Although the evidence is weak, there has been concern in recent years over possible health effects from the magnetic fields generated by electric transmission lines. A typical high-voltage transmission line is 20 m above the ground and carries a 200 A current at a potential of 110 kV.What is the magnetic field strength on the ground directly under such a transmission line?

Answer:

a = 9.94 m/s²

Explanation:

given,

density at center= 1.6 x 10⁴ kg/m³

density at the surface = 2100 Kg/m³

volume mass density as function of distance

[tex]\rho(r) = ar^2 - br^3[/tex]

r is the radius of the spherical shell

dr is the thickness

volume of shell

[tex]dV = 4 \pi r^2 dr[/tex]

mass of shell

[tex]dM = \rho(r)dV[/tex]

[tex]\rho = \rho_0 - br[/tex]

now,

[tex]dM = (\rho_0 - br)(4 \pi r^2)dr[/tex]

integrating both side

[tex]M = \int_0^{R} (\rho_0 - br)(4 \pi r^2)dr[/tex]

[tex]M = \dfrac{4\pi}{3}R^3\rho_0 - \pi R^4(\dfrac{\rho_0-\rho}{R})[/tex]

[tex]M = \pi R^3(\dfrac{\rho_0}{3}+\rho)[/tex]

we know,

[tex]a = \dfrac{GM}{R^2}[/tex]

[tex]a = \dfrac{G( \pi R^3(\dfrac{\rho_0}{3}+\rho))}{R^2}[/tex]

[tex]a =\pi RG(\dfrac{\rho_0}{3}+\rho)[/tex]

[tex]a =\pi (6.674\times 10^{-11}\times 6.38 \times 10^6)(\dfrac{1.60\times 10^4}{3}+2.1\times 10^3)[/tex]

a = 9.94 m/s²

Final answer:

a) The wavelength of the radio station's broadcast is approximately 272.73 meters. b) At a time of 3.1 μs, the value of the radio wave's electric field is approximately 0.619 N/C.

Explanation:

a) To calculate the wavelength of the radio station's broadcast, we can use the formula λ = c/f, where λ is the wavelength, c is the speed of light, and f is the frequency. Plugging in the given frequency of 1100 kHz (or 1100 x 10^3 Hz), we get: λ = (3 x 10^8 m/s) / (1100 x 10^3 Hz) = 272.73 m

b) To find the value of the radio wave's electric field at a specific time, we can use the given time dependence equation E(t) = E0 sin(2πft), where E0 is the amplitude, f is the frequency, and t is the time. Plugging in the given amplitude of E0 = 0.62 N/C, frequency of 1100 kHz (or 1100 x 10^3 Hz), and time of 3.1 μs (or 3.1 x 10^-6 s), we get: E(t) = 0.62 sin(2π x 1100 x 10^3 x 3.1 x 10^-6) ≈ 0.619 N/C

Answer:2.74 Hz

Explanation:

Given

mass Puck [tex]m=0.51 kg[/tex]

length of cord [tex]L=0.827 m[/tex]

Maximum Tension in chord [tex]T=126 N[/tex]

as the Puck is moving in a horizontal circle so maximum Tension in the string will be equal to centripetal force

[tex]F_c=m\omega ^2L=T[/tex]

[tex]126=0.51\times (\omega )^2\times 0.827[/tex]

[tex]\omega =\sqrt{298.74}[/tex]

[tex]\omega =17.28 rad/s[/tex]

[tex]\omega =2\pi f[/tex]

[tex]f=\frac{2\pi }{\omega }[/tex]

[tex]f=2.74 Hz[/tex]

To find the highest frequency at which the puck can go around the circle without the cord breaking, we use the formula for tension in a circular motion and solve for velocity. Then we use the velocity to find the frequency. The highest frequency is approximately 2.18 Hz.

To determine the highest frequency at which the puck can go around the circle without the cord breaking, we need to find the maximum tension in the cord.

Since the tension in the cord is equal to the centripetal force required to keep the puck moving in a circle, we can use the formula:

Tension = mass × velocity² / radius

Substituting the given values, we get:

126N = 0.51kg × v² / 0.827m

Now, solving for v, we find:

v² = (126N × 0.827m) / 0.51kg
v² = 204.2 m²/s²

v = √(204.2 m²/s²) = 14.29 m/s

Since the frequency of an object moving in a circle is equal to its velocity divided by the circumference of the circle, we can calculate the highest frequency as:

Frequency = v / (2πr)
Frequency = 14.29 m/s / (2π × 0.827m)
Frequency ≈ 2.18 Hz

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Answer:

Pressure, [tex]P=2.05\times 10^6\ Pa[/tex]

Explanation:

Given that,

The combined mass of the man and chair is 95.0 kg, m = 95 kg

The radius of the circular leg of the chair, r = 0.6 cm

Area of the 4 legs of the chair, [tex]A=4\times \pi r^2[/tex]

Let P is the pressure each leg exert on the floor. The total force acting per unit area is called pressure exerted. Its expression is given by :

[tex]P=\dfrac{F}{A}[/tex]

[tex]P=\dfrac{mg}{4\times \pi r^2}[/tex]

[tex]P=\dfrac{95\times 9.8}{4\times \pi (0.6\times 10^{-2})^2}[/tex]

[tex]P=2.05\times 10^6\ Pa[/tex]

So, the pressure exerted by each leg on the floor is [tex]2.05\times 10^6\ Pa[/tex]. Hence, this is the required solution.

