a conducting wire with conductance of 0.9s what is the conductivity of another wire of the same material and of the same length but the radius of its cross section is 3 times the radius of the cross section of the first wire

Answer:

First, we need to find the equivalent resistance of the parallel combination of 5.0 and 9.0 resistors:

1/R = 1/5.0 + 1/9.0

1/R = 0.4 + 0.1111

1/R = 0.5111

R = 1/0.5111

R ≈ 1.955 ohms

The equivalent resistance of the parallel combination is approximately 1.955 ohms.

Next, we need to find the total resistance of the circuit:

R_total = 4.0 + 1.955

R_total = 5.955 ohms

The total resistance of the circuit is approximately 5.955 ohms.

Using Ohm's Law, we can find the current through the circuit:

I = V/R_total

I = 6.0/5.955

I ≈ 1.006 A

The current through the circuit is approximately 1.006 A.

Finally, we can use the current divider rule to find the current through the 5.0-ohm resistor:

I_5 = (R_parallel / (R_parallel + R_series)) * I_total

I_5 = (1.955 / (1.955 + 4.0)) * 1.006

I_5 ≈ 0.383 A

The current through the 5.0-ohm resistor is approximately 0.383 A.

Horizontal lines are referred to as being parallel to a x-axis in coordinate geometry. A line is referred to as horizontal if two points on the line share the same y-coordinate points.

The intersection of the vertical and horizontal (X axis) real number lines is shown on the axis graph (Y axis). The Y axis is known as the dependent variable of the data set, whereas the X axis is typically used to refer to the independent variable of the data set.

The horizontal x-axis and vertical y-axis are the two axes of a line graph (vertical). A different type of data is indicated at each of the points where the axes connect, and (0,0). The x-axis is referred to as an independent axis since the numbers it represents are not reliant on any of the variables being assessed.

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The frequency of vibration of the block supported by two identical parallel vertical springs with spring stiffness constant k and mass m is [tex](1 / 2\pi) * \sqrt(2k / m).[/tex]

In physics, frequency is the number of waves that pass a fixed point in a unit of time as well as the number of cycles or vibrations that a body in periodic motion experiences in a unit of time.

The frequency of vibration of a mass-spring system is given by the formua:

[tex]f = (1 / 2\pi) * \sqrt(k / m)[/tex]

where f is the frequency of vibration, k is the spring constant, and m is the mass of the object.

In this case, the block is supported by two identical parallel vertical springs, each with spring stiffness constant k. So, the effective spring constant is the sum of the individual spring constants:

k_eff = 2k

The mass of the block is given as m.

So, the frequency of vibration of the block supported by two identical parallel vertical springs can be calculated as:

[tex]f = (1 / 2\pi) * \sqrt(k_eff / m) = (1 / 2\pi) * \sqrt((2k) / m)\\f = (1 / 2\pi) * \sqrt(2k / m)[/tex]

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Answer:

A) The maximum height above the roof reached by the rock can be found using the formula:

h = (v₀²sin²θ)/(2g)

where v₀ is the initial velocity (32.0 m/s), θ is the angle of the initial velocity (29.0°), and g is the acceleration due to gravity (9.81 m/s²).

Plugging in the values, we get:

h = (32.0²sin²29.0)/(2(9.81)) = 31.1 m

Therefore, the maximum height above the roof reached by the rock is 31.1 meters.

B) The vertical component of the velocity just before the rock strikes the ground is:

vᵥ = v₀sinθ - gt

where t is the time it takes for the rock to reach the ground.

We can find t by using the formula:

h = v₀sinθt - (1/2)gt²

where h is the height of the building (14.0 m). Rearranging this formula and solving for t, we get:

t = (v₀sinθ + sqrt((v₀sinθ)² + 2gh))/g

Plugging in the values, we get:

t = (32.0sin29.0 + sqrt((32.0sin29.0)² + 2(9.81)(14.0)))/9.81 = 4.01 s

Therefore, the vertical component of the velocity just before the rock strikes the ground is:

vᵥ = 32.0sin29.0 - 9.81(4.01) = -14.3 m/s

Note that the negative sign indicates that the velocity is directed downwards.

