3.) A boat is moving south at 15.0, a passenger is walking north along the boat deck at 4.0 and throws a
rock north at 3.0, what is the velocity of the rock when viewed from the boat and the shore? (2)

The magnitude of the resultant force exerted on the barge is approximately 3.6 kN.

To calculate the magnitude of the resultant force exerted on the barge, we can use the law of cosines:

c^2 = a^2 + b^2 - 2ab cos(C)

where c is the magnitude of the resultant force, a and b are the magnitudes of T1 and T2, respectively, and C is the angle between T1 and T2 (which is 30° in this case).

Substituting the given values, we get:

c^2 = (3 kN)^2 + (4 kN)^2 - 2(3 kN)(4 kN) cos(30°)

c^2 = 9 kN^2 + 16 kN^2 - 24 kN^2 cos(30°)

c^2 = 25 kN^2 - 24 kN^2 cos(30°)

c^2 = 25 kN^2 - 12 kN^2

c^2 = 13 kN^2

c = sqrt(13) kN

c ≈ 3.6 kN

Therefore, the magnitude of the resultant force exerted on the barge is approximately 3.6 kN.

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Overall, it is important to consider the individual differences and cultural context when applying Erikson's theory to adolescence and young adulthood.

What is Erikson's theory?

Erikson's theory is a psychoanalytic theory that describes the development of the human personality across eight stages throughout the lifespan. Each stage is characterized by a particular crisis or conflict that must be resolved in order for the individual to develop a healthy personality.

Erik Erikson proposed a theory of psychosocial development that includes eight stages spanning from infancy to old age. In adolescence and young adulthood, the stage is identity versus role confusion. Erikson believed that during this stage, individuals explore and experiment with different identities and roles as they seek to establish a sense of self and their place in the world.

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Answer:

the net force is 40 N to the left.

Explanation:

To find the net force, we need to subtract the force going to the right from the force going to the left:

Net force = 60 N - 20 N

Net force = 40 N to the left

Therefore, the net force is 40 N to the left.

Answer:

Human development option D

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 correct answer is C.  The kinetic energy of the proton will remain the same if it enters a uniform magnetic field that is perpendicular to its velocity . Therefore, it stays the same.

Kinetic energy is a fundamental concept in physics that refers to the energy of an object in motion. When an object moves, it possesses kinetic energy that is proportional to its mass and the square of its velocity. The formula for kinetic energy is K = 1/2mv^2, where m is the mass of the object and v is its velocity.

The concept of kinetic energy is important in many areas of physics, including mechanics, thermodynamics, and relativity. It is used to describe the motion of particles in a gas, the motion of planets in a solar system, and the motion of subatomic particles in a particle accelerator. The concept of kinetic energy is closely related to other concepts in physics, such as potential energy, work, and momentum. Understanding kinetic energy is essential for understanding the behavior of physical systems in motion.

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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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Answer:

a) To find the acceleration at the bottom of the pipe, we can use the formula for centripetal acceleration: a = v^2 / r where v is the velocity, and r is the radius (half of the diameter) of the pipe. Since the velocity is constant and equal to 5 m/s, and the radius of the pipe is 4 m, the acceleration at the bottom is:

a = (5 m/s)^2 / 4 m a = 6.25 m/s^2

b) To find the force on the bike at an angle of 30° up from the bottom, we need to use the formula for centripetal force: F = m * a where m is the mass of the bike and rider (given as 400 kg) and a is the centripetal acceleration calculated in part (a). The force on the bike at an angle of 30° up from the bottom is:

F = 400 kg * 6.25 m/s^2 * cos(30°) F = 3,464 N

c) To find the minimum velocity at the top for the bike and rider to stay moving in a circle, we can use the same formula for centripetal acceleration and solve for velocity: a = v^2 / r v = sqrt(a * r) where r is the radius of the pipe (again, 4 m) and a is the centripetal acceleration required to keep the bike and rider moving in a circle, which is equal to the acceleration due to gravity at the top of the pipe:

a = g = 9.81 m/s^2 v = sqrt(9.81 m/s^2 * 4 m) v = 6.26 m/s

d) Comparing the minimum velocity calculated in part (c) to the constant speed of 5 m/s given in the question, we can see that the bike and rider do have sufficient velocity to stay moving on a circle at the top of the pipe.

