The graph below shows how the velocity of a motorbike
varies with time during the final 10s of a race.
(i)Describe the motion shown by the graph.
(ii)Calculate the acceleration in the first phase of the motion.
(iii)Show that during the final 10 s the motorbike travels a distance of approximately 800 m

equal in magnitude but opposite in direction

The string moves to the right, as it restores its original position with the median plane of the bow. As a result, the string "pulls" on the arrow with a force F2. 2. The tip of the arrow T moves slightly to the left.

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2 conditioner are

Explanation:

force applied and displacement produced

The force acting on the system of two blocks connected by a rope passing over a pulley is -13.26 N.

The system of two blocks connected by a rope passing over a pulley are M1 and M2, where M1 rests on the triangle edge to the left of the pulley, which makes an angle of theta subscript 1 with the base of the triangle. The coefficient of friction between box M1 and the surface is mu subscript 1. M2 rests on the triangle edge to the right of the pulley, which makes an angle of theta subscript 2 with the base of the triangle.

The coefficient of friction between box M2 and the surface is mu subscript 2. The system sits atop a scalene triangle whose long edge forms the base. The pulley is attached to the apex of the triangle.M1 has a mass of 2.25 kg and is on an incline of angle 1=42.5∘ that has a coefficient of kinetic friction 1=0.205. M2 has a mass of 5.55 kg and is on an incline of angle 2=33.5∘ that has a coefficient of kinetic friction 2=0.105.The free-body diagram of M1 shows that the weight of M1 acts straight downwards (vertically) and the normal force acts perpendicular to the slope.

The force of friction opposes the motion and acts opposite to the direction of motion.M1 = 2.25 kgTheta subscript 1 = 42.5 degreesMu subscript 1 = 0.205g = 9.81 m/s²In the free-body diagram of M2, the normal force acts perpendicular to the incline of the slope, the weight of the object acts vertically downwards and parallel to the incline, and the force of friction opposes the motion and acts opposite to the direction of motion.M2 = 5.55 kgTheta subscript 2 = 33.5 degreesMu subscript 2 = 0.105g = 9.81 m/s²The tension in the string is the same throughout the rope. Since the masses are being pulled by the same rope, the acceleration of the objects is the same as the acceleration of the rope.

The tension in the string is directly proportional to the acceleration of the objects and the rope.A system of two blocks connected by a rope passing over a pulley has a total mass of M. The acceleration of the system is given by the formula below:a = [(m1-m2)gsin(θ1) - μ1(m1+m2)gcos(θ1)] / (m1 + m2)Where, μ1 = 0.205 is the coefficient of friction of block M1θ1 = 42.5 degrees is the angle of the incline of block M1M1 = 2.25 kg is the mass of block M1M2 = 5.55 kg is the mass of block M2g = 9.81 m/s² is the acceleration due to gravitysinθ1 = sin 42.5 = 0.67cosθ1 = cos 42.5 = 0.75The acceleration of the system is:a = [(2.25-5.55)(9.81)(0.67) - (0.205)(2.25+5.55)(9.81)(0.75)] / (2.25 + 5.55)a = -1.7 m/s² (the negative sign indicates that the system is accelerating in the opposite direction).

The force acting on the system is given by:F = MaWhere M is the total mass of the system and a is the acceleration of the system. The total mass of the system is:M = m1 + m2M = 2.25 + 5.55M = 7.8 kgThe force acting on the system is:F = 7.8(-1.7)F = -13.26 N (the negative sign indicates that the force is acting in the opposite direction).

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We have the next information

T=2.00 x 10^-1 s=0.2s

r=3.5 inch

For the angular speed

where omega is the angular speed, T is the period

We substitute

For the linear speed on the rim of the disc, we will use the next formula

in this case r= 3.5/2=1.75 inch

Then for the linear speed on the point at 0.750 inches from the center of the disk.

