Workmen are trying to free an SUV stuck in the mud. To extricate the vehicle, they use three horizontal ropes, producing the force vectors shown in the figure.(Figure 1) Take F1 = 914N ,F2 = 704N , and F3 = 399N . Part A Find the x components of each of the three pulls. Enter your answer as three numbers, separated with commas. F1x,F2x,F3x = N SubmitMy AnswersGive Up Try Again Part B Find the y components of each of the three pulls. Enter your answer as three numbers, separated with commas. F1y,F2y,F3y = N SubmitMy AnswersGive Up Try Again Part C Use the components to find the magnitude of the resultant of the three pulls. F? = N SubmitMy AnswersGive Up Try Again Part D Use the components to find the direction of the resultant of the three pulls. Enter your answer as the angle counted from +x axis in the counterclockwise direction.

Answer:b

Explanation:

Given

Force of attraction is F when charges are d distance apart.

Electrostatic force is given by

[tex]F=\frac{kq_1q_2}{d^2}---1[/tex]

where k=constant

[tex]q_1[/tex] and [tex]q_2[/tex] are charges

d=distance between them

In order to double the force i.e. 2F

[tex]2F=\frac{kq_1q_2}{d'^2}----2[/tex]

divide 1 and 2 we get

[tex]\frac{F}{2F}=\frac{d'^2}{d^2}[/tex]

[tex]d'=\frac{d}{\sqrt{2}}[/tex]

Answer:

1.686 m

Explanation:

From coulomb's law,

F = kq1q2/r² ...................................... Equation 1

Where F = electrostatic force  between the two charges, q1 = first charge, q2 = second charge, r = distance between the charges.

making r the subject of the equation,

r = √(kq1q2/F).......................... Equation 2

Given: F = 5.05 N, q1 = 28.0 μC = 28×10⁻⁶ C, q2 = 57.0 μC = 57.0×10⁻⁶ C

Constant: k = 9.0×10⁹ Nm²/C².

Substituting into equation 2

r = √(9.0×10⁹×28×10⁻⁶×57.0×10⁻⁶/5.05)

r = √(14364×10⁻³/5.05)

r = √(14.364/5.05)

r = √2.844

r = 1.686 m

r = 1.686 m.

Thus the distance must be 1.686 m

Answer: q = 7.06 × 10^-21 C

the magnitude of the charges that make up the dipole is 7.06 × 10^-21 C

Explanation:

Given:

Torque, τ = 6.60×10^−26N⋅m

Angle made by p with a uniform electric field, θ = 90° (perpendicular)

Electric field, E = 8.50×10^4N/C

Length between dipole r = 1.10×10^−10 m

Torque acting on the dipole is given by the relation,

τ = pE sinθ....1

But,

p = qr .....2

Substituting equation 1 to 2

τ= qrEsinθ ....3

Making q the subject of formula

q = τ/rEsinθ .....4

Where;

q = magnitude of the charges that make up the dipole.

Substituting the given values into equation 4:

q = 6.60×10^−26N⋅m/(1.10×10^−10 m × 8.50×10^4N/C × sin90°)

q = 0.70588 × 10^-20 C

q = 7.06 × 10^-21 C

The magnitude of the charge making up the electric dipole, given a torque of 6.60×10−26N⋅m when the dipole moment is perpendicular to an electric field of 8.50×104N/C, and a separation of 1.10×10−10m between the charges, is approximately 7.05×10−21 C.

The torque (τ) on a dipole in a uniform electric field is given by the equation τ = pEsinθ, where p is the dipole moment, E is the electric field strength, and θ is the angle between the dipole moment and the electric field. In this case, the dipole moment is perpendicular to the electric field, so θ= 90 degrees, and sinθ= 1. The dipole moment, p, is the product of the magnitude of the charge (q) and the separation (d) between the charges, so p = qd.

Given τ = 6.60×10−26N⋅m, E = 8.50×104N/C, and d = 1.10×10−10m, we can first use the torque equation to find the dipole moment: p = τ / E = 6.60×10−26N⋅m / 8.50×104N/C = 7.76×10−31 C⋅m. Then, use p = qd to calculate the charge: q = p / d = 7.76×10−31 C⋅m / 1.10×10−10m = 7.05×10−21 C.

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Answer:

t = 0.0105 s

Explanation:

given,

mass of the baseball = 435 g = 0.435 Kg

initial speed of ball, u = 36 m/s

final speed of ball, v = 49 m/s

contact force , F = 3500 N

we know,

Impulse is equal to change in momentum

I = m v - m u

I = 0.435 (49-(-36))

I = 0.435 x 85

I = 36.975 kg.m/s

we also know that

I = F x t

36.975 = 3500 x t

t = 0.0105 s

time of contact is equal to 0.0105 s

Answer:

2630250 N/C, horizontally left

0

[tex]1.67\times 10^{-8}\ s[/tex]

1434.825 N/C, horizontally left

Explanation:

m = Mass of particle

u = Initial velocity = [tex]4.2\times 10^6\ m/s[/tex]

v = Final velocity = 0

t = Time taken

s = Displacement = 3.5 cm

q = Charge of particle = [tex]1.6\times 10^{-19}\ C[/tex]

Force is given by

[tex]F=qE[/tex]

Acceleration is given by

[tex]a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times E}{1.67\times 10^{-27}}\\\Rightarrow a=95808383.23353E[/tex]

[tex]v^2-u^2=2as\\\Rightarrow v^2-u^2=2\times 95808383.23353Es\\\Rightarrow E=\dfrac{0^2-(4.2\times 10^6)^2}{2\times 95808383.23353\times 0.035}\\\Rightarrow E=-2630250\ N/C[/tex]

Magnitude of electric field is 2630250 N/C

Direction is horizontally to the left

The angle counterclockwise from left is zero.