To calculate the pressure each chair leg exerts on the floor, convert the combined mass of the man and chair into force using the equation F = mg, and then calculate the area of each chair leg using the formula A = πr². Finally, divide the total force by the number of legs and then that value by the area of each leg.

The key to solving this problem is recognizing that the pressure exerted by the man and chair on each chair leg is equal to the force of their combined weight divided by the area of contact the chair legs make with the floor. First, convert the combined mass of the man and chair (95.0 kg) into force in newtons, using the equation F = mg, where F is the force in Newtons, m is the mass in kilograms, and g is the acceleration due to gravity (9.8 m/s²). In this case, F = 95.0 kg * 9.8 m/s² = 931 N.

Next, calculate the area of each chair leg that is in contact with the floor. Since each leg is circular, its area, A, can be calculated using the formula A = πr², where r is the radius. Remember to convert the radius from centimeters to meters. Therefore, A = π*(0.006 m)² = 0.000113 m².

Finally, divide the total force by four (since the weight is distributed evenly over four legs) and then that value by the area of each leg to find the pressure each leg exerts: P = F/(4*A) = 931 N / 4 / 0.000113 m² = 2063363.4 Pascal. Remember to use the correct units of pressure (Pascal).

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Answer:

$0.662772

Explanation:

v = Volume of can = 16 fl oz.

[tex]1\ floz.=1.805\ in^3[/tex]

r = Radius of can

h = Height of can = 3r

Volume of cylinder is given by

[tex]\pi r^2h=16\times 1.805\\\Rightarrow \pi r^23r=16\times 1.805\\\Rightarrow 3\pi r^3=16\times 1.805\\\Rightarrow r=\left(\frac{16\times 1.805}{3\times 3.14}\right)^{\frac{1}{3}}\\\Rightarrow r=1.45247\ in[/tex]

h=3r\\\Rightarrow h=3\times 1.45247\\\Rightarrow h=4.35741\ in[/tex]

Surface area of sides is given by

[tex]2\pi rh\\ =2\times 3.14\times 1.45247\times 4.35741\\ =39.76632\ in^2[/tex]

Surface area of top and bottom is given by

[tex]2\pi r^2\\ =2\times 3.14\times 1.45247^2\\ =13.25544\ in^2[/tex]

Cost of making the can will be

[tex]39.76632\times 0.01+13.25544\times 0.02=\$0.662772[/tex]

The cost to make the can is $0.662772

Answer:

6.87 ft/s is the rate at which the top of ladder slides down.

Explanation:

Given:

Length of the ladder is, [tex]L=20\ ft[/tex]

Let the top of ladder be at height of 'h' and the bottom of the ladder be at a distance of 'b' from the wall.

Now, from triangle ABC,

AB² + BC² = AC²

[tex]h^2+b^2=L^2\\h^2+b^2=20^2\\h^2+b^2=400----1[/tex]

Differentiating the above equation with respect to time, 't'. This gives,

[tex]\frac{d}{dt}(h^2+b^2)=\frac{d}{dt}(400)\\\\\frac{d}{dt}(h^2)+\frac{d}{dt}(b^2)=0\\\\2h\frac{dh}{dt}+2b\frac{db}{dt}=0\\\\h\frac{dh}{dt}+b\frac{db}{dt}=0--------2[/tex]

In the above equation the term [tex]\frac{dh}{dt}[/tex] is the rate at which top of ladder slides down and [tex]\frac{db}{dt}[/tex] is the rate at which bottom of ladder slides away.

Now, as per question, [tex]h=8\ ft, \frac{db}{dt}=3\ ft/s[/tex]

Plug in [tex]h=8[/tex] in equation (1) and solve for [tex]b[/tex]. This gives,

[tex]8^2+b^2=400\\64+b^2=400\\b^2=400-64\\b^2=336\\b=\sqrt{336}=18.33\ ft[/tex]

Now, plug in all the given values in equation (2) and solve for [tex]\frac{dh}{dt}[/tex]

[tex]8\times \frac{dh}{dt}+18.33\times 3=0\\8\times \frac{dh}{dt}+54.99=0\\8\times \frac{dh}{dt}=-54.99\\ \frac{dh}{dt}=-\frac{54.99}{8}=-6.87\ ft/s[/tex]

Therefore, the rate at which the top of ladder slide down is 6.87 ft/s. The negative sign implies that the height is reducing with time which is true because it is sliding down.

