difference across the 1.76 µF capacitor.

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Answer:

72 Nm²

Explanation:

work = F . d

work = (12i ,0j) . (6i ,-8j )

work = 72 J

The 12N force is a vector acting in the direction of the positive x axis, so its vector notation is 12i + 0j.

It's not exactly clear what the point's location is but I interpret it to be x= m*6, y= -m*8 (I hope the "m" wasn't a typo). The direction vector from the origin is then m*6i -m*8j

When a force vector F acts in a nonparallel direction d, the work is given by:

W = |F|*|d|*cos(theta) where theta is the angle between the vectors.

alternatively you can use the dot product of the two vectors to get:

W = (12N)*(m*6) + (0N)*(-m*8) = 72 Nm²

Final answer:

The work done on the block by the constant force of 12 N in the +x direction is 24 J.

Explanation:

The work done on the block can be calculated by multiplying the force applied to the block by the displacement of the block in the direction of the force. In this case, the force applied is 12 N in the +x direction and the displacement is (6 - 8) m = -2 m. Since the force and displacement are in opposite directions, the work done will be negative. The formula for work is given by:

Work = Force * Displacement * cos(theta)

where theta is the angle between the force and the displacement.

In this case, the angle between the force and the displacement is 180 degrees, so cos(theta) = -1. Substituting the values into the formula:

Work = 12 N * -2 m * (-1) = 24 J

The cylinder will descend a distance of f/k before coming momentarily to rest.

When the cylinder is released, it will initially accelerate downwards due to the force of gravity. As it accelerates, the spring attached to it will begin to stretch, exerting an upward force on the cylinder. The cylinder will continue to descend until the force exerted by the spring is equal in magnitude to the force of kinetic friction between the cylinder and the walls of the tube. At this point, the net force on the cylinder will be zero, and it will come momentarily to rest.

To find the distance the cylinder descends before coming to rest, we can equate the force exerted by the spring with the force of kinetic friction:

kx = f

Where k is the force constant of the spring and x is the distance the cylinder has descended. Solving for x gives:

x = f/k

Therefore, the vertical distance the cylinder descends before it comes momentarily to rest is given by x = f/k.

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Answer:

Explanation:

Acceleration without friction on an inclined plane = g sinθ

Acceleration with friction on inclined plane = g sinθ - μ g cosθ

s = 1/2 a t²

For motion on friction-less surface

s = 1/2 g sinθ t₁²

For motion on frictional surface

s = 1/2( g sinθ - μ g cosθ)  t₂²

t₂ = 2t₁

(  sinθ - μ  cosθ)  t₂² =  sinθ t₁²

(  sinθ - μ  cosθ)  4t₁² =  sinθ t₁²

(  sinθ - μ  cosθ)  4 =  sinθ

4sinθ - sinθ = 4μ  cosθ

3sinθ = 4μ  cosθ

3 / 4 tanθ =μ  

μ  = .75  tanθ

Answer:

The centripetal acceleration at highway speed is greater.

Explanation:

We assume the motion of the car is uniformly accelerated. Let the highway speed be v.

By the equation of motion,

[tex]v=u+at[/tex]

[tex]a=\dfrac{v-u}{t}[/tex]

u is the initial velocity, a is acceleration and t is time

Because the car starts from rest, u = 0.

[tex]a_T=\dfrac{v}{t}[/tex]

This is the tangential acceleration of the thread of the tire.

The centripetal acceleration is given by

[tex]a_C=\dfrac{v^2}{r}[/tex]

r is the radius of the tire.

Comparing both accelerations and applying commonly expected values to r and t, the centripetal acceleration is seen to be greater. The radius of a tyre is, on the average, less than 0.4 m. Then the centripetal acceleration is about

[tex]a_C=\dfrac{v^2}{0.3}=2.5v^2[/tex]

The tangential acceleration can only be greater in the near impossible condition that the time to attain the speed is on the order of microseconds.

The minimum coefficient of static friction between the pavement and the tires is 0.156.

The minimum coefficient of static friction is calculated by applying Newton's second law of motion in determining the net force.