ANSWER

ω=31.41 rad/sec

v on the rim= 54.97 inches/sec

v on the point=23.56 inches/sec

Answer:

T = 80√3 N ≈ 139 N

W = 160 N

Explanation:

Sum of forces on B in the x direction:

∑F = ma

80 N sin 60° − T sin 30° = 0

T = 80 N sin 60° / sin 30°

T = 80√3 N

T ≈ 139 N

Sum of forces on B in the y direction:

∑F = ma

80 N cos 60° + T cos 30° − W = 0

W = 80 N cos 60° + T cos 30°

W = 40 N + 120 N

W = 160 N

Answer:

24 m/s

Explanation:

V=Vi+at

given initial velocity is 0, acceleration is 2 and time is 12 sec

find final velocity by plugging in the variables into the equation above.

V=0+2x12=24m/s

Answer:

920000

Explanation:

Each kg contains 1,000 grams

Answer:

a

Explanation:

away from

The direction the swimmer accelerates when she pushes against the wall is away from the wall.

Newton's third law states that, for every action, there is an equal and opposite reaction.

Here,

Given that, the swimmer is making a turn at the pool wall.

In order to take a turn, the swimmer have to exert a push against the wall with her feet, such that it is the action and she turning back is the reaction of it.

According to Newton's third law, for every action, there is an equal and opposite reaction. So, the wall exerts equal force back to the swimmer. As a result, the swimmer will be accelerated towards the opposite direction of the wall, thus making a turn.

Hence,

The direction the swimmer accelerates when she pushes against the wall is away from the wall.

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1. The waterfall is 8.54 m high

2. The velocity with which the swimmer enters the water is 14.836 m/s

s = ut

2.5 = 1.9 × t

Divide both sides by 1.9

t = 2.5 / 1.9

t = 1.32 s

h = ½gt²

h = ½ × 9.8 × 1.32²

h = 8.54 m

The velocity with which the swimmer enters the water can be obtained as follow:

v = u + gt

v = 1.9 + (9.8 × 1.32)

v = 1.9 + 12.936

v = 14.836 m/s

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The skater's final angular velocity is approximately 9.86 rad/s.

The skater's final angular velocity can be calculated using the principle of conservation of angular momentum. The equation for angular momentum is given by:

L = Iω

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Initially, the skater has an angular momentum of:

L_initial = I_initial * ω_initial

Substituting the given values:

L_initial = 2.12 kg m² * 3.25 rad/s

The skater's final angular momentum remains the same, as angular momentum is conserved:

L_final = L_initial

The final moment of inertia is given as 0.699 kg m². Therefore, the final angular velocity can be calculated as:

L_final = I_final * ω_final

0.699 kg m² * ω_final = 2.12 kg m² * 3.25 rad/s

Solving for ω_final:

ω_final = (2.12 kg m² * 3.25 rad/s) / 0.699 kg m²

Hence, the skater's final angular velocity is approximately 9.86 rad/s.

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The average time it took for the tablet pieces to dissolve in room-temperature water is 42.00 seconds.

Room temperature refers to the typical temperature range that is comfortable for humans in an indoor environment. It is generally considered to be between 68°F (20°C) and 77°F (25°C). However, the exact definition of room temperature can vary depending on the context and the standards of a particular region or industry. In scientific experiments or industrial settings, room temperature may be defined more precisely and may range from 20°C to 25°C, or even narrower ranges such as 22°C to 24°C.

To calculate the average time it took for the tablet pieces to dissolve in room temperature water, we need to add up the times from all three trials and divide by the number of trials (3):

(41.00 + 44.00 + 41.00) / 3 = 42.00 seconds

Therefore, the average time it took for the tablet pieces to dissolve in room-temperature water is 42.00 seconds.

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Answer: 1968 cm3

Explanation: You first need to convert all of your dimensions to the same units.

240 mm = 24 cm

20.5 cm = 20.5 cm

0.04 m = 4 cm

Then volume is Length x width x height = 24 cm x 20.5 cm x 4 cm =

The velocity of the ball just before it hits the ground is 24.29m/s.

What is acceleration due to gravity?

