when the mass of water that an iceberg displaces is equal to the mass of the iceberg, it floats. this is an example of group of answer choices upwelling. isostacy. tomography. gravity.

Wire's resistivity is approximately 3.72 × 10^−8 Ωm.

To find the wire's resistivity, we can follow these steps:

1. Calculate the wire's cross-sectional area.
2. Use Ohm's Law to find the resistance.
3. Use the resistance and cross-sectional area to find the resistivity.

Step 1: Calculate the wire's cross-sectional area.
The wire has a diameter of 3.1 mm, so its radius is 1.55 mm. Convert the radius to meters (1.55 × 10⁻³ m). The cross-sectional area (A) of the wire can be calculated using the formula A = πr², where r is the radius.

A = π(1.55 × 10⁻³ m)² ≈ 7.54 × 10⁻⁶ m²

Step 2: Use Ohm's Law to find the resistance.
Ohm's Law states that V = IR, where V is voltage, I is current, and R is resistance. We know the electric field (E) is 6.9 × 10⁻² V/m and the current (I) is 14 A. To find the voltage (V), we can multiply the electric field by the length of the wire (L). However, we don't know the length of the wire. Instead, we can find the resistance per unit length (R/L) by dividing both sides of Ohm's Law by L:

(V/L) = I(R/L) → E = I(R/L)

Now, we can solve for (R/L):

(R/L) = E/I = (6.9 × 10⁻² V/m) / 14 A ≈ 4.93 × 10⁻³ Ω/m

Step 3: Use the resistance and cross-sectional area to find the resistivity.
Resistivity (ρ) can be calculated using the formula ρ = (R × A) / L. Since we have the resistance per unit length (R/L) and the cross-sectional area (A), we can write the formula as:

ρ = (R/L) × A

Now, plug in the values:

ρ = (4.93 × 10⁻³ Ω/m) × (7.54 × 10⁻⁶ m²) ≈ 3.72 × 10⁻⁸ Ωm

So, the wire's resistivity is approximately 3.72 × 10⁻⁸ Ωm.

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The fermi energy for silver, assuming the effective mass equals the free electron mass is 5.58 x 10⁻¹⁹ J.

To calculate the Fermi energy for silver, use the following formula:

E_F = (h² / 2m) × (3π² × n)^(2/3)

where:

h = Planck's constant = 6.626 x 10⁻³⁴ J s

m = effective mass of electron = mass of free electron = 9.109 x 10⁻³¹ kg

n = number density of free electrons = 6.1 x 10²² electrons/cm³

Substituting the given values into the formula:

E_F = (6.626 x 10⁻³⁴ J s)² / (2 × 9.109 x 10⁻³¹ kg) × (3π² × 6.1 x 10²² cm⁻³)^(2/3)

E_F = 5.58 x 10⁻¹⁹  J

Therefore, the Fermi energy for silver is approximately 5.58 x 10⁻¹⁹ J.

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When placed near a negatively charged metal sphere.The electric field strength at this location is 1.25 × 10⁴ N/C directed toward the sphere .

Electric field strength = 6.00 ×10⁻² N -   4 .80 × 10⁶ C

                                                     =   1.25 × 10⁴N/C

The intensity of an electric field at a specific location is quantified by its electric field strength. The volt per meter (V/m or Vm-1) is the most common measurement. A potential difference of one V between two points separated by one meter is represented by a field strength of one V/m. The intensity of an electric field at a specific location is quantified by its electric field strength.

Where electric field lines are closer together, the field is stronger, and where they are further apart, the field is weaker. As you get further away from a point charge, the electric field gets weaker.

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A full water tank in the shape of a hemispherical bowl of radius 5 m. 5 m. 8.6875 × [tex]10^5[/tex] J work is required to pump all the water to a height of 5 m 5 m above the tank

The amount of work required to pump all the water in the full water tank of the shape of a hemispherical bowl of radius 5m to a height of 5m above the tank is 8.6875 x [tex]10^5[/tex] J.

Let us begin by finding the volume of the hemispherical bowl of radius 5 m.

V = (2/3)πr³

where r is the radius of the hemispherical bowl.

Substituting r = 5 m in the above formula for volume,

we get

V = (2/3) × π × 5³ m³ = 523.6 m³

Since the bowl is full of water, the volume of water it contains is also 523.6 m³.