[tex]F_c = Wsin(\theta) + \mu_s W cos(\theta)\\\\\frac{mv^2}{r} = mg sin(\theta) + \mu_s mg cos(\theta)\\\\\mu_s mg cos(\theta)\ = \frac{mv^2}{r} - mg sin(\theta) \\\\\mu _ s = \frac{mv^2 \ - \ mgr sin(\theta)}{mg rcos(\theta)}[/tex]

where;

[tex]\mu _ s = \frac{1200 \times 18.1^2 \ - \ 1200 \times 9.8 \times 100 sin(10.4)}{1200 \times 9.8 \times 100 \times cos(10.4)}\\\\\mu_s = 0.156[/tex]

Thus, the minimum coefficient of static friction between the pavement and the tires is 0.156.

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To solve this problem we will apply the concepts related to energy conservation. We know that potential energy is transformed into kinetic and vice versa energy. Since the energy accumulated in the upper part is conserved as potential energy, when the object is thrown all that energy will be converted into kinetic energy. Therefore we will have the following relation,

[tex]KE = PE[/tex]

[tex]KE = mgh[/tex]

Here,

m = mass

g = Gravitational acceleration

h = Height

Replacing,

[tex]KE = (6)(9.8)(5)[/tex]

[tex]KE = 294J[/tex]

Therefore the kinetic energy at the bottom is 294J

Answer:

Explanation:

Given that,

Sphere of radius R

At radial distance ¼R it has an electric field of Eo

Magnitude of electric field E at r=2R

Solution

electric field is given as

E=KQ/r²

Now at R,

E=KQ/R²

When, at r=R/4 inside the sphere is given as

Eo=kQ/R³ • R/4

Eo=kQ/4R²

When, r =2R

Then,

E=kQ/(2R)²

E=kQ/4R²

Comparing this to Eo

Then, E=Eo

Then, the electric field at r=2R is equal to the electric field at r=R/4

Answer:

    a = 4.8 10⁻⁵ m

Explanation:

The diffraction phenomenon is described by the expression

        a sin θ = m λ

How the pattern is observed on a distant screen

        tan θ = y / L = sin θ / cos θ

Since the angle is very small in these experiments ’we can approximate the tangent function

    tan θ = sin θ = y / L

We substitute

           a y / L = m λ

The first minimum occurs for m = 1

        a = λ L / y

        a = 680 10⁻⁹ 5.5 / 0.078

        a = 4.8 10⁻⁵ m

Answer:

correct answer is (b)  Longitudinal lines remain straight

Explanation:

solution

As we know that Deformation of circular shaft in the torsion is associate with twisting of  shaft more than an specify with the yielding limit.

so when any angle of twist is obtain in the torsion and that is beyond the specified safety limit of shaft

than that shaft will be fail.

but it does not regain its original shape and it will cause permanent deformation

so that we can say longitudinal lines which is twist, they will not regain to original back position as straight

but they will remain in curved shape.

so here incorrect statement is b Longitudinal lines remain straight

Though torsion theory assumes that cross sections of a torsionally loaded shaft remain flat, in reality, under heavy loading, they can warp and so this statement is not entirely accurate.

When a circular shaft deforms under torsion, certain assumptions are made about its deformation according to the torsion theory. These assumptions include:

1) cross sections remain flat and perpendicular to the axis of the shaft,

2) longitudinal lines remain straight,

3) circular sections remain circular, and

4) radial lines on the sections remain straight. However, the statement that is not entirely accurate is that cross sections remain flat. In reality, under severe torsional loading, the cross sections might warp and not remain entirely flat.

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Answer:

The average current that this cell phone draws when turned on is 0.451 A.

Explanation:

Given;

voltage of the phone, V = 3.7 V

electrical energy of the phone battery, E = 3.15 x 10⁴ J

duration of battery energy, t = 5.25 h

The power the cell phone draws when turned on, is the rate of energy consumption, and this is calculated as follows;

[tex]P = \frac{E}{t}[/tex]

where;

P is power in watts

E is energy in Joules

t is time in seconds

[tex]P = \frac{3.15*10^4}{5.25*3600s} = 1.667 \ W[/tex]

The average current that this cell phone draws when turned on:

P = IV

[tex]I = \frac{P}{V} =\frac{1.667}{3.7} = 0.451 \ A[/tex]

Therefore, the average current that this cell phone draws when turned on is 0.451 A.

The average current that the cell phone draws when turned on is approximately [tex]\(0.450 \, \text{A}\)[/tex].

