Two conductors are made of the same material and have the same length. conductor a is a solid wire of diameter 1.3 mm. conductor b is a hollow tube of outside diameter 8.0 mm and inside diameter 7.0 mm. what is the resistance ratio ra/rb , measured between their ends?

r1 = 5*10^10 m , r2 = 6*10^12 m

v1 = 9*10^4 m/s

From conservation of energy

K1 +U1 = K2 +U2

0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2

0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2

M is mass of sun = 1.98*10^30 kg

G = 6.67*10^-11 N.m^2/kg^2

0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))

v2 = 5.35*10^4 m/s

The speed of a comet in an elliptical orbit around the sun can be calculated using Kepler's second law, which states that a planet sweeps out equal areas in equal amounts of time. This means that when a comet is closer to the sun, it moves faster, and when it is farther away, it moves slower.

The speed of a comet in an elliptical orbit around the sun can be calculated using Kepler's second law, which states that a planet sweeps out equal areas in equal amounts of time. This means that when a comet is closer to the sun, it moves faster, and when it is farther away, it moves slower. To find the speed of the comet when it is 6 x 10^12 m from the sun, we can use the fact that the area swept out by the comet is the same at both points. Using the distances given, we can calculate the speed of the comet when it is 6 x 10^12 m from the sun.

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The maximum speed with which this car can round a second unbanked curve is 40 m/s

[tex]\texttt{ }[/tex]

Centripetal Acceleration can be formulated as follows:

[tex]\large {\boxed {a = \frac{ v^2 } { R } }[/tex]

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

[tex]\texttt{ }[/tex]

Centripetal Force can be formulated as follows:

[tex]\large {\boxed {F = m \frac{ v^2 } { R } }[/tex]

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

initial speed of the car = v₁ = 20 m/s

radius of first curve = R₁ = 80 m

radius of second curve = R₂ = 320 m

Asked:

final speed of the car = v₂ = ?

Solution:

Firstly , we will derive the formula to calculate the maximum speed of the car:

[tex]\Sigma F = ma[/tex]

[tex]f = m \frac{v^2}{R}[/tex]

[tex]\mu N = m \frac{v^2}{R}[/tex]

[tex]\mu m g = m \frac{v^2}{R}[/tex]

[tex]\mu g = \frac{v^2}{R}[/tex]

[tex]v^2 = \mu g R[/tex]

[tex]\boxed {v = \sqrt { \mu g R } }[/tex]

[tex]\texttt{ }[/tex]

Next , we will compare the maximum speed of the car on the first curve and on the second curve:

[tex]v_1 : v_2 = \sqrt { \mu g R_1 } : \sqrt { \mu g R_2 }[/tex]

[tex]v_1 : v_2 = \sqrt { R_1 } : \sqrt { R_2 }[/tex]

[tex]20 : v_2 = \sqrt { 80 } : \sqrt { 320 }[/tex]

[tex]20 : v_2 = 1 : 2[/tex]

[tex]v_2 = 2(20)[/tex]

[tex]\boxed {v_2 = 40 \texttt{ m/s}}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Circular Motion

The work done due to friction is [tex]\boxed{59.58{\text{ N}}}[/tex].

Further explanation:

Here, we have to calculate the total work done due to frictional force.

Frictional force is a non-conservative force and the work done by a non-conservative force is path dependent.

Given:

The mass of the physics book is [tex]1.9{\text{ kg}}[/tex].

Coefficient of friction between the book and floor is [tex]0.2[/tex].

The distance between observer that is (you) and Beth is [tex]8{\text{ m}}[/tex]

The distance between Beth and Carlos is [tex]8{\text{ m}}[/tex].

Formula and concept used:

The work done by the conservative force is equal to the product of force and the displacement.

Now, we will know about displacement,

Displacement: The shortest distance between the initial and final position of the object is known as displacement.

But, in case of the non-conservative forces, the work done is equal to the product of the force and the total distance travel by the object.

So, friction force can be calculated as,

[tex]{F_r} = \mu mg[/tex]

Here, [tex]m[/tex] is the mass of the book, [tex]\mu[/tex] is the coefficient of friction.

So, work done due to friction when observer slides the book to the Beth and Beth slides the book to Carlos.

[tex]\boxed{W = {F_r} \cdot d}[/tex]

Here, [tex]{F_r}[/tex] is the friction force, [tex]d[/tex] is the total distance travel against friction force.

