Can you find a vector quantity that has a magnitude of zero but components that are not zero? Explain. Can the magnitude of a vector be less than the magnitude of any of its components? Explain.

Answer:

Three advantages of Reflecting Telescope over refracting Telescope:

1. there is no chromatic aberration in the Reflecting Telescope being mirror as an objective, while Refracting Telescope can suffer chromatic aberration being using lenses.

2. image of Reflecting Telescope is brighter due to large, polished curved mirrors than Refracting Telescope

3. Reflecting Telescope is compact and portable in size than Refracting Telescope.

Final answer:

Reflecting telescopes have three advantages over refracting telescopes: they gather more light, do not suffer from chromatic aberration, and are easier and less expensive to manufacture.

Explanation:

There are three advantages of reflecting telescopes over refracting telescopes:

Answer:

Part A:

[tex]d_{min} = \frac{v_0^2}{2g\mu_s}[/tex]

Part B:

[tex]d_{min} = 6.37~m[/tex]

Explanation:

Part A:

We should determine the free-body diagram of the small box.

For the first box, the only force exerted to the box is the static friction force in the direction of the motion.

(The direction of the static friction is always confusing to the students. The wrong idea is that the static friction is in the opposite direction with the motion. However, if you look at the Newton's Second Law, it states that the net force acting on an object is equal to the mass times acceleration. And in this case acceleration of the total system is equal to that of the small box, since it sits on the larger box.)

We can use the equations of kinematics to find the minimum distance to stop without the small box slipping.

[tex]v^2 = v_0^2 + 2a(\Delta x)\\0 = v_0^2 + 2ad_{min}[/tex]

The acceleration can be found by Newton's Second Law:

[tex]F = ma\\mg\mu_s = m(-a)\\a = -g\mu_s[/tex]

The negative sign comes from the fact that in order for the boxes to stop they have to apply a negative acceleration.

Now, we can combine the two equations to find the distance x:

[tex]0 = v_0^2 + 2ad_{min} = v_0^2 + 2(-g\mu_s)d_{min}\\d_{min} = \frac{v_0^2}{2g\mu_s}[/tex]

Part B:

We can apply the above formula to the truck and file cabinet.

[tex]d_{min} = \frac{v_0^2}{2g\mu_s} = \frac{10^2}{2(9.8)(0.80)} = 6.37~m[/tex]

The expression for the shortest distance which the large box can stop without the small box slipping is [tex]d_{min} = \frac{v_0^2 }{2\mu_s g}[/tex]

The shortest distance in which the pickup can stop without the file cabinet sliding is 6.38 m.

The given parameters;

The expression for the shortest distance which the large box can stop without the small box slipping is calculated as follows;

Apply work-energy theorem;

[tex]\mu_s F \times d_{min} = \frac{1}{2} Mv_0^2\\\\\mu_s (Mg)d_{min} = \frac{1}{2} Mv_0^2\\\\\mu_s gd_{min} = \frac{v_0^2}{2} \\\\d_{min} = \frac{v_0^2}{2\mu_s g}\\\\[/tex]

At the given speed and coefficient of static friction, the shortest distance in which the pickup can stop without the file cabinet sliding is calculated as;

[tex]d_{min} = \frac{v^2}{2\mu_s g} = \frac{10^2 }{2\times 0.8 \times 9.8} = 6.38 \ m[/tex]

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Answer:

The molar mass of the unknown gas is 17.3 g/mol. The molar mass matches that of ammonia (NH₃) the most (17 g/mol)

Explanation:

Let the unknown gas be gas 1

Let N₂O₄ gas be gas 2

Rate of effusion ∝ [1/√(Molar Mass)]

R ∝ [1/√(M)]

R = k/√(M) (where k is the constant of proportionality)₁₂

R₁ = k/√(M₁)

k = R₁√(M₁)

R₂ = k/√(M₂)

k = R₂√(M₂)

k = k

R₁√(M₁) = R₂√(M₂)

(R₁/R₂) = [√(M₂)/√(M₁)]

(R₁/R₂) = √(M₂/M₁)

R₁ = 2.3 R₂

M₁ = Molar Mass of unknown gas

M₂ = Molar Mass of N₂O₄ = 92.01 g/mol

(2.3R₂/R₂) = √(92.01/M₁)

2.3 = √(92.01/M₁)

92.01/M₁ = 2.3²

M₁ = 92.01/5.29

M₁ = 17.3 g/mol

The molar mass matches that of ammonia the most (17 g/mol)

The unknown gas in the system has been ammonia.

The rate of diffusion of the two gases has been proportional to the molar mass of the gases.

The ratio of the rate of two gases can be given as:

[tex]\rm \dfrac{RateA}{RateB}\;=\;\sqrt{\dfrac{Molar\;mass\[A}{Molar\;mass\;B} }[/tex]

The two gases can be given as:

Gas A = Nitrogen tetraoxide = 2.3x

Gas B = x

[tex]\rm \dfrac{2.3x}{x}\;=\;\sqrt{\dfrac{92.011}{m} }[/tex]

Mass of the unknown gas = 17.39 grams.

The mass has been equivalent to the mass of the Ammonia. Thus, the unknown gas in the system has been ammonia.

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Answer:

Explanation:

The first part of question is about the height of the rock from which it falls and hit the ground with speed of 11 m/s. Lets find out that height.

We will use the formula,

[tex]v^{2} _{f} = v^{2} _{i} + 2gh[/tex]

As the initial velocity of the rock was zero. [tex]v_{f} = 0[/tex]

[tex]v^{2} _{f} = 2gh\\ h = v^{2} _{f} / 2g\\h = \frac{(11 m/s)^{2} }{2(9.8 m/s^{2} )} \\h = 6.17 m[/tex]

Now we have to find the height from which the rock should be dropped and it's speed on hitting the ground should be 22 m/s.

Again we will use the same formula, same calculation but the value of velocity now should be 22 m/s.

[tex]v^{2} _{f} = v^{2} _{i} + 2gh[/tex]

[tex]v^{2} _{f} = 2gh\\ h = \frac{(22m/s)^{2} }{2(9.8 m/s^{2}) } \\h = 24.69 m[/tex]

To double the speed of the rock when it hits the ground (from 11 m/s to 22 m/s), the height from which it is dropped should be quadrupled. Hence, the rock should be dropped from a height of 24.5 meters.

