The place you get your hair cut has two nearly parallel mirrors 6.50 m apart. As you sit in the chair, your head is 3.00 m from the nearer mirror. Looking toward this mirror, you first see your face and then, farther away, the back of your head. (The mirrors need to be slightly nonparallel for you to be able to see the back of your head, but you can treat them as parallel in this problem.) How far away does the back of your head appear to be?

Answer:

The partial pressure of H2 is 0.375 atm

The partial pressure of Ne is also 0.375 atm

Explanation:

Mass of H2 = 1 g

Mass of Ne = 1 g

Mass of Ar = 1 g

Mass of Kr = 1 g

Total mass of gas mixture = 1 + 1 + 1 + 1 = 4 g

Pressure of sealed container = 1.5 atm

Partial pressure of H2 = (mass of H2/total mass of gas mixture) × pressure of sealed container = 1/4 × 1.5 = 0.375 atm

Partial pressure of Ne = (mass of Ne/total mass of gas mixture) × pressure of sealed container = 1/4 × 1.5 = 0.375 atm

The partial pressures of H2 and Ne in the mixture under ideal gas behavior are approximately 1.278 atm and 0.128 atm respectively. This is calculated using their mole fractions and the total pressure.

In this scenario, we're essentially dealing with ideal gas behavior. The partial pressure of a gas in a mixture can be determined by the mole fraction of the gas times the total pressure. The moles of each gas can be calcuated using the equation: moles = mass/molar mass. The molar masses of H2, Ne, Ar, and Kr are approximately 2 g/mol, 20 g/mol, 40 g/mol, and 84 g/mol, respectively.

For example, moles of H2 ≈ 1.000 g / 2 g/mol = 0.5 mol. Likewise, moles of Ne = 1.000 g / 20 g/mol = 0.05 mol, moles of Ar = 1.000 g / 40 g/mol = 0.025 mol, and moles of Kr = 1.000 g / 84 g/mol = 0.012 mol. The total moles of gas = 0.5 mol + 0.05 mol + 0.025 mol + 0.012 mol = 0.587 mol.

The mole fraction of H2 = 0.5 mol / 0.587 mol = 0.852. Therefore, the partial pressure of H2 = 0.852 * 1.5 atm = 1.278 atm. Applying the same calculations for Ne, the mole fraction of Ne = 0.05 mol / 0.587 mol = 0.085. Therefore, the partial pressure of Ne = 0.085 * 1.5 atm = 0.128 atm.

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Answer:

The coefficient of kinetic friction between the puck and the ice is 0.11

Explanation:

Given;

initial speed, u = 9.3 m/s

sliding distance, S = 42 m

From equation of motion we determine the acceleration;

v² = u² + 2as

0 = (9.3)² + (2x42)a

- 84a = 86.49

a = -86.49/84

|a| = 1.0296

[tex]F_k = \mu_k N[/tex] = ma

where;

Fk is the frictional force

μk is the coefficient of kinetic friction

N is the normal reaction = mg

μkmg = ma

μkg = a

μk = a/g

where;

g is the gravitational constant = 9.8 m/s²

μk = a/g

μk = 1.0296/9.8

μk = 0.11

Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11

Answer:

(c) 10.29 J

(d) 113.19 J

(e) 113.19 J

(f) 10061 N/m

Explanation:

15 cm = 0.15 m

Let g = 9.8 m/s2

(c) The work done by gravitational force is the product of gravity force and the distance compressed

[tex]E_p = mgx = 7*9.8*0.15 = 10.29 J[/tex]

(d) By using law of energy conservation with potential energy reference being 0 at the maximum compression point. As the ball falls and come to a stop at the compression point, its potential energy is transferred to elastic energy, which is the work that the mattress does on the ball:

[tex]E_p = E_e[/tex]

[tex]E_e = mgh[/tex]

where h = 1.5 + 0.15 = 1.65 m is the vertical distance that it falls.