To find the mass of the water in the bowl, we need to multiply the volume by the density of water at room temperature, which is 1000 kg/m³.

Mass of water in the bowl = 523.6 × 1000 kg = 523600 kg

The potential energy of the water when it is lifted to a height of 5m above the tank is given by

mgh

where m is the mass of water, g is the acceleration due to gravity and h is the height of the water above the tank.

Substituting the given values, we get

Potential energy of water = 523600 × 9.81 × 5 J = 25690800 J

Therefore, the amount of work required to pump all the water to a height of 5m above the tank is equal to the potential energy of the water, which is 25690800 J, rounded off to 8.6875 × 10^5 J.

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Wegener's theory of continental drift was an important early step in our understanding of how the continents move, but it lacked a complete explanation for the forces driving their movement.

Alfred Wegener's theory of continental drift proposed that the continents were once joined together as a single landmass, which he called Pangaea, and that they gradually moved apart over time. He suggested that the continents moved due to the forces of tides, centrifugal forces, and other factors. Although Wegener's theory was met with skepticism when it was first proposed in the early 20th century, it is now widely accepted as a valid explanation for the movement of continents.

Wegener based his theory on several lines of evidence, including the fit of the continents, the distribution of rocks and fossils, and the presence of glacial deposits in areas that are now too warm for glaciers. However, his explanation for the forces driving the movement of the continents was incomplete and lacked a plausible mechanism. It was not until the discovery of plate tectonics in the mid-20th century that a more complete explanation for continental drift was provided.

Plate tectonics explains the movement of the continents by the movement of the plates that make up the Earth's crust. The plates are driven by convection currents in the mantle, which move material up from the hot interior of the Earth, causing the plates to move apart, slide past each other, or collide. As the plates move, they carry the continents with them, causing them to drift apart or come together.

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Indeed, Alfred Wegener's assertions make logical. His hypothesis of continental drift, which is backed by a variety of geological and paleontological data, postulated that the continents were originally united together in a single landmass before drifting apart.

According to Alfred Wegener's hypothesis of continental drift, Pangaea, all the continents previously belonged to a single supercontinent. He thought the continents drifted to their present locations over millions of years after the supercontinent started to disintegrate some 200 million years ago. The matching of geological characteristics and fossils across continents, the resemblance of rock formations and mountain ranges, and the discovery of magnetic stripes on the ocean floor have all been used to support this idea, which was previously viewed with suspicion. In light of this, Wegener's assertions that the continents were originally linked and then drifted apart make sense, and his hypothesis has completely changed how we think about the Earth's past and current dynamics.

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The given statement "Before rotating the platform, the hanging mass is disconnected from the test mass and removed from the platform." is True because the concept of a rotating platform involves a disc that rotates about its central axis with a pendulum suspended from its edge.

The plane of rotation and the plane of the pendulum oscillation are separated by a tiny angle. A rotating platform is utilized to generate an artificial gravity environment in space. The centrifugal acceleration produced by rotation is used to imitate the gravitational pull of Earth's mass on objects.

The pendulum is an instrument that measures acceleration, and it functions by oscillating with a period that is dependent on the magnitude of the acceleration it experiences. Besides, it has a number of other applications, including scientific research, engineering tests, and astronaut training in simulated gravity. The test mass is left alone on the platform while the hanging mass is disconnected and removed from the platform before the platform is rotated.

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The block is moving at 6.0 m/s when the spring is compressed 15 cm.

We can use conservation of energy to solve this problem. The initial kinetic energy of the block will be converted to potential energy when the spring is compressed. At this point, the block will momentarily come to a stop before bouncing back, so its velocity will be zero.

Let's first calculate the potential energy stored in the spring when it is compressed 15 cm:

Δx = 15 cm = 0.15 m (conversion from cm to m)

k = 2.0 kN/m = 2000 N/m (conversion from kN/m to N/m)

x = Δx = 0.15 m

[tex]U = (1/2) k x^2 = (1/2) * 2000 * (0.15)^2 = 22.5 J[/tex]

Now, we can equate this potential energy to the initial kinetic energy of the block:

[tex]K = (1/2) mv^2 = (1/2) * m * (6.0)^2 = 18 m J[/tex]

where m is the mass of the block.

