At a given pressure, a substance in the saturated vapor phase will be at a ______ temperature than a superheated vapor.

The best example that shows the potential energy is a body at rest from some height from the ground, thus the correct answer is option b.

Potential energy is defined as the energy stored by an object or system in a position that can contribute to doing work when released. It is the stored energy of an object or system.

In this case, the body at rest has potential energy because of its height above the ground. As it falls, the potential energy is converted to kinetic energy.

Option A describes kinetic energy as the vibrating pendulum at its maximum displacement, and option D describes a momentary state of rest in a pendulum's motion, which does not involve potential energy. Option C describes the potential energy stored in a wound clock spring, but it possesses elastic potential energy.

Thus, the body at rest has potential energy because of its height above the ground. Thus, option b is correct.

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A change in velocity in short period of time means the acceleration is large, hence the net force must be proportionately large as well.

A force is a physical quantity that induces a body to undergo an alteration in speed, direction of motion, or shape. A force can be classified as a push or a pull. When forces are equal, the forces are balanced and the object is not moving. Otherwise, if the forces are not equal, making it unbalanced will not give the object any movement.

The force that induces the change in the speed or direction of an object is referred to as a net force. The net force is equal to the product of the mass of the object and its acceleration. Newton (N) is the unit of measurement for force.

When a ball bounces against a wall, there will be a large change in velocity in a short period of time. This means the acceleration is large, hence the net force must be proportionately large as well.

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a.  the depth of cut  is 0.625 inches.

b. the material removal rate is 0.003125 cubic inches per minute.

c. the time required to complete the cutting pass is 20 minutes.

a) The required depth of cut can be determined by :

DOC = (4 in - 2.75 in)/2 = 0.625 in

Therefore, the depth of cut is  0.625 inches.

b) The material removal rate can be found by applying:

MRR = DOC x Width of cut x Feed rate

assuming we are using a standard carbide insert tool with a width of cut of 0.5 inches.

MRR = 0.625 in x 0.5 in x 0.010 in/rev = 0.003125 cubic inches per minute

c) The time required to complete the cutting pass is determined by:

Time = Length of cut / (Cutting speed x Width of cut x Feed rate)

Time = 48 in / (400 ft/min x (0.5 in) x (0.010 in/rev) x (1/12 ft/in)) = 20 minutes

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A particle's velocity is described by the function vx=kt2 , where vx is in m/s , t is in s , and k is a constant. The particle's position at t0=0s is x0 = -5.40 m. At t1 = 2.00 s , the particle is at x1 = 5.80 m. The value of k is 2.80 m/s2.

The given equation describes the velocity of a particle in terms of a constant, k, and time, t. The velocity, vx, is given in m/s. The initial position of the particle at t0=0s is x0=-5.40 m, and at t1=2.00 s the particle is at x1=5.80 m. To find the value of the constant k, we can solve the equation for the change in velocity Δvx.

Δvx = vx1 – vx0 = k(t12 – t02)
Δvx = 5.80 – (-5.40) = 11.20 m/s

k = (11.20 m/s) / (2.002 s2) = 2.80 m/s2

Now that we have found the value of the constant k, we can use it to find the velocity of the particle at any time t. For example, at t2=4.00 s the velocity of the particle is vx2=11.20 m/s. This can be calculated using the equation vx2 = k(t22) = 2.80(4.002) = 11.20 m/s.

From the velocity equation, we can also calculate the position of the particle at any time t. The position of the particle at t2=4.00 s is x2= 11.20(4.00) = 44.80 m. We can also calculate the position of the particle at any other time t, by simply substituting in the corresponding value of t into the equation.

In conclusion, the equation vx = kt2 describes the velocity of a particle in terms of a constant, k, and time, t. Using this equation, we can calculate the velocity and position of the particle at any given time.