[tex]E_e = 7*9.8*1.65 = 113.19 J[/tex]

(e) Before the compression, the potential energy of the mattress is 0. After the compression, the potential energy is 113.19J. So it has increased by 113.19J due to the potential energy transferred from the falling ball.

(f) [tex]E_e = 113.19 = kx^2/2[/tex]

[tex]k0.15^2/2 = 113.19[/tex]

[tex]k = 10061 N/m[/tex]

Answer:

(C) Wg = 113.2J

(D) Wm = 10.3J

(E) E = 1/2kx² + mgh where h is the height above the mattress and x is the compressed distance in the mattress.

(F) k = 457N/m.

Explanation:

See attachment below.

Answer:

The magnetic field intensity at the point P, equidistant from the wires is

9.566 x  [tex]10^{-5}[/tex] T

Explanation:

The magnetic field intensity can be obtained by resolving the vertical and horizontal components of the magnetic field. The step by step calculation is contained in the pictures attached;

Answer:

Squids = 450 - 490 nm (Moderate Frequency) (Blue)

Bees = 300 - 650 nm (Lower Frequency Bands)

Frogs = 280 - 580 nm (Very Low Frequency)

Explanation:

All of the above mentioned ranges are compared to that of humans.

I'm just surprised a little bit in the imagination that how these organisms see the world through their unique eyes. On the other hands, they are evolved like this just like we do so that may not be surprising enough. SIKE

Answer:

t=63s

tempersture gradient -2.36 x 10⁴

Explanation:

Answer:

The time required to penetrate the 40 cm clay layer is 126.82 years

Explanation:

Given:

Hydraulic gradient = 1 [tex]\frac{cm}{cm}[/tex]

Permeability of clay [tex]K = 1 \times 10^{-8}[/tex] [tex]\frac{cm}{sec}[/tex]

Time required to penetrate 40 cm clay layer,

  [tex]= \frac{40}{K}[/tex]

  [tex]= \frac{40}{1 \times 10^{-8} }[/tex]

  [tex]= 40 \times 10^{8}[/tex] sec

But we have to find time in years,

  [tex]= \frac{40 \times 10^{8} }{24 \times 60 \times 60 \times 365}[/tex] years

  [tex]= 126.82[/tex] years

Therefore, the time required to penetrate the 40 cm clay layer is 126.82 years

Answer:

its a

Explanation:

Answer:

(d) 14.7 J.

Explanation:

Using the equation,

W = mghsin∅.................  equation 1

Where W = work  done on  the cart by the force of gravity, g = acceleration due to gravity, h = height, ∅ = angle between the displacement and the force of gravity.

Given: m = 2 kg, g = 9.8 m/s², h = 1.5 m, ∅ = 30°

Substitute into equation 1

W = 2(9.8)(1.5)sin30

W = 2×9.8×1.5×0.5

W = 14.7 J.

The right option is (d) 14.7 J

The work done by gravity on a 2-kg cart rolling down a 1.5 m, 30° incline is 14.7 J, corresponding to option (d).

When an object slides down an inclined plane, the work done by gravity can be calculated using the formula W = mgh, where W is the work done, m is the mass of the object, g is the acceleration due to gravity (9.8 m/s2), and h is the vertical height. The vertical height can be determined from the incline's length and the angle of inclination using the sine function (h = l · sin(θ)). For a 2-kg cart rolling a distance of 1.5 m down a 30° incline, the work done by gravity would be W = m · g · l · sin(θ) = 2 kg · 9.8 m/s2 · 1.5 m · sin(30°) = 2 · 9.8 · 1.5 · 0.5 = 14.7 J, which corresponds to option (d).

Answer:

n = 1.7*10²² electrons.