[tex]a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times -2630250}{1.67\times 10^{-27}}\\\Rightarrow a=-2.52\times 10^{14}\ m/s^2[/tex]

[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-4.2\times 10^6}{-2.52\times 10^{14}}\\\Rightarrow t=1.67\times 10^{-8}\ s[/tex]

The time taken is [tex]1.67\times 10^{-8}\ s[/tex]

Acceleration is given by

[tex]a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times E}{9.11\times 10^{-31}}\\\Rightarrow a=175631174533.4797E[/tex]

[tex]v^2-u^2=2as\\\Rightarrow v^2-u^2=2\times 175631174533.4797Es\\\Rightarrow E=\dfrac{0^2-(4.2\times 10^6)^2}{2\times 175631174533.4797\times 0.035}\\\Rightarrow E=-1434.825\ N/C[/tex]

Magnitude of electric field is 1434.825 N/C

Direction is horizontally to the left

Answer:

F' = 169.45N

This is a vector addition involving two vectors. In order to do this correctly, we need to resolve each of those forces into their vertical and horizon components and sum them up accordingly (all vertical components summed together and all horizontal components summed together). Then the magnitude of the summation is found by taking the square root of the sun of the squares of the summations along the vertical and the horizontal.

Explanation:

See the attachment below for the full solution to the problem.

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Answer: Electron affinity of F equals

275.8kJ/mol

Explanation: Electron affinity is the energy change when an atom gains an electron.

Let's first calculate the energy required -E(r)to dissociate KF into ions not neutral atom which is given.

E(r) = {z1*z2*e²}/{4π*permitivity of space*r}

z1 is -1 for flourine

z2 is +1 for potassium

e is magnitude of charge 1.602*EXP{-9}C

r is ionic bond length of KF(is a constant for KF 0.217nm)

permitivity of free space 8.854*EXP{-12}.

Now let's solve

E(r)= {(-1)*(1)*(1.602*EXP{-9})²} /

{4*3.142*8.854*EXP(-12)*0.217*EXP(-9)

E(r) = - 1.063*EXP{-18}J

But the energy is released out that is exothermic so we find - E(r)

Which is +1.603*EXP{-18}J

Let's now convert this into kJ/mol

By multiplying by Avogadro constant 6.022*EXP(23) for the mole and diving by 1000 for the kilo

So we have,

1.603*EXP(-18) *6.022*EXP(23)/1000

-E(r) = 640.2kJ/mol.

Now let's obtain our electron affinity for F

We use this equation

Energy of dissociation (nuetral atom)= electron affinity of F +(-E(r)) + ionization energy of K.

498kJ/mol

=e affinity of F + 640.2kJ/mol

+(-418kJ/mol)

(Notice the negative sign in ionization energy for K. since it ionize by losing an electron)

Making electron affinity of F subject of formula we have

Electron affinity (F)=498+418-640.2

=275.8kJ/mol.

Final answer:

The cake's temperature decreases according to Newton's Law of Cooling, and we can use this law to calculate the time it takes for the cake to reach a certain temperature.

Explanation:

The cake is removed from a 350°F oven and placed in a 70°F room. After 30 minutes, the cake's temperature decreases to 200°F. To find out when it will be 100°F, we can use Newton's Law of Cooling.

According to Newton's Law of Cooling, the rate of change of the temperature of an object is proportional to the difference between the object's temperature and the temperature of its surroundings. This can be expressed as:

T' = -k(T - Ts)

where T' is the rate of change of temperature with respect to time, T is the temperature of the object, Ts is the temperature of the surroundings, and k is the cooling constant. In this case, we can rearrange the equation to solve for time:

t = (1/k) * ln((T - Ts) / (T0 - Ts))

where t is the time, T0 is the initial temperature of the object, and ln is the natural logarithm.

Plugging in the values from the problem:

t = (1/k) * ln((200 - 70) / (350 - 70))

We can find the value of the cooling constant, k, by using the given information. Since we know the temperature dropped from 350°F to 200°F in 30 minutes, we can use this to find k:

-k = (T' / (T - Ts)) = (200 - 70) / (350 - 70) / 30

simplifying, we get:

k = -((200 - 70) / (350 - 70)) / 30

Now we can substitute the value of k into the equation for time:

t = (1 / -((200 - 70) / (350 - 70)))) * ln((200 - 70) / (350 - 70))

Calculating this equation will give us the approximate time it takes for the cake to reach 100°F.

Answer:

Vector C = -1.56i^ +1.07j^

This question requires that we use the properties of the scalar product of two vectors to find the required x and y component.

The dot product of two perpendicular vectors is equal to and the product of two vectors that are not parallel is equal to a nonzero value.

These are the properties that have been used in solving this problem alongside solving the simultaneous questions generator.

Explanation:

The full solution can be found in the attachment below.