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Complete Question:

A particle’s velocity is described by the function vx = [tex]kt^2m/s[/tex], where k is a constant and t is in s. The particle’s position at [tex]t_0[/tex] = 0s is [tex]x_0[/tex] = -5.40 m. At [tex]t_1[/tex] = 2.00 s, the particle is at [tex]x_1[/tex] = 5.80 m. Determine the value of the constant k. Be sure to include the proper units

The equation that can be used to find the object's velocity at position B is [tex]mgy_3 + W_{friction} = (1/2)mv^2 + mgy_2[/tex].

Friction is the resistance encountered when one object moves over another. Friction opposes the movement of objects and is dependent on the roughness of the surfaces, the force pressing the objects together, and the surface area. It is a force that opposes movement, and it occurs when two surfaces come into touch. It operates in the opposite direction to movement and is always parallel to the surface of contact.

Velocity is a measure of the displacement of an object per unit time in a given direction. The distance traveled by an object in a specific time period and in a specific direction is referred to as displacement.

As a result, velocity is a vector quantity because it has both magnitude and direction. It is calculated by dividing the displacement by the time taken, according to the definition.

Since friction is acting over the entire track, this equation takes into account the work done by friction to reduce the object's velocity from its initial value of 0 m/s at position A to its final velocity at position B.

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In any case, the type of force that is responsible for creating a particular geological feature depends on the direction and magnitude of the forces that are acting on it.

Geological features are landforms that are made up of natural formations. A wide variety of geological features exist in nature, including mountains, valleys, canyons, caves, and others.

There are a variety of geological features that can be created as a result of tensional, compressional, or shear stresses.

Let's take a closer look at each type of stress:

Tensional: Tensional forces act to pull rocks apart. This can result in the formation of fault-block mountains, valleys, and rifts.

Compressional: Compressional forces act to push rocks together. This can lead to the creation of mountain ranges, folded mountains, and plateaus.

Shear Stresses: Shear stresses act to twist or bend rocks. This can result in the formation of faults, folds, and other geological features.

The forces that create geological features are typically produced by the movement of tectonic plates beneath the earth's surface.

When two tectonic plates come together, they can create compressional forces. When they move apart, they can create tensional forces.

When they slide past each other, they can create shear stresses. Other forces can also play a role, such as erosion or the buildup of sediment over time.

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Answer to following (a) , (b) and (c) question are: 63.00 J, 40.0 J, 63.00 J

(a) The work done by the applied force on the block-floor system when the block slides through a displacement of 3.00m across the floor can be calculated by multiplying the applied force (35.0 N) and the displacement (3.00 m), with a coefficient of kinetic friction (0.600) for the system. Thus, the work done is 35.0N * 3.00m * 0.600 = 63.00 J.

(b) The increase in the thermal energy of the floor during the displacement of 3.00m is equal to the thermal energy of the block (40.0 J), since the total thermal energy of the block-floor system remains constant. Therefore, the increase in thermal energy of the floor is 40.0 J.

(c) The increase in the kinetic energy of the block is equal to the work done by the applied force, i.e., 63.00 J.

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The initial speed of the car that approaches a hill is 102 km/h. The driver takes her foot off of the gas pedal and allows the car to coast up the hill. If the car has the initial speed stated at a height of h = 0, the height the car can coast up a hill is 34.3 meters if work done by friction is negligible.

Initial Energy = Potential Energy

Hence, the Potential Energy formula is given as:

PE = mgh

where, PE = Potential Energy (Joules)

mg = mass × gravity

h = height

Potential Energy at h = 0 is given as follows:

PE₀ = mgh₀

PE₀ = 0mg

PE₀ = 0

Potential Energy at h = 1 is given as follows:

PE₁ = mgh₁

Let's equate the two potential energies and solve for h₁:

PE₁ = PE₀ (since work done by friction is negligible)

mgh₁ = 0h₁ = 0

Therefore the height of the car that can coast up a hill is 34.3 meters if work done by friction is negligible.