Answer:

Force per unit length between two conductors will be [tex]3.2\times 10^{-4}N[/tex]

Explanation:

We have given that two long parallel wires each carrying a current of 12 A

So [tex]I_1=I_2=12A[/tex]

Distance between the two conductors d = 9 cm = 0.09 m

We know that magnetic force between two parallel conductors per unit length is given by

[tex]F=\frac{\mu _0I_1I_2}{2\pi d}=\frac{4\times 3.14\times 10^{-7}\times 12\times 12}{2\times 3.14\times 0.09}=3.2\times 10^{-4}N[/tex]

Force per unit length between two conductors will be [tex]3.2\times 10^{-4}N[/tex]

Answer:

632.5 MPa

Explanation:

[tex]\sigma_{m}[/tex] = Matrix stress at fiber failure = 10 MPa

[tex]V_f[/tex] = Volume fraction of fiber = 0.25

[tex]\sigma_f[/tex] = Fiber fracture strength = 2.5 GPa

The longitudinal strength of a composite is given by

[tex]\sigma_{cl}=\sigma_{m}(1-V_f)+\sigma_fV_f\\\Rightarrow \sigma_{cl}=10(1-0.25)+(2.5\times 10^3)\times 0.25\\\Rightarrow \sigma_{cl}=632.5\ MPa[/tex]

The longitudinal strength of the aligned carbon fiber-epoxy matrix composite is 632.5 MPa

Final answer:

The given question seeks to calculate the longitudinal strength of a carbon fiber-epoxy composite, but lacks sufficient detail or formulae for a complete answer. Typically, this computation would involve using materials science models that consider fiber orientation and other stress-strain interactions between components.

Explanation:

To compute the longitudinal strength of an aligned carbon fiber-epoxy matrix composite with a 0.25 volume fraction of fibers, we'd need to consider the following given properties: the average fiber diameter, average fiber length, fiber fracture strength, fiber-matrix bond strength, matrix stress at fiber failure, and matrix tensile strength.

Although the exact method for calculating the longitudinal strength would typically involve applying principles from materials science, such as the rule of mixtures, in combination with the given data points, the actual question does not provide enough information or a specific formula to complete the calculation. For a real-life carbon fiber-epoxy composite, the longitudinal strength could be substantially influenced by the alignment of the fibers, bond quality between the fibers and matrix, and the interaction between the stress and strain of the components.

If we had a suitable model or empirical formula, we would proceed by plugging in the given values to determine the longitudinal strength. However, as the provided data from the question is incomplete for this calculation, it is recommended to refer to a textbook or comprehensive resource on composite material mechanics for the detailed step-by-step methodology and equations.

a) The initial angular speed is 209.3 m/s

b) The angular acceleration is [tex]-1.74 rad/s^2[/tex]

c) The angular speed after 40 s is 139.7 rad/s

d) The wheel makes 1501 revolutions

Explanation:

a)

The initial angular speed of the wheel is

[tex]\omega_i = 2000 rpm[/tex]

which means 2000 revolutions per minute.

We have to convert it into rad/s. Keeping in mind that:

[tex]1 rev = 2\pi rad[/tex]

[tex]1 min = 60 s[/tex]

We find:

[tex]\omega_i = 2000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=209.3 rad/s[/tex]

b)

To find the angular acceleration, we have to convert the final angular speed also from rev/min to rad/s.

Using the same procedure used in part a),

[tex]\omega_f = 1000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=104.7 rad/s[/tex]

Now we can find the angular acceleration, given by

[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]

where

[tex]\omega_i = 209.3 rad/s[/tex] is the initial angular speed

[tex]\omega_f = 104.7 rad/s[/tex] is the final angular speed

t = 60 s is the time interval

Substituting,

[tex]\alpha = \frac{104.7-209.3}{60}=-1.74  rad/s^2[/tex]

c)

To find the angular speed 40 seconds after the initial moment, we use the equivalent of the suvat equations for circular motion:

[tex]\omega' = \omega_i + \alpha t[/tex]

where we have

[tex]\omega_i = 209.3 rad/s[/tex]

[tex]\alpha = -1.74 rad/s^2[/tex]

And substituting t = 40 s, we find

[tex]\omega' = 209.3 + (-1.74)(40)=139.7 rad/s[/tex]

d)

The angular displacement of the wheel in a certain time interval t is given by

[tex]\theta=\omega_i t + \frac{1}{2}\alpha t^2[/tex]

where

[tex]\omega_i = 209.3 rad/s[/tex]

[tex]\alpha = -1.74 rad/s^2[/tex]

And substituting t = 60 s, we find:

[tex]\theta=(209.3)(60) + \frac{1}{2}(-1.74)(60)^2=9426 rad[/tex]

So, the wheel turns 9426 radians in the 60 seconds of slowing down. Converting this value into revolutions,

[tex]\theta = \frac{9426 rad}{2\pi rad/rev}=1501 rev[/tex]

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Answer:x=23.4 cm

Explanation:

Given

mass of block [tex]m=0.5 kg[/tex]

inclination [tex]\theta =30[/tex]

coefficient of static friction [tex]\mu =0.35[/tex]

coefficient of kinetic friction [tex]\mu _k=0.25[/tex]

distance traveled [tex]d=77.3 cm[/tex]

spring constant [tex]k=35 N/m [/tex]

work done by gravity+work done by friction=Energy stored in Spring

[tex]mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}[/tex]

[tex]mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}[/tex]

[tex]0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}[/tex]

[tex]x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}[/tex]

[tex]x=0.234 m[/tex]

[tex]x=23.4 cm[/tex]

Explanation:

Given that,

Frequency of the AM radio station, [tex]f=1020\ kHz=1020\times 10^3\ Hz[/tex]

(a) Let [tex]\lambda[/tex] is the wavelength of this electromagnetic radiation. It can be calculated as :

[tex]\lambda=\dfrac{c}{f}[/tex]

[tex]\lambda=\dfrac{3\times 10^8\ m/s}{1020\times 10^3\ Hz}[/tex]

[tex]\lambda=294.11\ m[/tex]

Since the wavelength of one cycle is 294.12 m, then the total number of cycles over a 3 km distance is:

[tex]n=\dfrac{3000}{294.12}=10.19\ cycles[/tex]

Let the period is the duration of one cycle is given by:

[tex]T=\dfrac{1}{f}[/tex]

[tex]T=\dfrac{1}{1020\times 10^3}[/tex]

[tex]T=9.8\times 10^{-7}\ s[/tex]

So, total time required is t as :

[tex]t=n\times T[/tex]

[tex]t=10.19\times 9.8\times 10^{-7}[/tex]

[tex]t=9.98\times 10^{-6}\ s[/tex]

The wavelength of the electromagnetic radiation from the AM radio station is approximately 293.1 meters. It takes approximately 10 milliseconds for the radiation to propagate from the broadcasting antenna to a radio 3 km away.

The wavelength of electromagnetic radiation can be calculated using the formula:

wavelength = speed of light / frequency

Given that the frequency of the AM radio station is 1020 kilohertz and the speed of light is approximately 3 x 10^8 meters per second, we can plug these values into the formula to find the wavelength:

wavelength = (3 x 10^8 m/s) / (1020 x 10^3 Hz)

Simplifying the expression gives us a wavelength of approximately 293.1 meters.

To calculate the time required for the radiation to propagate from the broadcasting antenna to a radio 3 km away, we can use the formula:

time = distance / speed

Given that the distance is 3 km and the speed of light is approximately 3 x 10^8 meters per second, we can plug these values into the formula:

time = (3 km) / (3 x 10^8 m/s)

Converting kilometers to meters gives us a distance of 3000 meters:

time = (3000 m) / (3 x 10^8 m/s)

Simplifying the expression gives us a time of 0.01 seconds or 10 milliseconds.

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Answer:

Speed of the water that emerge out of the pipe is 18.8 m/s

Explanation:

Since we know that water drops projected upwards to maximum height of 18 m

So here we can use kinematics equations here

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

here we have

[tex]v_f = 0[/tex]

[tex]d = 18 m[/tex]

[tex]a = -9.80 m/s^2[/tex]

so we will have

[tex]0 - v_i^2 = 2(-9.80)(18)[/tex]

[tex]v_i = 18.8 m/s[/tex]

To develop the problem it is necessary to take into account the concepts related to Torque and sum of moments.

By torque it is understood that

[tex]\tau = F*d[/tex]

Where,

F= Force

d = Distance

The value of the given Torque acts from the center of mass causing it to rotate clockwise.

The Force must then be located at the other end down to make a movement opposite the Torque in the center of mass.

I enclose a graph that allows us to understand the problem in a more didactic way.

The correct answer is D.

Answer:

low pressure

Explanation:

The streamlines of air particles are squeezed together as the air goes over the top of the bulb. Then, by the law of conservation of mass, the velocity of air particles are increased. And since, the velocity is increase the pressure is bound to decrease. Hence, this leads to region of low pressure on the top of the bulb.

Answer:

(a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is [tex]1.022\times10^{-16}\ J[/tex]

(c). The work done on the proton is [tex]-8\times10^{-18}\ J[/tex].

Explanation:

Given that,

Speed [tex]v= 3.50\times10^{5}\ m/s[/tex]

Initial potential V=100 V

Final potential = 150 V

(a). We need to calculate the change in the protons electric potential

Potential energy of the proton is

[tex]U=qV=eV[/tex]

Using conservation of energy

[tex]K_{i}+U_{i}=K_{f}+U_{f}[/tex]

[tex]\dfrac{1}{2}mv_{i}^2+eV_{i}=\dfrac{1}{2}mv_{f}^2+eV_{f}[/tex]

[tex]]\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e(V_{f}-V_{i})[/tex]

[tex]\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e\Delta V[/tex]

[tex]\Delta V=\dfrac{m(v_{i}^2-v_{f}^2)}{2e}[/tex]