C) The horizontal distance from the base of the building to the point where the rock strikes the ground can be found using the formula:

d = v₀cosθt

Plugging in the values, we get:

d = 32.0cos29.0(4.01) = 96.4 m

Therefore, the horizontal distance from the base of the building to the point where the rock strikes the ground is 96.4 meters.

Answer:

Explanation:

The 33 N force is at a 90 degree angle, whereas the 44 N force is at a 60 degree angle with the x-axis.

Assume that the third force makes a theta-angle contact with the x-axis.

Since the object is in balance, the total force acting on it will equal zero.

Find the accumulation of the x-axis forces.

[tex]\begin{aligned} 33\cos 90{}^\circ +44\cos 60{}^\circ +{{F}_{3}}\cos \theta &=0 \\ 0+22\text{ N}+{{F}_{3}}\cos \theta &=0 \\ {{F}_{3}}\cos \theta &=-22\text{ N }......\text{ }\left( 1 \right) \end{aligned}[/tex]

Find accumulation of the y-axis forces.

[tex]\begin{aligned} 33\sin 90{}^\circ +44\sin 60{}^\circ +{{F}_{3}}\sin \theta &=0 \\ 33\text{ N}+38.11\text{ N}+{{F}_{3}}\sin \theta &=0 \\ {{F}_{3}}\sin \theta &=-71.11\text{ N }......\text{ }\left( 2 \right) \end{aligned}[/tex]

Identify the magnitude.

[tex]\begin{aligned} F&=\sqrt{{{\left( {{F}_{3}}\cos \theta \right)}^{2}}+{{\left( {{F}_{3}}\sin \theta \right)}^{2}}} \\ &=\sqrt{{{\left( -22\text{ N} \right)}^{2}}+{{\left( -71.11\text{ N} \right)}^{2}}} \\ &=74.43\text{ N} \end{aligned}[/tex]

Identify the direction.

[tex]\begin{aligned} \tan \theta &=\left( \frac{{{F}_{3}}\sin \theta }{{{F}_{3}}\cos \theta } \right) \\ \theta &={{\tan }^{-1}}\left( \frac{{{F}_{3}}\sin \theta }{{{F}_{3}}\cos \theta } \right) \\ \theta &={{\tan }^{-1}}\left( \frac{-71.11\text{ N}}{-22\text{ N}} \right) \\ \theta &=72.8{}^\circ \end{aligned}[/tex]

To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum of a system remains constant if there are no external forces acting on it. We can set the momentum of the asteroid before the collision equal to the momentum of the asteroid-Silly Putty system after the collision.

The momentum of the asteroid before the collision is:

P_before = m_ast * v_ast

where m_ast is the mass of the asteroid and v_ast is its velocity.

The momentum of the asteroid-Silly Putty system after the collision is:

P_after = (m_ast + m_sp) * v_final

where m_sp is the mass of the Silly Putty, v_final is the velocity of the asteroid-Silly Putty system after the collision, which we assume to be zero.

We can equate these two expressions for momentum and solve for the mass of the Silly Putty:

m_sp = (m_ast * v_ast) / (-v_final)

Substituting the given values:

m_ast = 25,000 kg

v_ast = 1500.0 m/s

v_final = -100.0 m/s

m_sp = (25,000 kg * 1500.0 m/s) / (-(-100.0 m/s))

m_sp = 375,000 kg m/s / 100.0 m/s

m_sp = 3750 kg

Therefore, we would need 3750 kg of Silly Putty to stop the asteroid if we launched it at a velocity of -100.0 m/s.

This is the pressure at the top of the container due to the height of the water. The pressure at the bottom of the container due to the height of the water is 0 Pa since the datum was set at the bottom.