Explanation:

       [tex]I =\frac{\Delta q}{\Delta t} \\ \\ \Delta q = I* \Delta t \\ \\ \Delta t = 5hs*\frac{3600s}{1h} = 18000 s[/tex]

       ⇒ Δq = I * Δt = 0.151 A * 18000 s = 2718 C

         Δq = n*e

        [tex]n = \frac{\Delta q}{e} =\frac{2718C}{1.6-19C} = 1.7e22 electrons.[/tex]

Explanation:

Below is an attachment containing the solution.

Answer:

1. Charge on plates (plates remain connected to battery) increases.

2. Electric potential energy (plates isolated from battery before inserting dielectric) decreases.

3.Capacitance (plates isolated from battery before inserting dielectric) increases.

4. Voltage between plates (plates remain connected to battery) remains the same.

5. Charge on plates (plates isolated from battery before inserting dielectric) remains the same.

6. Capacitance (plates remain connected to battery) increases.

7. Electric potential energy (plates remain connected to battery) increases.

8. Voltage between plates (plates isolated from battery before inserting dielectric) decreases.

Explanation:

When a dielectric material is inserted between the plates of a capacitor, the capacitance is increase by a factor K, the dielectric constant.

[tex]C = KC_0[/tex]

By the capacitance formula, the other factors change accordingly.

1. Charge on plates (plates remain connected to battery) increases, because charge and capacitance are directly proportional.

2. Electric potential energy (plates isolated from battery before inserting dielectric) decreases, because potential is inversely proportional to capacitance, and potential energy is given by the following formula

[tex]U = \frac{1}{2}CV^2[/tex]

3.Capacitance (plates isolated from battery before inserting dielectric) increases.

4. Voltage between plates (plates remain connected to battery) stays the same, because the voltage is applied by the battery.

5. Charge on plates (plates isolated from battery before inserting dielectric) remains constant. If the plates isolated from the battery, then the total charge is conserved.

6. Capacitance (plates remain connected to battery)  increases when a dielectric is inserted.

7. Electric potential energy (plates remain connected to battery)  increases, because when plates remain connected to battery the voltage remains the same. But the capacitance increases. Therefore, electric potential energy increases.

8. Voltage between plates (plates isolated from battery before inserting dielectric) decreases, because voltage is inversely proportional to capacitance.

Final answer:

When a dielectric is inserted into a capacitor, the capacitance increases regardless of whether the plates are connected to a battery or isolated. If connected to a battery, charge on the plates increases while voltage remains the same; if isolated, charge remains the same while voltage decreases.

Explanation:

When inserting a dielectric material into a parallel-plate capacitor, the effect on different parameters depends on whether the capacitor is connected to a battery or isolated:

Charge on plates (plates remain connected to battery): The charge on the plates will increase.

Electric potential energy (plates isolated from  the battery before inserting the dielectric): The electric potential energy will decrease due to the polarization of the dielectric.

Capacitance (plates isolated from the battery before inserting the dielectric): The capacitance will increase.

Voltage between plates (plates remain connected to battery): The voltage will remain the same, as it is maintained by the battery.

Charge on plates (plates isolated from battery before inserting dielectric): The charge will remain the same as isolation prevents any charge exchange.

Capacitance (plates remain connected to battery): The capacitance will increase.

Electric potential energy (plates remain connected to the battery): The electric potential energy will increase, as more charge is accumulated due to the dielectric.

Voltage between plates (plates isolated from battery before inserting dielectric): The voltage will decrease because the same charge is now across a larger capacitance.

It is important to note that the introduction of a dielectric material affects the capacitance of a parallel-plate capacitor by reducing the effective electric field and permitting greater charge storage capability for a given voltage.

Answer: The statement is TRUE.

Explanation: The Newton's third law states that a body that experiments an external force responds with an opposite force with same magnitude. Therefore, the statement is true.