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Answer:

0.69s

Explanation:

10 cm = 0.1 m

Let t be the time that radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude. At that time we have the angular velocity would be

[tex]\omega = \alpha t = 2.1 t[/tex]

And so the radial acceleration is

[tex]a_r = \omega^2 r = (2.1t)^2 r = 2.1^2 t^2 * 0.1= 0.441 t^2 m/s^2[/tex]

The tangential acceleration is always the same since angular acceleration is constant:

[tex]a_t = \alpha * r = 2.1 * 0.1 = 0.21 m/s^2[/tex]

For these 2 quantities to be the same

[tex]a_r = a_t[/tex]

[tex]0.441 t^2 = 0.21[/tex]

[tex]t^2 = 0.21/0.441 = 0.4762[/tex]

[tex]t = \sqrt{0.4762} = 0.69 s[/tex]

The value of t whereby the radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude is; t = 0.69 s

Let the time that radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude be denoted as t. Thus, angular velocity at time (t) is;

ω = αt

where;

α is angular acceleration

We are given;

Thus, ω = 2.1t

Now, we can find the radial acceleration from the formula;

α_r = ω²r

Thus;

α_r =  (2.1t)² × 0.1

α_r = 0.441 t² m/s²

The tangential acceleration is gotten from the formula is;

α_t = α × r

α_t = 2.1 × 0.1

α_t = 0.21 m/s²

The condition in the question implies that the tangential acceleration is equal to the radial acceleration. Thus;

α_t = α_r

0.21 = 0.441 t²

t = √(0.21/0.441)

t = 0.69 s

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To solve this problem we will start by calculating the distance traveled while relating the first acceleration in the given time. From that acceleration we will calculate its final speed with which we will calculate the distance traveled in the second segment. With this speed and the acceleration given, we will proceed to calculate the last leg of its route.

Expression for the first distance is

[tex]s_1 = ut +\frac{1}{2} at^2[/tex]

[tex]s_1 = 0+\frac{1}{2} (3.8)(6)^2[/tex]

[tex]s_1 = 68.4m[/tex]

The expression for the final speed is

[tex]v = v_0 +at[/tex]

[tex]v = 0+(3.8)(6)[/tex]

[tex]v = 22.8m/s[/tex]

Then the distance becomes as follows

[tex]s_2 = vt[/tex]

[tex]s_2 = (22.8)(1.6)[/tex]

[tex]s_2 = 36.48m[/tex]

The expression for the distance at last sop is

[tex]v_1^2=v_0^2 +2as_3[/tex]

[tex]22.8^2 = 0+2(3.3)s_3[/tex]

[tex]s_3 =78.7636m[/tex]

Therefore the required distance between the signs is,

[tex]S = s_1+s_2+s_3[/tex]

[tex]S = 68.4+36.48+78.76[/tex]

[tex]S = 183.64m[/tex]

Therefore the total distance between signs is 183.54m

Answer :

New force becomes, F' = 1.83 N

Explanation:

Let two point charges exert a force of 7.35 N force on each other. The electric force between two charges is given by :

[tex]F=\dfrac{kq_1q_2}{r^2}[/tex]

[tex]q_1\ and\ q_2[/tex] are charges

r is the distance between charges if the distance between them is increased by a factor of 2, r' = 2r

New force is given by :

[tex]F'=\dfrac{kq^2}{r'^2}[/tex]

[tex]F'=\dfrac{kq^2}{(2r)^2}[/tex]

[tex]F'=\dfrac{1}{4}\dfrac{kq^2}{r^2}[/tex]

[tex]F'=\dfrac{1}{4}\times 7.35[/tex]

F' = 1.83 N

So, the new force between charges will be 1.83 N. Therefore, this is the required solution.          

To find the net electric force on an electron due to two point charges at the origin and at x=0.8m, one has to first calculate the electric field due to each charge at the location of the electron. After finding the total electric field it's multiplied with the charge of the electron to obtain the force.

The subject of this question is Physics, specifically the concept of Electric Force. In this problem, we need to find the net force on an electron located on the x-axis at two different points because of two point charges at the origin and at x=0.8m.

Step 1: Calculate the electric field at the given point due to each point charge by using the formula E= KQ/r² where k = 9 x 10⁹ N m²/C²(Q is the charge and r is the distance from the charge to the point in question).

Step 2: Once the electric field due to each charge is found, sum these together to get the total electric field at the point.

Step 3: Electric force that an electron experiences in the field can be found using the electric field (E) and the charge of an electron (e) by the formula F= eE where e is 1.602 x 10⁻¹⁹ Coulomb.

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The net electric force on an electron at x = 0.200 m is [tex]1.66 * 10^{-17}[/tex] N towards the origin, due to both charges. At x = 1.20 m, the net force is [tex]6.95 * 10^{-18}[/tex] N towards q1, away from the origin.

A -4.00 nC point charge is at the origin, and a second -5.50 nC point charge is on the x-axis at x = 0.800 m.

1. Find the net electric force that the two charges would exert on an electron placed at a point on the x-axis at x = 0.200 m.

To find the net electric force on an electron at x = 0.200 m, we will use Coulomb's Law:

[tex]F = (k * |q_1 * q_2| )/ r^2[/tex]

Here, [tex]q_1[/tex] and [tex]q_2[/tex] are the charges, r is the distance between them, and k is Coulomb's constant,[tex]8.99 * 10^9 N.m^2/C^2.[/tex]

The electron at x = 0.200 m is 0.200 m from q1 and 0.600 m from [tex]q_2[/tex]. Both forces are attractive since all charges are negative.