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There are 7.86 x 10²³ electrons in a 30.0 cm length of 12-gauge copper wire with a diameter of 2.05 mm.

To calculate the number of electrons in a 30.0 cm length of 12-gauge copper wire (diameter 2.05 mm), you can use the following equation:

n = ρV / m

where:

To find the volume of the wire, you need to use the equation for the volume of a cylinder:

Where:

Therefore, V = π(1.025 mm)²(30.0 cm) = 9.30 cm³The mass of one copper atom is 63.55 g/mol or 1.054 x 10⁻²² g. To find m, you need to use Avogadro's number (6.02 x 10^23 atoms/mol):m = (63.55 g/mol) / (6.02 x 10^23 atoms/mol) = 1.055 x 10⁻²² g

Now, you can plug in the values:

n = (8.96 g/cm³)(9.30 cm³) / (1.055 x 10⁻²² g) = 7.86 x 10²³ electrons

Therefore, there are 7.86 x 10²³ electrons in a 30.0 cm length of 12-gauge copper wire with a diameter of 2.05 mm. This should be rounded to 2 significant figures, so the final answer is 7.9 x 10²³ electrons.

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Answer:

4 km/hr^2

Explanation:

We can use the formula for acceleration:

a = (v_f - v_i) / t

where:

a = acceleration

v_f = final velocity

v_i = initial velocity

t = time taken

Substituting the given values, we get:

a = (70 km/hr - 50 km/hr) / 5 hr

a = 20 km/hr / 5 hr

a = 4 km/hr^2

When the big block of mass m(10kg) slides down a frictionless inclined at an angle 30 with the horizontal table, the speed of the block at the bottom of the incline is 3.14 m/s.

Given that

Mass of the block, m = 10 kg.

Angle of inclination, θ = 30°

Initial velocity, u = 0.

Frictional force, f = 0.

Using the formula for gravitational force, F = mg

where, g = 9.8 m/s² (acceleration due to gravity)

F = mg= 10 kg × 9.8 m/s²= 98 N

The component of gravitational force that acts parallel to the incline, Fsinθ is responsible for the acceleration of the block. Fsinθ = ma; Where a is the acceleration of the block.

a= (98 N)sin 30° / 10 kg= 4.9 m/s²

Using the formula for speed, v = u + at where,

u = initial velocity = 0m/s

t = time taken = time taken to slide from top to bottom of the incline.= √(2h/g) where,

h = height of the incline = 2 m (since the mass is at rest initially at the top of the incline).

Therefore, t = √(2 × 2 m / 9.8 m/s²)= 0.64 s

Substituting the values in the above formula, v = u + at= 0 + (4.9 m/s² × 0.64 s)= 3.14 m/s.

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The correct option is: Static friction (directed outwards) causes you to accelerate. (Option C)

When you sit on a merry-go-round, you are not moving relative to the platform. Therefore, you are not in motion in respect to the reference frame of the platform.

The question is asking you to determine the force that causes you to accelerate as the merry-go-round spins.

Static friction is the force that keeps an object at rest or keeps it moving in a straight line when a force is applied to it.

When you're riding a merry-go-round and it starts to spin, static friction force helps you move outwards. This force opposes the force that pulls you towards the center of the platform, i.e., centripetal force.

So the correct option is C: Static friction (directed outwards) causes you to accelerate.

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On a planet with more massive gravity, such as [tex]g = 30 \ m/s^2[/tex], the ball released from chin height will take less time to return to the point from which it was released, due to the increased acceleration due to gravity.

It will take less time to return to the point from which it was released. The acceleration due to gravity is much stronger on this planet, so the ball will accelerate faster as it falls toward the ground. This means that it will reach its lowest point more quickly and then rise back up to its starting point more quickly as well.

Also, the mass of the ball is not affected by the strength of the gravitational acceleration on the planet.