The acceleration an object experiences as a result of gravitational force is known as acceleration due to gravity. M/s2 is its SI unit. Its vector nature—which includes both magnitude and direction—makes it a quantity. The unit g stands for gravitational acceleration. At sea level, the standard value of g on earth's surface is 9.8 m/s2.

Given:

Initial velocity u = 10m/s

height h = 25m

we know that the acceleration due to gravity g is : 9.8m/s

Therefore using the formula,

v² - u² = 2gs   where s is the height in this case and v is the final velocity.

v² - 10² = 2 x 9.8 x 25

=> v² = 490 + 100

=> v = √590

=> v = 24.29m/s

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Hi there!

We can begin by breaking the problem up into two directions, horizontal and vertical.

The collision is perfectly inelastic since the two objects stick, so:

1 = car

2 = truck

Momentum in the east direction (positive and horizontal, car NOT going in this direction has a velocity of 0 m/s):

m2v2 = (m1 + m2)v

(650 · 11) / (350 + 650) = 7.15 m/s

Momentum in the north directions (only initial is the northward velocity of the car):

m1v1 = (m1 + m2)v

(350 · 16)/(350 + 650) = 5.6 m/s

Find the overall velocity using the pythagorean theorem:

√(5.6²) + (7.15²) = 9.08 m/s

Find the direction using trigonometry. Use tangent in this situation:

tan Ф = O/A

tan⁻¹(5.6/7.15) = 38° north of east

The theorem showing that △ ABC ≅ △ def is Density functional theorem.

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When a typical comet in the Oort cloud is only 1 km in diameter, then the mass of the Oort cloud is changes to  4.18×10⁹ kg.

The mass of each comet int he Oort cloud is,

mass = density × Volume

volume of sphere = 4/3(π×r²) where, r = radius of the sphere

From the given,

diameter = 1 km

radius = Diameter / 2 = 1/2(D)

Volume of the comet = 4/3 (π×r²) = 4/3 (π× (1/2 (10)³)²)

                                   = 4/3×(π) ×(1/8×10⁶)

Mass = density  × volume, where density = 1000 kg/m³

         = 1000  × 4/3×(π) ×(1/8×10⁶)

        = 4.18 ×10⁹ kg

When the typical comet in the Oort cloud is only 12 km in diameter the mass of Oort cloud is 4.18×10⁹ kg.

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Jonathan needs to maintain a separation of 0.543 mm between the plates to get the desired charge, and a dielectric constant of 92.6 to achieve a separation of 5 mm with a dielectric.

(a) Using the equation Q = CV, where Q is the charge, C is the capacitance, and V is the voltage, we can solve for the capacitance: C = Q/V =\((8.15 x 10^-9 C) / (50 V) = 1.63 x 10^-10 F.\)

Then, using the formula for capacitance of parallel plate capacitors: C = ε0A/d, where ε0 is the permittivity of free space, A is the area of the plates, and d is the separation distance, we can solve for the separation distance: d =\(_{3}OA/C = (8.85 x 10^-12 F/m) x (0.01 m^2) / (1.63 x 10^-10 F) = 0.543 mm.\)

(b) To find the dielectric constant, we can use the formula for capacitance of a parallel plate capacitor with a dielectric: C = εrε0A/d, where εr is the relative permittivity or dielectric constant of the material. Solving for εr, we get: εr = Cd / ε0A = \((1.63 x 10^-10 F)\) x (0.005 m) / \((8.85 x 10^-12 F/m)\) x \((0.01 m^2)\) = 92.6.

Therefore, Jonathan should use a dielectric with a relative permittivity of 92.6 to achieve a separation of 5 mm.

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Answer:

F' = F/12

Therefore, the electrostatic force is reduced to one-twelve of its original value.