Force due to [tex]q_1[/tex]:

[tex]F_1 = k * |e * q_1| / (0.200)^2 = 8.99 * 10^9 N.m^2/C^2 * |-1.6 x 10^-{19} C * -4.00 * 10^{-9} C| / (0.200 m)^2 = 1.44 * 10^{-17} N[/tex]

Force due to [tex]q_2[/tex]:

[tex]F_2 = k * |e * q_2| / (0.600)^2 = 8.99 * 10^9 N.m^2/C^2 * |-1.6 x 10^{-19} C * -5.50 x 10^{-9} C| / (0.600 m)^2 = 2.20 * 10^{-18} N[/tex]

The net force will be the sum of these forces, taking directions into account.

Net force = F1 (towards -x) + F2 (towards -x) [tex]= 1.44 * 10^{-17} N + 2.20 * 10^{-18} N = 1.66 * 10^{-17} N[/tex]

This force is towards the origin.

2. Find the net electric force that the two charges would exert on an electron placed at a point on the x-axis at x = 1.20 m.

We repeat the procedure for the electron at x = 1.20 m:

The electron at x = 1.20 m is 1.20 m from q1 and 0.400 m from q2.

Force due to q1:

[tex]F1 = k * |e * q1| / (1.20)^2 = 8.99 * 10^9 N.m^2/C^2 * |-1.6 x 10^{-19} C * -4.00 * 10^-9 C| / (1.20 m)^2 = 2.00 * 10^{-18} N[/tex]

Force due to q2:

[tex]F2 = k * |e * q2| / (0.400)^2 = 8.99 * 10^9 N.m^2/C^2 * |-1.6 x 10^{-19} C * -5.50 * 10^{-9} C| / (0.400 m)^2 = 4.95 * 10^{-18} N[/tex]

The net force is:

Net force = F1 (towards +x) - F2 (towards -x) [tex]= 2.00 * 10^{-18} N + 4.95 * 10^{-18} N = 6.95 * 10^{-18}[/tex] N towards q1, away from the origin

To solve this problem it is necessary to apply the concepts related to the voltage depending on the electric field and the distance, as well as the load depending on the capacitance and the voltage. For the first part we will use the first mentioned relationship, for the second part, we will not only define the load as the capacitance by the voltage but also place it in terms of the Area, the permittivity in free space, the voltage and the distance.

PART A ) Voltage in function of electric field and distance can be defined as,

[tex]V = Ed[/tex]

Our values are,

[tex]E = 1.2*10^5 V/m[/tex]

[tex]d = 3.0mm = 3*10^{-3}[/tex]

Replacing,

[tex]V = (1.2*10^5)(3*10^{-3})[/tex]

[tex]V = 360v[/tex]

Therefore the potential difference across the capacitor is 360V

PART B) The charge can be defined as,

[tex]Q = CV = \frac{\epsilon AV}{d}[/tex]

Here,

[tex]\epsilon = 8.85*10^{-12} F/m[/tex], Permittivity of free space

[tex]A = s^2[/tex], area of each capacitor plate

s = Length of capacitor plate

Replacing,

[tex]Q = \frac{\epsilon AV}{d}[/tex]

[tex]Q = \frac{(8.85*10^{-12})(0.03)^2(240)}{2.0*10^{-8}m}[/tex]

[tex]Q = 9.558*10^{-10}C[/tex]

Therefore the charge on each plate is [tex]9.558*10^{-10}C[/tex]

The potential difference across the parallel-plate capacitor is 360 V. The charge on each plate of the capacitor is approximately 0.95 x 10^-8 C.

The potential difference across a parallel-plate capacitor is calculated using the formula V = Ed, where E is the electric field strength and d is the distance (or spacing) between the plates. As given, E is 1.2 x 10^5 V/m, and d is 3.0 mm (or 3.0 x 10^-3 m). Therefore, the potential difference V across the plates is given by V = 1.2 x 10^5 V/m * 3.0 x 10^-3 m = 360 V.

The amount of charge Q on each plate of the capacitor can be found using the formula Q = εEA, where ε is the permittivity of free space (ε = 8.85 x 10^-12 F/m), E is the electric field strength, and A is the area of the plate. Substituting the values given, we have A = 3.0 cm * 3.0 cm = 9 cm^2 = 9 x 10^-4 m^2, E = 1.2 x 10^5 V/m, and ε = 8.85 x 10^-12 F/m. Therefore, Q = εEA = 8.85 x 10^-12 F/m * 1.2 x 10^5 V/m * 9 x 10^-4 m^2 ≈ 0.95 x 10^-8 C.

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The procedures involves using short pulses of high intensity laser beams to repair detached parts of the retina. The energy delivered per pulse, average pressure exerted by laser, new wavelength and frequency of laser light inside the eye, and the maximum electric and magnetic field in the laser beam are calculated. The details of these calculations have been given.

A) The energy delivered to the retina with each pulse can be calculated using the formula Energy = Power x Time. Here, power is 250mW or 0.250 joules/second and time is 1.50 ms or 0.00150 seconds. So, the energy is 0.250 joules/second x 0.00150 seconds = 0.000375 joules or 375 μJ per pulse.

