At which point are two objects said to be at thermal equilibrium with each other?
Group of answer choices

When both their temperature and thermal energy are equal

When their thermal energy is equal

When their heat capacities are equal

When their temperatures are equal

Answer:

The answer to your problem is, D. In an orbit, the satellite and the central body are the two foci of the ellipse

Explanation:

Our solar system contains the sun and eight planets revolving around the sun. Planets like earth, mars, mercury etc. are revolving around sun in their own orbits. Every planet's orbit around the Sun is an ellipse, according to Kepler's First Law.

Two focal points, or foci, make up an ellipse. The overall distance of a planet from these 2 focus points is constant during its orbit. Additionally, an ellipse has two symmetry lines.

The orbital ellipse's central focus is always where the sun is positioned. The sun is centered. The planet's orbit is an ellipse, thus as it revolves around its axis, the distance from the sun changes continuously.

Thus the answer to your problem is, D. In an orbit, the satellite and the central body are the two foci of the ellipse

The speed of the car after the collision is determined as 0.2 m/s.

The speed of the car after the collision is calculated by applying the principle of conservation of linear momentum as follows;

m1u1 + m2u2 = v(m1 + m2)

where;

20000(3) - 40000(1.2) = v (20000 + 40000)

12,000 = 60,000v

v = 12,0000 / 60,000

v = 0.2 m/s

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The orbital radius of the satellite above Webb13's equator is around 11,360 kilometres.

There are several rotating periods (of an astronomic object) the length of time needed for it to revolve in relation to the nearby stars. (of an object revolving on Earth) the duration of its axis rotation in relation to the earth (assumed fixed).

Since 1 hour equals 60 minutes x 60 seconds, or 3600 seconds, the orbital period of the satellite is the same as the planet's rotational period, which is 22.9 hours, or 82,440 seconds. The following formula may be used to determine the radius of the satellite's orbit:

[tex]r = (G * M * T^2 / 4π^2)^(1/3)[/tex]

where r is the orbit's radius, G is the gravitational constant, M is the planet's mass, and T is the satellite's orbital period.

Using the specified values:

[tex]G = 6.67 × 10^-11 m^3 kg^-1 s^-2 (gravitational constant)[/tex]

[tex]M = 2.75 × 10^25 kg (mass of Webb13)[/tex]

T = 82,440 s (orbital period of satellite)

The units can be changed to kilometres and then entered into the formula as follows:

[tex]r = (6.67 × 10^-11 m^3 kg^-1 s^-2 * 2.75 × 10^25 kg * (82,440 s)^2 / (4π^2))^(1/3)[/tex]

[tex]r = 1.136 × 10^7 m[/tex]

[tex]r = 11,360 km[/tex]

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Answer:

Human development option D

The velocity of ball B after impact would be2.7147 i + 2.6987 j ft/s

To solve this problem, we can use the conservation of momentum and energy.

First, let's find the momentum of ball A before the collision. The momentum is given by:

p = mv

The mass of each ball is the same, so we can write:

p_A = mV_A

where V_A is the velocity vector of ball A.

We can break V_A into its x and y components as follows:

V_Ax = V_A cos(θ)

V_Ay = V_A sin(θ)

where θ is the angle between the velocity vector and the x-axis.

Substituting in the given values, we get:

V_Ax = 3 cos(70°) = 0.9063 ft/s

V_Ay = 3 sin(70°) = 2.8830 ft/s

So, the momentum of ball A before the collision is:

p_A = mV_A = m (V_Ax i + V_Ay j) = m (0.9063 i + 2.8830 j) lb·ft/s

Next, we need to find the velocity of ball A after the collision. We can use conservation of momentum and energy to do this.

p_A + p_B = p_A' + p_B'

where p_B is the momentum of ball B before the collision, and p_A', p_B' are their respective momenta after the collision.

Since ball B is initially at rest, its momentum before the collision is zero:

p_B = 0

Conservation of energy tells us that the total kinetic energy of the system before the collision is equal to the total kinetic energy of the system after the collision:

1/2 m V_A^2 = 1/2 m V_A'^2 + 1/2 m V_B'^2

where V_A' and V_B' are the velocities of the balls after the collision.