Pressure is a physical quantity used to measure the force applied by an object to another object or by an object to a surface. It can be expressed as the force per unit area and is typically measured in units such as pounds per square inch (psi) or pascals (Pa). Pressure can be applied to gases, liquids, and solids, and is generated by the weight of the atmosphere, the force of gravity, and by the movement of air or liquids. Pressure can also be created by mechanical devices such as pumps and compressors, as well as by chemical reactions.

ρ = density of water = 1000 kg/m3

g = acceleration due to gravity = 9.81 m/s2

h = height of the container = 7.7 in = 0.198 m

Using the equation ρgh = density x gravity x height, we can calculate the pressure at the top of the container to be:

ρgh = (1000 kg/m3)(9.81 m/s2)(0.198 m) = 1960.38 Pa

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Answer:

The car has approximately 1.42 × 10^6 J of energy when it moves at 101 miles per hour.

Explanation:

First, we need to convert the initial velocity and kinetic energy to SI units:

Initial velocity: 32 miles per hour = 14.3 meters per second (rounded to 2 decimal places)

Kinetic energy: 5 × 10^5 J (given)

Next, we can use the formula for kinetic energy:

KE = (1/2)mv^2

where KE is the kinetic energy, m is the mass of the car, and v is the velocity of the car.

Solving for mass:

m = 2KE/v^2

Substituting the given values:

m = 2(5 × 10^5 J) / (14.3 m/s)^2 ≈ 1569.93 kg (rounded to 2 decimal places)

Now, we can use the same formula to calculate the kinetic energy of the car when it moves at 101 miles per hour (rounded to 2 decimal places):

KE = (1/2)mv^2 = (1/2)(1569.93 kg)(45.06 m/s)^2 ≈ 1.42 × 10^6 J

Therefore, the car has approximately 1.42 × 10^6 J of energy when it moves at 101 miles per hour.

To solve this problem, we can use the kinematic equation for vertical motion under constant acceleration, which is given by:

[tex]y = vi*t + (1/2)at^2[/tex]

What is displacement?

Displacement is a vector quantity that refers to the overall change in position of an object from its initial position to its final position.

It is a straight line distance between the initial and final position, in a specific direction.

where y is the displacement (in this case, the vertical distance the ball falls), vi is the initial velοcity (0 in this case since the ball is thrοwn hοrizοntally), a is the acceleratiοn due tο gravity[tex](-32.2 ft/s^2)[/tex], and t is the time it takes fοr the ball tο reach hοme plate. We can find t by dividing the distance tο hοme plate by the hοrizοntal velοcity:

[tex]t = 60.5 ft / (102.5 mi/h * 5280 ft/mi * 1 h/3600 s) = 0.400 s[/tex]

Now we can use the kinematic equation to find y:

[tex]y = 0 + (1/2)(-32.2 ft/s^2)(0.400 s)^2 = -2.57 ft[/tex]

Note that the negative sign indicates that the displacement is downward. Therefore, the ball falls about 2.57 feet vertically by the time it reaches home plate.

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The rocket travels a horizontal distance of 57.4 m before hitting the ground.

The initial speed of the rocket is v0=√(v0x^2+v0y^2)=7.28 m/s.

The rocket's velocity at the maximum height is zero, as it momentarily stops moving vertically and starts falling back down.

To solve this problem, we can use the kinematic equations of motion. Let's first find the velocity and position as a function of time:

vx(t) = v0x + ∫ax(t) dt = v0x + (1/3)αt^3

vy(t) = v0y + ∫ay(t) dt = v0y + βt - (1/2)γt^2

x(t) = ∫vx(t) dt = v0x t + (1/12)αt^4

y(t) = ∫vy(t) dt = v0y t + (1/2)βt^2 - (1/6)γt^3

Now, let's find the time t1 when the rocket reaches its maximum height:

ay(t1) = 0

β - γt1 = 0

t1 = β/γ = 6.43 s

At t1, the rocket's height is:

y(t1) = v0y t1 + (1/2)βt1^2 - (1/6)γt1^3

y(t1) = 7.00 m/s × 6.43 s + (1/2) × 9.00 m/s2 × (6.43 s)^2 - (1/6) × 1.40 m/s3 × (6.43 s)^3

y(t1) = 92.5 m

Now, let's find the time t2 when the rocket hits the ground. We can do this by solving for the positive root of the quadratic equation:

y(t) = 0

(1/2)γt^2 - βt - v0y = 0

Using the quadratic formula, we get:

t2 = (β + √(β^2 + 2γv0y))/γ = 8.01 s

Finally, let's find the horizontal distance traveled by the rocket:

x(t2) = v0x t2 + (1/12)αt2^4

x(t2) = 1.00 m/s × 8.01 s + (1/12) × 2.50 m/s4 × (8.01 s)^4

x(t2) = 57.4 m

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Ball A will have the greater density because it has twice as much mass as ball B for the same volume. Ball C will have the greater density because it has 3 times the volume of ball D and only 1/3 the mass.

Volume is the quantity of three-dimensional space occupied by an object or a substance. It is measured in cubic units, such as liters or gallons. Volume is an important concept in mathematics, physics, chemistry, and engineering, and is often used to calculate the amount of material needed for a certain project. For example, in architecture, engineers may use volume to determine the amount of concrete needed to build a bridge. In cooking, cooks use volume to measure the amount of ingredients needed for a recipe. In physics, volume is used to measure the amount of space an object occupies, or the amount of space within an object, such as a liquid or gas.

Ball P will have the greater density because it is twice as large as ball Q for the same mass. Ball X will have the greater density because it is twice as big as ball Y but weighs only half as much.

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Answer:

To find the height of the hill, we can use the conservation of energy principle, which states that the initial potential energy of the child and sled at the top of the hill is equal to their final kinetic energy at the bottom of the hill. The potential energy is given by:

PE = mgh

where m is the mass of the child and sled, g is the acceleration due to gravity (9.81 m/s^2), and h is the height of the hill.

The kinetic energy is given by:

KE = (1/2)mv^2

where v is the speed of the child and sled at the bottom of the hill.

Equating the potential and kinetic energies, we have:

mgh = (1/2)mv^2

Canceling the mass, we get:

gh = (1/2)v^2

Solving for h, we have:

h = (1/2) v^2 / g

Substituting the given values, we get:

h = (1/2) (9.94 m/s)^2 / 9.81 m/s^2

h = 5.06 m

Therefore, the height of the hill is 5.06 meters.

The intensity of sound are- For distance of 25 m: I = 0.108 W/m² and For distance of 45 m: I = 0.034 W/m².

Your comprehension of the inverse square law, which states that a sound's intensity varies inversely to the square of its distance from its source, will be put to the test in this challenge.

I = k • (1/R²)

when,  distance R is 15m ,  intensity of a sound source in open air to be 0.3 W/m².

So,

0.3 = k • (1/15²)

k = 225 * 0.3

k = 67.5

For distance of 25 m:

I = k • (1/R²)

I = 67.5 • (1/25²)

I = 0.108 W/m².

For distance of 45 m:

I = k • (1/R²)

I = 67.5 • (1/45²)

I = 0.034 W/m².

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Answer:

search it up

Explanation:

The intensity of the force F is calculated to be approximately 2.83 N.

Coefficient of sliding friction is a value that measures force of sliding friction for particular surface type.

Weight = mg = 1 kg x 10 m/s² = 10 N (acting downwards)

F_vertical - Weight = 0

F_vertical = 10 N

As Frictional force = coefficient of friction x normal force

Frictional force = 0.2 x 10 N = 2 N (acting to the left)

F_horizontal = F x cos(45) = F / √2

F_horizontal - frictional force = 0

F_horizontal = frictional force = 2 N

F / √2 = 2 N

F = 2 N x √2

F ≈ 2.83 N

Therefore, the intensity of the force F is approximately 2.83 N.