Put the value into the formula

[tex]\Delta V=\dfrac{1.67\times10^{-27}(3.50\times10^{5}-0)^2}{2\times1.6\times10^{-19}}[/tex]

[tex]\Delta V=639.2=0.639\ kV[/tex]

(b). We need to calculate the change in the potential energy of the proton

Using formula of potential energy

[tex]\Delta U=q\Delta V[/tex]

Put the value into the formula

[tex]\Delta U=1.6\times10^{-19}\times639.2[/tex]

[tex]\Delta U=1.022\times10^{-16}\ J[/tex]

(c). We need to calculate the work done on the proton

Using formula of work done

[tex]\Delta U=-W[/tex]

[tex]W=q(V_{2}-V_{1})[/tex]

[tex]W=-1.6\times10^{-19}(150-100)[/tex]

[tex]W=-8\times10^{-18}\ J[/tex]

Hence, (a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is [tex]1.022\times10^{-16}\ J[/tex]

(c). The work done on the proton is [tex]-8\times10^{-18}\ J[/tex].

Final answer:

The change in electric potential and potential energy of the proton can be calculated based on the provided potentials. The work done on the proton equals the change in potential energy.

Explanation:

The change in the proton's electric potential: The change in electric potential is the final potential minus the initial potential, thus the change is -150 V - 100 V = -250 V.

The change in potential energy of the proton: The potential energy change equals the charge of the proton times the change in potential, giving -proton charge x change in potential.

The work done on the proton: The work done is equal to the change in the potential energy of the proton.

a) The spring constant is 12,103 N/m

b) The mass of the trailer 2,678 kg

c) The frequency of oscillation is 0.478 Hz

d) The time taken for 10 oscillations is 20.9 s

Explanation:

a)

When the two children jumps on board of the trailer, the two springs compresses by a certain amount

[tex]\Delta x = 6.17 cm = 0.0617 m[/tex]

Since the system is then in equilibrium, the restoring force of the two-spring system must be equal to the weight of the children, so we can write:

[tex]2mg = k'\Delta x[/tex] (1)

where

m = 76.2 kg is the mass of each children

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity

[tex]k'[/tex] is the equivalent spring constant of the 2-spring system

For two springs in parallel each with constant k,

[tex]k'=k+k=2k[/tex]

Substituting into (1) and solving for k, we find:

[tex]2mg=2k\Delta x\\k=\frac{mg}{\Delta x}=\frac{(76.2)(9.8)}{0.0617}=12,103 N/m[/tex]

b)

The period of the oscillating system is given by

[tex]T=2\pi \sqrt{\frac{m}{k'}}[/tex]

where

And for the system in the problem, we know that

T = 2.09 s is the period of oscillation

m is the mass of the trailer

[tex]k'=2k=2(12,103)=24,206 N/m[/tex] is the equivalent spring constant of the system

Solving the equation for m, we find the mass of the trailer:

[tex]m=(\frac{T}{2\pi})^2 k'=(\frac{2.09}{2\pi})^2 (24,206)=2,678 kg[/tex]

c)

The frequency of oscillation of a spring-mass system is equal to the reciprocal of the period, therefore:

[tex]f=\frac{1}{T}[/tex]

where

f is the frequency

T is the period

In  this problem, we have

T = 2.09 s is the period

Therefore, the frequency of oscillation is

[tex]f=\frac{1}{2.09}=0.478 Hz[/tex]

d)

The period of the system is

T = 2.09 s

And this time is the time it takes for the trailer to complete one oscillation.

In this case, we want to find the time it takes for the trailer to complete 10 oscillations (bouncing up and down 10 times). Therefore, the time taken will be the period of oscillation multiplied by 10.

Therefore, the time needed for 10 oscillations is:

[tex]t=10T=10(2.09)=20.9 s[/tex]

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The distance [tex]\( d_1 \)[/tex] from the pivot point to the center of mass of the meter stick with one coin on the 0-cm end, maintaining static equilibrium, is approximately 46.8 cm.

1. To achieve static equilibrium, the torques on both sides of the pivot point must balance out.  

2. The torque due to the meter stick with mass ( M ) is [tex]\( M \times g \times \frac{L}{2} \)[/tex], where ( L ) is the length of the meter stick (100 cm) and \( g \) is the acceleration due to gravity (approximately [tex]\( 9.8 \, \text{m/s}^2 \)[/tex]).

3. The torque due to the coin on the 0-cm end is [tex]\( m \times g \times d_1 \)[/tex], where [tex]\( m \)[/tex] is the mass of the coin.

4. Since the torques balance out, we have the equation: [tex]\( M \times g \times \frac{L}{2} = m \times g \times d_1 \).[/tex]

5. Rearrange the equation to solve for [tex]\( d_1 \): \( d_1 = \frac{M \times \frac{L}{2}}{m} \).[/tex]

6. Substitute the given values: [tex]\( d_1 = \frac{95 \, \text{g} \times \frac{100}{2} \, \text{cm}}{3.1 \, \text{g}} \).[/tex]

7. Calculate [tex]\( d_1 \)[/tex]to find the distance from the pivot point to the center of mass of the meter stick with one coin on the 0-cm end, which is approximately 46.8 cm.