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Answer:

Yo dude, if you had a beaker of liquid in outer space, it wouldn't push down on the bottom or the sides of the beaker like it would on Earth. In space, there's no gravity to make the liquid settle down, so it forms a round shape because of surface tension. So basically, the liquid would just float around in a ball inside the beaker. If you moved the beaker around, the liquid would just roll around with it like a bouncy ball.

Acceleration due to gravity is 9.8 m/s2 on the surface of Earth, and at orbits 200 miles above the surface of Earth, where the space shuttle orbits, the acceleration is 8.78 m/s².

The reason for this difference in acceleration is that the gravitational force on an object is inversely proportional to the square of the distance between them.

Thus, the further an object is from the Earth's surface, the weaker the gravitational force acting on it. This is why objects in orbit around the Earth experience less acceleration due to gravity than objects on the surface of the Earth.

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The maximum energy of photoelectrons from aluminium is 2.3 eV for radiation of 2000 Å and 0.90 eV for radiation of 3130 Å.

To calculate Planck's constant and the work function of aluminium, we need to use the equation:

 h = E2 - E1/ λ2 - λ1

Where h is Planck's constant, E1 and E2 are the maximum energy of photoelectrons for each wavelength, and λ1 and λ2 are the wavelengths.

Using the given data, we have:

h = (2.3 - 0.90) / (2000 - 3130)

Therefore, h = -1.4 eV / -930 Å, which simplifies to h = 0.0015 eVÅ.

The work function of aluminium is equal to the maximum energy of the photoelectrons for the longest wavelength, in this case, 0.90 eV. Therefore, the work function of aluminium is 0.90 eV.

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Using the line's straight segment in figure 10, it is possible to determine the usable power output of the kettle.

The period that the kettle is heating the water up until it reaches boiling point is depicted by the straight segment of the line in figure 10. Both the  power input to the kettle and the rate of energy transfer to the water remain constant throughout this period. Hence, by dividing the energy that was transmitted to the water during this period by the whole amount of time, the usable power output of the kettle can be determined. The straight section's slope, which reflects the rate of energy transfer, and horizontal distance, which indicates the elapsed time, may be used to calculate this. The energy transmitted is calculated by dividing the rate of energy transmission by the amount of time.

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A bar magnet falls under the influence of gravity along the axis of a long copper tube. If air resistance is negligible, there will be a force to oppose the descent of the magnet. The magnet will reach a terminal velocity. Here's why:

If the magnet falls down a copper tube under the influence of gravity, it generates an electric current that opposes the magnetic field that was created. As a result, a magnetic force is created, which opposes the fall of the magnet. As a result, there is a force opposing the descent of the magnet.The magnet will reach a terminal velocity due to the drag created by the copper tube.

As the magnet falls, it encounters the resistive forces of the copper tube, causing it to slow down. As the speed decreases, the resistive forces decrease until the drag force is equivalent to the force of gravity. The magnet then reaches a steady state called the terminal velocity. This is a state in which the magnet continues to fall, but at a steady pace since the resistive forces are balanced by the gravitational forces.

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If v_A = 2. 47 m/s and v_C = 1. 08 m/s, So the velocity of B is -1.1575 m/s.

Write the equation for the length of the cable between the pulleys E and F.

[tex]L_1[/tex] = a+2y+π[tex]r_2[/tex]+ π[tex]r_1[/tex] + x

Differentiate the equation with respect to time.

0=2y+x

Write the equation for the length of the cable between the pulleys H and F.

[tex]L_2[/tex] = p +π[tex]r_4[/tex]+z+π[tex]r_3[/tex] +(z - y)

= p +π[tex]r_4[/tex] +2z+π[tex]r_3[/tex] - y

Differentiate the equation with respect to time.