Answer:

[tex]F=800N[/tex]

Explanation:

The average force is defined as the mass of the body multiplied by its average velocity over the contact time. According to the third Newton's law,  the magnitude of the average force exerted on the wall by the ball is equal to  the magnitude of the average force exerted on the ball by the wall. Thus:

[tex]F=m\frac{v_f-v_i}{t}\\F=0.80kg\frac{25\frac{m}{s}-(-25\frac{m}{s})}{0.05s}\\F=800N[/tex]

Answer:

800 N

Explanation:

Force: This can be defined as the product of mass and acceleration. The S.I unit of force is Newton(N).

From the question,

F = m(v-u)/t ....................... Equation 1

Where F = force exerted on the wall by the ball, m = mass of the ball, v = final velocity, u = initial velocity, t = time.

Note: Assuming the direction of the initial motion to be negative

Given: m = 0.80 kg, v = 25 m/s ( bounce back), u = -25 m/s, t = 0.05 s

Substitute into equation 1

F = 0.8[25-(25)]/0.05

F = 0.8(50)/0.05

F = 0.8(1000)

F = 800 N

Answer: [tex]V_{\epsilon}\propto \frac{d\phi_{B}}{dt}[/tex]

Explanation:

A direct proportionality means a linear relationship between two variables and rate of change means an application of derivatives. Hence, the mathematical model is:

[tex]V_{\epsilon}\propto \frac{d\phi_{B}}{dt}[/tex]

Faraday's Law of Induction, primarily focused on Physics, reveals that the induced electromotive force in a coil is proportional to the rate of change of magnetic flux through the coil and is also dependent on the number of coil turns.

Understanding Faraday's Law of Induction

Faraday's Law of Induction is a critical concept in electromagnetism, underlying the working principles of various electrical devices. The induced electromotive force (emf) is directly proportional to the time rate of change of magnetic flux. In simpler terms, when the magnetic environment of a circuit changes, an emf is induced. Faraday's Law is given by the equation EMF = -N(ΔΦ/Δt), where 'N' represents the number of turns in a coil, and (ΔΦ/Δt) is the rate of change of magnetic flux over time. This equation highlights three main factors influencing the induced EMF:

The negative sign in Faraday's Law is a reflection of Lenz's Law, which indicates that the induced emf always works to oppose the change in magnetic flux that produces it.

Answer:

[tex]CPP=52mm \ Hg[/tex]

Explanation:

Cerebral Perfusion Pressure is obtained by subtracting IntraCranial Pressure(ICP) from the Mean Arterial Pressure(MAP). Adequate cerebral perfusion requires a minimum goal of [tex]70mm \ Hg[/tex]. MAP is obtained using the formula:-

[tex]MAP=\frac{(diastolic \ blood \ pressure\times2)+\ systolic \ blood \ pressure)}{3}\\\\MAP=\frac{2\times60+90}{3}\\\\MAP=70mm \ Hg\\CPP=MAP-ICP\\CPP=70-18\\CPP=52mm \ Hg[/tex]

Answer:

1.6*10⁻¹⁹ J = 1 eV

Explanation:

        [tex]\Delta U_{e} = e*\Delta V[/tex]

        ⇒ [tex]\Delta U_{e} = e*\Delta V = 1.6e-19 C* 1.0 V = 1.6e-19 J = 1 eV[/tex]

Explanation:

Below is an attachment containing the solution.

Answer:

(a) L = 149.2 cm

(b) d = 0.033 cm

Explanation:

Note that the resistivity of copper is

[tex]\rho = 1.72\times 10^{-8}~{\rm \Omega . m} = 1.72\times 10^{-6}~{\rm \Omega.cm}[/tex]

and the mass density of copper is

[tex]d = 8.96~{\rm g/cm^3}[/tex]

We will use the following formula to relate the resistance to other parameters

[tex]R = \frac{\rho L}{A} = \frac{\rho L}{\pi r^2}[/tex]

If all the copper with 1.15 g is used, then

[tex]m = dV\\1.15 = 8.96 \times (L\pi r^2)\\L\pi r^2 = 0.128[/tex]

Back to the resistance:

[tex]0.3 = \frac{1.72\times 10^{-6} L}{\pi r^2}\\L = \pi r^2 (1.74\times 10^5)\\L = \frac{0.128}{L}(1.74\times 10^5)\\L = 149.2~{\rm cm}[/tex]

Then, the diameter is

[tex]149.2\times \pi r^2 = 0.128\\r = 0.0165~{\rm cm}\\d = 2r = 0.033~{\rm cm}[/tex]

Answer:

h = 4.281 m

Explanation

Given Data,

Weight of disk = 2kg

Speed of water jet (V) = 10 m/s

diameter of water jet = 25 mm

Cal. the velocity of the water jet as function height by applying Bernoulli's  eqtn of water surface to the jet

[tex]\frac{P}{\rho } +\frac{V^{2}}{2} + gz[/tex] = Constant

[tex]\frac{V_{0}^{2}}{2} + g(0) = \frac{V^{2}}{2} + g[/tex]

[tex]V=\sqrt{V_{0}^{2}-2gh}[/tex]

Relation between  [tex]V_{0}[/tex] & V

[tex]m=\rho V_{0}A_{0}[/tex]

[tex]\rho VA=\rho V_{0}A_{0}[/tex]

[tex]VA= V_{0}A_{0}[/tex]

Momentum

[tex]F_{w}+F_{d}= \frac{\partial }{\partial t}\int_{cv} w\rho dA + \int_{cs} w\rho V dA[/tex]

[tex]-mg = w_{1}[-\rho VA] +w_{2}[\rho VA][/tex]

[tex]w_{1}[/tex] = V

[tex]w_{2}[/tex] = 0

[tex]mg = \rho V^{2}A[/tex]

[tex]mg = \rho VV_{0} A[/tex]

[tex]mg = \rho VV_{0} A_{0}[/tex]

[tex]mg = \rho V_{0} A_{0} \sqrt{V_{0} ^{2}-2gh }[/tex]

Solving for h

[tex]h = \frac{1}{2g}[V_{0}^{2}-\frac{mg}{\rho V_{0}A_{0}}][/tex]

g is gravitational acc.

[tex]= \frac{1}{2\times 9.81}[10^{2}-(\frac{2\times 9.81}{999\times10\times\frac{\Pi }{4}\times(0.025)^{2}})^{2} ][/tex]

[tex]= \frac{1}{19.62}[100-(\frac{19.68}{4.9038})^{2}][/tex]

[tex]= \frac{100-16.0078}{19.62}[/tex]

h = 4.281 m

h of disk on which it remains stationary.

Answer:

Total drag force is 88.74 N

Rate of heat transfer over the entire plate per unit width is 26451.6 W

Explanation:

The rate of heat transfer over the entire plate per unit width can be obtained by applying the Newton's law of cooling. The pictures attached are step by step solution to the question;

Answer:

Please find attached file for complete answer solution and explanation of same question.

Explanation:

Answer:

[tex] m = \rho V[/tex]

Since we have an ball we can consider this like a sphere and the volume is given by [tex] V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1.69cm)^3 = 20.218 cm^3 = 0.00002022 m^3[/tex]

The density for the copper is approximately [tex] \rho = 8940 kg/m^3[/tex]

So then the mass is :

[tex] m =8940 kg/m^3 * 0.00002022m^3 = 0.1808 Kg[/tex]

And now we have everything in order to replace into the formula for F, like this:[tex] F = 0.1808 Kg *9.8 m/s^2 + 0.884 kg/s * 0.093 m/s= 1.772 +0.975 N = 2.747 N[/tex]And that would be the final answer for this case.