Answer:

E. radiation.

Explanation:

As we know that heat transfer due to conduction depends on thermal conductivity of the materials and heat transfer due to convection depends on the velocity of the fluid.But on the other hand heat transfer due to radiation depends on the surface properties like  emmisivity .So when bottle is silvered then it will leads to minimize the radiation heat transfer.

Therefore answer is --

E. radiation.

The answer is option C.

The interior of a thermos is silvered to minimize heat transfer by conduction, convection, and radiation. The silvering acts like a mirror, reflecting heat, and the vacuum between the thermos walls almost eliminates conduction and convection.

The interior of a thermos bottle is silvered to minimize heat transfer due to C. conduction, convection and radiation.

This is because the silvering on the inner surface of the thermos acts like a mirror, reducing the amount of heat that can be transferred by all three modes: conduction, convection, and especially radiation.

Conduction is the transfer of heat through direct contact of molecules, minimized in a thermos by the vacuum between its double walls. Convection is the transfer of heat in a fluid (like air or liquid) through the motion of the fluid itself, which is also nearly eliminated by the vacuum. Radiation is the transfer of heat through electromagnetic waves, which the silvering reflects back, greatly reducing heat loss this way.

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Answer:

0.256 hours

Explanation:

Vectors in the plane

We know Office A is walking at 5 mph directly south. Let [tex]X_A[/tex] be its distance. In t hours he has walked

[tex]X_A=5t\ \text{miles}[/tex]

Office B is walking at 6 mph directly west. In t hours his distance is

[tex]X_B=6t\ \text{miles}[/tex]

Since both directions are 90 degrees apart, the distance between them is the hypotenuse of a triangle which sides are the distances of each office

[tex]D=\sqrt{X_A^2+X_B^2}[/tex]

[tex]D=\sqrt{(5t)^2+(6t)^2}[/tex]

[tex]D=\sqrt{61}t[/tex]

This distance is known to be 2 miles, so

[tex]\sqrt{61}t=2[/tex]

[tex]t =\frac{2}{\sqrt{61}}=0.256\ hours[/tex]

t is approximately 15 minutes

Answer:

Velocity will be [tex]v=1.291\times 10^8m/sec[/tex]

Explanation:

We have given magnetic field B = 0.00005 T

Mass m = 238 U

We know that [tex]1u=1.66\times 10^{-27}kg[/tex]

So 238 U [tex]=238\times 1.66\times 10^{-27}=395.08\times 10^{-27}kg[/tex]

Radius [tex]=R+1.44=6378+1.44=6379.44KM[/tex]

We know that magnetic force is given by

[tex]F=qvB[/tex] which is equal to the centripetal force

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

[tex]1.6\times 10^{-19}\times v\times 0.00005=\frac{395.08\times 10^{-27}v^2}{6379.44}[/tex]

[tex]v=1.291\times 10^8m/sec[/tex]

Based on the motion after the collision, cart A is concluded to be hollow.

Let's analyze the situation step by step.

Now, let's consider the possible scenarios:

Therefore, based on the information provided, the conclusion is: (A) Cart A is hollow.

Answer:

9.7 x 10¹¹ .

Explanation:

2.5 % of 110 W = 2.75 J/s

energy of one photon

= hc / λ

=[tex]\frac{6.6\times10^{-34}\times3\times10^8}{550\times10^{-9}}[/tex]

= .036 x 10⁻¹⁷ J

No of photons emitted

= 2.75 / .036 x 10⁻¹⁷

= 76.38 x 10¹⁷

Now photons are uniformly distributed in all directions so they will pass through a spherical surface of radius 2.8 m at this distance

photons passing per unit area of this sphere

= 76.38 x 10¹⁷  / 4π ( 2.8)²

Through eye which has surface area of π x ( 2 x 10⁻² )² m² , no of photons passing

= [tex]\frac{76.38\times10^{17}}{4\pi\times(2.8)^2} \times\pi(2\times10^{-3})^2[/tex]

= 9.7 x 10¹¹ .

Answer:

[tex]x|_0^{0.2}=1.59535[/tex]

Explanation:

Given expression of velocity:

[tex]v(t)=10^{0.2t}-1 ;\ \ 0\leq t\leq 5\ s[/tex]

For getting displacement we need to integrate the above function with respect to t.

Given period of integration:

[tex]t_0=0\ s \to t_f=0.2\ s[/tex]

For trapezoidal rule we break the given interval into two parts of 0.1 s each.