We can use the coefficient of restitution (e) to relate the velocities of the balls before and after the collision:

e = (V_B' - V_A') / (V_A - V_B)

Substituting in the given values, we get:

e = (V_B' - V_A') / (3 - 0)

Solving for V_B', we get:

V_B' = e (V_A - V_B) + V_A'

Substituting in the known values, we get:

V_A' = (0.9063 i + 2.8830 j) ft/s

e = 0.9

Solving for V_B', we get:

V_B' = e (V_A - V_B) + V_A'

= 0.9 (3 i + 0 j) + (0.9063 i + 2.8830 j)

= 2.7147 i + 2.6987 j ft/s

So, the velocity of ball B after the collision is:

V_B' = 2.7147 i + 2.6987 j ft/s

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Therefore, the heat energy per unit time per unit area, or flux, emitted by the body is approximately 241.7 W/m².

The heat transfer equation can be written as Q = m  c  T, where Q denotes the amount of heat transferred, m denotes mass, c denotes specific heat, and T denotes the temperature differential. Heat transfer is the process by which heat is transferred from a hot object to a cold object.

The Stefan-Boltzmann Law can be used to determine the thermal energy per unit time per unit area, or flux, released by a body:

F = σ * € * T⁴

To convert 20.0°C to Kelvin, we add 273.15 K to get 293.15 K.

Now we can plug in the values:

F = 5.67 x 10⁻⁸ W/m² K⁴ * 0.500 * (293.15 K)⁴

F ≈ 241.7 W/m²

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I = M g sin / (L  B), where M, L,, and B are the provided variables and g is the acceleration due to gravity (9.81 m/s2), determines the amount of current in the wire that will keep it at rest.

The magnitude of the magnetic field is inversely related to the perpendicular distance from the wire and proportionate to the current.

We need to use the equation for the force on a current-carrying wire in a magnetic field to estimate the size of the current in the wire that will keep it at rest:

F = I L × B

The weight of the wire's component moving down the incline must be balanced by the force acting on the wire:

F = M g sinθ

For the wire to remain at rest, these two forces must be equal:

I L × B = M g sinθ

Solving for I, we get:

I = M g sinθ / (L × B)

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Answer:

The answer to your problem is, 28 kg x m/s

Explanation:

m = 4

v = 7

P = 4 x 7

= 28

There is no other way to do it.

Thus the answer to your problem is, 28 kg x m/s

We can use Faraday's Law of Electromagnetic Induction to find the induced emf (electromotive force) in the secondary coil:

emf = -N(dΦ/dt)

where

emf = induced emf

N = number of turns in the coil

dΦ/dt = rate of change of magnetic flux through the coil.

In this problem,

N = 12,000 turns, and the change in magnetic flux is:

dΦ = 740 uWb - 40 uWb = 700 uWb

The time interval for this change =180 us, or 180 x 10^-6 seconds. Therefore, the rate of change of magnetic flux is:

dΦ/dt = (700 uWb) / (180 x 10^-6 s) = 3.89 mWb/s

Now we can find the induced emf:

emf = -N(dΦ/dt) = -(12,000 turns)(3.89 mWb/s) = -46.68 volts

Note that the negative sign indicates that the induced emf will produce a current that opposes the change in magnetic flux that caused it.

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if in a binary star system a white dwarf exceeds this limit through mass transfer, it will explode and a Type Ia Supernova will be the end result.

The clock will run faster in Pluto

The force constant is 81 N/m

The period is   3.5 s

Clocks on Jupiter, which has a stronger gravitational field than Pluto, would run slower than clocks on Pluto. This means that clocks on Pluto would appear to run faster when compared to clocks on Jupiter.

We know that;

F = Ke

K = F/e

K = 68.28738 N/84.81007 * 10^-2 m

K = 81 N/m

Then;

T = 2π√L/g

T = 2(3.14)√0.5/1.6

T = 3.5 s

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Marcel should place Gilles at about 0.828 m to the right of the pivot in order to balance the seesaw.

For the seesaw to be balanced, the clockwise moment caused by Gilles sitting on the right side of the pivot must be equal to the counterclockwise moment caused by Jacques sitting on the left side of the pivot. The moment (M) is given by the weight of the child multiplied by the distance from the pivot:

M = w × L1 = W × L

where w is the weight of Gilles.