Equating the two equations, we get:

18 m J = 22.5 J

[tex]m = 22.5 J / 18 (m/s)^2 = 1.25 kg[/tex]

Now that we know the mass of the block, we can use conservation of momentum to find its velocity when the spring is compressed:

Before the collision:

[tex]m_1 = 1.25 kg[/tex]

[tex]v_1 = 6.0 m/s[/tex]

After the collision:

[tex]m_2 = 1.25 kg[/tex]

[tex]v_2 = ?[/tex]

Conservation of momentum:

[tex]m_1v_1 = m_2v_2[/tex]

[tex]v_2 = m_1v_1/m_2 = (1.25 kg) * (6.0 m/s) / (1.25 kg) = 6.0 m/s[/tex]

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The increase in potential by 1260 V would take t = 1260V * (2π * 0.15)/(μ * 0.0000020) seconds.

The potential difference across a conducting sphere of radius r is given by V = (μ I)/2πr, where μ is the magnetic permeability and I is the current passing through the sphere.

In this case, the radius of the sphere is 15 cm, the current passing into the sphere is 1.0000020 A, and the current passing out of the sphere is 1.0000000 A.

Substituting the given values into the equation for potential difference, we get:

V = (μ (1.0000020 - 1.0000000)) / (2π * 0.15)

Therefore, the increase in potential by 1260 V would take t = 1260V * (2π * 0.15)/(μ * 0.0000020) seconds.

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The time could it sustain a current of 1.3 A is 3.96 minutes.

To calculate the time it can sustain a current of 1.3 A, we divide the Power = Voltage x Current

in the battery (in Joules) by the current (in Amps).

The voltage of the battery is 1.3 V

The energy stored in the battery is 6.7 kJ

The current drawn by the battery is 1.3 A

We know that,

Power = Voltage x Current

P = 1.3 x 1.3

P = 1.69 W

Now,

Energy = Power x Time

6.7 kJ = 1.69 W x Time

Time = 6.7 kJ/1.69 W

Time = 3.962 min ≈ 3.96 min

Hence, the battery could sustain a current of 1.3 A for about 3.96 minutes.

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A farsighted person has a near point of 50 cm, and to bring their near point to 25 cm, we will need a converging lens.

Let the required lens' focal length be f, and the near point be p1. We will now use the lens formula to determine the strength of the lens (in diopters).

Formula: 1/f = 1/p1 + 1/p2

Since the lens formula is expressed in meters, we must first convert the near point to meters: 50 cm = 0.5 mp1 = 0.5 m - 0.25 m (because we want the image to be formed 25 cm away from the eye) = 0.25 m

Putting in these values in the above formula, we get:1/f = 1/0.25 - 1/0.5 1/f = 4 - 2 f = 1/2 f = 0.5 m Diopters are defined as the reciprocal of the focal length in meters, and we get the strength of the lens as:

Strength of the lens = 1/f = 1/0.5 = 2 diopters. Therefore, a lens of 2 diopters strength will be required to bring the farsighted person's near point to 25 cm.

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The movement of crustal plates is best described as a continuing long-term process.

Plate tectonics is a theory of geology that describes the movement and interactions of lithospheric plates that cover the Earth's surface. There are three types of plate boundaries: divergent, convergent, and transform. They all have varying rates of movement and result in various geological phenomena.

As a result, the movement of crustal plates is best described as a continuing long-term process. Crustal plates move slowly over time as they are pushed and pulled by the Earth's internal forces. The direction and speed of movement vary from one plate to another, and the rate of movement is determined by several factors, including the thickness and composition of the lithosphere.

Plate tectonics is a process that operates on a global scale and has a significant impact on the Earth's surface features and geological history.

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Brody initially hears the frequency of the toy as 1071 Hz (approximately).

To solve this problem, we need to use the Doppler effect formula, which relates the frequency of a wave to the relative motion between the source and the observer.

The frequency heard by Brody initially:

[tex]f' = f * (v + vo) / (v + vs)[/tex]

where f is the frequency of the toy (950 Hz), v is the speed of sound (assumed to be 343 m/s at room temperature), vo is the velocity of the observer (Brody, running at 7 m/s), and vs is the velocity of the source (the toy, moving away from Brody at 34 m/s).

[tex]f' = 950 Hz * (343 m/s + 7 m/s) / (343 m/s - 34 m/s)\\\f' = 950 Hz * 350 / 309\\\\\f' = 1071 Hz (approximately)[/tex]

Therefore, Brody initially hears the frequency of the toy as 1071 Hz (approximately).