To find the average current drawn by the cell phone, we need to use the relationship between power, voltage, and current. Let's go step by step.

Step 1: Calculate the power consumed by the cell phone

We already have:

- The total energy provided by the battery: [tex]\( \Delta W = 3.15 \times 10^4 \, \text{J} \)[/tex]

- The total time of operation: [tex]\( \Delta t = 5.25 \, \text{hours} \)[/tex]

First, convert the operation time from hours to seconds:

[tex]\[\Delta t = 5.25 \, \text{hours} \times 3600 \, \text{seconds/hour} = 5.25 \times 3600 = 18900 \, \text{s}\][/tex]

Now, calculate the power P using the energy and time:

[tex]\[P = \frac{\Delta W}{\Delta t} = \frac{3.15 \times 10^4 \, \text{J}}{18900 \, \text{s}} \approx 1.666 \, \text{W}\][/tex]

Step 2: Use the power to find the average current

We know that power P, voltage V, and current I are related by the equation:

[tex]\[P = VI\][/tex]

We can solve for the current I

[tex]\[I = \frac{P}{V}\][/tex]

Given the battery voltage [tex]\(V = 3.70 \, \text{V}\)[/tex] and the power [tex]\(P \approx 1.666 \, \text{W}\)[/tex], we can calculate the current:

[tex]\[I = \frac{1.666 \, \text{W}}{3.70 \, \text{V}} \approx 0.450 \, \text{A}\][/tex]

The complete question is this:

The battery for a certain cell phone is rated at 3.70 V. According to the manufacturer it can produce 3.15 x 104 J of electrical energy, enough for 5.25 h of operation, before needing to be recharged. Find the average current that this cell phone draws when turned on.

Ok, so what I did so far was convert time into seconds and found Power:

t = 18900 s

P = ΔW/Δt =student submitted image, transcription available below= 1.6666 W

I think you have to use the problem : P = VabI = I2R = εI - I2R

Answer:

Φ = 2.26 10⁶ N m² / C

Explanation:

The electric flow is

           Ф = E .ds = [tex]q_{int}[/tex] / ε₀

Since we have two components for this flow. The uniform electric field and the flux created by the point charge.

The flux of the uniform electric field is zero since the flux entering is equal to the flux leaving the Gaussian surface

The flow created by the point load at the origin is

              Φ = q_{int} /ε₀

Let's calculate

              Φ = 2.00 10⁻⁶ /8.85 10¹²

              Φ = 2.26 10⁶ N m² / C

Answer:

Explanation:

Since , resistances are joined in parallel , potential difference across either of R₁ or R₂ will be same , and it will be equal to emf of the cell provided no internal resistance exists in the cell.

Since potential difference across R₁ is equal to emf and resistance is R₁ , current through it too will be same as that in the previous case.

The potential difference across resistor R1 remains the same when R2 is added in parallel, and the current through R1 also remains unchanged because the voltage across it does not change.

When R2 is added in parallel to R1, the potential difference (voltage) across R1 remains the same because in a parallel circuit, all components have the same potential difference across them. However, the current i1 through R1 remains the same as previously as well since the potential difference across it hasn't changed and Ohm's Law (I = V/R) dictates that the current through a resistor will depend only on the potential difference across it and its resistance, which remains unchanged.

Adding R2 in parallel with R1 has the effect of decreasing the overall resistance of the circuit as found using the equation 1/Rp = 1/R1 + 1/R2. This means that the total current drawn from the battery will increase, but it will be divided among R1 and R2. Since the voltage across R1 remains constant, the current through R1 is unaffected by the addition of R2.

Answer:

Th = 50°C

Explanation:

See the attachment below.

K = 7.75

Tc = 13.5°C = 273+13.5 = 286.5K

Answer:

292.984 K or 19.98 °C

Explanation:

Using,

η = Tc/(Th-Tc)................ Equation 1

Where η = coefficient of performance, Tc = cold temperature of the refrigerator, Th = hot temperature of the refrigerator.