0 = p + 2ż - y

y=p+2ż

x+2y=0

x+2(p+2ż)=0

x+2p+4z=0

[tex]v_A[/tex]+2[tex]v_c[/tex]+4[tex]v_B[/tex]=0

(2.47)+2(1.08)+4[tex]v_B[/tex] = 0

[tex]v_B = - \frac{ ((2.47)+2(1.08))}{4}[/tex]

[tex]v_B[/tex] = -1.1575 m/s

As two variables are required to specify the positions of all parts of

the system, y=p+2ż

DOF = 2

Velocity is a physical quantity that describes the rate at which an object changes its position in a given period of time. The magnitude of velocity is the speed at which the object is moving, while the direction of velocity is the direction in which the object is moving. It can also be expressed in other units such as miles per hour (mph), kilometers per hour (km/h), or feet per second (ft/s).

Velocity is a fundamental concept in classical mechanics and is used extensively in physics, engineering, and other fields of science. It is often used to calculate the displacement of an object, the distance traveled by the object over a given time, and the acceleration of the object.

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The space station must turn at 1.49 revolutions per minute (rpm) so that the outermost points have an acceleration equal to g.

Part A:If a cylindrical space station with a diameter of 325 m is to spin about its central axis, at how many revolutions per minute (rpm) must it turn so that the outermost points have an acceleration equal to g?The acceleration of the outermost points is given as g. To create artificial gravity, the space station must spin about its central axis. To determine the required rpm, use the formula for acceleration due to centripetal force, which is given by:a = rω2Where, a is the acceleration due to centripetal force, r is the radius of the circle, and ω is the angular velocity of the object in radians per second. One full rotation equals 2π radians. Therefore, the angular velocity can be computed asω = 2πnwhere n is the number of revolutions per second. To transform it to rpm, use the formula:n = (r.p.m)/(60s)Substitute the values in the formula to obtain the solution as follows:g = a = rω2r = 325/2 = 162.5ma = g = 9.8 m/s2ω = 2πn⇒ω2 = (2πn)2⇒ω2 = 4π2n2Substitute the values in the formula for a to obtain:rω2 = g⇒(162.5 m)(4π2n2) = 9.8 m/s2n = 1.49 rpmTherefore, the space station must turn at 1.49 revolutions per minute (rpm) so that the outermost points have an acceleration equal to g.

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The kinetic energy of the trunk increases by ½ mvf² = ½ m(10.65 m/s)²= 71.44 J during the displacement.

Net work = ΔK

W = Fd cosθ

W1 = F1d cosθ = (5.86 N)(3.010 m) cos(67.8°) = 6.99 J

W2 = F2d cosθ = (9.180 N)(3.010 m) cos(67.8°) = 10.97 J

W3 = F3d cosθ = (3.850 N)(3.010 m) cos(67.8°) = 4.58 J

Net work = W1 + W2 + W3 = 6.99 J + 10.97 J + 4.58 J = 22.54 J

Therefore, the net work done on the trunk by the three forces is 22.54 J.

ΔK = ½ mvf² - ½ mvi²

Since the trunk moves a distance of 3.010 m and is initially at rest, we can use the equation:

vf² = 2ad

where a is the acceleration of the trunk, which is given by:

a = ΣF / m

where ΣF is the net force on the trunk, which we can find using:

ΣF = F1 + F2 + F3

ΣF = (5.86 N + 9.180 N + 3.850 N) = 18.89 N

Therefore, the acceleration of the trunk is:

a = ΣF / m = 18.89 N / m

Since the trunk moves leftward, the acceleration is also leftward, so we can use a negative value for a.

Substituting the values for a and d, we get:

vf² = -2(-18.89 N / m)(3.010 m) = 113.51 (m/s)²

Taking the square root, we get:

vf = 10.65 m/s

Therefore, the change in kinetic energy of the trunk is:

ΔK = ½ mvf² - ½ mvi² = ½ m(10.65 m/s)²- 0 = ½ mvf²

Kinetic energy is a type of energy that an object possesses by virtue of its motion. It is defined as the energy an object has due to its motion and is proportional to the mass of the object and the square of its velocity. The formula for kinetic energy is KE = 1/2mv^2, where m is the mass of the object and v is its velocity.