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The magnification produced by the convex mirror when the object distance is (a) 14 cm is 0.63 and (b) 16 cm is 0.6.To find the magnification produced by a convex mirror with a focal length of -24 cm when the object distance is  14 cm and  16 cm, we'll first need to calculate the image distance for each case using the mirror formula.

Mirror formula = 1/f = 1/v + 1/u, Where f is the focal length, v is the image distance, and u is the object distance.
(a) When the object distance (u) is 14 cm:
1/f = 1/v + 1/u
1/-24 = 1/v + 1/14
Now, solve for the image distance (v):
1/v = 1/-24 - 1/14
1/v = -38/336
v = 336/38
v = -8.84 cm Now we can find the magnification (M) using the formula: M = -v/u

For the object distance of 14 cm:
M = -(-8.84)/14
M = 0.63

(b) When the object distance (u) is 16 cm:
1/f = 1/v + 1/u
1/-24 = 1/v + 1/16
Solve for the image distance (v) again:
1/v = 1/-24 - 1/16
1/v = -40/384
v = 384/40
v = -9.6 cm. we can find the magnification (M) using the formula: M = -v/u

For the object distance of 16 cm:
M = -(-9.6)/16
M = 0.6.

Therefore the magnification produced by the mirror when the object distance is 14 cm is 0.63 and 16 cm is 0.6.

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a) Using energy methods and assuming negligible air resistance, the speed of the ski at the base of the incline is approximately 34.1 m/s. b) The ski will travel approximately 110.6 m along the level snow.

a) When the ski descends the gradient, its original potential energy is transformed into kinetic energy. Friction causes part of this energy to be wasted, which reduces the ski's speed at the bottom of the hill. The ski's ultimate kinetic energy may be calculated by subtracting the work done by friction from its starting potential energy using the laws of energy conservation. b) The ski only has kinetic energy left as it reaches the bottom of the gradient since all of its potential energy has been expelled. The ski's kinetic energy is transformed into thermal energy as it travels down the flat snow by the frictional force pressing on it, slowing it down until it ultimately comes to a stop. making use of energy saving once more.

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Answer:

p1v1=p2v2

v2=p1v1/p2

=(1 atm ×12 L)/2 atm

v2= 6 L

The total current drawn in the circuit was calculated to be 21.7 A.

Given the appliances are in parallel So the voltage across all the appliances is the same i.e. V = 120 V

We know that P = IΔV

The current in the hair dryer is

I₁ = 1350/120

I₁ = 11.25 A

The current in the curling iron

I₂ = 700/120

I₂ = 5.83 A

The current inside the electric light fixture

I₃ = 550/120

I₃= 4.58 A

So the total current drawn (I) is equal to

I = I₁+ I₂ +I₃

I = 11.25 A +5.83 A +4.58 A

I = 21.7 A

Since all components in a parallel circuit have the same electrical junctions, the voltage across parallel components is the same. The total current is equal to the sum of each individual branch current.

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Answer:

To calculate the change in gravitational potential energy of the book as it falls from different starting points, we need to use the formula:

ΔPE = mgh

where ΔPE is the change in gravitational potential energy, m is the mass of the book, g is the gravitational acceleration (9.81 m/s^2), and h is the change in height.

A) If the book is released from a point on the floor of the room to the left of the hatch, it will fall towards the hatch and then fall further towards the basement floor. The change in height is the distance from the floor to the basement floor, which is 2.2 meters. Therefore, the change in gravitational potential energy is:

ΔPE = (7.5 N) x (9.81 m/s^2) x (2.2 m) = 162.8 Joules

B) If the book is released from a point on the floor of the room to the right of the hatch, it will fall towards the hatch and then fall further towards the basement floor as well. Therefore, the change in gravitational potential energy is the same as in Part A:

ΔPE = (7.5 N) x (9.81 m/s^2) x (2.2 m) = 162.8 Joules

C) If the book is released from a point at the height of the chair seat (0.45 m above the floor), it will fall towards the hatch and then fall further towards the basement floor. The change in height is the distance from the chair seat to the basement floor, which is:

h = 0.45 m + 2.2 m = 2.65 m

Therefore, the change in gravitational potential energy is:

ΔPE = (7.5 N) x (9.81 m/s^2) x (2.65 m) = 187.8 Joules

D) If the book is released from a point at the height of the bookshelf (1.5 m above the floor), it will fall directly towards the basement floor. The change in height is the distance from the bookshelf to the basement floor, which is:

h = 1.5 m + 2.2 m = 3.7 m

Therefore, the change in gravitational potential energy is:

ΔPE = (7.5 N) x (9.81 m/s^2) x (3.7 m) = 268.4

The optical power of eyeglass lens required to correct the vision of a nearsighted person with a far point of 30cm is -3.33 diopters (D).