Make Th the subject of the equation

η (Th-Tc) = Tc

η Th-η Tc = Tc

Th = (Tc+ηTc)/η ....................... Equation 2

Given: Tc = -13.5°C+273 = 259.5 °C, η  = 7.75

Substitute into equation 2

Th = (259.5+7.75×259.5)/7.75

Th = (259.5+2011.125)/7.75

Th = 2270.625/7.75

Th = 292.984 °C or 19.98 °C

Hence the maximum hot reservoir temperature = 292.984 °C or 19.98 °C

Answer:23 N

Explanation:

Given

Weight of box [tex]W=60\ N[/tex]

Coefficient of static friction is [tex]\mu _s=0.5[/tex]

Applied force [tex]F=23\ N[/tex]

When Force is applied box is observed to be at rest i.e. static friction is overcoming the applied force.

Thus Static friction [tex]F_s[/tex]=applied force

[tex]F_s=23\ N[/tex]

Although it maximum value can go up to [tex](F_s)_{max}=\mu _sN[/tex]

[tex]F_s=0.5\times 60[/tex]

[tex]F_s=30\ N[/tex]

Answer:

6104 N/C.

Explanation:

Given:

k = 8.99 × 10^9 Nm2/C^2

Qx = 1.3 × 10^-5 C

rx = 7 m

Qy = 1 × 10−5 C

ry = 4 m

E = F/Q

= kQ/r^2

Ex = (8.99 × 10^9 × 1.3 × 10^−5) ÷ 7^2

= 2385.1 N/C.

Ey = (8.99 × 10^9 × 1.0 × 10^−5) ÷ 4^2

= 5618.75 N/C

Eo = sqrt(Ex^2 + Ey^2)

= sqrt(3.157 × 10^7 + 5.69 × 10^6)

= 6104 N/C.

The magnitude of the electric field at the origin due to the charges fixed at (7 m, 0) and (0, 4 m) is approximately 6107.83 N/C. This was calculated by determining the electric fields from each charge and then using vector addition to find the resultant electric field. The calculation involved using the formula E = k |q| / r² for each charge, followed by determining the resultant using Pythagoras' theorem.

Let's determine the electric field at the origin due to two fixed charges. The first charge is -1.3 × 10⁻⁵ C located at (7 m, 0), and the second charge is 1 × 10⁻⁵C located at (0, 4 m).

The electric field due to a point charge q at a distance r is given by E = k |q| / r², where k = 8.99 × 10⁹ N·m²/C² is the Coulomb constant.

Step-by-Step Calculation

Calculate the distance from each charge to the origin:

Charge 1 at (7 m, 0): r₁ = 7 m

Charge 2 at (0, 4 m): r₂ = 4 m

Compute the electric field due to each charge:

For q₁ = -1.3 × 10⁻⁵ C:
E₁ = k |q₁| / r₁² = (8.99 × 10⁹ N·m²/C²) (1.3 × 10⁻⁵ C) / (7 m)² ≈ 2381.57 N/C along the negative x-direction.

For q₂ = 1 × 10⁻⁵ C:
E₂ = k |q₂| / r₂² = (8.99 × 10⁹ N·m²/C²) (1 × 10⁻⁵ C) / (4 m)² ≈ 5621.88 N/C along the positive y-direction.

Determine the resultant electric field at the origin using vector addition. Since the fields are perpendicular:

Resultant E = √(E₁² + E₂²)
≈ √((2381.57 N/C)² + (5621.88 N/C)²)
≈ √(5662179.6649 + 31643689.7344)
≈ √(37305869.3993)
Resultant E ≈ 6107.83 N/C

The magnitude of the resultant electric field at the origin is therefore approximately 6107.83 N/C.

The angular velocity of the merry-go-round increases as a child walks towards the center, due to the conservation of angular momentum and a decrease in the system's moment of inertia.

When a child walks from the outer edge of a rotating merry-go-round towards the center, the system's angular momentum remains constant due to the conservation of angular momentum. However, the moment of inertia of the system decreases as the child moves closer to the center. Since angular momentum is the product of the moment of inertia and angular velocity (L = I∙ω), and it is conserved, the angular velocity of the merry-go-round must increase to compensate for the decrease in moment of inertia.

This concept can be compared to a figure skater pulling in their arms to spin faster. As the skater's arms come closer to the body, their moment of inertia decreases, causing an increase in their spinning speed, or angular velocity, with no external torques acting on the system.

The same principle applies to the merry-go-round: as the child moves inward, the system's angular velocity increases to conserve angular momentum. This can be observed in playgrounds and is a practical example of conservation laws in rotational motion.