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:

As a result, the ultimate internal diameter is D = 300 mm + D,D = 309.18 mm, and the ring's change in breadth is 1.53 mm.

Thermal expansion is the process through which an item enlarges and expands as a result of a change in temperature. The molecules take up more space because they move more quickly on average at higher temperatures. As a result, when anything is heated up, it gets bigger.

We must apply the thermal expansion formula to this issue in order to find a solution:

ΔL = α L ΔT

where L is the length change, is the thermal expansion coefficient, L is the starting length, and T is the temperature change.

a) The final internal diameter:

ΔD = α D ΔT

Substituting the values given, we get:

ΔD = (51 x 10^-6/°C) x 300 mm x 600°C

ΔD = 9.18 mm

The final internal diameter is therefore:

D = 300 mm + ΔD

D = 309.18 mm

b) The change in width of the ring:

The original width of the ring is 50 mm. We can use the same formula to find the change in width:

ΔW = α W ΔT

Substituting the values given, we get:

ΔW = (51 x 10^-6/°C) x 50 mm x 600°C

ΔW = 1.53 mm

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The position of the center of mass of the system is 21 cm from the origin.

The position of the center of mass of a two-particle system can be calculated using the formula,

xcm = (m1x1 + m2x2) / (m1 + m2)

where x1 and x2 are the positions of particles m1 and m2, respectively, and m1 and m2 are their masses.

Substituting values into the formula for xcm,

xcm = (m1x1 + m2x2) / (m1 + m2)

xcm = (m15 cm + m225 cm) / (m1 + m2)

xcm = (m15 cm + 4m125 cm) / (m1 + 4m1)

xcm = (5 cm + 4*25 cm) / (1 + 4)

xcm = 105 cm / 5

xcm = 21 cm

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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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a) The takeoff speed of the froghopper can be found using the following equation:

v^2 = 2gh/(1 - cos^2(theta))

where:

v = takeoff speed

g = acceleration due to gravity (9.81 m/s^2)

h = maximum height (293 mm = 0.293 m)

theta = takeoff angle (62.0 degrees)

Substituting the given values into the equation, we get:

v^2 = 2(9.81)(0.293)/(1 - cos^2(62.0))

v^2 = 0.571

v = sqrt(0.571)

v ≈ 0.756 m/s

Therefore, the takeoff speed of the froghopper is approximately 0.756 m/s.

b) The kinetic energy generated by the froghopper can be found using the following equation:

KE = 0.5mv^2

where:

m = mass (12.8 mg = 0.0128 g)

v = takeoff speed (0.756 m/s)

Substituting the given values into the equation, we get:

KE = 0.5(0.0128)(0.756)^2

KE ≈ 0.00346 J

(1 J = 10^6 microjoules)

Therefore, the kinetic energy generated by the froghopper for this jump is approximately 0.00346 microjoules.

c) The energy per unit body mass required for this jump can be found by dividing the kinetic energy by the mass of the froghopper:

energy per unit body mass = KE/m

Substituting the values we obtained earlier, we get:

energy per unit body mass = 0.00346/0.0128

energy per unit body mass ≈ 0.270 J/kg

Therefore, the energy per unit body mass required for this jump is approximately 0.270 joules per kilogram of body mass.

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.

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

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:

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

Answer: Min: 35

Max: 100

Mean: 73.53

Standard deviation (SX): 20.17

Mode: 72

Median: 72.5

Explanation:

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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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 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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Answer: Parietal lobe

Explanation: The parietal lobe controls the ability to read, write, and understand spatial concepts. Therefore, you gain the ability to process language through the left hemisphere of your brain.

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]

[tex]\\[/tex]

[tex]{===========================================}[/tex]

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

[tex]{===========================================}[/tex]

[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]

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:

Star C (with a mass of 20 solar masses) will become a giant first, after about 14 million years.

Star A (with a mass of 2 solar masses) will become a giant next, after about 85 million years.

Star B (with a mass of 0.5 solar masses) will become a giant last, after about 12.5 billion years.

Explanation:

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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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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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.