The optical power of an eyeglass lens required to correct the vision of a nearsighted individual is determined by their far point. The far point is the maximum distance at which the person can see objects clearly without any visual aids. A nearsighted individual has a far point that is closer than the normal far point of infinity. To determine the optical power needed to correct the nearsightedness, the far point distance is converted to meters and then the reciprocal of that value is calculated to obtain the optical power in diopters. For example, if the far point is 30 cm, the optical power needed to correct the nearsightedness is approximately 3.33 diopters. This optical power can be achieved by wearing eyeglass lenses with the appropriate refractive power.

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The moment of inertia of the flywheel is approximately 0.0337 kg·m^2.

To solve this problem, we can use the formula for rotational kinetic energy:

KE = (1/2) I ω²

where KE is the kinetic energy of the flywheel, I is its moment of inertia, and ω is its angular velocity.

We can first convert the angular velocity from revolutions per minute (rpm) to radians per second (rad/s):

ω = (11000 rpm) * (2π rad/rev) * (1 min/60 s) = 1146.13 rad/s

Next, we can plug in the values for KE and ω and solve for I:

KE = 4.9 MJ = 4.9 × 10⁶ J

ω = 1146.13 rad/s

(1/2) I ω² = KE

(1/2) I (1146.13 rad/s)² = 4.9 × 10⁶ J

I = (2 * 4.9 × 10⁶J) / (1146.13 rad/s)²

I = 0.0337 kg·m²

Therefore, the moment of inertia of the flywheel is approximately 0.0337 kg·m².

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The quality factor of a series resonance band-pass filter when the same resistor, capacitor, and inductor are used is 20.

The quality factor of a series resonance band-pass filter when the same resistor, capacitor, and inductor are used is the reciprocal of the quality factor of the parallel resonance band-pass filter.
In this case, the quality factor of a parallel resonance band-pass filter is found to be 0.05. If the same resistor, capacitor, and inductor are used to construct a series resonance band-pass filter, the quality factor of the filter is given by;

Qs = 1/QR

Where, Qs = Quality factor of a series resonance band-pass filter, QR = Quality factor of a parallel resonance band-pass filter.

Therefore,

Qs = 1/0.05 = 20

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If a person weighs 670 N on Earth, their weight on the surface of a neutron star that has the same mass as our sun and a diameter of 16.0 km would be 9.26 x 10^12 N.

To find out, we'll need to use the formula for surface gravity: g = (GM)/r²

where: g is surface gravity in N/kg, G is gravitational constant in Nm²/kg², M is the mass in kilograms, r is the radius in meters.

First, we'll need to find the mass of the neutron star using the mass of the sun, which is Ms = 1.989 x 10³⁰ kg.

M = Ms = 1.989 x 10³⁰ kg

The radius of the neutron star is 16.0 km or 16,000 m. r = 16,000 m

Now we can plug these values into the surface gravity formula:

g = (GM)/r²g = [(6.674 x 10^-11 Nm²/kg²)(1.989 x 10³⁰ kg)]/(16,000 m)²g = 1.15 x 10^12 N/kg

Finally, to find the weight of the person on the surface of the neutron star, we'll multiply their mass by the surface gravity:

weight = mgweight = (670 N)/(9.81 m/s²)weight = 68.27 kg

weight on neutron star = (68.27 kg)(1.15 x 10^12 N/kg)

weight on neutron star = 9.26 x 10^12 N

Therefore, the weight of the person on the surface of a neutron star that has the same mass as our sun and a diameter of 16.0 km would be 9.26 x 10^12 N.

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The planet's average distance from the star is approximately 0.78 astronomical units (AU).

Based on the information given, the velocity curve obtained using the Doppler method has a period of 6 months. This means that the planet completes one full orbit around the star in 6 months.