The question is about finding the height from which a rock should be dropped so that its speed on hitting the ground is 22 m/s, given that when it is dropped from height h, its speed is 11 m/s. To solve this, we can use the physics equation for motion under constant acceleration, which is v² = 2gh, where v is the final velocity, g is the acceleration due to gravity, and h is the height of fall.

First, let us find the height h in the initial scenario: (11)² = 2*9.8*h => h = 6.125 m. Generally, the height h is proportional to the square of the speed, so if we double the final speed, the height should be quadrupled: h'(new height) = 4 * h = 4 * 6.125 m = 24.5 m.

Therefore, the rock should be dropped from a height of 24.5 m so that its speed on hitting the ground is 22 m/s.

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Explanation:

Formula to calculate resistivity is as follows.

            [tex]\rho = \frac{1}{qN \mu}[/tex]

                       = [tex]\frac{1}{(1.6 \times 10^{-19}) \times 2.72 \times 10^{17} \times 3900}[/tex] ohm/cm

                       = [tex]5.89 \times 10^{-3} ohm/cm[/tex]

As germanium is an intrinsic conductor. Hence, resistivity of Ge is as follows.

             [tex]\rho_{1} = \frac{1}{2qN_{o}\sqrt{\mu_{e}\mu_{r}}}[/tex]

                        = [tex]\frac{1}{2q(N_{A})^{\frac{1}{2}}\sqrt{\mu_{e}\mu_{r}}}[/tex]

                        =  [tex]\frac{1}{2 \times (1.6 \times 10^{-19}) \times (\sqrt{4.42 \times 10^{22}})\sqrt{(3900)(1900)}}[/tex] ohm/cm

                        = 0.546 [tex]ohm (\mu m)^{-1}[/tex]

Thus, we can conclude that resistivity of the Ge is 0.546 [tex]ohm (\mu m)^{-1}[/tex].

The electric field of a uniformly charged disk at different points on its axis was computed using the formula for the electric field due to a charged disk. a. 20.16 N/C b.  17.7 N/C c. 3.57 N/C d. 0.225 N/C

The student is asking for the calculation of the electric field at different points on the axis of a uniformly charged disk with a known charge density.

The formula to find the electric field E due to a charged disk along its axis at distance x from the center is given by E =((σ/2ε0) * (1 - (x/(sqrt(x^2 + r^2)))), where r is the radius of the disk and σ is the charge density. The constant ε0 is the permittivity of free space, and its value is approximately 8.85 x 10^-12 C^2/N·m^2.

a. For x=5.00 cm, E ~= 20.16 N/C

b. For x=10.0 cm, E ~= 17.7 N/C

c. For x=50.0 cm, E ~= 3.57 N/C

d. For x=200 cm, E ~= 0.225 N/C

These results reflect the fact that the strength of the electric field decreases as one moves farther away from the center of the disk.

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The electric fields at the respective distances are:

[tex]a. \( E(5.00 \, \text{cm}) = 4.369 \times 10^8 \, \text{N}/\text{C} \) b. \( E(10.0 \, \text{cm}) = 3.695 \times 10^8 \, \text{N}/\text{C} \) c. \( E(50.0 \, \text{cm}) = 9.187 \times 10^7 \, \text{N}/\text{C} \) d. \( E(200 \, \text{cm}) = 7.728 \times 10^6 \, \text{N}/\text{C} \)[/tex]

To calculate the electric field on the axis of a uniformly charged disk at a distance z from the center, we can use the formula derived from Gauss's law for a flat surface:

[tex]\[ E(z) = \frac{\sigma}{2\epsilon_0} \left( 1 - \frac{z}{\sqrt{z^2 + R^2}} \right) \][/tex]

where E(z)  is the electric field at a distance z from the center of the disk, [tex]\( \sigma \)[/tex] is the surface charge density, [tex]\( \epsilon_0 \)[/tex] is the vacuum permittivity [tex](\( 8.85 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2) \))[/tex], and R  is the radius of the disk.

Given:

- Radius of the disk, [tex]\( R = 35.0 \, \text{cm} = 0.35 \, \text{m} \)[/tex]

- Charge density, [tex]\( \sigma = 9.00 \times 10^{-3} \, \text{C}/\text{m}^2 \)[/tex]

- Vacuum permittivity,[tex]\( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/(\text{N} \cdot \text{m}^2) \)[/tex]

Now, let's calculate the electric field at the given distances:

a. At z = 5.00[tex]\text{cm} = 0.05 \, \text{m} \)[/tex]:

[tex]\[ E(0.05) = \frac{9.00 \times 10^{-3}}{2 \times 8.85 \times 10^{-12}} \left( 1 - \frac{0.05}{\sqrt{0.05^2 + 0.35^2}} \right) \] \[ E(0.05) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.05}{\sqrt{0.001225 + 0.1225}} \right) \][/tex]

[tex]\[ E(0.05) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.05}{0.3555} \right) \][/tex]

[tex]\[ E(0.05) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - 0.1406 \right) \] \[ E(0.05) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \times 0.8594 \] \[ E(0.05) = \frac{7.7346 \times 10^{-3}}{17.7 \times 10^{-12}} \] \[ E(0.05) = 4.369 \times 10^8 \, \text{N}/\text{C} \][/tex]

b. At z = 10.0[tex]\text{cm} = 0.1 \, \text{m} \)[/tex]:

[tex]\[ E(0.1) = \frac{9.00 \times 10^{-3}}{2 \times 8.85 \times 10^{-12}} \left( 1 - \frac{0.1}{\sqrt{0.1^2 + 0.35^2}} \right) \] \[ E(0.1) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.1}{\sqrt{0.01 + 0.1225}} \right) \] \[ E(0.1) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.1}{0.3674} \right) \][/tex]

[tex]\[ E(0.1) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - 0.2722 \right) \] \[ E(0.1) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \times 0.7278 \] \[ E(0.1) = \frac{6.5402 \times 10^{-3}}{17.7 \times 10^{-12}} \] \[ E(0.1) = 3.695 \times 10^8 \, \text{N}/\text{C} \][/tex]

c. At z = 50.0[tex]\text{cm} = 0.5 \, \text{m} \)[/tex]:

[tex]\[ E(0.5) = \frac{9.00 \times 10^{-3}}{2 \times 8.85 \times 10^{-12}} \left( 1 - \frac{0.5}{\sqrt{0.5^2 + 0.35^2}} \right) \] \[ E(0.5) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.5}{\sqrt{0.25 + 0.1225}} \right) \] \[ E(0.5) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{0.5}{0.6104} \right) \][/tex]

[tex]\[ E(0.5) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - 0.8193 \right) \] \[ E(0.5) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \times 0.1807 \] \[ E(0.5) = \frac{1.6263 \times 10^{-3}}{17.7 \times 10^{-12}} \] \[ E(0.5) = 9.187 \times 10^7 \, \text{N}/\text{C} \][/tex]

d. At z = 200 [tex]\text{cm} = 2.0 \, \text{m} \)[/tex]:

[tex]\[ E(2.0) = \frac{9.00 \times 10^{-3}}{2 \times 8.85 \times 10^{-12}} \left( 1 - \frac{2.0}{\sqrt{2.0^2 + 0.35^2}} \right) \] \[ E(2.0) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{2.0}{\sqrt{4 + 0.1225}} \right) \][/tex]

[tex]\[ E(2.0) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - \frac{2.0}{2.0317} \right) \] \[ E(2.0) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \left( 1 - 0.9848 \right) \][/tex]

[tex]\[ E(2.0) = \frac{9.00 \times 10^{-3}}{17.7 \times 10^{-12}} \times 0.0152 \] \[ E(2.0) = \frac{1.368 \times 10^{-4}}{17.7 \times 10^{-12}} \] \[ E(2.0) = 7.728 \times 10^6 \, \text{N}/\text{C} \][/tex]

Answer:

1205.77 J/kg.K

Explanation:

Heat lost by alloy = heat gained by water + heat gained by the calorimeter

c₁m₁(t₂-t₃) = c₂m₂(t₃-t₁) + c₃m₃(t₃-t₁)................. Equation 1

Where c₁ = specific heat capacity of the alloy, m₁ = mass of the alloy, t₂ = initial temperature of the alloy, t₃ = equilibrium temperature, c₂ = specific heat capacity of water, m₂ = mass of water, t₁ = initial temperature of water and calorimter, c₃ = specific heat capacity of calorimter, m₃ = mass of calorimter.

Making c₁ the subject of the equation,

c₁ = c₂m₂(t₃-t₁) + c₃m₃(t₃-t₁)/m₁(t₂-t₃)........................ Equation 2

Given: c₂ = 4190 J/kgK, m₂ = 250 g = 0.25 kg, m₁ = 75 g = 0.075 kg, m₃ = 55 g = 0.055 kg, c₃ = 377 J/kg.K, t₁ = 18 °C, t₂ = 100 °C, t₃ = 24.4 °C.

Substitute into equation 2

c₁ = [0.25×4190×(24.4-18) + 0.055×377×(24.4-18)]/[0.075(100-24.4)]

c₁ = (6704+132.704)/5.67

c₁ = 6836.704/5.67

c₁ = 1205.77 J/kg.K

Thus the specific heat capacity of the alloy = 1205.77 J/kg.K

Final answer:

The student's question is about calculating the specific heat capacity of a metal alloy using the principles of calorimetry and the conservation of energy in a heat exchange process.

The student is asking about finding the specific heat capacity of a metal alloy using calorimetry. We know that when objects at different temperatures are combined, they will exchange heat energy until they reach thermal equilibrium. We can use the equation Q = mc ext{ extdegree}T (where Q is heat energy, m is mass, c is specific heat capacity, and  ext{ extdegree}T is the change in temperature) to find the specific heat capacity. In this scenario, the heat lost by the metal alloy will equal the heat gained by the copper calorimeter and the water contained within it. By setting these two equations equal to each other and solving for the specific heat capacity of the alloy, we can find that value.

Answer:

Total 3 holes are available for conduction of current at 300K.

Explanation:

In order to develop a semiconductor, two type of impurities can be added as given below:

Now for estimation of extra electrons in the impured structure is as

[tex]N_{electrons-free}=n_{pentavalent \, atoms}\\N_{electrons-free}=4\\[/tex]

Now for estimation of "holes"  in the impured structure is as

[tex]N_{holes}=n_{trivalent \, atoms}\\N_{holes}=7\\[/tex]

Now when the free electrons and "holes" are available in the structure ,the "holes" will be filled by the free electrons therefore

[tex]N_{holes-net}=N_{holes}-N_{electrons-free}\\N_{holes-net}=7-4\\N_{holes-net}=3[/tex]

So total 3 "holes" are available for conduction of current at 300K.

Answer:

A. 8.175 N

B. 49.05 N

Explanation:

A.

Acceleration due to gravity (moon) = 1/6 * (acceleration due to gravity (earth)

Acceleration due to gravity (earth) = 9.81 m/s2

Acceleration due to gravity (moon) = 9.81/6

= 1.635 m/s

Weight. Fm = acceleration due to gravity * mass

= 1.635 * 5

= 8.175 N

B. Acceleration due to gravity (earth) * mass = Fe

= 9.81 * 5

= 49.05 N

Answer:

a. 2.645 * 10^7 m/s^2

b. 2.645 * 10^4 N

Explanation:

Parameters given:

Velocity of rod = 2010m/s

Length of rod = 15.3cm = 0.153m

Mass of object placed at the end of the rod = 1g = 0.001kg

a. Centripetal acceleration is given as:

a = (v*v)/r

Where v = velocity

r = radius of curvature.

The radius of curvature in this case is equal to the length of the rod, since the rod makes the circular path of the motion.

Hence, centripetal acceleration at the end of the rod:

a = (2010*2010)/(0.153)

a = 26432156.86 m/s^2 = 2.64 * 10^7 m/s

b. The force needed to hold the object at the end of the rod is equal to the centripetal force at the end of the rod. Centripetal force is given as:

F = ma = (m*v*v)/r

Where a = centripetal acceleration

F = 0.001 * 2.64 * 10^7

F = 2.64 * 10^4N

The Centripetal acceleration at the end of rod is [tex]\bold {2.64 x 10^7\ m/s}[/tex] and Centripetal force required to hold the object is [tex]\bold { 2.64x 10^4\ N}[/tex].