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Answer: 0.1

Explanation:

To find the tension on the stone, we can use the centripetal force formula, which is given by F = (mv^2)/r, where m is the mass of the object, v is its velocity, and r is the radius of the circular path.

In this case, the stone is tied to a string and is moving in a circle of radius 50 cm (or 0.5 m), so we have:

F = (0.05 kg) x (1 m/s)^2 / 0.5 m

F = 0.1 N

Therefore, the tension on the string is 0.1 N.

So, the correct answer is 0.1.

Answer:  240 inch-lbs.

Explanation:

The spring constant, k, is given by:

k = (F/x)

where F is the force, x is the displacement, and k is the spring constant.

Using the given information, we can calculate the spring constant as:

k = (80 lbs / 2 in) = 40 lbs/in

To find the work done in stretching the spring from its relaxed state to a total of 4 inches, we need to integrate the force over the displacement:

W = ∫F dx

Since the force required to stretch the spring varies with displacement, we need to break up the integration into two parts: one from 0 to 2 inches, and another from 2 to 4 inches.

W = ∫0^2 F dx + ∫2^4 F dx

The first integral is:

∫0^2 F dx = ∫0^2 (kx) dx = (1/2)kx^2 = (1/2)(40 lbs/in)(2 in)^2 = 80 inch-lbs

The second integral is:

∫2^4 F dx = ∫2^4 (2k) dx = 2kx = 2(40 lbs/in)(4 in - 2 in) = 160 inch-lbs

Therefore, the total work done in stretching the spring from its relaxed state to a total of 4 inches is:

W = ∫0^2 F dx + ∫2^4 F dx = 80 inch-lbs + 160 inch-lbs = 240 inch-lbs

So, the work done in stretching the spring from its relaxed state a total of 4 inches is 240 inch-lbs.

The rate at which οil is flοwing is 0.494 m³/s. This means that the prοduct οf the fluid's density (ρ), crοss-sectiοnal area (A), and velοcity (v) is cοnstant.

Density is a physical prοperty οf matter that represents hοw much mass is cοntained within a given vοlume οf a substance. It is defined as the amοunt οf mass per unit vοlume and is typically expressed in units οf kilοgrams per cubic meter (kg/m³) οr grams per cubic centimeter (g/cm³).

We can use the principle οf cοntinuity tο sοlve this prοblem. Accοrding tο this principle, the mass flοw rate οf a fluid remains cοnstant as it flοws thrοugh a pipe οf varying diameter.

Therefore, we can write:

[tex]\rho_1A_1v_1 = \rho_2A_2v_2[/tex]

where the subscripts 1 and 2 refer to the sections of the pipe before and after the constriction, respectively.

We can rearrange this equation to solve for the velocity of the oil in the pipe:

[tex]v_2 = (A_1/A_2) \times (v_1 \times (\rho_1/\rho_2))[/tex]

where A1 and A2 are the cross-sectional areas of the pipe before and after the constriction, respectively.

Using the given values, we get:

[tex]v_2 = (\pi/4) \times (1.11 m)^2 \times (8130 N/m^2 / 821 kg/m^3) / [(\pi/4) \times (0.666 m)^2] \times (6097.5 N/m^2 / 821 kg/m^3)[/tex]

v₂ = 1.74 m/s

Finally, we can calculate the rate at which oil is flowing using the formula:

Q = A₂ × v₂

Using the given values, we get:

Q = (π/4)  × (0.666 m)² × 1.74 m/s

Q = 0.494 m³/s

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The metaphor "sawing himself in two" highlights the basketball player's agility as his legs and upper body move in opposing directions before coming together at the rim to hit a shot.

Essentially, employing figurative language involves distorting the meaning of words to convey a point, sound clever, or create a joke. Figurative language is a common technique used in narrative writing when the author wants to make the reader feel strongly about something.