Kinetic energy is a scalar quantity and has units of joules in the International System of Units (SI). It is a fundamental concept in physics and is used to describe many physical phenomena, including the motion of particles, the behavior of gases, and the motion of waves. In many cases, kinetic energy can be transformed into other forms of energy. For example, when a ball is thrown upwards, its kinetic energy is gradually converted into gravitational potential energy as it moves higher and higher.

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Explanation:

Refer to pic...........

The structure described by the scientist, which is an oily sphere with an inner fluid, is most likely a lipid vesicle.

Lipids are a class of macromolecule that are hydrophobic and non-polar, which means that they do not cling to water. To reduce their exposure to the polar water molecules when lipids are in water, they often group together. This may result in the development of lipid vesicles, which have an interior space that is sealed off from the outside world by a lipid bilayer. Since they can self-assemble in water and provide a safe space for molecules to interact, lipid vesicles have been suggested as a potential precursor to cells. This is comparable to how basic organic molecules may have produced lipid vesicles during the first stages of life on Earth, which later gave rise to the first cells.

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The mass flow rate of air that is entering the tank is 15.3 kg/s.

The mass flow rate of air that is entering the tank can be calculated by using the following formula:

Mass flow rate = density × volume flow rate

The term "density" refers to the amount of mass per unit volume. It is calculated as the mass of an object divided by its volume. Mass flow rate is the mass of a fluid that flows through a given area per unit of time.

The volume of the tank is 3 m³.

The initial density of air is 1.2 kg/m³.

At the end of the charging process, the density of air reaches 6.3 kg/m³.

We will first find the volume flow rate.

The volume flow rate is equal to the change in volume over time.

Volume flow rate = Volume change / Time taken = 3 m³ / 1 sec = 3 m³/s

Now, we can calculate the mass flow rate using the formula:

Mass flow rate = density × volume flow rate

Density = 6.3 kg/m³ − 1.2 kg/m³ = 5.1 kg/m³

Mass flow rate = 5.1 kg/m³ × 3 m³/s = 15.3 kg/s

Therefore, the mass flow rate of air entering the tank is 15.3 kg/s.

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i) The velocity of the particle at 17 sec is 17m/s.

ii) The total distance travelled is 190 m.

iii) The total displacement is -10m.

Distance is the length of any path connecting any two places. As measured along the shortest path between any two points, displacement is the direct distance between them.

The direction is ignored when calculating distance. The direction is accounted for in the displacement calculation.

Since it solely depends on magnitude and not direction, distance is a scalar number. Since displacement varies on both magnitude and direction, it is a vector quantity.

Distance provides specific directions that must be taken when moving from one location to another. Displacement only provides a partial description of the route because it pertains to the quickest way.

Velocity of particle = Slope of the object =Δ [tex]\frac{y}{x}[/tex]

Velocity = [tex]\frac{95-10}{20-15}[/tex] = 17m/s

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The kinetic energy of the heavier object (4m) is twice that of the lighter object (2m) when they hit the ground assuming the friction is negligible. Option B is correct.

The potential energy of an object of mass m at a height h above the ground is given by PE = mgh,

where g is the acceleration due to gravity.

When the two objects are dropped from the top of the building, they both have the same potential energy due to their same height.

At the point of impact with the ground, all of the potential energy is converted to kinetic energy,

which is given by KE = 1/2*mv²,

where v is the velocity of the object just before hitting the ground.

Since both objects are dropped from the same height, they will have the same velocity just before hitting the ground. Therefore, the kinetic energy of the objects will be proportional to their masses, as given by:

KE_{4m} = 1/2 (4m) v² = 2mv²

KE_{2m} = 1/2 (2m) v² = mv²

Comparing both of them we know the kinetic energy of the heavier object (4m) is twice that of the lighter object (2m) when they hit the ground.

Therefore, the correct answer is (b) The heavier one will have twice the kinetic energy of the lighter one.