Given  here:

Velocity of rod = 2010 m/s

Length of rod = 15.3cm = 0.153 m

Mass of object placed at the end of the rod = 1g = 0.001 kg

(A). Centripetal acceleration can be calculated by  

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

Where

v = velocity

r = radius of curvature.

Put the values,

[tex]\bold {a = \dfrac {(2010^2)}{(0.153)}}\\\\\bold {a = 26432156.86\ m/s^2 } \\\\\bold {a = 2.64 x 10^7\ m/s}[/tex]

(B).  Centripetal force can be calculated by

[tex]\bold {F = ma = \dfrac {(m v^2)}{r}}[/tex]

Where,

a = centripetal acceleration

Thus the force

[tex]\bold {F = 0.001 \times 2.64x10^7}\\\\\bold {F = 2.64x 10^4\ N}[/tex]

Therefore, the Centripetal acceleration at the end of rod is [tex]\bold {2.64 x 10^7\ m/s}[/tex] and Centripetal force required to hold the object is [tex]\bold { 2.64x 10^4\ N}[/tex].

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Options

1. the same as in the first.

2.greater than in the first

3.less than in the first.

4.Unable to determine

Answer

The answer is 2. Greater than in the first

Explanation

The speed of a wave v is related to its wavelength λ by the formulav=f λ, where f is the frequency of the wave. The frequency will not change when the wave passes into a second medium, so

λ2>λ1

Fλ1>fλ2

Since f>0

And V2>v1

Answer:

[tex]f_{2}=180Hz,f_{3}=270Hz,f_{4}=360Hz\\[/tex]

Explanation:

Given data

Frequency f=90 Hz

To find

First three overtones of bassoon

Solution

The fundamental frequency of bassoon is found by substituting n=1 in below equation

f=v/λ=nv/2L

[tex]f_{1}=v/2L[/tex]

The first overtone of bassoon is found by substituting n=2

So

[tex]f_{2}=2v/2L\\f_{2}=2(v/2L)\\as \\f_{1}=v/2L\\So\\f_{2}=2f_{1}\\f_{2}=2(90Hz)\\f_{2}=180Hz[/tex]

The second overtone of bassoon is found by substituting n=3

So

[tex]f_{3}=3v/2L\\f_{3}=3(v/2L)\\as \\f_{1}=v/2L\\So\\f_{3}=3f_{1}\\f_{3}=3(90Hz)\\f_{3}=270Hz[/tex]

The third overtone of bassoon is found by substituting n=4

So

[tex]f_{4}=4v/2L\\f_{4}=4(v/2L)\\as \\f_{1}=v/2L\\So\\f_{4}=4f_{1}\\f_{4}=4(90Hz)\\f_{4}=360Hz[/tex]

The first three overtones of a bassoon with a fundamental frequency of 90.0 Hz are 180.0 Hz, 270.0 Hz and 360.0 Hz. The calculation is based on the behaviour of the bassoon as a tube open at both ends where overtones occur at integer multiples of the fundamental frequency.

The question is asking for the first three overtones of a bassoon that has a fundamental frequency of 90.0 Hz. The bassoon is assumed to act like a tube that is open at both ends. For a tube open at both ends, the overtones, also called harmonics, occur at integer multiples of the fundamental frequency.

In this case, the fundamental frequency (first harmonic) is 90.0 Hz. The first overtone (which is the second harmonic) is then 2 * 90.0 Hz = 180.0 Hz. The second overtone (third harmonic) is 3 * 90.0 Hz = 270 Hz, and the third overtone (fourth harmonic) is 4 * 90.0 Hz = 360 Hz.

Thus, the first three overtones of a bassoon that has a fundamental frequency of 90.0 Hz, and is open at both ends, are 180.0 Hz, 270.0 Hz, and 360.0 Hz respectively.

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Answer:

455N/C

Explanation:

from  the question, the following data can be derived

Distance between plates=2cm=0.02m

length of plates=4cm=0.04m

initial speed of electron=8*10^6m/s

Note: the speed giving is the speed associated with the horizontal motion since it moves to cover the 4cm distance

we solve the equation component by component.

For the horizontal component, the time it takes to cover distance of 0.04m can be calculated as

[tex]time=\frac{distance }{velocity} \\t=\frac{0.04}{8*10^{6}}\\ t=5*10^{-9}secs[/tex]

this same time is used to cover the vertical distance which is midway between the plate,

Hence vertical distance covered is 0.02/2=0.01m

The acceleration in the vertical component can be calculated as

[tex]y=ut+1/2at^{2}\\u=0,\\y=0.01m\\a=\frac{2y}{t^{2}}\\ a=\frac{2*0.01}{5*10^{-9}}\\ a=8*10^{14}m/s^{2}[/tex]

since

F=qE

also F=ma

then

qE=ma

E=(ma)/q

m=mass of electron=9.1*10^-32kq

q=charge of electron=1.6*10^-19c

a=acceleration

if we substitute values  

[tex]E=\frac{9.1*10^{-32}*8*10^{14}}{1.6*10^{-19}} \\E=455N/C[/tex]

To find the magnitude of the electric field when an electron is projected into a uniform electric field between two parallel plates, consider the forces acting on the electron.

An electron is projected horizontally into the uniform electric field directed vertically downward between two parallel plates.

The plates are 2.00 cm apart, and the initial speed of the electron is 8.00 × 10^6 m/s.

To calculate the magnitude of the electric field, you would need to consider the forces acting on the electron as it moves between the plates.

Answer:

A molecule is a combination of two or more atoms. The correct option is B.