Using Kepler's third law, we can relate the orbital period of the planet to its distance from the star:

(T^2 / a^3) = (4π^2 / GM)

where;

Since we know the orbital period of the planet (6 months) and the mass of the star (similar to the Sun), we can solve for the semi-major axis:

a = (T^2 GM / 4π^2)^(1/3)

Substituting the given values, we get:

a = [(6 months)^2 * (1 solar mass) * (6.67 x 10^-11 N m^2/kg^2) / (4π^2)]^(1/3)

Simplifying the expression, we get:

a ≈ 0.78 AU

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The string is under 4.82N of tension.

Tension is the pulling force exerted on a string, rope, cable, or wire when it is stretched or pulled. The tension on a string is equal to the amount of force applied to it divided by its cross-sectional area.

The tension on the string of a conical pendulum can be calculated using the equation:

[tex]T =\frac{ (4\pi^2m)}{(L^2T^2)},[/tex]

where T is the string's tension, m is the pendulum's mass, L is the string's length, and T is the pendulum's period of motion.

m=mass of pendulum=0.400kg

L=length of string =0.9m

T=pendulum's period of motion=1.4s

Plugging in the given values, we get:

[tex]T =\frac{ (4 \pi^2 * 0.400 kg) }{ (0.9 m^2 * 1.4 s^2)}\\\\T = 4.82 N[/tex]

Therefore,The Tension on the string is 4.82N

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The capacitance is needed to store charge Q in a capacitor with a potential difference between its plates of 9.5 is Q/9.5

A capacitor is a two-terminal electrical device that can store energy in the form of an electric charge. It consists of two electrical conductors that are separated by a distance. The space between the conductors may be filled by vacuum or with an insulating material known as a dielectric.

Capacitance is expressed as the ratio of the electric charge on each conductor to the potential difference (i.e., voltage) between them.

The capacitance needed to store charge Q in a capacitor with a potential difference between its plates of 9.5 is given by the formula:

C= Q/V

Where, C is the capacitance, Q is the charge stored, and V is the potential difference between the plates.

Therefore,

C= Q/9.5

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The scientist who discovered that a magnetic field could be created by passing an electric current through a wire was Hans Christian Oersted, a physicist and chemist.

Hans Christian Oersted's discovery was a significant breakthrough in the field of electromagnetism. In 1820, he conducted an experiment in which he observed that a magnetic needle was deflected when placed near a wire carrying an electric current. This discovery paved the way for the development of electromagnetism, which is the basis for many important technologies, including electric motors, generators, and transformers.

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False, Fiber optic cable is thinner and lighter than unshielded twisted pair (UTP) cable.

Fiber optic cables are made up of strands of glass or plastic fibers that transmit data through light pulses. They are immune to electromagnetic interference and can transmit signals over longer distances than UTP cables, which are made up of copper wires.The content loaded Fiber optic cable is used in high-speed internet connections because it can transmit large amounts of data at very high speeds. Additionally, fiber optic cables are more secure than UTP cables because they are much more difficult to tap into without being detected. Therefore, the statement that fiber optic cable is thicker and heavier than UTP cable is false.

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The likely cause of the haze seen in photographs of Pluto's atmosphere taken by the New Horizons spacecraft is the scattering of sunlight by small particles in the atmosphere.

Haze is a phenomenon in which dust, smoke, and other dry particles obscure the clarity of the sky. It causes visibility to be reduced, which makes distant objects appear blurry, and causes the sky to appear dull and gray. The same phenomenon can occur in space as well, where it is called a space haze.

Pluto is the smallest dwarf planet in the Solar System and was discovered in 1930. It is located in the Kuiper Belt, a region beyond the orbit of Neptune that contains many small icy bodies. Pluto is the largest object in the Kuiper Belt, with a diameter of 2377 km. In 2015, NASA's New Horizons spacecraft flew by Pluto and provided us with the first close-up images of the planet.

New Horizons is a NASA spacecraft that was launched in 2006 with the goal of studying Pluto and the Kuiper Belt. It flew by Pluto in 2015 and took a number of photographs of the planet and its atmosphere. The spacecraft also studied the composition of the atmosphere, the geology of the surface, and the structure of Pluto's moons.

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Answer:

No.

Explanation:

It will be soo painful that the person won't be able to feel it any,ore the person will feel nothing but dieing and can't talk