Rearranging this equation, we get:

L1 = (W × L) / w

Substituting the given values, we get:

L1 = (72.0 N × 0.80 m) / w

We don't know the weight of Gilles, so we cannot solve for L1. However, we can set up an equation to find the weight w needed to balance the seesaw:

W × L = w × L1

Substituting the given values, we get:

72.0 N × 0.80 m = w × L1

Solving for w, we get:

w = (72.0 N × 0.80 m) / L1

Now we can substitute this expression for w into the earlier equation for L1, giving:

L1 = (W × L) / [(72.0 N × 0.80 m) / L1]

Simplifying, we get:

L1^2 = (W × L × L1) / (72.0 N × 0.80 m)

L1^2 = (W × L) / (72.0 N × 0.80 m)

Substituting the given values and solving, we get:

L1 = sqrt[(72.0 N × 0.80 m) / (76.8 N)]

L1 ≈ 0.828 m

Therefore, Marcel should place Gilles at a distance of approximately 0.828 m to the right of the pivot to balance the seesaw.

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To solve the problem, we can use the principle of conservation of heat, which states that the heat lost by the glass is gained by the alcohol.Let T be the initial temperature of the alcohol in degrees Celsius.The heat lost by the glass is given by:Q1 = mcΔT1where m is the mass of the glass, c is the specific heat capacity of glass, and ΔT1 is the change in temperature of the glass.Substituting the given values, we get:Q1 = (250 g) × (840 J/kg°C) × (60°C - T)The heat gained by the alcohol is given by:Q2 = mcΔT2where m is the mass of the alcohol, c is the specific heat capacity of alcohol, and ΔT2 is the change in temperature of the alcohol.Substituting the given values, we get:Q2 = (700 g) × (2500 J/kg°C) × (20°C - T)Since the total heat lost by the glass is equal to the total heat gained by the alcohol, we can set Q1 equal to Q2 and solve for T:Q1 = Q2(250 g) × (840 J/kg°C) × (60°C - T) = (700 g) × (2500 J/kg°C) × (20°C - T)Simplifying and solving for T, we get:T = 32.5°CTherefore, the initial temperature of the alcohol was 32.5°C.

Explanation:

When a car is stopped, facing upwards on a hill, the friction acts in the opposite direction to the motion that the car would naturally take if it were not stopped.

In this case, the car would roll backwards down the hill due to the force of gravity. The friction between the tires and the road surface acts in the opposite direction to this motion, providing a force that opposes the car's tendency to roll backwards. Therefore, the friction acts in the forward direction, up the hill, to prevent the car from rolling backwards.

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Answer:

work = (37.0 kcal) - (82.5 kcal) = -45.5 kcal

Explanation:

The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In this case, the refrigerator is removing heat from the freezer and releasing it through the condenser on the back. Therefore, the work done by the compressor is:

work = heat removed - heat released

The negative sign indicates that work was done on the refrigerator by an external agent (e.g., an electric motor) to remove heat from the freezer and release it through the condenser.

I got you Bruv :)

The two parts of the model that are missing are as shown in the diagram:

Oxygen is essential for cellular respiration, which is the process by which cells convert food molecules (such as glucose) into energy in the form of ATP (adenosine triphosphate). Cellular respiration occurs in the mitochondria of cells and is a complex series of metabolic reactions that requires oxygen as the final electron acceptor in the electron transport chain.

Without oxygen, cellular respiration cannot proceed beyond glycolysis, which is the first step in the process. Glycolysis breaks down glucose into two molecules of pyruvate and produces a small amount of ATP. However, this process is not very efficient, and it cannot sustain cellular activity for very long.

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A coal power station generates electricity by first burning coal to heat water into steam, then passing that steam through a turbine to make it spin.

A. Energy stores:

Chemical potential energy in the coal

Thermal energy in the water

Kinetic energy in the steam

Kinetic energy in the turbine

Electrical energy in the generator

Energy transfers:

Chemical potential energy in the coal to thermal energy in the water (by burning the coal)

Thermal energy in the water to kinetic energy in the steam (by boiling the water)

Kinetic energy in the steam to kinetic energy in the turbine (by passing through the turbine)

Kinetic energy in the turbine to electrical energy in the generator (by driving the generator)

B. Energy transferred into heat store of water: 20,000 J x 0.9 = 18,000 J

Energy dissipated in the turbine: 18,000 J x 0.3 = 5,400 J

Remaining energy after turbine: 18,000 J - 5,400 J = 12,600 J

Energy transferred as electrical energy: 12,600 J x 0.85 = 10,710 J

C. Power output = energy input per second = 10,000 J/s

So, the power output of the power station is 10,710 J/s (since 85% of the remaining energy is transferred as electrical energy).