Explanation:

For this case if we assume that we have a damping motion the force action on the vertical direction would be:

[tex] F = mg + bv[/tex]

Where F represent the upward force on the copper ball

m represent the mass

g = 9.8 m/s^2 represent the gravity

b = 0.884 kg/s represent the proportionality constant

v = 9.3 cm/s = 0.093 m/s represent the velocity

We can solve for the mass from the following expression:

[tex] m = \rho V[/tex]

Since we have an ball we can consider this like a sphere and the volume is given by [tex] V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1.69cm)^3 = 20.218 cm^3 = 0.00002022 m^3[/tex]

The density for the copper is approximately [tex] \rho = 8940 kg/m^3[/tex]

So then the mass is :

[tex] m =8940 kg/m^3 * 0.00002022m^3 = 0.1808 Kg[/tex]

And now we have everything in order to replace into the formula for F, like this:[tex] F = 0.1808 Kg *9.8 m/s^2 + 0.884 kg/s * 0.093 m/s= 1.772 +0.975 N = 2.747 N[/tex]And that would be the final answer for this case.

Answer: x, y (0, 1)

Explanation:

the X coordinate of the center mass is

X(c) = Σm(i)*x(i) / Σx(i)

X(c) = (0 + 0 + 0 + 0) / (1.93 + 3.06 + 2.41 + 3.96)

X(c) = 0 / 11.36

X(c) = 0

The y coordinate of the center mass is

Y(c) = Σm(i)y(i) / Σm(i)

Y(c) = [(1.93)(2.93) + (3.06)(2.58) + (2.41)(0) + (3.96)(-0.498)] / (1.93 + 3.06 + 2.41 + 3.96)

Y(c) = (5.6549 + 7.8948 + 0 - 1.97208) / 11.36

Y(c) = 11.57762 / 11.36

Y(c) = 1.02

Therefore, the center of masses is at x, y (0, 1)

Answer:

Explanation:

electric potential of electron

= k Qq / r

= - 9 x 10⁹ x 1.4 x 10⁻¹⁵ x 1.6 x 10⁻¹⁹ / ( 1.2 x 10⁻² )

= - 16.8 x 10⁻²³ J

If v be the escape velocity

1/2 m v² = potential energy of electron

= 1/2 x 9.1 x 10⁻³¹ x v² = 16.8 x 10⁻²³

v² = 3.69 x 10⁸

v = 1.92 x 10⁴ m /s

Answer:

d. Test the rods one pair at a time in each possible end‑to‑end orientation until finding a pair that repels; the leftover rod is unmagnetized.

Explanation:

Repulsion is the surest test of magnetisation . If two objects repel each other , they must be magnets . If two objects attract each other , they may not be magnets. In later case , one may be non - magnet. So the last test will help us identify the unmagnetised rod.

Answer:

21.5 °C.

Explanation:

Given:

α = −5.00 × 10−4 °C−1

To = 4°C

Ro = 217 Ω

Rt = 215.1 Ω

Rt/Ro = 1 + α(T - To)

215.1/217 = 1 + (-5 × 10^−4) × (T - 4)

-0.00876 = -5 × 10^−4 × (T - 4)

17.5 = (T - 4)

T = 21.5 °C.

The temperature on a spring day can be calculated by using the relationship between temperature and resistance. Given the initial resistance, the change in resistance and the temperature coefficient of resistivity for carbon, the spring day temperature can be deduced to be 21.51 °C.

The change in temperature can be determined by using the equation ΔR = R0αΔT, where ΔR is the change in resistance, R0 is the initial resistance, α is the temperature coefficient of resistivity and ΔT is the change in temperature. The initial resistance of the carbon resistor is given as 217.0 Ω, the resistance on a spring day was 215.1 Ω, so the change in resistance ΔR is 215.1 - 217.0 = -1.9 Ω. The temperature coefficient for carbon is given as -5.00×10-4 C-1, and α is -5.00×10-4 C-1. We can rearrange the above equation to solve for ΔT: ΔT = ΔR / (R0 * α) = -1.9 Ω / (217.0 Ω * -5.00×10-4 C-1) = 17.51 C.