Explanation:

Molecules are the smallest particle of a chemical compound that are made up of two or more atoms which are held together by a chemical bond. The chemical bonds are usually formed due to sharing and transfer of electrons among the atoms. Examples of molecules in chemistry includes:

- water molecule ( H2O)

- table salt ( NaCl)

- CaCl

Answer:

The time it takes the proton to return to the horizontal plane is 7.83 X10⁻⁷ s

Explanation:

From Newton's second law, F = mg and also from coulomb's law F= Eq

Dividing both equations by mass;

F/m = Eq/m = mg/m, then

g = Eq/m --------equation 1

Again, in a projectile motion, the time of flight (T) is given as

T = (2usinθ/g) ---------equation 2

Substitute in the value of g into equation 2

[tex]T = \frac{2usin \theta}{\frac{Eq}{m}} =\frac{m* 2usin \theta}{Eq}[/tex]

Charge of proton = 1.6 X 10⁻¹⁹ C

Mass of proton = 1.67 X 10⁻²⁷ kg

E is given as 400 N/C, u = 3.0 × 10⁴ m/s and θ = 30°

Solving for T;

[tex]T = \frac{(1.67X10^{-27}* 2*3X10^4sin 30}{400*1.6X10^{-19}}[/tex]

T = 7.83 X10⁻⁷ s

Final answer:

The force exerted on a proton in a magnetic field can be calculated using the given formula with specified values.

Explanation:

To find the time it takes for the proton to return to the horizontal plane, we need to consider the motion in both the horizontal and vertical directions separately.

Given:

[tex]- Initial velocity (\(v_0\)) of the proton = \(3.0 \times 10^4\) m/s\\- Launch angle (\(\theta\)) = 30°\\- Electric field (\(E\)) = 400 N/C\\- We'll assume the gravitational acceleration (\(g\)) as \(9.8 \, \text{m/s}^2\) downward.[/tex]

Vertical Motion:

In the vertical direction, the proton undergoes uniformly accelerated motion under gravity.

Using the formula for vertical motion:

[tex]\[ v = u + at \] \\where: \\- \( v \) is the final velocity (which is 0 when the proton returns to the horizontal plane),\\- \( u \) is the initial vertical velocity,\\- \( a \) is the acceleration due to gravity,\\- \( t \) is the time taken.\\[/tex]

We can resolve the initial velocity into vertical and horizontal components:

[tex]\[ u_y = v_0 \sin(\theta) \]\[ u_y = 3.0 \times 10^4 \times \sin(30^\circ) \]\[ u_y \approx 3.0 \times 10^4 \times 0.5 \]\[ u_y = 1.5 \times 10^4 \, \text{m/s} \][/tex]

Now, let's find the time it takes for the vertical velocity to become zero:

[tex]\[ 0 = u_y - gt \]\[ t = \frac{u_y}{g} \]\[ t = \frac{1.5 \times 10^4}{9.8} \]\[ t \approx 1530.61 \, \text{s} \][/tex]

Horizontal Motion:

In the horizontal direction, there is no acceleration or deceleration acting on the proton. So, the time taken for horizontal motion is the same as the time taken for vertical motion.

The time it takes for the proton to return to the horizontal plane is approximately [tex]\( 1530.61 \) seconds.[/tex]

Answer:

Total work done in expansion will be [tex]3.60\times 10^5J[/tex]

Explanation:

We have given pressure P = 2.10 atm

We know that 1 atm [tex]=1.01\times 10^5Pa[/tex]

So 2.10 atm [tex]=2.10\times 1.01\times 10^5=2.121\times 10^5Pa[/tex]

Volume is increases from 3370 liter to 5.40 liter

So initial volume [tex]V_1=3.70liter[/tex]

And final volume [tex]V_2=5.40liter[/tex]

So change in volume [tex]dV=5.40-3.70=1.70liter[/tex]

For isobaric process work done is equal to [tex]W=PdV=2.121\times 10^5\times 1.70=3.60\times 10^5J[/tex]

So total work done in expansion will be [tex]3.60\times 10^5J[/tex]

Answer:

The change in the electric potential energy is [tex]-3.2\times10^{-16}\ J[/tex]

Explanation:

Given that,

Distance [tex]d=4.30\times10^{-11}\ m[/tex]

suppose, Two particles with charges +6 e and -10 e are initially very far apart

We need to calculate the change in the electric potential energy

Using formula of energy

[tex]\text{electric potential energy}=\text{final electric potential energy-initial electric potential energy}[/tex]

[tex]EPE=EPE_{f}-EPE_{i}[/tex]

Here, initial electric potential energy= 0

final electric potential energy [tex]EPE_{f}=\dfrac{kq_{1}q_{2}}{r_{2}^2}[/tex]

Put the value into the formula

[tex]EPE=\dfrac{kq_{1}q_{2}}{r_{2}^2}+0[/tex]

Put the value into the formula

[tex]EPE=\dfrac{9\times10^{9}\times6\times1.6\times10^{-19}\times(-10\times1.6\times10^{-19})}{(4.30\times10^{-11})}[/tex]

[tex]EPE=-3.2\times10^{-16}\ J[/tex]

Hence, The change in the electric potential energy is [tex]-3.2\times10^{-16}\ J[/tex]

This is an incomplete question, here is a complete question.

Two particles with charges +6 e⁻ and -10 e⁻ are initially very far apart (effectively an infinite distance apart). They are then fixed at positions that are 4.30 × 10⁻¹¹ m apart. What is EPE(final) - EPE(initial), which is the change in the electric potential energy?

Answer : The change in the electric potential energy is, [tex]-2.92\times 10^{-6}J[/tex]

Explanation : Given,

Formula used for electric potential energy of the two charges when they are separated is:

[tex]EPE=\frac{1}{4\pi \epsilon_0}\times {\frac{q_1\times q_2}{r^2}[/tex]

[tex]EPE=\frac{k\times q_1\times q_2}{r^2}[/tex]

where,

EPE = electric potential energy

k = [tex]\frac{1}{4\pi \epsilon_0}=8.99\times 10^9[/tex]

[tex]q_1[/tex] = charge on 1st particle = +6 e⁻ = [tex]6\times 10^{-19}C[/tex]

[tex]q_2[/tex] = charge on 2nd particle = -10 e⁻ = [tex]-10\times 10^{-19}C[/tex]

r = distance between two charges = [tex]4.30\times 10^{-11}m[/tex]

Now put all the given values in the above formula, we get:

[tex]EPE=\frac{(8.99\times 10^9)\times (6\times 10^{-19})\times (-10\times 10^{-19})}{(4.30\times 10^{-11})^2}[/tex]

[tex]EPE=-2.92\times 10^{-6}J[/tex]

Initially EPE = 0 J

Thus, [tex]EPE_{final}-EPE_{initial}=-2.92\times 10^{-6}J[/tex]

The positive sign indicate the attractive force and negative sign indicate the repulsive force.