B) The average pressure exerted by the laser beam can be calculated using the formula Pressure = Power/Area where the area of the laser spot is the area of a circle of diameter 510 μm. The laser light is completely absorbed upon hitting a surface, therefore the pressure the beam exerts is 2* Power/Area of beam.

C) Inside the vitreous humor of the eye, the wavelength of the laser light decreases as it is inversely proportional to the refractive index of the medium. So the new wavelength in the vitreous humor is 810 nm/ 1.34 = 604.48 nm.

D) The frequency of light doesn't change when it enters a different medium so it remains the same in the vitreous humor as it was when in air. It can be calculated using the formula Frequency = Speed of Light / Wavelength.

E) The maximum electric field in the laser beam can be computed using the formula for energy density.

F) The maximum value of the magnetic field in the laser beam is proportional to the maximum electric field.

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The energy delivered to the retina per pulse is 0.375 mJ. The pressure exerted is approximately 3.68 Pa, and the wavelength inside the vitreous humor is 604.5 nm. The frequency is 3.70 × 10¹⁴ Hz, with maximum electric and magnetic field values of 1.53 × 10⁵ V/m and 5.10 × 10⁻⁴ T, respectively.

A) Energy Delivered to the Retina Per Pulse

The power of the laser beam is 250 mW, which is 0.250 W. The duration of each pulse is 1.50 ms, which is 1.50 × 10⁻³ s.

Calculate the energy delivered during each pulse using the formula:

Energy (J) = Power (W) × Time (s).

Energy = 0.250 W × 1.50 × 10⁻³ s

= 3.75 × 10⁻⁴ J (or 0.375 mJ).

B) Average Pressure Exerted by the Pulse

If the beam is fully absorbed, the pressure exerted is given by:

Pressure (P) = Power (P) / Area (A) × c,

where c is the speed of light (approximately 3 × 10⁸ m/s).

Calculate the area of the circular spot:

Area (A) = π (d/2)²

where diameter

d = 510 μm = 510 × 10⁻⁶ m.

Area = π (510 × 10⁻⁶ / 2)²

≈ 2.04 × 10⁻⁷ m².

P = 0.250 W / (2.04 × 10⁻⁷ m²) × 3 × 10⁸ m/s

≈ 3.68 N/m² (Pa).

C) Wavelength of Laser Light Inside the Vitreous Humor

The wavelength inside the medium is given by: λ' = λ / n, where λ is the wavelength in air (810 nm) and n is the refractive index (1.34).

λ' = 810 nm / 1.34

≈ 604.5 nm.

D) Frequency of the Laser Light Inside the Vitreous Humor

The frequency of light remains constant through different media and is given by:

f = c / λ,

where c is the speed of light

λ is the wavelength in air.

Frequency f = 3 × 10⁸ m/s / 810 × 10⁻⁹ m

≈ 3.70 × 10¹⁴ Hz.

E) Maximum Value of the Electric Field in the Laser Beam

The maximum electric field E₀ is given by:

E₀ = √(2I/ε₀c),

where I is the intensity,

ε₀ is the permittivity of free space,

c is the speed of light.

Intensity I = P / A

= 0.250 W / 2.04 × 10⁻⁷ m²

≈ 1.23 × 10⁶ W/m².

Using ε₀ = 8.85 × 10⁻¹² F/m: E₀

≈ √(2 × 1.23 × 10⁶ W/m² / 8.85 × 10⁻¹² F/m × 3 × 10⁸ m/s)

≈ 1.53 × 10⁵ V/m.

F) Maximum Value of the Magnetic Field in the Laser Beam

The magnetic field B₀ is related to the electric field E₀ by:

B₀ = E₀ / c.

B₀ = 1.53 × 10⁵ V/m / 3 × 10⁸ m/s

≈ 5.10 × 10⁻⁴ T.

These calculations help us understand how laser pulses interact with biological tissues, aiding in precision medical procedures.

Therefore, the energy delivered to the retina per pulse is 0.375 mJ. The pressure exerted is approximately 3.68 Pa, and the wavelength inside the vitreous humor is 604.5 nm. The frequency is 3.70 × 10¹⁴ Hz, with maximum electric and magnetic field values of 1.53 × 10⁵ V/m and 5.10 × 10⁻⁴ T, respectively.

There are some placeholders in the expression, but they can be safely assumed

Answer:

(a) [tex]f=1617.9\ Hz[/tex]

(b) [tex]T=0.618\ ms[/tex]

(c) [tex]A=20 \ Volts[/tex]

(d) [tex]\varphi=60^o[/tex]

Explanation:

Sinusoidal Waves

An oscillating wave can be expressed as a sinusoidal function as follows

[tex]V(t)&=A\cdot \sin(2\pi ft+\varphi )[/tex]

Where

[tex]A=Amplitude[/tex]

[tex]f=frequency[/tex]

[tex]\varphi=Phase\ angle[/tex]

The voltage of the question is the sinusoid expression  

[tex]V(t)=20cos(5\pi\times 103t+60^o)[/tex]

(a) By comparing with the general formula we have

[tex]f=5\pi\times 103=1617.9\ Hz[/tex]

[tex]\boxed{f=1617.9\ Hz}[/tex]

(b) The period is the reciprocal of the frequency:

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

[tex]\displaystyle T=\frac{1}{1617.9\ Hz}=0.000618\ sec[/tex]

Converting to milliseconds

[tex]\boxed{T=0.618\ ms}[/tex]

(c) The amplitude is

[tex]\boxed{A=20 \ Volts}[/tex]

(d) Phase angle:

[tex]\boxed{\varphi=60^o}[/tex]

To solve this problem we will apply the concept related to kinetic energy based on the ideal gas constant and temperature. From there and with the given values we will find the temperature of the system. As the temperature is the same it will be possible to apply the root mean square speed formula that is dependent on the element's molar mass, the ideal gas constant and the temperature, this would be:

[tex]KE = \frac{3}{2} RT[/tex]

Where,

KE = Average kinetic energy of an ideal gas

[tex]R = 8.314JK^{-1}mol^{-1}[/tex]= Ideal gas constant

T = Temperature

Replacing we have,

[tex]KE = \frac{3}{2} RT[/tex]

[tex]5930J/mol = \frac{3}{2}(8.314JK^{-1}mol^{-1})T[/tex]

[tex]T = 475.503K[/tex]

Therefore the temperature is 475.5K

RMS velocity of [tex]F_2[/tex] gas is

[tex]v_{rms} = \sqrt{\frac{3RT}{M}}[/tex]

Where,

M = Molar mass of [tex]F_2[/tex]

[tex]M = 38.00g/mol[/tex]

[tex]M = 38.00*10^{-3} kg/mol[/tex]

[tex]T = 475.5K[/tex]

[tex]R = 8.314JK^{-1}mol^{-1}[/tex]

Replacing we have,

[tex]v_{rms} = \sqrt{\frac{3RT}{M}}[/tex]

[tex]v_{rms} = \sqrt{\frac{3(8.314JK^{-1}mol^{-1})(475.5K )}{38.00*10^{-3} kg/mol}}[/tex]

[tex]v_{rms} = 558.662m/s[/tex]

Therefore, the RMS velocity of [tex]F_2[/tex] gas is 558.6m/s

Answer:

Air, water, rock and soil

Explanation:

The environment where the general properties of wave motion can be observed is the wave medium which is any substance or particle that carries the wave, or through which the wave travels.

Answer:

a. Electric field lines can never cross each other.

d. Wider spacing between electric field lines indicates a lower magnitude of electric field.

Explanation:

a. The electric field lines cannot be crossed, since this would mean that there would be more than one electric field vector for the same point at the place where the crossing occurs.

d. The space between the field lines is inversely proportional to the intensity of the electric field.

Answer: "a" and "d" are correct

Explanation:

(a) Field lines can never cross. Since a field line represents the direction of the field at a given point, if two field lines crossed at some point, that would imply that the electric field was pointing in two different directions at a single point.

(b) Wider spacing between electric field lines indicates a lower magnitude of electric field.

Answer:

Explanation:

Given

Three blocks are placed over each other at a certain distance.

Center of gravity of each block is at distance of 0.5 L from one end of block.

First We consider block 1 and 2

Block 1 center of gravity will try to tumble the block 1 if center of gravity torque goes beyond 0.5 L of second block.

i.e. maximum distance up to which block 1 is placed over block 2 is [tex]x=0.5 L[/tex]

combined center of gravity of 1 and 2 is

Center of gravity [tex]x=\frac{0.5L+L}{2}=\frac{3L}{4}[/tex]

Now consider block 2 and 3

Combined center of gravity of  block 1 and 2 will tumble over when their Center of gravity goes beyond edge of block 1

i.e. maximum value of [tex]d=\frac{3L}{4}[/tex]

(b) As the no of blocks increases center of gravity increases so maximum value of [tex]d\rightarrow \infty[/tex]

In the static equilibrium physics problem, the maximum overhang distance d for three stacked bricks without toppling is initially calculated for two bricks, then for three bricks. The problem scales with an infinite number of bricks to reveal an overhang approaching half the brick's length.

The question pertains to the physics concept of static equilibrium, specifically torque and center of mass in systems with multiple stacked objects. For part (a), considering the top two bricks first, the maximum overhang achievable without the bricks tumbling is one-fourth of the length of one brick. When adding the third brick, the maximum overhang distance d will increase, but calculation requires a step-by-step process taking into account the center of mass of the bricks in the stack and the fulcrum point. For part (b), with an infinite number of bricks, optimal stacking achieves a maximum distance that approaches half the length of a single brick, as per the harmonic series solution to this classic physics problem.

Explanation:

Using Newtons second law on each block

F = m*a

Block 1

[tex]T_{1} - u*g*M_{1} = M_{1} *a \\\\T_{1} = M_{1}*(a + u*g) ... Eq1[/tex]

Block 2

[tex]T_{2} - u*g*M_{2} = M_{2} *a \\\\T_{2} = M_{2}*(a + u*g) ... Eq2[/tex]

Block 3

[tex]- (T_{1} + T_{2} ) + g*M_{3} = M_{3} *a \\\\T_{1} + T_{2} = M_{3}*( -a + g) ... Eq3[/tex]