[tex]\boxed{\begin{aligned}d&=8+8\\&=16\text{ m}\end{aligned}}[/tex]

Substitute the value of [tex]{F_r}[/tex] in above equation.

[tex]W = \mu mgd[/tex]                                                                       …… (1)

Calculation:

Substitute [tex]0.2[/tex] for [tex]\mu[/tex], [tex]1.9{\text{ kg}}[/tex] for [tex]m[/tex] and [tex]16{\text{ m}}[/tex] for [tex]d[/tex] in equation (1).

[tex]\begin{aligned}W&=\left( {0.2} \right)\left( {1.9} \right)\left( {9.8} \right)\left({16}\right)\\&=59.58{\text{ N}}\\\end{gathered}[/tex]

The work done due to friction is [tex]\boxed{59.58{\text{ N}}}[/tex].

Learn more:

1. Acceleration against friction: https://brainly.com/question/7031524

2. Water is a compound because: https://brainly.com/question/4636675

3. Conservation of momentum during collision https://brainly.com/question/9484203

Answer detail:

Grade: Senior School

Subject: Physics

Chapter: Work and Energy

Keywords:

Book slide, Beth, Carlos, work done, friction, four corners, square, 8m length, 59.58 N, connecting these people, displacement.

Answer:

The horizontal and vertical velocity of the skier are both 29.4 m/s.

Explanation:

If no forces are acting in the horizontal direction, then the horizontal component of the velocity remains constant in time:

Vx = 29.4 m/s

For the vertical component the gravity acceleration must be considered, and the vertical velocity can be written as:

Vy = g*t

Here the value for g is 9,8 m/s2. Note that the initial vertical velocity is zero.  

Evaluating in the velocity’s expressions for a time of 3 seconds we obtain:

Vx = 29.4 m/s

Vy = 9.8 m/s2 * 3 s = 29.4 m/s

Answer:

Interference pattern- alternate bright and dark fringes

Explanation:

When a monochromatic light is incident on a thin barrier having two slits, an interference pattern can be observed.

Dark and bright fringes would appear on the screen placed in front of it. It would not be random pattern. It would be a smooth pattern with distinctive dark fringe and bright fringe alternatively.

When the light from two source would superimpose, a constructive interference would lead to bright fringe and destructive interference would lead to dark fringe.

Final answer:

The magnitude of the acceleration of the larger block across the table is 2.45 m/s², calculated using the gravitational force acting on the hanging mass and applying Newton's second law to the system of both blocks.

Explanation:

To calculate the magnitude of the acceleration of the larger block across the table, we need to apply Newton's second law of motion which states that the force equals the mass times the acceleration (F = ma). In this scenario, only the gravitational force acting on the 10 kg hanging mass needs to be considered for the system of both blocks, as the table is frictionless and there are no other external forces acting on the horizontal block.
We calculate the gravitational force acting on the 10 kg block, which is the weight of the block (W = mg), where m is the mass and g is the gravitational acceleration (approximately 9.8 m/s2). For the 10 kg block, W = 10 kg × 9.8 m/s2 = 98 N. This force is the net force causing both blocks to accelerate, and because the rope is massless and the pulley is frictionless, the tension in the rope is uniform throughout.
The total mass of the system is 30 kg + 10 kg = 40 kg. Using the equation F = ma, rearrange to find acceleration (a = F/m). So the acceleration of the system is a = 98 N / 40 kg = 2.45 m/s2. Therefore, the magnitude of the acceleration of the larger block across the table is 2.45 m/s2.

or an object with mass  moving at a velocity , the angular momentum  with respect to a reference point is defined using the cross product as:

where  is the position vector of the object that describes the object’s position with respect to the reference point. The units for measuring angular momentum is kg m2 s-1. Since angular momentum is defined in terms of a cross product, the direction of the angular momentum vector is taken to be in a direction perpendicular to both the particle’s position vector  and its velocity vector .

Imma answer so other dude can have brainliest.  

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The rate of heat loss from a hot-water pipe by natural convection is calculated using the formula Q = h * A * ΔT. After plugging in the given values and conducting the appropriate calculations, the rate of heat loss turns out to be 2512 W.

Let's determine the rate of heat loss from a hot-water pipe by natural convection. The formula to calculate the rate of heat loss though natural convection is: Q = h * A * ΔT, where:

Let's plug the numbers in:

First calculate the surface area, A = π * 0.05 m * 10 m = 1.57 m2. Then, to find Q, we use the formula Q = 25 w/m2·k * 1.57 m2 * 64 K = 2512 W. Therefore, the rate of heat loss from the pipe by natural convection is 2512 W.