D. Three specific improvements that could be made to the power station to make it more efficient are:

Implementing better combustion techniques to increase the efficiency of burning coal.

Using better insulation materials to minimize heat loss in the power station.

Using more efficient turbines and generators to convert kinetic energy to electrical energy.

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Velocity of block B when it is at half of its path is approximately 10.17 m/s.

Law of conservation of energy is a fundamental principle in physics that states that energy cannot be created or destroyed, but can only be transformed from one form to another.

Potential energy of block B is : PE = mgh

m is mass of the block, g is acceleration due to gravity, and h is height of the block above the ground.

Initial potential energy of block B is: PEi = mgh = 70 kg × 9.81 m/s² × 15 m = 10290 J

When block B is at half of its path, its height above the ground is: h/2 = 15 m / 2 = 7.5 m

Final potential energy of block B at this height is: PEf = mgh/2 = 70 kg × 9.81 m/s² × 7.5 m = 5143.5 J

Change in potential energy is:

ΔPE = PEf - PEi = 5143.5 J - 10290 J = -5146.5 J

Kinetic energy of block B at half of its path is: KE = -ΔPE = 5146.5 J

Kinetic energy of block B is given by: KE = (1/2)mv²

v = √(2KE/m) = √(2ΔPE/m)

v = √(2 × 5146.5 J / 70 kg) = 10.17 m/s

Therefore,  velocity of block B when it is at half of its path is approximately 10.17 m/s.

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Answer: H=5.1m

Explanation:

Given:

Ball 1 height= 10m

Ball 2 initial velocity=10m/s

use the kinematic equation:

S=(vi)t+12at2

I choose my sign convention to be up=positive, down=negative, so  a=−9.81ms2 (a is taken as the value of gravity)

Ball 1 dropped from 10m :

−(10−H)=0+12(−9.81)t2

Note that (10-S) is negative because that displacement is *below* the starting point.

12(9.81)t2=10−H

 ——- equation (1)

Ball 2 thrown upward at 10 m/s :

H=(10)t+12(−9.81)t2

or

12(9.81)t2=10t−H

 ——- equation (2)

equation (1) minus equation (2):

0=(10−H)−(10t−H)

t=1       equation (1):

12(9.81)12=10−H

H=5.1m

Given Information:

Ball One:

[tex]\vec y_{0} = 10 \ m[/tex] (Indicating the initial position)

We also know the ball was dropped from rest. So, [tex]\vec v_{0_{1} } = 0 \ m/s[/tex].

Ball Two:

[tex]\vec v_{0} = 10 \ m/s[/tex] (Indicating the initial velocity)

We also know the ball was throw from the ground. So, [tex]\vec y_{0_{1} } = 0 \ m[/tex].

The Information we want to Find:

[tex]\vec y_{c} = ?? \ m[/tex] (Indicating the position the two projectiles collide)

Using the Following Kinematic Equation to Solve:

[tex]\Delta \vec x = \vec v_{0}t + \frac{1}{2} \vec at[/tex]

For ball one...

[tex]\Delta \vec y = \vec v_{0}t + \frac{1}{2} \vec a_{y} t[/tex]

[tex]\Longrightarrow \vec y_{c}- \vec y_{0} = \vec v_{0_{1} }t + \frac{1}{2} \vec a_{y}t[/tex]

[tex]\Longrightarrow \vec y_{c}- \vec y_{0} = (0)t + \frac{1}{2} \vec a_{y}t[/tex]

[tex]\Longrightarrow \vec y_{c}- \vec y_{0} = \frac{1}{2} \vec a_{y}t[/tex]

[tex]\Longrightarrow \vec y_{c} = \frac{1}{2} \vec a_{y}t +\vec y_{0}[/tex] => Equation 1

For ball two...

[tex]\Delta \vec y = \vec v_{0}t + \frac{1}{2} \vec a_{y}t[/tex]

[tex]\Longrightarrow \vec y_{c}- \vec y_{0_{1} } = \vec v_{0}t + \frac{1}{2} \vec a_{y}t[/tex]

[tex]\Longrightarrow \vec y_{c} = \vec v_{0}t + \frac{1}{2} \vec a_{y}t +\vec y_{0_{1} }[/tex]

[tex]\Longrightarrow \vec y_{c} = \vec v_{0}t + \frac{1}{2} \vec a_{y}t + 0[/tex]

[tex]\Longrightarrow \vec y_{c} = \vec v_{0}t + \frac{1}{2} \vec a_{y}t[/tex] => Equation 2

Set equations 1 and 2 equal to each other and solve for the time that they collide.