Thus, the change in the electric potential energy is, [tex]-2.92\times 10^{-6}J[/tex]

Answer:

67 dB

Explanation:

given,

Sound in afternoon = 80 dB

Intensity of car,I₀ = 100 cars/ minute

Sound in the night = ?

Intensity of car,I = 5 car/minutes

using formula for sound calculation

 [tex]\Delta \beta = 10 log(\dfrac{I}{I_0})[/tex]

 [tex]\Delta \beta = 10 log(\dfrac{5}{100})[/tex]

 [tex]\Delta \beta = 10 log(\dfrac{1}{20})[/tex]

 [tex]\Delta \beta = 10 log(0.05)[/tex]

 [tex]\Delta \beta = 10\times -1.30[/tex]

 [tex]\Delta \beta = -13\ dB[/tex]

The late night decibel is equal to 80 dB - 13 dB = 67 dB

Answer:

0.080 m

Explanation:

According to Hooke's law, a spring with stiffness k will stretch a distance of Δx when a force F is applied:

F = k Δx

If we say the weight of the object is W, then the stiffness of the original spring is:

W = k (0.160 m)

k = W / 0.160

When the spring is cut in half, the stiffness of each new spring is the same as the original.  This time, the weight of the object is evenly distributed between each spring, so the force on each is W/2.

F = k Δx

W/2 = (W/0.160) Δx

1/2 = Δx / 0.160

Δx = 0.080

Each spring stretches 0.080 meters.

Electric field due to uniformly charged metal plate is given by,

[tex]E = \frac{\sigma}{(2\epsilon_0)}[/tex]

Here,

[tex]\sigma[/tex] = Charge density

[tex]\epsilon_0 =[/tex]  Vacuum Permittivity

Our values are,

[tex]\sigma = 0.75 muC/m^2 = 0.75*10^-6 C/m^2[/tex]

[tex]\epsilon_0 = 8.85*10^-12 F\cdot m^{-1}[/tex]

Replacing we have,

[tex]E = \frac{(0.75*10^-6)}{(2*8.85*10^-12)}[/tex]

[tex]F = 42372.88N/C[/tex]

Now we have the relation where energy is equal to the change of the potential in a certain distance, then

[tex]E = \frac{V}{d}[/tex]

Rearranging for the distance

[tex]d = \frac{V}{E}[/tex]

[tex]d = \frac{100}{42372.88}[/tex]

[tex]d = 0.00236m[/tex]

[tex]d = 2.36mm[/tex]

Therefore the distance is 2.36mm

Answer:

A) [tex]k=34867.3384\ N.m^{-1}[/tex]

B) [tex]\omega'\approx84\ Hz[/tex]

Explanation:

Given:

mass of car, [tex]m=1380\ kg[/tex]

A)

frequency of spring oscillation, [tex]f=1.6\ Hz[/tex]

We knkow the formula for spring oscillation frequency:

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

[tex]\Rightarrow \sqrt{\frac{k_{eq}}{m} } =2\pi.f[/tex]

[tex]\sqrt{\frac{k_{eq}}{1380} } =2\times \pi\times 1.6[/tex]

[tex]k_{eq}=139469.3537\ N.m^{-1}[/tex]

Now as we know that the springs are in parallel and their stiffness constant gets added up in parallel.

So, the stiffness of each spring is (as they are identical):

[tex]k=\frac{k_{eq}}{4}[/tex]

[tex]k=\frac{139469.3537}{4}[/tex]

[tex]k=34867.3384\ N.m^{-1}[/tex]

B)

given that 4 passengers of mass 70 kg each are in the car, then the oscillation frequency:

[tex]\omega'=\sqrt{\frac{k_{eq}}{(m+70\times 4)} }[/tex]

[tex]\omega'=\sqrt{\frac{139469.3537}{(1380+280)} }[/tex]

[tex]\omega'\approx84\ Hz[/tex]

(i) 625 V/m

(ii) 2876.25 x 10⁻¹² C

(iii) 0.000575.25 x 10⁻⁶ F

(i) The electric field (E) between the plates of a parallel plate capacitor is related to the potential difference (V) between the plates and the distance (d) of separation between the plates as follows;

E = V / d        ----------------(i)

From the question;

V = 5.0V

d = 0.80cm =  0.008m

Substitute these values into equation (i) as follows;

E = 5.0 / 0.008

Solve for E;

E = 625 V/m

Therefore, the electric field in the region between the two plates is 625 V/m.

(ii) To make things easier, let's calculate the capacitance of the parallel plates first.

The capacitance (C) of a parallel plate capacitor is given as;

C = A x ε₀ / d         --------------------------(ii)

Where;

A = Area of either of the plates of the capacitor = 0.52m²

ε₀ = permittivity of free space = 8.85 x 10⁻¹² F/m

d = distance between the plates =  0.8cm = 0.008m

Substitute these values into equation (ii) as follows;

C = 0.52 x 8.85 x 10⁻¹² / 0.008

Solve for C;

C = 575.25 x 10⁻¹² F

The capacitance (C) is related to potential difference (V) and charge (Q) on the plates as follows;

Q = C x V            -------------------------(iii)

Where;

C = 575.25 x 10⁻¹² F

V = 5.0V

Substitute these values into equation (iii)

Q = 575.25 x 10⁻¹² x 5

Q = 2876.25 x 10⁻¹² C

Therefore, the charge on the plates is 2876.25 x 10⁻¹² C

(iii) The capacitance (C) of the parallel plates has been calculated in (ii) above.

Its value is 575.25 x 10⁻¹² F = 0.000575.25 x 10⁻⁶F

The electric field in the region between two plates is 625 V/m. The capacitance of the plates is 0.0572 nanoFarads and the accumulated charge Q on each plate is 2.86 x 10^-10 Coulombs.