Thus, the temperature on the spring day would be the winter day temperature plus the calculated change in temperature: T_spring = T_winter + ΔT = 4.00 °C + 17.51 °C = 21.51 °C.

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Answer:

[tex]3.2\times 10^{-7}\ m[/tex] or 0.32 μm.

Explanation:

Given:

The radiations are UV radiation.

The frequency of the radiations absorbed (f) = [tex]9.38\times 10^{14}\ Hz[/tex]

The wavelength of the radiations absorbed (λ) = ?

We know that, the speed of ultraviolet radiations is same as speed of light.

So, speed of UV radiation (v) = [tex]3\times 10^8\ m/s[/tex]

Now, we also know that, the speed of the electromagnetic radiation is related to its frequency and wavelength and is given as:

[tex]v=f\lambda[/tex]

Now, expressing the above equation in terms of wavelength 'λ', we have:

[tex]\lambda=\frac{v}{f}[/tex]

Now, plug in the given values and solve for 'λ'. This gives,

[tex]\lambda=\frac{3\times 10^8\ m/s}{9.38\times 10^{14}\ Hz}\\\\\lambda=3.2\times 10^{-7}\ m\\\\\lambda=3.2\times 10^{-7}\times 10^{6}\ \mu m\ [1\ m=10^6\ \mu m]\\\\\lambda=3.2\times 10^{-1}=0.32\ \mu m[/tex]

Therefore, the wavelength of the radiations absorbed by the ozone is nearly [tex]3.2\times 10^{-7}\ m[/tex] or 0.32 μm.

Answer:

The force developed in the link is 6.36 N.

Explanation:

Given that,

Mass of block A = 10 kg

Mass of block B = 6 kg

Coefficients of kinetic friction [tex]\mu_{A}= 0.1[/tex]

Coefficients of kinetic friction [tex]\mu_{B}= 0.3[/tex]

Suppose the angle is 30°

We need to calculate the acceleration

Using formula of acceleration

[tex]a=\dfrac{m_{A}g\sin\theta+m_{B}g\sin\theta-\mu_{A}m_{A}g\cos\theta-\mu_{A}m_{A}g\cos\theta}{m_{A}+m_{B}}[/tex]

Put the value into the formula

[tex]a=\dfrac{10\times9.8\sin30+6\times9.8\sin30-0.1\times10\times9.8\times\cos30-0.3\times6\times9.8\times\cos30}{16}[/tex]

[tex]a=3.415\ m/s^2[/tex]

We need to calculate the force developed in the link

For block A,

Using balance equation

[tex]ma=m_{A}g\sin\theta-\mu m_{A}g\cos\theta-T[/tex]

[tex]T=ma+\mu m_{A}g\cos\theta-m_{A}g\sin\theta[/tex]

Put the value into the formula

[tex]T=10\times3.415+0.1\times10\times9.8\times\cos30-10\times9.8\times\sin30[/tex]

[tex]T=-6.36\ N[/tex]

Negative sign shows the opposite direction of the force.

Hence, The force developed in the link is 6.36 N.

Answer:

w = 0.189 rad/ s

Explanation:

This exercise we work with the conservation of the moment, the system is made up of the merry-go-round and the child, for which we write the moment of two instants

Initial

         L₀ = I₀ w₀

Final

          [tex]L_{f}[/tex] = I w

         L₀ = L_{f}

         I₀ w₀ = I_{f} w

         .w = I₀/I_{f}   w₀

The initial moment of inertia is

        I₀ = 500 kg. m2

The final moment of inertia

         [tex]I_{f}[/tex] = 500 + m r²

         I_{f} = 500 + 20 1.5

         I_{f} = 530 kg m²

Initial angular velocity

         w₀ = 0.20 rad / s

Let's calculate

          w = 500/530 0.20

          w = 0.189 rad / s

Answer: 23.92 ns

Explanation:

The capacitance, C = (e(0) * A) / d

C = [8.85*10^-12 * 2*10^-2 * 10*10^-2] / 1*10^-3

C = (1.77*10^-14) / 1*10^-3

C = 1.77*10^-11

C = 17.7 pF

Vc = V * [1 - e^-(t/CR)]

75 = 100 * [1 - e^-(t/CR)]

75/100 = 1 - e^-(t/CR)

e^-(t/CR) = 1 - 0.75

e^-(t/CR) = 0.25

If we take the log of both sides, we then have

Log e^-(t/CR) = Log 0.25

-t/CR = In 0.25

t = -CR In 0.25

t = - 17.7*10^-12 * 975 * -(1.386)

t = 2.392*10^-8 s

t = 23.92 ns

Answer:

Voltage= 9.4 V

Current= 1.3 A

Explanation:

Equivalent resistance will be the sum of two resistors hence

[tex]R_{equ}[/tex]=2+7.2=9.2Ω

From Ohm's law, V=IR where I is current, R is equivalent resistance and V is voltage

Terminal voltage will be given by V=e-Ir where r is internal resistance, I is current and e is electromotive force.

From Ohm's law, current is given by dividing emf of the battery by the equivalent resistance hence

[tex]I=\frac {e}{R_{equ}}\\I=\frac {12}{9.2}=1.304347826A\approx 1.3A[/tex]

This is the current through external resistor

Terminal voltage will be given by V=e-Ir hence V=12-(1.3*2)=9.4 V

Final answer:

The terminal voltage across the 7.2
Ω resistor when connected to a 12.0 V battery with 2.0
Ω internal resistance is approximately 9.392 V, and the current flowing through the resistor is approximately 1.304 A.

Explanation:

The student is asking about the calculation of terminal voltage and current through a resistor when connected to a battery that has internal resistance. This is a problem related to the topic of electrical circuits in Physics.

To find the current through the 7.2
Ω resistor, use Ohm's law which states that V = IR, where V is voltage, I is current, and R is resistance. Since the battery has an internal resistance, we need to consider the total resistance in the circuit which is the sum of the internal resistance and the resistance of the 7.2
Ω resistor.

Total resistance (Rtotal) = Internal resistance (r) + External resistance (R) = 2.0
Ω + 7.2
Ω = 9.2
Ω

Using Ohm's Law, the current (I) through the circuit can be calculated as follows:
I = V / Rtotal = 12.0 V / 9.2
Ω = 1.304
A (approximately).

To find the terminal voltage (Vterminal) across the 7.2
Ω resistor when the current is flowing, we account for the voltage drop across the internal resistance:
Vterminal = emf - I × r = 12.0 V - (1.304
A × 2.0
Ω) = 12.0 V - 2.608 V = 9.392 V (approximately).

The terminal voltage across the 7.2
Ω resistor when it is connected to the battery is 9.392 V, and the current flowing through the resistor is 1.304
A.

Answer:

0.28 ft

Explanation:

We are given that

Strength=m=[tex]36ft^2/s[/tex]

Distance between source and sink=15 ft

Distance between the sink of the source and origin=[tex]a=\frac{15}{2}[/tex] ft

Uniform velocity, U=12 ft/s

We have to find the length of the oval.

Formula to find the half length of the body

[tex]\frac{l}{a}=(\frac{m}{\pi Ua}+1)^{\frac{1}{2}}[/tex]

Where a=Distance between sink of source and origin

U=Uniform velocity

m=Strength

l=Half length

Using the formula

[tex]\frac{l}{\frac{15}{2}}=(\frac{36}{\pi\times 12\times \frac{15}{2}}+1)^{\frac{1}{2}}[/tex]

[tex]l=\frac{2}{15}(\frac{36}{\pi\times 12\times \frac{15}{2}}+1)^{\frac{1}{2}}[/tex]

[tex]l=0.14[/tex]

Length of oval=[tex]2l=2(0.14)=0.28 ft[/tex]