[tex]\left \{ {{\vec y_{c} = \frac{1}{2} \vec a_{y}t +\vec y_{0}} \atop { \vec y_{c} = \vec v_{0}t + \frac{1}{2} \vec a_{y}t } \right.[/tex]

[tex]\Longrightarrow \frac{1}{2} \vec at +\vec y_{0}= \vec v_{0}t + \frac{1}{2} \vec at[/tex]

[tex]\Longrightarrow \vec y_{0}= \vec v_{0}t[/tex]

[tex]\Longrightarrow t=\frac{\vec y_{0}}{\vec v_{0}}[/tex]

[tex]\Longrightarrow t=\frac{10}{10}[/tex]

[tex]\Longrightarrow t=1 \ s[/tex]

Thus, the balls collide at time, t=1 s. We can now use this time to plug into equation 1 or 2 to find the height at which they collide. I will use equation 1.

[tex]\Longrightarrow \vec y_{c} = \frac{1}{2} \vec a_{y}t +\vec y_{0}[/tex]

[tex]\Longrightarrow \vec y_{c} = \frac{1}{2} (-9.8)(1) +10[/tex]

[tex]\Longrightarrow \vec y_{c} = 5.1 \ m \ \therefore \ Sol.[/tex]

*Note* [tex]\vec a_{y}[/tex] is the acceleration of gravity ([tex]-9.8 \ m/s^2 \ or \ -32 \ ft/s^2[/tex])

Final Answer: The balls collide at the height 5.1 m.

Artificial satellites are launched into Earth's orbit for various purposes, including communication, navigation, weather monitoring, scientific research, and military surveillance.

Artificial satellites sent into Earth's orbit the Earth, typically at an altitude between 200 and 22,000 miles, depending on its intended purpose. Satellites in low Earth orbit (LEO) travel at a speed of about 17,500 miles per hour, completing one orbit in about 90 minutes, while satellites in geostationary orbit (GEO) remain stationary above the equator at an altitude of about 22,236 miles.

Satellites can remain in orbit for many years, but eventually, they can fall out of orbit due to atmospheric drag or collisions with space debris. When a satellite falls out of orbit, it typically burns up in Earth's atmosphere, although larger satellites may leave debris that can pose a risk to other spacecraft.

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Answer: Min: 35

Max: 100

Mean: 73.53

Standard deviation (SX): 20.17

Mode: 72

Median: 72.5

Explanation:

The formula C = (20A) / d can be used to determine the capacitance of a circular parallel plate capacitor. As a result, the capacitor can store 0.000494 J of energy.

A charged parallel plate capacitor has 5 cm between its plates, and its internal electric field is 200 Vcm. The capacitor is totally submerged by a 2 cm wide uncharged metal bar. The metal bar's length is the same as the capacitor's.

C = (2πε₀A) / d

Where,

ε₀ = Permittivity of free space = [tex]8.854 × 10^−12 F/m[/tex]

A = Area of the plate =[tex]πr^2[/tex]

d = Distance between the plates

Therefore, the area of each plate (A) = [tex]πr^2 = π(0.05)^2 = 0.00785 m^2.[/tex]

[tex]C = (2πε₀A) / d = (2π × 8.854 × 10^−12 × 0.00785) / 0.002[/tex]

[tex]= 6.21 × 10^−11 F[/tex]So, the capacitance of the circular parallel plate capacitor is [tex]6.21 × 10^-11 F.[/tex]

[tex]U = 1/2 * C * V^2[/tex]

Substituting the values, we get:

[tex]U = 1/2 * 6.21 × 10^-11 * (200)^2[/tex]

= 0.000494 J

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An individual dropped the sandbag from a height of 63.504 metres.

The relationship between angular momentum L and moment of inertia I and angular speed, expressed in radians per second, is shown. Moment of inertia is different from mass in that it depends on both the form and location of the axis of rotation in addition to the amount of stuff present.

vf = vi + at

vf = 0 + (9.8 m/s²)(0.5 s)

vf = 4.9 m/s

vf = vi + at

0 = 12 m/s - (9.8 m/s²)t

t = 1.22 s

Δy = vi t + 1/2 a t²

Δy = (12 m/s)(1.22 s) + 1/2 (9.8 m/s²) (1.22 s)²

Δy = 7.33 m

Δy = vi t + 1/2 a t²

Δy = vi t + 1/2 a t²

0 = (4.2 m/s)(3.6 s) + 1/2 (9.8 m/s²) (3.6 s)² + Δy

Δy = -63.504 m

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B=0I2R(at centre of loop), B = 0 I 2 R(at centre of loop), where R is the loop's radius, gives the magnetic field strength at the loop's centre.