Literature forms like poetry, drama, prose, and even speeches use figurative language. Figurative language is a literary element that is employed throughout

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The signboard Rahul saw was likely a speed limit sign. This sign is used to inform drivers of the maximum speed they should be travelling at in that area.

Limit is a mathematical concept that describes the highest or lowest value that a given function or expression can reach. It is used as a tool to measure how a function or expression behaves as its variables approach a certain value. For example, the limit of a function as x approaches infinity is the highest value that the function can take on. It is often used to calculate the area under a curve, the rate of change of a function, and the slope of a line at a given point. Limits are also used to compare the relative sizes of different functions, and to determine the continuity of a function.

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According to the given statement The ring's final internal diameter would be 1218 mm.

Cubical expansion is the term for the phenomenon wherein the volume of a solid increases as it is heated. Also called volumetric expansion. The coefficient of volumetric expansion measures how much a material's volume expands as its temperature rises by one degree.

The following formula can be used to get the ring's ultimate interior diameter:

ΔL = αLΔT

where ΔL is the length increase, is the cubical growth coefficient, L is the starting length, and ΔT is the temp change.

In this instance, the change in width, which is double the change in length, is what we are searching for. Thus, the formula may be rewritten as follows:

ΔD = 2αDLΔT

where D represents the starting diameter and ΔD represents the diameter change.

By putting in the indicated values, we get:

ΔD = 2(51×10⁻⁶ /degree celsius)(300 mm)(600 degree celsius)

ΔD = 918 mm

As a result, the ring's final internal diameter would be:

Dfinal = Dinitial + ΔD = 300 mm + 918 mm = 1218 mm

Hence, the ring's final internal diameter would be 1218 mm.

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Answer:

Start:

Higher energy of reactants -> Transition state

Finish:

Lower energy of reactants -> Higher energy of products

Explanation:

Answer:

D. Thermoelectric thermometer

Explanation:

It preferred for rapidly changing temperature

Answer:

Answer:d. 109N/C

Explanation: The atomic electric field, the field between the atomic nucleus and the surrounding electron cloud, should possess information about the atomic species, local chemical bonding, and charge redistributions between bonded atoms.

Answer:

The energy produced by the sun when it converts 2 kg of mass into energy is given by Einstein's famous equation E = mc², where E is the energy produced, m is the mass converted, and c is the speed of light.

Substituting the values, we have:

E = (2 kg) x (299,792,458 m/s)²

E = 2 x 89,875,517,873,681,764 J

E ≈ 1.7975 x 10²⁰ J

Therefore, the sun produces approximately 1.7975 x 10²⁰ joules of energy when it converts 2 kg of mass into energy.

The moment of inertia of the wheel if the block rises to a height of h is 7.5 cm before momentarily coming to rest is 0.068 kg m².

We can use the conservation of mechanical energy to solve for the moment of inertia of the wheel. Initially, the system has kinetic energy, which is converted to potential energy at the highest point of the block's trajectory. Therefore, we can write:

Initial kinetic energy = Final potential energy

At the start, the wheel and block have kinetic energy due to the motion of the block:

KE i = 0.5 * m * v²

At the highest point of the block's trajectory, all of the kinetic energy is converted to potential energy due to the block's height above the ground:

PE_f = m * g * h

where g is the acceleration due to gravity.

Since the string is wrapped around the circumference of the wheel, the distance that the block moves upwards is equal to the distance that the string moves around the wheel, which is equal to the circumference of the wheel:

h = 2 * pi * R

Substituting this into the expression for potential energy:

PE_f = m * g * 2 * π * R

Equating initial kinetic energy with final potential energy:

0.5 * m * v² = m * g * 2 * π * R

Simplifying and solving for the moment of inertia of the wheel, I:

I = (m * v²) / (2 * g * π * R)

Substituting the given values:

I = (1.9 kg * 0.43 m/s)² / (2 * 9.81 m/s² * π * 0.081 m)

I = 0.068 kg m²

Therefore, the moment of inertia of the wheel is 0.068 kg m².

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