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As the child walks towards the center of the rotating merry-go-round, the system's angular momentum remains constant due to the conservation of angular momentum. The moment of inertia decreases, causing the angular velocity to increase.

A child is standing on the edge of a merry-go-round that is rotating with frequency f.

As the child walks towards the center of the merry-go-round, the angular momentum of the system (child plus merry-go-round) remains constant.

This is due to the conservation of angular momentum, which states that if no external torque acts on the system, the total angular momentum remains unchanged.

Angular momentum (L) is given by the formula:

where I is the moment of inertia and ω is the angular velocity.

complete question:

A child is standing on the edge of a merry-goround that is rotating with frequency f. The child then walks towards the center of the merry-go-round. For the system consisting of the child plus the merry-go-round, what remains constant as the child walks towards the center of the rotating merry-go-round?

Answer:

  B = -0.215 s⁻¹,  A₀ = 1.537º

Explanation:

To be able to make this graph we must linearize the data, the best procedure is to calculate the logarithm of the values

           A = A₀ [tex]e^{-Bt}[/tex] + C

The constant creates a uniform displacement of the graph if we start the graph at the angle of 1, the constant disappears, this is done by subtracting 1 from each angle, so the equation is

            (A-1) = A₀  e^{-Bt}

We do the logarithm

           Log (A-1) = log Ao –Bt log e

Make this graph because paper commercially comes in logarithm 10, if we use graph paper if we can calculate directly in base logarithm e, let's perform this calculation

              Ln (A-1) = Ln A₀ - Bt

To graph the points we subtract 1 from each angle and calculate the logarithm, the data to be plotted are

              θ’= ln (θ -1)

θ(º)    t(s)        θ'(º)

6.2      2     1.6487

4.7      4     1.3083

3.2      6    0.7885

2.4      8    0.3365

1.8     10   -0.2231

We can use a graph paper and graph on the axis and the primary angle (θ') and on the x-axis time, mark the points and this graph is a straight line, we see that the point for greater time has a linearity deviation , so we will use the first three for the calculations

To find the line described by the equation

              y -y₀ = m (x -x₀)

             m = (y₂ -y₁) / (x₂ -x₁)

Where m is the slope of the graph e (x₀, y₀) is any point, let's start as the first point of the series

            (x₀, y₀) = (2, 1.65)

           (x₂, y₂) = (2, 1.65)

           (x₁, y₁) = (6, 0.789)

We use the slope equation

           m = (1.65 - 0.789) / (2-6)

           m = -0.215  

The equation is

           y - 1.65 = -0.215 (x- 2)

           y = -0.215 x +0.4301

We buy the two equations and see that the slope is the constant B

            B = -0.215 s⁻¹

The independent term is

            b = ln A₀

            A₀ = [tex]e^{b}[/tex]

           A₀ = [tex]e^{0.43}[/tex]

           A₀ = 1.537º

Answer:

210,600πft-lb

Explanation:

Force is a function F(x) of position x then in moving from

x = a to x= b

Work done = [tex]\int\limits^b_a {Fx} \, dx[/tex]

Consider a water tank conical in shape

we will make small horizontal section of the water at depth h and thickness dh and also assume radius at depth h is w

we will have ,

[tex]\frac{w}{12} = \frac{(18-h)}{18} \\w = \frac{2}{3} (18-h)a[/tex]

weight of slice under construction

weight = volume × density × gravitational constant

[tex]weight = \pi \times w^2 \times dh \times 62.4\\= (62.4\pi w^2dh)lb[/tex]

Now we can find work done

[tex]W = \int\limits \, dw\\[/tex]

[tex]\int\limits^{18}_{3} {(62.4\pi } \, dx w^2dh)h\\= 62\pi \frac{4}{9} \int\limits^{18}_{3} {(18-h)^2hdh} \,[/tex]

= [tex]62.4\pi \times\frac{4}{9} (\frac{324}{2} h^2-\frac{36}{3} + \frac{h^4}{4})^1^8_3[/tex]

= 210,600πft-lb

Answer:

a) b) d)

Explanation:

The question is incomplete. The Complete question might be

In an inertial frame of reference, a series of experiments is conducted. In each experiment, two or three forces are applied to an object. The magnitudes of these forces are given. No other forces are acting on the object. In which cases may the object possibly remain at rest? The forces applied are as follows: Check all that apply.