The magnetic field strength diminishes with increasing radius. Radius of the loop has an inverse relationship with magnetic field intensity.

With its plane normal to an external field of magnitude 5.0102T, a circular coil with a radius of 10 cm and 16 turns that is carrying a current of 0.75 A is at rest. The coil is unrestricted in its ability to rotate around an axis in a plane perpendicular to the field direction.

Hence, the integral around any circle's diameter that is centred on a wire.

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The final velocity of the goalie is -0.613 m/s (indicating that he moves backwards), and the final velocity of the puck is 30.2 m/s (indicating that it moves forwards).

When particles, collections of particles, or solid bodies move in the same direction and get close enough to each other, they collide and generate mutual force.

This issue can be resolved by applying the laws of motion and kinetic energy conservation.

momentum conservation

Before the collision:

Total momentum = 0 (since the goalie is at rest)

After the collision:

Total momentum = m_goalie * v_goalie + m_puck * v_puck

Conservation of kinetic energy:

Before the collision:

Total kinetic energy = 0

After the collision:

Total kinetic energy = (1/2) * m_goalie * v_goalie² + (1/2) * m_puck * v_puck^2

We can use these two equations to solve for the final velocities of the goalie and the puck.

First, let's use the conservation of momentum equation to solve for v_goalie:

0 = m_goalie * v_goalie + m_puck * v_puck

v_goalie = - m_puck * v_puck / m_goalie

Now, we can substitute this expression for v_goalie into the conservation of kinetic energy equation:

(1/2) * m_puck * v_puck² = (1/2) * m_goalie * (- m_puck * v_puck / m_goalie)² + (1/2) * m_puck * 43.5²

Simplifying this equation and solving for v_puck, we get:

v_puck = 2 * (m_oalie / (m_goalie + m_puck)) * 43.5

v_puck = 30.2 m/s

Finally, we can substitute this value for v_puck back into the equation for v_goalie

v_goalie = - m_puck * v_puck / m_goalie

v_goalie = -0.613 m/s

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The collision is perfectly elastic the velocity of object 2 after the collision is 44.0 cm/s to the right.

Substituting the given values, we get:

KE = (1/2)(0.0131 kg)(0.201 m/s)^2 + (1/2)(0.0324 kg)(0 m/s)^2

KE = 0.000132 J

Since the collision is perfectly elastic, the total kinetic energy of the system after the collision is also equal to KE. Therefore:

KE' = [tex](1/2)m_1v_1'^2 + (1/2)m_2v_2'^2[/tex]

where KE' is the kinetic energy of the system after the collision.

We can arrange this equation to solve for v2':

v2' = [tex]\sqrt{((2/m2)(KE' - (1/2)m1v1'^2))}[/tex]

Substituting the given values , we get:

v2' = [tex]\sqrt{((2/0.0324 kg)(0.000132 J - (1/2)(0.0131 kg)(20.1 m/s)^2))}\\[/tex]

v2' = 0.440 m/s

Finally, we need to convert the velocity back to cm/s:

v2' = 44.0 cm/s (rounded to three significant figures)

A collision refers to an event where two or more objects come into contact with each other, often resulting in damage or a change in the objects' trajectories. Collisions can occur in various contexts, including physics, engineering, and traffic accidents. In physics, a collision involves the transfer of momentum and energy between objects. The types of collisions include elastic collisions, where there is no loss of kinetic energy, and inelastic collisions, where some of the kinetic energy is lost.

In engineering, collisions can occur when designing structures or machines to prevent them from failing under extreme loads. For example, cars are designed with crumple zones to absorb the impact of a collision and protect the passengers.

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Answer:

Approximately [tex]1.9\; {\rm kg}[/tex] (assuming that [tex]g = 9.81\; {\rm N\cdot kg^{-1}}[/tex].)

Explanation:

Let [tex]L = 0.9\; {\rm m}[/tex] denote the unextended length of each spring.