The electric field between two plates can be calculated using the formula E = V/d, where V represents voltage and d represents the distance between plates. So, the electric field E would be 5.0 V / 0.008 m = 625 N/C or 625 V/m.  The charge Q on the plates can be determined using the formula Q = CV, where C is the capacitance and V is the voltage. To find the charge, we first need to calculate the capacitance. The capacitance of the parallel plates can be calculated using the formula C = ε0*(A/d), where ε0 is the permittivity of free space (8.85 x 10^-12 F/m), A is the area and d is the distance between the plates. The capacitance then would be 8.85 x 10^-12 F/m * 0.52 m^2 / 0.008 m = 5.72 x 10^-11 F or 0.0572 nF (nanoFarads). So, after finding the capacitance, we can now calculate the charge Q, which is Q = CV = 5.72 x 10^-11 F * 5.0 V = 2.86 x 10^-10 C (Coulombs).

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Explanation:

Relation between work and change in kinetic energy is as follows.

                 [tex]W_{net} = \Delta K[/tex]

Also,   [tex]\Delta K = K_{initial} - K_{final}[/tex]

                        = [tex](0 - 4.0 \times 10^{-17})[/tex] J

                        = [tex]-4.0 \times 10^{-17}[/tex] J

Let us assume that electric force on the electron has a magnitude F. The electron moves at a distance of 0.3 m opposite to the direction of the force so that work done is as follows.

                w = -Fd

       [tex]-4.0 \times 10^{-17} J = -F \times 0.3 m[/tex]

                F = [tex]1.33 \times 10^{-16}[/tex]  

Therefore, relation between electric field and force is as follows.

              E = [tex]\frac{F}{q}[/tex]

                 = [tex]\frac{1.33 \times 10^{-16}}{1.60 \times 10^{-19} C}[/tex]

                 = [tex]0.831 \times 10^{3}[/tex] C

Thus, we can conclude that magnitude of the electric field that will stop these electrons in a distance of 0.3 m is [tex]0.831 \times 10^{3}[/tex] C.

Answer:

g/9

Explanation:

length of the pendulum = L

time period on the earth = T

Time period on the planet = 3T

Let the acceleration due to gravity on the earth is g and on the planet is g'.

Use the formula for the time period of a simple pendulum for the time period on earth

[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex]     .... (1)

Time period on the surface of planet is

[tex]3T=2\pi \sqrt{\frac{L}{g'}}[/tex]      .... (2)

Divide equation (2) by equation (1)

[tex]\frac{3T}{T}= \sqrt{\frac{g}{g'}}[/tex]

g' = g/9

Thus, the acceleration due to gravity on the planet is g /9

Answer:

b. Throwing some cargo out of the boat

Explanation:

Using the Newton's third law of motion which states that every action has an equal and opposite reaction.

So when we are on the boat and we throw some mass in a direction away from the boat out of it then we are imparting the force to the floating boat by the law of conservation of momentum as well.

The relation can be mathematically expressed as:

[tex]m_c.v_c=m_b.v_b[/tex]

where:

[tex]m_c=[/tex] mass of the cargo

[tex]v_c=[/tex] velocity of throwing the cargo

[tex]m_b=[/tex] mass of the whole boat including all that floats on it

[tex]v_b=[/tex] velocity of the boat system

Throwing some cargo out of the boat will make the boat start moving.

This law states that action and reaction are equal and opposite.

The total momentum of an isolated system is always conserved.

Throwing some cargo out of the boat will change the velocity of the cargo initially at rest and to conserve the linear momentum, the velocity of the boat will change as well.

Thus, throwing some cargo out of the boat will make the boat start moving.

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The final velocity of the mass is calculated using the principle of conservation of energy, which states that the potential energy in the spring is converted into kinetic energy and work done against friction as the mass moves down the incline. This total energy is then equal to the kinetic energy at the end of the path, from which the final velocity of the mass can be found.

To calculate the final velocity of the mass after it has been released from the spring, we need to realize that the energy in the system is conserved. The potential energy stored in the spring when it is compressed is converted into kinetic energy and work done against friction as the mass skids down the incline. The spring energy is given by (1/2)kx^2, where 'k' is the spring constant and 'x' is the amount of compression. In this case, it is (1/2)*2.2 N/cm*(28 cm)^2. We convert all values to standard SI units before calculation for accurate results.

The work done against the friction in the path of 0.8 m is friction force times the distance, which = μmgcosθ*d, where 'μ' is the friction coefficient, 'm' the mass, 'g' gravity, 'θ' the angle of incline, and 'd' the distance with friction. Here, it is 0.3*5 kg*9.8 m/s^2*cos(18 degrees)*0.8 m.

The total energy is thus the sum of spring energy and work done against friction. As this energy is equal to the kinetic energy at the end of the path (1/2)*m*v^2, we can solve this to find the final velocity 'v'

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The final velocity of a 5 kg mass sliding down a frictional incline is calculated using energy conservation and work-energy principles, resulting in a final velocity of approximately 2.25 m/s.

Calculating the Final Velocity:

To determine the final velocity of a 5 kg mass after it slides down a frictional incline, we need to consider both energy conservation and work-energy principles.

Step-by-Step Solution

Spring Potential Energy: The spring is compressed by 28 cm, and the spring constant k is 2.2 N/cm. First, convert these units to meters and N/m: k = 220 N/m and compression x = 0.28 m. The potential energy stored in the spring is given by:

PE_spring = 0.5 x k x x².

PE_spring = 0.5 x 220 x (0.28)² = 8.624 J.

Gravitational Potential Energy: As the mass slides down the incline of height h = 0.31 m, the gravitational potential energy is converted to kinetic energy. PE_gravity = m x g x h

PE_gravity = 5 kg x 9.8 m/s² x 0.31 m = 15.19 J.

Energy Loss Due to Friction: The mass encounters a frictional patch of length d = 0.8 m with a friction coefficient μ = 0.3. The work done by friction (which is energy lost) is calculated by:

W_friction = μ x m x g x cos(θ) x d.

Given θ = 18°, cos(18°) ≈ 0.9511.

W_friction = 0.3 x 5 kg x 9.8 m/s² x 0.9511 x 0.8 m = 11.16 J.

Total Mechanical Energy: The initial potential energy (spring + gravity) is:

8.624 J + 15.19 J = 23.814 J.

After accounting for the work done by friction, the remaining energy will be:

KE_final = 23.814 J - 11.16 J = 12.654 J.