Solving Eq1,2,3 simultaneously

Divide 1 and 2

[tex]\frac{T_{1} }{T_{2}} = \frac{M_{1}*(a+u*g)}{M_{2}*(a+u*g)} \\\\\frac{T_{1} }{T_{2}} = \frac{M_{1} }{M_{2} }\\\\ T_{1} = \frac{M_{1} *T_{2} }{M_{2} } .... Eq4[/tex]

Put Eq 4 into Eq3

[tex]T_{2} = \frac{M_{3}*(g-a) }{1+\frac{M_{1} }{M_{2} } } ...Eq5[/tex]

Put Eq 5 into Eq2 and solve for a

[tex]a = \frac{M_{3}*g -u*g*(M_{1} + M_{2}) }{M_{1} + M_{2} + M_{3} } .... Eq6[/tex]

Substitute back in Eq2 and use Eq4 and solve for T2 & T1

[tex]T_{2} = M_{2}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\T_{1} = M_{1}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\[/tex]

Answer:

For real gas the volume of a given mass of gas will increase with increase in temperature.

Explanation:

With the piston head locked in place and place above the fire,the volume of the gas will increase,because the volume of a given mass of gas increases with increase temperature.

Final answer:

With the piston head locked in place, the volume of the gas will stay the same when placed above a flame, but pressure will increase due to the rise in kinetic energy of the gas molecules.

Explanation:

When the piston head is locked in place and then placed above a flame, the volume of the gas will stay the same because the piston cannot move to allow for an increase in volume. According to Charles's Law, which describes the direct relationship between the temperature and volume of a gas at constant pressure, if a gas is heated and can expand, its volume will increase. However, since the piston in this scenario is locked and prevents the gas from expanding, the volume cannot change. The effect of heating the gas while the piston is locked will result in an increase in the gas pressure instead. This occurs because as the gas molecules are heated, they gain kinetic energy and move faster, colliding with the walls of the container more frequently and with greater force, which leads to increased pressure.

To find the volume flow rate in ideal flow between two horizontal tubes, use the equation A1v1 = A2v2, where A1 and A2 are the cross-sectional areas of the tubes and v1 and v2 are the velocities of the liquid. To find the volume flow rate as a function of DP, substitute the values of A1 and v1 into the equation Q = A1v1. For specific values of DP, substitute the values of A1 and v1 into the equation Q = A1v1.

For liquids, the volume flow rate can be determined using the equation A1v1 = A2v2. In this equation, A1 and A2 are the cross-sectional areas of the first and second tubes, and v1 and v2 are the velocities of the liquid flowing through the tubes. Since the first tube is larger in radius, it has a larger cross-sectional area. Therefore, the velocity of the liquid in the first tube is smaller than in the second tube. The volume flow rate can be expressed as:

Q = A1v1

where Q is the volume flow rate and is equal to the product of the cross-sectional area and velocity of the liquid in the first tube.

(a) The volume flow rate as a function of DP can be found by substituting the values of A1 and v1 into the equation Q = A1v1. Since the radii of the tubes are given, the cross-sectional areas can be calculated using the formula A = πr2.

(b) To evaluate the volume flow rate for DP = 6.00 kPa, substitute the values of A1 and v1 into the equation Q = A1v1.

(c) To evaluate the volume flow rate for DP = 12.0 kPa, substitute the values of A1 and v1 into the equation Q = A1v1.

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Answer:

d =  -26 mm

Explanation:

given,

initial position of paramecium, x = 65 mm

final position of paramecium, y = 39 mm

displacement of the paramecium's = ?

displacement  = final position - initial position

 d =  y - x                      

d = 39 - 65                

d =  -26 mm              

The paramecium's displacement comes out to be -26 mm

Answer:

      T = 6.0 N

Explanation:

given,

mass of the cord = 0.46 Kg

length of the supports = 7.2 m

time taken to travel = 0.74 s

tension in the chord = ?

using formula for tension calculation

[tex]T = \dfrac{v^2.m}{l}[/tex]

[tex]v = \dfrac{l}{s}[/tex]

[tex]v = \dfrac{7.2}{0.74}[/tex]

v = 9.73 m/s

now, calculation of tension

[tex]T = \dfrac{9.73^2\times 0.46}{7.2}[/tex]

      T = 6.0 N

The tension in the cord is equal to 6.0 N.

Answer:

<-16969.7, 0, 0> N

Explanation:

[tex]p_2[/tex] = Final momentum = < 33000, 0, 0> kg·m/s

[tex]p_1[/tex] = Initial momentum = <89000, 0, 0> kg·m/s

t = Time taken = 3.3 seconds

Impulse is given by

[tex]J=p_2-p_1\\\Rightarrow Ft=p_2-p_1\\\Rightarrow F=\dfrac{p_2-p_1}{t}\\\Rightarrow F=\dfrac{< 33000, 0, 0>-<89000, 0, 0>}{3.3}\\\Rightarrow F=\dfrac{<-56000, 0, 0>}{3.3}\\\Rightarrow F=<-16969.7, 0, 0>\ N[/tex]

The net force exerted on the truck is <-16969.7, 0, 0> N

The net force exerted on the truck by the surroundings is calculated using the change in momentum and the time interval. It is found to be approximately <-16970, 0, 0> Newtons.