The length of each spring is now [tex](L / \cos(\theta))[/tex]. The displacement of each spring would be [tex]x = L - (L / \cos(\theta)) = (1 - (1 / \cos(\theta)))\, L[/tex].

The tension in each spring would be [tex]T = k\, x[/tex], where [tex]k[/tex] is the spring constant.

Decompose the tension that each spring exerts on the block into two components:

The two horizontal components balance each other since they are equal in magnitude. The two vertical components add on to each other to exert a total upward force of [tex]2\, T\, \sin(\theta)[/tex] on the block.

Since the block is in equilibrium, the resultant force on the block will be [tex]0[/tex]. The sum of these two (upward) vertical components of tension should balance the (downward) weight of the block:

[tex]2\, T\, \sin(\theta) = m\, g[/tex], where [tex]m[/tex] is the mass of the block.

Rearrange this equation to find the mass of the block:

[tex]\begin{aligned} m &= \frac{2\, T\, \sin(\theta)}{g} \\ &= \frac{2\, k\, x\, \sin(\theta)}{g} \\ &= \frac{2\, k\, L\, (1 - (1 / \cos(\theta))\, \sin(\theta)}{g} \\ &= \frac{2\, (473)\, (0.9)\, (1 - (1 / \cos(20^{\circ})))\, \sin(20^{\circ})}{(9.81)}\; {\rm kg} \\ &\approx 1.9\; {\rm kg}\end{aligned}[/tex].

[tex]\blue{\huge {\mathrm{MASS \; OF \; THE \; BLOCK}}}[/tex]

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[tex]{\underline{\huge \mathbb{Q} {\large \mathrm {UESTION : }}}}[/tex]

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[tex] {\underline{\huge \mathbb{A} {\large \mathrm {NSWER : }}}} [/tex]

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[tex]{\underline{\huge \mathbb{S} {\large \mathrm {OLUTION : }}}}[/tex]

To solve for the mass of the block, we can use the forces acting on the block at equilibrium. We know that the force of gravity pulling down on the block is equal to the force of the springs pulling up.

The force of each spring can be found using Hooke's Law:

where:

In this case, the displacement is equal to the extension of the spring, which is given by:

where:

So the force of each spring is:

At equilibrium, the forces in the vertical direction must balance, so we have:

where

Substituting in the expression for [tex]\sf F_{spring}[/tex] and simplifying, we get:

[tex]\sf\qquad\implies 2kL(1-\cos\theta) = mg[/tex]

Solving for m, we obtain:

[tex]\sf\qquad\implies m = \dfrac{2kL(1-\cos\theta)}{g}[/tex]

Plugging in the given values, we get: [tex]\\\begin{aligned}\sf m&=\sf \dfrac{2(473\: N/m)(0.9\: m)[1-\cos(20^{\circ})][\sin(20^{\circ})]}{(9.81 m/s^2)}\\&=\boxed{\bold{\:1.9\: kg\:}}\end{aligned}[/tex]

Therefore, the mass of the block is 1.9 kg.

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[tex]- \large\sf\copyright \: \large\tt{AriesLaveau}\large\qquad\qquad\qquad\qquad\qquad\qquad\tt 04/02/2023[/tex]

Answer:C or B

Explanation:

Gases easily compress when pressure is applied, but liquids don't. So, option A.

The attracting and repellent forces that develop between the molecules of a substance are known as intermolecular forces.

Here,

The interactions between the individual molecules of a substance are mediated by the intermolecular forces. The majority of the physical and chemical features of matter are caused by intermolecular forces.

Compared to gases, which have relatively far-apart particles, liquids exhibit higher intermolecular forces due of the near proximity of their particles. The greater the intermolecular forces get as they get closer to one another since they are electrostatic in nature.

There is no force between particles in a gas. Particles have no restrictions on their movement. The container's walls colliding with one another provide the only forces that exist. which rely on the quantity of gas (number of collisions) and the momentum of each collision respectively. The pressure is nearly consistent throughout the entire mass of the gas because the molecules are free to move.

There is no compression of liquids. Since their volume is constant, changing pressure has no effect on it.

Hence,

Gases easily compress when pressure is applied, but liquids don't. So, option A.

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Answer:

Explanation:

The dimensions for the flow of electric current is amperes, often denoted by "A" in scientific notation.

In terms of the base SI units, one ampere is defined as the flow of one coulomb of electric charge per second through a conductor. It can also be expressed in milliamperes or macro amperes which is equal to 1/1000 and 1/1000000 of amperes respectively.