a)2 N; 2 N

b) 200 N; 200 N

c) 200 N; 201 N

d) 2 N; 2 N; 4 N

e) 2 N; 2 N; 2 N

f) 2 N; 2 N; 3 N

g) 2 N; 2 N; 5 N

h ) 200 N; 200 N; 5 N

For th object to remain at rest, sum of all forces must be equal to zero. Use minus sign to show opposing forces

a) 2+(-2)=0 here minus sign is to show the opposing firection of force

b) 200+(-200)=0

c) 200+(-201)[tex]\neq[/tex]0

d) 2+2+(-4)=0

e) 2+2+(-2)[tex]\neq[/tex]0

f) 2+2+(-3) [tex]\neq[/tex]0; 2+(-2)+3[tex]\neq[/tex]0

g) 2+2+(-5)[tex]\neq[/tex]0; 2+(-2)+5[tex]\neq[/tex]0

h)200 + 200 +(-5)[tex]\neq[/tex]0; 200+(-200)+5[tex]\neq[/tex]0

Explanation:

Given data:

G = 11.5×10⁶psi

d₂ = 2.0 inch

d₁ = 1.5 inch

ε[tex]_{max}[/tex] = 170 × 10⁻⁶

Y[tex]_{max}[/tex] = 2ε

T/J = τ[tex]_{max}[/tex] /R

[tex]\frac{Td_{2} }{2J}[/tex] = τ[tex]_{max}[/tex]         (1)

τ[tex]_{max}[/tex] = G Y

from 1 and 2

[tex]\frac{T d_{2} }{2J} = G Y_{max}[/tex]

T = [tex]\frac{2 G Y_{max}J }{d_{2} }[/tex]

[tex]\frac{2* 11.5*10^{6}*0.006895*10^{6}*340*10*^{-6} *\frac{\pi }{32}[2^{4}-1.5^{4}]*(0.0254)^{4} }{2* 0.0254}[/tex]

  = 474.14 Nm

Answer:

5.02 m/s

Explanation:

given,

frequency of sound emitted by bat, [tex]f_1[/tex] = 30 kHz = 30000 Hz

beat frequency detected  by bat[tex]f_b[/tex] = 900 Hz

Speed of sound, v = 340 m/s

[tex]f_b = f_1 -f_2[/tex]

[tex]900 = 30000-f_2[/tex]

[tex]f_2 = 30900[/tex]

Using Doppler's effect formula

[tex]f_2=f_1(\dfrac{v+v_b}{v-v_b})[/tex]

[tex]v_b = velocity\ of\ bat[/tex]

[tex]30900=30000(\dfrac{340+v_b}{340-v_b})[/tex]

[tex]1.03 =(\dfrac{340+v_b}{340-v_b})[/tex]

[tex]v_b = 5.02 m/s[/tex]

Hence, the speed of bat is equal to 5.02 m/s

Answer: The intensity level of sound in the bedroom is 80dB

Explanation:

Intensity of lawn mower at r=1m is 100dB

Beta1= 10dBlog(I1/Io)

100dB= 10dB log(I1/Io)

10^10= I1/Io

I1= Io(10^10)

10^12)×(10^10)= I1

I1=10^-2w/m^2

Intensity of lawn mower at r=20m

I2/I1=(r1/r2)^2 =(1/20)^2

I2= I1(1/400)

I2=2.5×10^-3W_m^2

Intensity of 4 lown mowers at 20m fro. Window

= 10dBlog(4I2/Io)

= 10^-4/10^-12

=80dB

The pilot's maximum endurance speed in a circle of 85 m diameter without blacking out can be calculated using the centripetal acceleration formula, taking into account the acceleration limit of 7 g, where 1 g equals 9.80 m/s².

The question requires the application of concepts from physics, specifically the idea of centripetal force and acceleration when a body moves in a circular path.

Given that the maximum acceleration a pilot can endure without blacking out is 7 g, and knowing that 1 g is equivalent to 9.80 m/s², we can calculate the maximum speed under these conditions using the formula for centripetal acceleration, which is `a = v² / r`, where `a` is the acceleration, `v` is the speed, and `r` is the radius of the circle.