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The rate of heat loss from the pipe by natural convection is 2512 W.

To determine the rate of heat loss from a hot-water pipe, we can use the formula :

Q = h ×A×ΔT

Where:

First, we calculate the surface area of the pipe:

The external diameter of the pipe is given as 0.05 m, and the length is 10 m. The surface area of a pipe is :

A = π ×D×L

Substituting the values:

A = π × 0.05 m × 10 m = 1.57 m²

Next, we calculating the temperature difference:

ΔT = T(pipe) - T(air) = 80°C - 16°C = 64°C

Finally, using heat transfer formula:

Q = h×A×ΔT = 25 W/m²·K × 1.57 m² × 64 = 2512 W

The rate of heat loss from the pipe by natural convection is  2512 W.

Final answer:

To find the cat's speed at the top of the incline, we use the Work-Energy Theorem. The work done moving the cat up the ramp increases her kinetic energy, resulting in a final speed of 5.1 m/s.

Explanation:

To determine the speed of the cat at the top of the incline, we can apply the Work-Energy Theorem. The theorem states that the work done on an object is equal to the change in its kinetic energy. In mathematical terms, Work = ΔKE = KEfinal - KEinitial, where KE is kinetic energy given by ½mv2. The work done on the cat by pushing her up the ramp can be calculated as the force applied (parallel to the ramp) times the distance moved along the ramp, which is Work = Force × Distance = 41.0 N × 2.00 m = 82.0 J.

The initial kinetic energy of the cat can be calculated using her initial speed at the bottom of the ramp, KEinitial = ½ × 8.50 kg × (1.90 m/s)2. Plugging in the values, we get KEinitial = 15.33 J. The final kinetic energy at the top of the incline can be found by adding the work done to the initial kinetic energy: KEfinal = KEinitial + Work = 15.33 J + 82.0 J = 97.33 J. Solving for the final speed, we set KEfinal = ½ × 8.50 kg × v2 equal to 97.33 J and solve for v, finding that the cat's speed at the top of the ramp is 5.1 m/s.

Using the law of conservation of momentum, which states that initial momentum equals final momentum in a closed system, the initial speed of the truck before it collided with the stationary car is calculated to have been 12 m/s.

This question pertains to the concept of momentum in physics. The law of conservation of momentum states that the total momentum of a system of objects is constant if no external forces are acting on it. In this case, the truck and the car form the system of objects.

The initial momentum of the system before the collision is given by the momentum of the truck, since the car is stationary. The final momentum after the collision is given by the combined mass of the truck and car, multiplied by their common speed.

So, if we denote the initial speed of the truck as V, we have: 3000kg * V (initial momentum of truck) = (3000kg + 1000kg) * 9m/s (final momentum of the system)

Solving for V, we find the initial speed of the truck to have been 12 m/s.

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Sample Response: If the height of the cylinder increases, the temperature of the water increases, because a greater height means the cylinder has more potential energy that can be converted to thermal energy, increasing the temperature of the water.

The current carried by the wire and the power rating of the heater are 20 A and 2400 W respectively.

Given data:

The potential difference across the electric heater is, V' = 120 V.

The total resistance of the nichrome wire is, [tex]R= 6.00 \;\rm \Omega[/tex].

First we need to apply the Ohm's law to find the current through the wire. The expression for the Ohm's law is given as,

[tex]V'= I \times R\\\\I =\dfrac{V'}{R}\\\\I =\dfrac{120}{6}\\\\I=20 \;\rm A[/tex]

Now, the expression for the electric power through the heater is given as,

[tex]P= V \times I\\\\P= 120 \times 20\\\\P =2400 \;\rm W[/tex]

Thus, we can conclude that the current carried by the wire and the power rating of the heater are 20 A and 2400 W respectively.

Learn more about the Ohm's law here:

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Answer:

Four conditions need to be met for an object to be seen-an object, an eye, a source of light, and a direct, unblocked path between the object and the eye

hope this helps

Answer:

4560m=4.560km

Explanation:

Analysis conceptual

We know that:

1km=1000m

Then, the conversion factor from meters to kilometers is: 1km/1000m

Known data:

4560 m

problem development

We multiply the amount given in m by the conversion factor (1km/1000m ):

[tex]4560 m(\frac{1km}{1000m} )=4.560km[/tex]

4560m=4.560km

Answer:

Should be same

Explanation:

An experiment is conducted to test a theory. It is expected that the result of the experiment would match with the theoretical result. Also, if the experiment is repeated by the same or another person, it should give the same result. If not, then either there is error in conduction of experiment or in the theory.