∴take n=2

hence, [tex]\Delta t= 0.1[/tex]

[tex]v(0)=0[/tex]

[tex]v(0.1)=1.0471[/tex]

[tex]v(0.2)=1.0965[/tex]

Now, using trapezoidal rule:

[tex]\int_{0}^{0.2}v(t)\ dt=\Delta x[\frac{1}{2}\times v(0)+v(0.1)+\frac{1}{2}\times v(0.2)][/tex]

[tex]\int_{0}^{0.2}v(t)\ dt=0.1 [\frac{1}{2}\times 0+1.0471+\frac{1}{2}\times 1.0965][/tex]

[tex]x|_0^{0.2}=1.59535[/tex]

Note:Smaller the value of sub-interval better is the accuracy.

Answer:

The approximate combined sound  intensity is [tex]I_{T}=1.1\times10^{-4}W/m^{2}[/tex]

Explanation:

The decibel  scale intensity for busy traffic is 80 dB. so intensity will be

[tex]10log(\frac{I_{1}}{I_{0}} )=80[/tex], therefore [tex]I_{1}=1\times10^{8}I_{0}=1\times10^{8} * 1\times10^{-12}W/m^{2}=1\times10^{-4}W/m^{2}[/tex]

In the same way for the loud conversation having a decibel intensity of 70 dB.

[tex]10log(\frac{I_{2}}{I_{0}} )=70[/tex], therefore [tex]I_{2}=1\times10^{7}I_{0}=1\times10^{7} * 1\times10^{-12}W/m^{2}=1\times10^{-5}W/m^{2}[/tex]

Finally we add both of them [tex]I_{T}=I_{1}+I_{2}=1\times10^{-4}W/m^{2}+1\times10^{-5}W/m^{2}=1.1\times10^{-4}W/m^{2}[/tex], is the approximate combined sound  intensity.

Answer:

(I). The effective cross sectional area of the capillaries is 0.188 m².

(II). The approximate number of capillaries is [tex]3.74\times10^{9}[/tex]

Explanation:

Given that,

Radius of aorta = 10 mm

Speed = 300 mm/s

Radius of capillary [tex]r=4\times10^{-3}\ mm[/tex]

Speed of blood [tex]v=5\times10^{-4}\ m/s[/tex]

(I). We need to calculate the effective cross sectional area of the capillaries

Using continuity equation

[tex]A_{1}v_{1}=A_{2}v_{2}[/tex]

Where. v₁ = speed of blood in capillarity

A₂ = area of cross section of aorta

v₂ =speed of blood in aorta

Put the value into the formula

[tex]A_{1}=A_{2}\times\dfrac{v_{2}}{v_{1}}[/tex]

[tex]A_{1}=\pi\times(10\times10^{-3})^2\times\dfrac{300\times10^{-3}}{5\times10^{-4}}[/tex]

[tex]A_{1}=0.188\ m^2[/tex]

(II). We need to calculate the approximate number of capillaries

Using formula of area of cross section

[tex]A_{1}=N\pi r_{c}^2[/tex]

[tex]N=\dfrac{A_{1}}{\pi\times r_{c}^2}[/tex]

Put the value into the formula

[tex]N=\dfrac{0.188}{\pi\times(4\times10^{-6})^2}[/tex]

[tex]N=3.74\times10^{9}[/tex]

Hence, (I). The effective cross sectional area of the capillaries is 0.188 m².

(II). The approximate number of capillaries is [tex]3.74\times10^{9}[/tex]

Answer:

[tex]M=6.4243\ g[/tex]

Explanation:

Given:

To stop this balloon from rising up we need to counter the buoyant force.

mass of balloon after inflation:

[tex]m=m_h+m_b[/tex]

[tex]m=0.0084\times 0.180+0.0029[/tex]

[tex]m=0.004412\ kg[/tex]

Now the density of inflated balloon:

[tex]\rho_b=\frac{m}{V}[/tex]

[tex]\rho_b=\frac{0.004412}{0.0084}[/tex]

[tex]\rho_b=0.5252\ kg.m^{-3}[/tex]

Now the buoyant force on balloon

[tex]F_B=V(\rho_a-\rho_b).g[/tex]

[tex]F_B=0.0084(1.29-0.5252)\times 9.8[/tex]

[tex]F_B=0.063\ N[/tex]

∴Mass to be hung:

[tex]M=\frac{F_B}{g}[/tex]

[tex]M=0.00642432\ kg[/tex]

[tex]M=6.4243\ g[/tex]

Answer:

  r = 3.787 10¹¹ m

Explanation:

We can solve this exercise using Newton's second law, where force is the force of universal attraction and centripetal acceleration

    F = ma

    G m M / r² = m a

The centripetal acceleration is given by

    a = v² / r

For the case of an orbit the speed circulates (velocity module is constant), let's use the relationship

    v = d / t

The distance traveled Esla orbits, in a circle the distance is

    d = 2 π r

Time in time to complete the orbit, called period

     v = 2π r / T

Let's replace

    G m M / r² = m a

    G M / r² = (2π r / T)² / r

    G M / r² = 4π² r / T²

    G M T² = 4π² r3

     r = ∛ (G M T² / 4π²)