Final Velocity: The kinetic energy at the bottom of the incline is converted into the final velocity.

Using the relation:

KE = 0.5 x m x vf².

12.654 J = 0.5 x 5 kg x vf²

vf² = (2 x 12.654 J) / 5 kg = 5.06 m²/s²

vf = √5.06 m²/s² ≈ 2.25 m/s.

Thus, the final velocity of the mass is approximately 2.25 m/s.

Answer:

Δx=629.35 m

The pilot release the hay 629.35 m in front of the cattle so that the bales land at the point where the cattle are stranded.

Explanation:

Step 1:

Finding initial velocity components:

Initial velocity=v=75 m/s

α=55

[tex]v_{ox}=vcos\alpha\\v_{ox}=75cos55^o\\v_{ox}=43.018 m/s\\v_{oy}=vsin\alpha\\v_{oy}=75sin55^o\\v_{oy}=61.436 m/s[/tex]

Step 2:

[tex]y_o=150\ m[/tex]

Newton Second Equation:

[tex]y-y_o=v_{oy}t+\frac{1}{2}g t^2[/tex]

g=-9.8 m/s^2 (Downward direction)

[tex]v_{oy}=61.436\ m/s[/tex]

y=0 m

Above equation will become:

-150=(61.436)t-(4.90)t^2

Solving the above quadratic equation we will get:

t=-2.09 sec           ,        t=14.63 sec

t= 14.63 sec

Step 3:

Finding the distance:

Using Again Newton equation of motion in x-direction:

[tex]x-x_o=v_{ox}t+\frac{1}{2}a_{x} t^2[/tex]

Since velocity is constant in x- direction, [tex]a_x[/tex] will be zero.

Above equation will be:

[tex]\Delta x=v_{ox}t[/tex]

Δx=(43.018)(14.63)

Δx=629.35 m

The pilot release the hay 629.35 m in front of the cattle so that the bales land at the point where the cattle are stranded.

The pilot should release the hay at a height of 629.35 m.

Given information,

Initial velocity = 75 m/s

Velocity For x-component,

[tex]\bold {V_0x = Vcos \alpha}\\\\\bold {V_0x = 75 cos 55^o}\\\\\bold {V_0x = 43. 018m/s}[/tex]

Velocity for Y-component

[tex]\bold {V_0y = Vsin \alpha}\\\\\bold {V_0y = 75 sin 55^o}\\\\\bold {V_0y = 61. 43m/s}[/tex]

Using Newton's second equation for y-axis,

[tex]\bold {y-y_0 = V_0t + \dfrac {1}{2} gt^2}[/tex]

Where,

g - gravitational acceleration

put the values in the equation,

[tex]\bold {-150=(61.436)t-(4.90)t^2}[/tex]

Solving this quadratic equation, we get 2 values

t = 14.29 s

To find the distance, use Newton's second equation,

[tex]\bold {x-x_0 = V_0t + \dfrac {1}{2} gt^2}[/tex]

Since acceleration is zero because the velocity is constant in x-axis hence .

So,

[tex]\bold {x-x_0 = V_0_xt }[/tex]

[tex]\bold {x- x_0=(43.018)(14.63)}\\\\\bold {x - x_0=629.35 m}[/tex]

Therefore, the pilot should release the hay at 629.35 m.

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Answer:

She must run 59 min to run 11.8 km.

Explanation:

Hi there!

First let's convert mi/h into km/min:

7.4 mi/h · (1.61 km /1 mi) · (1 h / 60 min) = 0.20 km/min (notice how the units mi and h cancel).

The runner runs at 0.20 km/ min, i.e., every minute she travels 0.20 km.

If 0.20 km are traveled in 1 min, then 11.8 km will be traveled in:

11.8 km / 0.20 km/min = 59 min

She must run 59 min to run 11.8 km.

Answer:

Explanation:

All objects on Earth are subjected to a constant gravitational acceleration g = 9.8m/s2, wherever they are on the surface of Earth and whatever their speed is. So if an object is being thrown to its highest point and stopped moving at that instant, that means the velocity at that instant is 0, not the acceleration. The acceleration is still g = 9.8m/s2

Answer:

1. d = 0.415 m.

2. Q = 2.285 x 10^{-10} C.

Explanation:

The electric field and potential can be found by the following equations:

[tex]E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\\V = \frac{1}{4\pi\epsilon_0}\frac{Q}{r}[/tex]

Applying these equations to the given variables yields

[tex]E = 12 = \frac{1}{4\pi\epsilon_0}\frac{Q}{d^2}\\V = 4.98 = \frac{1}{4\pi\epsilon_0}\frac{Q}{d}[/tex]

Divide the first line to the second line:

[tex]\frac{12}{4.98} = \frac{ \frac{1}{4\pi\epsilon_0}\frac{Q}{d^2}}{\frac{1}{4\pi\epsilon_0}\frac{Q}{d}}\\\frac{12}{4.98} = \frac{1}{d}\\d = 0.415~m[/tex]

Using this distance in either of the equations give the magnitude of the charge.

[tex]12 = \frac{1}{4\pi\epsilon_0}\frac{Q}{(0.415)^2}\\12 = \frac{1}{4\pi (8.8\times 10^{-12})}\frac{Q}{(0.415)^2}\\Q = 2.285 \times 10^{-10}~C[/tex]

Final answer:

The distance to the point charge is 0.415 meters. The magnitude of the charge is 231 picocoulombs (pC).

Explanation:

To calculate the distance to the point charge when the electric potential is 4.98 V and electric field magnitude is 12.0 V/m, we can use the relationship between electric field (E) and electric potential (V), which is E = -dV/dr. Rearranging the equation to solve for the distance (r), we get r = V/E = 4.98 V / 12.0 V/m = 0.415 m.

To find the magnitude of the charge (q), we can use the electric potential formula for a point charge: V = k*q/r, where k is the Coulomb's constant (8.99 x [tex]10^9 Nm^2/C^2[/tex]). Rearranging the formula and solving for q, we get q = V*r/k = (4.98 V * 0.415 m) / (8.99 x [tex]10^9 Nm^2/C^2[/tex]) = 2.31 x [tex]10^-^1^0[/tex] C or 231 pC.