Explanation:

The electrostatic force of attraction or repulsion between to charges is given by Coulomb's Law:

F = kq₁q₂/r²   ---------- equation 1

where,

F = Electrostatic Force

k = Coulomb's Constant

q₁ = magnitude of first charge

q₂ = magnitude of 2nd charge

r = distance between charges

Now, if we double the distance between charges and reduce one charge to one-third value, then the force will become:

F' = kq₁'q₂'/r'²

where,

q₁' = (1/3)q₁

q₂' = q₂

r' = 2r

Therefore,

F' = k(1/3 q₁)(q₂)/(2r)²

F' = (1/12)kq₁q₂/r²

using equation 1:

F' = F/12

Therefore, the electrostatic force is reduced to one-twelve of its original value.

Answer: I=0.52 A

Explanation:

Given

Voltage is \(2.5\ V\)

Resistance \(R=4.8\ \Omega\)

Current is given by

\(I=\dfrac{V}{R}\)

\(\Rightarrow I=\dfrac{2.5}{4.8}=0.52\ A\)

To calculate the torque required to create an angular acceleration, we can use the formula:

Torque = Moment of Inertia × Angular Acceleration

The moment of inertia of a disk can be calculated using the formula:

Moment of Inertia = (1/2) × Mass × Radius^2

Given:

Mass = 15,000 kg

Radius = 6.14 m

Angular Acceleration = 0.0500 rad/s^2

First, calculate the moment of inertia:

Moment of Inertia = (1/2) × Mass × Radius^2

Moment of Inertia = (1/2) × 15,000 kg × (6.14 m)^2

Next, calculate the torque:

Torque = Moment of Inertia × Angular Acceleration

Torque = Moment of Inertia × 0.0500 rad/s^2

Now, let's plug in the values and calculate:

Moment of Inertia = (1/2) × 15,000 kg × (6.14 m)^2

Moment of Inertia ≈ 283,594.13 kg·m^2

Torque = 283,594.13 kg·m^2 × 0.0500 rad/s^2

Torque ≈ 14,179.71 N·m

Rounding to three significant figures, the torque required to create an angular acceleration of 0.0500 rad/s^2 is approximately 14,180 N·m.

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Answer:

Her average speed is: 100 m/min

Explanation:

Recall that the formula for average speed is given by:

Speed = Distance / time

Then in our case, this is

Speed = 300 m / 3 min = 100 m/min

Answer:

The final temperature of the bullets is 327.3 ºC.

Explanation:

Let suppose that a phase change does not occur during collision and collided bullets stop at the end. We represent the phenomenon by the First Law of Thermodynamics:

\(K_{A, o} + K_{B, o}-K_{A}-K_{B}+U_{A,o} + U_{B,o}-U_{A}-U_{B} = 0\) (1)

Where:

\(K_{A,o}\), \(K_{A}\) - Initial and final translational kinetic energies of the 15-g bullet, measured in joules.

\(K_{B,o}\), \(K_{B}\) - Initial and final translational kinetic energies of the 7.75-g bullet, measured in joules.

\(U_{A,o}\), \(U_{A}\) - Initial and final internal energies of the 15-g bullet, measured in joules.

\(U_{B,o}\), \(U_{B}\) - Initial and final internal energies of the 7.75-g bullet, measured in joules.

By definitions of translational kinetic energy and sensible heat we expand and simplify the equation above:

\(\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T) = 0\) (2)

Where:

\(m_{A}\), \(m_{B}\) - Masses of the 15-g and 7.75-g bullets, measured in kilograms.

\(v_{A,o}\), \(v_{A}\) - Initial and final speeds of the 15-g bullet, measured in meters per second.

\(v_{B,o}\), \(v_{B}\) - Initial and final speeds of the 7.75-g bullet, measured in meters per second.

\(c\) - Specific heat of lead, measured in joules per kilogram-Celsius degree.

\(T_{o}\), \(T\) - Initial and final temperatures of the bullets, measured in Celsius degree.