To find the net force exerted on the truck by the surroundings, we need to use the change in momentum (Δp) and the time (Δt) over which the change occurs. The change in momentum is obtained by subtracting the final momentum vector from the initial momentum vector. In this case:

Δp = <33000, 0, 0> kg·m/s - <89000, 0, 0> kg·m/s = <-56000, 0, 0> kg·m/s

The time interval Δt is given as 3.3 seconds. The net force (F) can be calculated using Newton's second law in its impulse-momentum form:

F = Δp / Δt

The net force vector is:

F = <-56000, 0, 0> kg·m/s / 3.3 s = <-16969.7, 0, 0> N

Therefore, the net force exerted on the truck by the surroundings as a vector is approximately <-16970, 0, 0> Newtons.

Answer:

m=57.65 kg

Explanation:

Given Data

Ricardo mass m₁=80 kg

Canoe mass m₂=30 kg

Canoe Length L= 3 m

Canoe moves x=40 cm

When Canoe was at rest the net total torque is zero.

Let the center of mass is at x distance from the canoe center and it will be towards the Ricardo cause. So the toque around the center of mass is given as

[tex]m_{1}(L/2-x)=m_{2}x+m_{2}(L/2-x)[/tex]

We have to find m₂.To find the value of m₂ first we need figure out the value of.As they changed their positions the center of mass moved to other side by distance 2x.

so

2x=40

x=40/2

x=20 cm

Substitute in the above equation we get

[tex]m_{x}=\frac{m_{1}(L/2-x)-m_{2}x }{L/2+x}\\m_{x}=\frac{80(\frac{3}{2}-0.2 )-30*0.2}{3/2+0.2}\\m_{x}=57.65 kg[/tex]

Answer:

The magnitude of the tension in the rope on the right is 500 N.

Explanation:

Given that,

Weight of scaffold= 400 N

Weight of first painter = 500 N

Weight of second painter = 400 N

Tension in the rope = 800 N

According to figure,

We need to calculate the magnitude of the tension in the rope on the right

Using balance equation

[tex]T_{r}+T=W_{p}+W_{sp}+W_{s}[/tex]

Where, [tex]T_{r}[/tex]=Tension on left side

[tex]W_{r}[/tex]=weight of first painter

[tex]W_{sr}[/tex]=weight of second painter

[tex]W_{r}[/tex]=weight of scaffold

Put the value in the equation

[tex]800+T=400+500+400[/tex]

[tex]T=1300-800[/tex]

[tex]T=500\ N[/tex]

Hence, The magnitude of the tension in the rope on the right is 500 N.

The tension in the right rope supporting the scaffold and the painters can be found by using the equilibrium principle, which results to be 500N.

In physics, we often deal with problems like this using the principle of equilibrium, stating that the sum of all the forces in a system is zero. In this case, the scaffold is stationary, not moving up or down. Therefore, the sum of the forces on it should be zero.

The total weight of the scaffold and the painters is 400N (scaffold) + 500N (first painter) + 400N (second painter) = 1300N. It's being held by two ropes, the left having tension 800N and the right of which we are seeking the tension.

Using the principle of equilibrium, the sum of the tensions in the ropes should equal to total weight. So, if 800N (left rope) + Attention (right rope) = 1300N. This yields the equation, Tension (right rope) = 1300N - 800N, which equates to 500N, the tension in the right rope.

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The wave speed and wavelength will differ across strings of different thicknesses held at the same tension; the wave frequency remains the same.

When two strings of different thicknesses are made of the same material and held at the same tension, the wave speed and wavelength will differ, whereas the wave frequency will remain the same. This is because the speed of a wave on a string is determined by the tension (T) and the linear mass density (μ), where v = (T/μ)¹/2; as the tension is constant and the material is the same, it is the difference in thickness (hence, different densities) that causes a variance in wave speed. Since wave frequency (f) is related to the speed (v) and wavelength (λ) by the equation v = fλ, and the oscillator creates waves at a fixed frequency, a change in wave speed inherently impacts the wavelength.

Answer:

9.25 x 10^-4 Nm

Explanation:

number of turns, N = 8

major axis = 40 cm

semi major axis, a = 20 cm = 0.2 m

minor axis = 30 cm

semi minor axis, b = 15 cm = 0.15 m

current, i = 6.2 A

Magnetic field, B = 1.98 x 10^-4 T

Angle between the normal and the magnetic field is 90°.

Torque is given by

τ = N i A B SinФ

Where, A be the area of the coil.

Area of ellipse, A = π ab = 3.14 x 0.20 x 0.15 = 0.0942 m²

τ = 8 x 6.20 x 0.0942 x 1.98 x 10^-4 x Sin 90°

τ = 9.25 x 10^-4 Nm

thus, the torque is 9.25 x 10^-4 Nm.

The magnitude of the torque on the coil is 9.25 x 10⁻⁴ Nm

According to the question we have the following data:

Number of turns of the coil N = 8

Semi-major axis of the ellipse a = 40/2 cm = 0.2 m

Semi-minor axis, b = 30/2 cm = 0.15 m

Current in the coil, i = 6.2 A

Magnetic field, B = 1.98 x 10⁻⁴ T

The angle between the normal and the magnetic field is 90°.

So the torque on the coil is given by:

τ = NiABsinθ

Now, the area of ellipse:

A = πab

A = 3.14 x 0.20 x 0.15 = 0.0942 m²

Thus,

τ = 8 x 6.20 x 0.0942 x 1.98 x 10⁻⁴ x sin 90°

τ = 9.25 x 10⁻⁴ Nm

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