The diameter of the circle is given as 85 meters, which means the radius `r` will be half of that, 42.5 meters. The maximum acceleration of 7 g translates to 7 × 9.80 m/s², which is 68.6 m/s². We can rearrange the formula to solve for `v`, leading to `v = √(a × r)`.

Substituting the values provided, we get `v = √(68.6 m/s² × 42.5 m)

v = √2923.175 m²/s² ≈ 54.1 m/s

Therefore, the maximum speed the pilot can tolerate while spinning in the horizontal circle is approximately 54.1 m/s, which is equivalent to about 192 km/h.

Incomplete question.The complete one is here

Spacecraft have been sent to Mars in recent years. Mars is smaller than Earth and has correspondingly weaker surface gravity. On Mars, the free-fall acceleration is only 3.8m/s2. What is the orbital period of a spacecraft in a low orbit near the surface of Mars?

Answer:

[tex]T=5900s=99min[/tex]

Explanation:

Given

[tex]r_{satelite}=r_{mars}=3.37*10^{6}m\\ g_{mars}=3.8m/s^{2}\\[/tex]

To find

orbital period of a spacecraft T

Solution

An the initial calculating is computing the angular velocity of satellite :

[tex]w=\frac{2\pi }{T}\\ w=\frac{2\pi }{110min}(1min/60s)\\ w=9.52*10^{-4}rad/s[/tex]

Computing T

[tex]T=\frac{2\pi }{w}\\ as\\w=\sqrt{\frac{a}{r} }\\ So\\T=\frac{2\pi }{\sqrt{\frac{a}{r} }} \\T=\frac{2\pi }{\sqrt{\frac{3.8m/s^{2} }{3.37*10^{6} m} }}\\T=5900s=99min[/tex]

Answer:

11.2 Km/Sec

Explanation:

Escape velocity is the velocity that one has to achieve to go out of the gravitational effect of a body. Here they are talking about escaping from the Earth's gravity. To travel to Mars, first we will have to go far enough so that the impact of Earth's gravity is negligible.

The escape velocity ([tex]v_{e}[/tex]) can be calculate using the given formula:

[tex]v_{e} = \sqrt{\frac{2GM}{r}}[/tex]

where,

G = Gravitational Constant

M = Mass of the object from which we need to escape (Here it is the mass of the Earth)

r = distance from the center of that object (In this case it is equal to the radius of Earth)

For Earth, this comes out to be 11.2 KM/Sec

Answer:

Normal Conversation: i=106i0

i(dB)=60

Power saw a 3 feet: i=1011i0

i(dB)=110

Jet engine at 100 feet: i=1018i0

i(dB)=180

Explanation:

if these are the same as edge, then these are the answers! :)

Normal Conversation: i=106i0

The intensity in decibels is 60,110,180, respectively.

⇒ normal conversation: i =60(dB)

⇒power saw 3 feet: i =110(dB)

⇒jet engine at 100 feet: i = 180(dB)

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Answer:

The correct answer is the number 4. A wave on a pond is a mechanical wave which requires a medium to travel.

Explanation:

Mechanical waves are those that need a material medium to propagate. The waves of the sea and the waves that we produce on a guitar string, the sound, are examples of mechanical waves. Electromagnetic waves are energetic pulses capable of propagating in a vacuum. This way, a wave on a pond is a mechanical wave which requires a medium to travel.

Final answer:

A wave on a pond is a mechanical wave requiring water to propagate, while electromagnetic waves, such as light, don't require a medium and can travel through a vacuum. Therefore, statement 4 is correct. This reflects fundamental differences in how mechanical and electromagnetic waves propagate and their need for a medium.

Explanation:

To compare waves on a pond with electromagnetic waves, we should understand that a wave on a pond is a mechanical wave which requires a medium to travel (option 4). This is because mechanical waves are disturbances that propagate through a material medium and the pond water serves as this medium. On the other hand, electromagnetic waves do not need a medium; they can travel through a vacuum because they consist of oscillating electric and magnetic fields.

So, the correct statement would be that a wave on a pond is a mechanical wave which requires a medium to travel. Therefore, the waves on a pond are similar to sound waves since both are mechanical and need a medium. In contrast, light waves are an example of electromagnetic waves, which means they can travel through the vacuum of space without any medium.