Now we clear the final temperature of the bullets:

\((m_{A}+m_{B})\cdot c \cdot (T-T_{o}) = \frac{1}{2}\cdot [m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})]\)

\(T-T_{o} = \frac{m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})}{(m_{A}+m_{B})\cdot c}\)

\(T= T_{o}+\frac{m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})}{(m_{A}+m_{B})\cdot c}\) (3)

If we know that \(T_{o} = 30\,^{\circ}C\), \(m_{A} = 15\times 10^{-3}\,kg\), \(m_{B} = 7.75\times 10^{-3}\,kg\), \(v_{A,o} = 295\,\frac{m}{s}\), \(v_{B,o} = 375\,\frac{m}{s}\), \(v_{A} = 0\,\frac{m}{s}\), \(v_{B} = 0\,\frac{m}{s}\) and \(c = 128\,\frac{J}{kg\cdot ^{\circ}C}\), then the final temperature of the collided bullets is:

\(T = 30\,^{\circ}C+\frac{(15\times 10^{-3}\,kg)\cdot \left[\left(295\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]+(7.75\times 10^{-3}\,kg)\cdot \left[\left(375\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{(15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(128\,\frac{J}{kg\cdot ^{\circ}C} \right)}\)

\(T = 852.534\,^{\circ}C\)

Given that found temperature is greater than melting point, then we conclude that supposition was false. If we add the component of latent heat of fussion, then the resulting equation is:

\(\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T)-U = 0\) (4)

\(U=\frac{1}{2}\cdot m_{A}\cdot (v_{A,o}^{2}-v_{A}^{2})+ \frac{1}{2}\cdot m_{B}\cdot (v_{B,o}^{2}-v_{B}^{2})+(m_{A}+m_{B})\cdot c\cdot (T_{o}-T)\)

If we know that \(T_{o} = 30\,^{\circ}C\), \(T = 327.3\,^{\circ}C\), \(m_{A} = 15\times 10^{-3}\,kg\), \(m_{B} = 7.75\times 10^{-3}\,kg\), \(v_{A,o} = 295\,\frac{m}{s}\), \(v_{B,o} = 375\,\frac{m}{s}\), \(v_{A} = 0\,\frac{m}{s}\), \(v_{B} = 0\,\frac{m}{s}\) and \(c = 128\,\frac{J}{kg\cdot ^{\circ}C}\), then latent heat received by the bullets during impact is:

\(U =\frac{1}{2}\cdot (15\times 10^{-3}\,kg)\cdot \left[\left(295\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right] + \frac{1}{2}\cdot (7.75\times 10^{-3}\,kg)\cdot \left[\left(375\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]+(15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(128\,\frac{J}{kg\cdot ^{\circ}C} \right) \cdot (30\,^{\circ}C-327.3\,^{\circ}C)\)\(U = 331.872\,J\)

The maximum possible latent heat (\(U_{max}\)), measured in joules, that both bullets can receive during collision is:

\(U_{max} = (m_{A}+m_{B})\cdot L_{f}\) (5)

Where \(L_{f}\) is the latent heat of fusion of lead, measured in joules per kilogram.

If we know that  \(m_{A} = 15\times 10^{-3}\,kg\), \(m_{B} = 7.75\times 10^{-3}\,kg\) and \(L_{f} = 2.45\times 10^{4}\,\frac{J}{kg}\), then the maximum possible latent heat is:

\(U_{max} = (15\times 10^{-3}\,kg+7.75\times 10^{-3}\,kg)\cdot \left(2.45\times 10^{4}\,\frac{J}{kg} \right)\)

\(U_{max} = 557.375\,J\)

Given that \(U < U_{max}\), the final temperature of the bullets is 327.3 ºC.

In fluid dynamics, the terms "lines of flow" and "streamlines" are used to describe the path followed by individual fluid particles as they move within a flow. While both terms are closely related, they have slightly different meanings.

Lines of flow refer to the imaginary lines that are drawn in a fluid flow field to represent the direction in which the fluid particles move. These lines can be fluid dynamics,as the path that a fluid particle would trace out over time. Lines of flow can provide information about the overall flow pattern and direction in a fluid system.

On the other hand, streamlines are the actual curves or lines that are tangent to the velocity vector at every point in the flow field.

These streamlines are defined as the locus of points through which a fluid particle passes at a given instant. Streamlines are a mathematical representation of the flow pattern and can be determined through computational fluid dynamics (CFD) or experimental methods.