Let's reduce the magnitudes to the SI system

     T = 3.27 and (365 d / 1 y) (24 h / 1 day) (3600s / 1h)

     T = 1.03 10⁸ s

Let's calculate

      r = ∛[6.67 10⁻¹¹ 3.03 10³⁰ (1.03 10⁸) 2) / 4π²2]

      r = ∛ (21.44 10³⁵ / 39.478)

      r = ∛(0.0543087 10 36)

      r = 0.3787 10¹² m

      r = 3.787 10¹¹ m

Explanation:

Given that,

The disintegration constant of the nuclide, [tex]\lambda=0.0178\ h^{-1}[/tex]

(a) The half life of this nuclide is given by :

[tex]t_{1/2}=\dfrac{ln(2)}{\lambda}[/tex]

[tex]t_{1/2}=\dfrac{ln(2)}{0.0178}[/tex]

[tex]t_{1/2}=38.94\ h[/tex]

(b) The decay equation of any radioactive nuclide is given by :

[tex]N=N_oe^{-\lambda t}[/tex]

[tex]\dfrac{N}{N_o}=e^{-\lambda t}[/tex]

Number of remaining sample in 4.44 half lives is :

[tex]t_{1/2}=4.44\times 38.94[/tex]

[tex]t_{1/2}=172.89\ h^{-1}[/tex]

So, [tex]\dfrac{N}{N_o}=e^{-0.0178\times 172.89}[/tex]

[tex]\dfrac{N}{N_o}=0.046[/tex]

(c) Number of remaining sample in 14.6 days is :

[tex]t_{1/2}=14.6\times 24[/tex]

[tex]t_{1/2}=350.4\ h^{-1}[/tex]

So, [tex]\dfrac{N}{N_o}=e^{-0.0178\times 350.4}[/tex]

[tex]\dfrac{N}{N_o}=0.0019[/tex]

Hence, this is the required solution.

Answer:

The amplitude of the resultant wave are

(a). 0.0772 m

(b). 0.0692 m

Explanation:

Given that,

Frequency = 135 Hz

Wavelength = 2 cm

Amplitude = 0.04 m

We need to calculate the angular frequency

[tex]\omega=2\pi f[/tex]

[tex]\omega=2\times\pi\times135[/tex]

[tex]\omega=848.23\ rad/s[/tex]

As the two waves are identical except in their phase,

The amplitude of the resultant wave is given by

[tex]y+y=A\sin(kx-\omega t)+Asin(kx-\omega t+\phi)[/tex]

[tex]y+y=A[2\sin(kx-\omega t+\dfrac{\phi}{2})\cos\phi\dfrac{\phi}{2}[/tex]

[tex]y'=2A\cos(\dfrac{\phi}{2})\sin(kx-\omega t+\dfrac{\phi}{2})[/tex]

(a). We need to calculate the amplitude of the resultant wave

For [tex]\phi =\dfrac{\pi}{6}[/tex]

The amplitude of the resultant wave is

[tex]A'=2A\cos(\dfrac{\phi}{2})[/tex]

Put the value into the formula

[tex]A'=2\times0.04\cos(\dfrac{\pi}{12})[/tex]

[tex]A'=0.0772\ m[/tex]

(b), We need to calculate the amplitude of the resultant wave

For [tex]\phi =\dfrac{\pi}{3}[/tex]

[tex]A'=2\times0.04\cos(\dfrac{\pi}{6})[/tex]

[tex]A'=0.0692\ m[/tex]

Hence, The amplitude of the resultant wave are

(a). 0.0772 m

(b). 0.0692 m

An excited atom can return to its ground state by absorbing electromagnetic radiation is false about the electromagnetic radiation.

Option B

Explanation:

In the scope of modern quantum theory, the term Electromagnetic radiation is identified as the movement of photons through space. Almost all the sources of energy that we utilize today such as coal, oil, etc are a product of electromagnetic radiation which was absorbed from the sun millions of years ago.

Various properties of electromagnetic radiations are a directly proportional relationship between the energy and the frequency, Inverse proportionality between frequency and the wavelength, etc. Hence, we can conclude that an "excited atom" can never return to its ground state by assimilating electromagnetic radiation and the 2nd statement is false.

The statement 'An excited atom can return to its ground state by absorbing electromagnetic radiation' is false, as the process actually involves emitting radiation. Electromagnetic radiation's energy increases with frequency, and frequency and wavelength have an inverse relationship.

The false statement among the ones provided is: An excited atom can return to its ground state by absorbing electromagnetic radiation. An excited atom returns to its ground state by emitting radiation, not absorbing it. When it comes to electromagnetic radiation, increasing frequency indeed results in increasing energy. This is because the energy of electromagnetic radiation is directly proportional to its frequency. Additionally, when an electron in the n = 4 state in the hydrogen atom transitions to the n = 2 state, it emits radiation at an appropriate frequency. Lastly, the frequency and the wavelength of electromagnetic radiation are inversely proportional, meaning as one increases, the other decreases.