Furthermore, all electromagnetic waves move at the same speed in a vacuum, which is known as the speed of light. This is a fundamental difference between mechanical and electromagnetic waves since the speed of mechanical waves depends on the medium in which they are traveling. The discovery that electromagnetic waves do not require a medium led to the abandonment of the aether theory, which was invented to provide a medium for light and other electromagnetic wave propagation.

Answer:

(a) the initially stationary spelunker is accelerated to a speed of 4.70 m/s - 6106 J

(b) he is then lifted at the constant speed of 4.70 m/s - 5488 J

(c) finally he is decelerated to zero speed. How much work is done on the 56.0 kg rescue by the force lifting him during each stage - 4869 J

Explanation:

knowing

d = 10 m

m = 56 kg

The work done by the applied force to pull the spelunker is given by

Wa + Wg = Kf - Ki

Wg = -mgd

First

Wa = mgd + 0.5 [tex]mv^{2}_{f}[/tex] - 0.5 [tex]mv^{2}_{i}[/tex]

Wa = (56*9.8*10) + (0.5*56*[tex]4.7^{2}[/tex])

Wa = 6106 J

Second

Kf = Ki

Wa = mgd + 0.5 [tex]mv^{2}_{f}[/tex] - 0.5 [tex]mv^{2}_{i}[/tex]

Wa = 56*9.8*10

Wa = 5488 J

Third

Kf = 0

Wa = mgd + 0.5 [tex]mv^{2}_{f}[/tex] - 0.5 [tex]mv^{2}_{i}[/tex]

Wa = (56*9.8*10) - (0.5*56*[tex]4.7^{2}[/tex])

Wa = 4869 J

Answer:

Explanation:

Length, l = 35.5 cm = 0.355 m

y = 2 x

Slope, y / x = 2

tan θ = 2

θ = 63.4°

i = 22.5 A

B = 0.318 i Tesla

[tex]\overrightarrow{l}=0.355\widehat{k}+0.355 Cos63.4\widehat{i}+0.355 Sin63.4\widehat{j}[/tex]

[tex]\overrightarrow{l}=0.16\widehat{i} + 0.32\widehat{j}+0.355\widehat{k}[/tex]

[tex]\overrightarrow{B}=0.318\widehat{i}[/tex]

The magnetic force is given by

[tex]\overrightarrow{F}=i(\overrightarrow{l}\times \overrightarrow{B})[/tex]

[tex]\overrightarrow{F}=22.5\left ( 0.16\widehat{i}+0.32\widehat{j}+0.355\widehat{k} \right )\times 0.318\widehat{i}[/tex]

[tex]\overrightarrow{F}=2.54\widehat{j}-2.29\widehat{k}[/tex]

Magnitude of force

[tex]F=\sqrt{2.54^{2}+2.29^{2}}[/tex]

F = 3.42 N

The angle is Ф

tanФ = -2.29/2.54

Ф = 42° below y axis

The total force experienced by the wire in the magnetic field has a magnitude of 145 N and is directed -30 degrees relative to the z-axis, according to the right-hand rule and Lorentz force law.

In order to determine the direction of the total force on the wire, you would need to use the right-hand rule. First, you know the current is flowing down the z-axis and out along the y=2x line in the xy-plane. You also know that the magnetic field, B, is given by (.318 i)T which lies along the x-axis.

According to the Lorentz force law, the force exerted on a current-carrying conductor in a magnetic field is given by F = I * (L x B), where I represents the current, L is the length vector of the wire, and x is the cross product. Because the wire is bent at a right angle, the force will have two components: one for each section of the wire.

The force along the z-axis is given by F = I*LB*sin(theta), where theta is the angle between L and B. Here, L = 35.5/2 = 17.75 cm, B = 0.318 T, I = 22.5 A, and theta = 90 degrees (since B is along x-axis and L is along z-axis). Calculating gives Fz = 22.5 A * 17.75 cm * 0.318 T * sin(90) = 126.6 N.

The force along the y-axis is similar, but with the length vector along y=2x in the xy plane and B still along the x-axis, so theta = 180 degrees - arctan(2). Substituting the values we get Fy = - 22.5 A * 17.75 cm * 0.318 T * sin(180 - arctan(2)) = -73.8 N.

Therefore, the total force is given by the vector sum of Fz and Fy, which has magnitude sqrt(Fz^2 + Fy^2) = 145 N, in the direction of atan2(Fy, Fz) = -30 degrees relative to the z-axis.

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