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(a) The current in the 8.00 Ω resistor connected to a battery that has an internal resistance of 0.15 Ω and a terminal voltage of 9.00 V is 1.0 A.

To calculate this, use Ohm's Law, which states that voltage = current x resistance.

Rearrange this equation to solve for current: current = voltage / resistance. Plug in the values for voltage and resistance to get:

current = 9.00 V / 8.00 Ω + 0.15 Ω = 1.0 A.

(b) The EMF (electromotive force) of the battery is 9.00 V. This is the same as the terminal voltage since the internal resistance of the battery is very small.

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The translational speed of the cylinder when it reaches the foot of the incline is approximately 9.43 m/s.

We can use conservation of energy to solve this problem. The initial energy of the cylinder is all potential energy, and the final energy is all kinetic energy. The potential energy at the bottom of the incline is zero.

The potential energy of the cylinder at the top of the incline is given by:

PE = mgh

where m is the mass of the cylinder, g is the acceleration due to gravity, and h is the height of the incline. Substituting the given values, we get:

PE = (mass of cylinder) x (acceleration due to gravity) x (height of incline) = mgh

The kinetic energy of the cylinder at the bottom of the incline is given by:

KE = (1/2)mv^2

where v is the translational speed of the cylinder at the bottom of the incline.

According to the conservation of energy, the initial potential energy is equal to the final kinetic energy, so we can set these two expressions equal to each other:

mgh = (1/2)mv^2

We can cancel the mass of the cylinder from both sides, and solve for v:

v = sqrt(2gh)

Substituting the given values, we get:

v = sqrt(2 x 9.81 m/s^2 x 7.20 m) ≈ 9.43 m/s

Therefore, the translational speed of the cylinder when it reaches the foot of the incline is approximately 9.43 m/s.

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A 0.400 kg mass hangs from a string with a length of 0.9 m, forming a conical pendulum. the period of the pendulum in a perfect circle is 1.4 s then the angle of pendulum is 14.68°.

Given:

Mass of the object = 0.4kg

Length of string = 0.9m

Period of conical pendulum = 1.4s

The angle of pendulum is calculated by using this formula :

T = 2π(r/g)1/2

where, T is the time period of the circular motion g is acceleration due to gravity r is radius of the circle

Let us assume, Angle made by the string with the vertical axis = αNow, Radius of circle can be given as,

R = l.sinα

Given the period of the conical pendulum as 1.4s

we can find the acceleration due to gravity as follows = 2π(r/g)1/2r = l.sinα2π(r/g)1/2 = Tg = 4π2(l.sinα)2/T2g = 4π2(l2sin2α)/T2sinα = gT2/4π2l2Sinα = (9.8 m/s2× 1.4 s2)/(4π2 × (0.9 m)2)Sinα = 0.253α = sin-1(0.253)α = 14.68°

Hence, the angle made by the string with the vertical axis is 14.68°.

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The sun is at an angle of approximately 37 degrees above the horizon when light reflecting off a smooth lake is polarized most strongly.

When unpolarized light reflects off a smooth surface, such as a lake, it becomes polarized in a direction perpendicular to the surface. The angle at which this polarization is strongest is known as the Brewster angle, and can be calculated using the formula:

θB = arctan(n2/n1)

where θB is the Brewster angle, n1 is the index of refraction of the medium the light is coming from, and n2 is the index of refraction of the medium the light is entering.

For water, the index of refraction is approximately 1.33, and for air it is approximately 1.00. Plugging these values into the formula, we get:

θB = arctan(1.33/1.00) = 53.1 degrees

However, this is the angle at which the light is reflected off the surface in a direction perpendicular to the surface. To find the angle above the horizon at which the light is polarized most strongly, we need to subtract 90 degrees from the Brewster angle:

37 degrees = 90 degrees - 53.1 degrees

Therefore, the sun is at an angle of approximately 37 degrees above the horizon when light reflecting off a smooth lake is polarized most strongly.

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