In the 1960s, the West Coast saw the rise of Surf Rock, a dance style associated with the beach culture and the Black Arts Movement which promoted a blend of traditional and contemporary African American dance styles.

In the 1960s, on the West Coast, particularly in California, a new dance style known as Surf Rock emerged. This musical and dance movement was a product of the beach culture prevalent there. Bands like the Kingsmen and the Ventures popularized this genre, and the music often emphasized guitar riffs and had less focus on the lyrics, a shift influenced by the noisy dance halls where the bands performed. The dance associated with Surf Rock was an energetic response to the instrumental tunes and was part of the broader youth culture's expression at the time.

While Surf Rock mainly dominated, there were also important contributions to dance from African American communities. The Black Arts Movement arose during the 1960s and celebrated "Black Dance," which was a blend of different dance styles that included historical and traditional elements of African heritage. This movement was part of a broader spectrum of cultural expression during the decade which sought to empower and reconnect with African American roots.

Final answer:

During the 1960s, the West Coast saw the rise of Surf Rock, a dance genre closely connected to the laid-back, free-spirited surfing lifestyle and characterized by instrumental rock music.

Explanation:

In the 1960s, on the West Coast, particularly in California, a dance movement that emerged was Surf Rock. This genre of music created a unique dance vibe and culture, mostly due to its instrumental focus and connection with the surf scene. Surf Rock music was linked to a style of guitar-led rock music that often had a strong association with the surfing culture and lifestyle associated with Southern California. It helped define a youthful, beach-oriented counterculture that was free-spirited and often synonymous with the freedom and allure of the West Coast. Bands like the Kingsmen, famous for 'Louie Louie', and the Ventures with 'Walk Don't Run', became the soundscape of this era, influencing the dances that accompanied their music.

The minimum speed the rock must have at the top of the circular path to always stay in contact with the bottom of the bucket is approximately 5.04 m/s, based on principles of circular motion and gravitational force.

This problem can be solved using the principles of circular motion and Newton's laws of motion. When the rock is at the top of the vertical circle, the minimum speed it should have is equal to the speed at the point where the gravitational force acting downwards equals the required centripetal force to maintain circular motion. To achieve this, you can use the formula v² = 2gy to find out the minimum velocity needed.

Given:
Mass of the rock (m) = 500g = 0.5kg
Gravitational acceleration (g) = 9.81m/s²
Radius of the circle (r), since diameter = 2.6m, r = d/2 = 1.3m

The minimum speed needed at the top of the circle, v, is found by rearranging the equation to v = √(2gr). Substituting the given values, we get: v = √(2*9.81m/s²*1.3m) = √(25.386) ≈ 5.04 m/s

So, the rock must have a minimum speed of 5.04 m/s at the top of its circular path to ensure it stays in the bucket.

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The scenario presented is a physics problem that involves the concepts of forces and circular motion. It necessitates determining the minimum velocity a rock must have when at the highest point of its circular path to maintain contact with the bucket. The calculated velocity required at the highest point of the trajectory, based on the provided diameter of the circle and the force of gravity, is approximately 3.5 m/s.

The scenario is a physics problem that deals with centripetal force and gravity. The minimum speed the rock needs to have at the top of the circle to stay in contact with the bucket can be calculated using the centripetal acceleration equation, ac = v²/r. Here, we know that the centripetal acceleration must be at least equal to gravity (9.8 m/s²) to maintain contact. We also know that the radius (r) of the circle is the diameter divided by two, which is 2.6m/2 = 1.3m. So, we can set up the equation 9.8m/s² = v²/1.3m, which when solved, gives us v = √(9.8m/s² * 1.3m) = √12.74 m²/s², resulting in v ≈ 3.5 m/s. So, the rock needs to travel at a speed of about 3.5 m/s at the top of the circle to stay in contact with the bucket.

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You are facing North. Turn 90 degrees left. W.

Turn 180 degrees right. E.

Reverse direction. W.

Turn 45 degrees left. S.W.

Reverse direction. N.E.

In which direction are you now facing? North East