In laminar flow, where the fluid flows smoothly in layers with minimal turbulence, the lines of flow and streamlines coincide with each other. This means that the lines of flow and streamlines are identical, and fluid particles move along these lines without crossing each other.

The streamlines do not intersect or cross paths in laminar flow, indicating that the flow is well-ordered and predictable.

In summary, lines of flow and streamlines both describe the path followed by fluid particles in a flow field, with the key difference being that lines of flow are conceptually drawn paths, while streamlines are the actual curves tangent to the velocity vector at each point. In laminar flow, the lines of flow and streamlines coincide, reflecting the organized and smooth nature of the flow.

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Answer:

Types of tidal turbines

Axial turbines.

Crossflow turbines.

Flow augmented turbines.

Oscillating devices.

Venturi effect.

Tidal kite turbines.

Turbine power.

Resource assessment.

Answer:

Axial turbines

Crossflow turbines

flow augmented turbines

Answer:

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 5000 × 15

We have the final answer as

Hope this helps you

SeaWorld is a chain of theme parks and oceanariums that showcase marine life through educational exhibits, live shows, and thrilling rides. While it has faced criticism for its treatment of animals, SeaWorld has made changes to prioritize conservation and phased out its orca breeding program.

SeaWorld is a chain of theme parks and oceanariums that primarily focuses on marine life and entertainment. The company operates various parks across the United States, including SeaWorld parks in Orlando, San Diego, and San Antonio, as well as Busch Gardens parks in Tampa and Williamsburg. SeaWorld offers a combination of educational exhibits, live shows, and thrilling rides, with a special emphasis on marine animals such as dolphins, whales, sea lions, penguins, and sharks.

SeaWorld parks provide visitors with opportunities to observe and interact with marine creatures up close, while also offering educational programs that aim to raise awareness about marine conservation and preservation. The parks feature captivating shows featuring trained animals, where they perform impressive behaviors and stunts, showcasing their intelligence and natural abilities.

Over the years, SeaWorld has faced criticism from animal rights activists and environmentalists, who argue that the captivity and use of marine animals for entertainment purposes is unethical and harmful to the animals' well-being. These concerns have led to significant changes in the company's practices, including the phasing out of its orca breeding program and the introduction of more educational and conservation-focused initiatives.

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The Moon was formed from the collision between Earth and another Mars-sized body- giant impact theory

The Moon was formed from collisions and combining- The Colliding Planetesimals Theory

The Moon was locked into orbit by Earth's gravity as it was passing by it- capture hypothesis

The Moon was formed from the same cloud of material as Earth- accretion hypothesis

Define The Colliding Planetesimals Theory.

Early in the Solar System's existence, planetesimals (extremely massive rocks like asteroids) orbiting the Sun and the Earth came into contact with one another, breaking them apart. Condensing from this material was the Moon.

According to the capture hypothesis, put forth by Michael Mark Woolfson in 1964, the Solar System came into being as a result of tidal interactions between the Sun and a protostar with a low density. The Sun's gravity would have attracted material from the protostar's diffuse atmosphere, which would have subsequently collapsed to create the planets.

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The potential difference between points A and B in the circuit is 12 V.

To find the potential difference between points A and B in the given circuit, determine the total voltage across the circuit.

From the information provided, there are four resistors labeled as 12 Ω each. Assuming these resistors are connected in series, the total resistance in the circuit is 12 Ω + 12 Ω + 12 Ω + 12 Ω = 48 Ω.

Next, determine the current flowing through the circuit. The voltage is given as 24 V.

Using Ohm's Law (V = IR), calculate the current:

I = V / R

I = 24 V / 48 Ω

I = 0.5 A

Since the resistors are connected in series, the current remains the same throughout the circuit.

Finally, to find the potential difference between points A and B, multiply the current by the resistance between those points:

V_AB = I × R_AB

V_AB = 0.5 A * 24 Ω

V_AB = 12 V

Therefore, the potential difference between points A and B in the circuit is 12 V.

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