Would you expect the tensile strength of polychlorotrifluoroethylene to be greater, the same as, or less than that of a polytetrafluoroethylene specimen having the same molecular weight and degree of crystallinity?

Answer:

360N

Explanation:

The frictional force = centripetal force = mass * centripetal acceleration = ma

Since centripetal acceleration = V^2/ r

Frictional force = mv^2 / r

Substitute the values of m = 80kg, radius = 2m and tangential speed = 3m/s

Frictional force = 80 * (3)^2/ 2 = 360N

The force of friction that keeps the person sitting on the edge of the rotating platform is 360 N.

To calculate the force of friction that keeps an 80-kg person sitting on the edge of a horizontal rotating platform, we need to find the centripetal force required to keep the person moving in a circular path. The centripetal force is provided by the frictional force between the person and the platform.

 The formula for centripetal force [tex](\(F_c\))[/tex] is given by:

[tex]\[ F_c = \frac{mv^2}{r} \][/tex]

Let's plug in the values:

[tex]\[ F_c = \frac{(80\text{ kg})(3\text{ m/s})^2}{2\text{ m}} \][/tex]

[tex]\[ F_c = \frac{(80\text{ kg})(9\text{ m}^2/\text{s}^2)}{2\text{ m}} \][/tex]

[tex]\[ F_c = \frac{720\text{ kg}\cdot\text{m}/\text{s}^2}{2} \][/tex]

[tex]\[ F_c = 360\text{ N} \][/tex]

 Therefore, the force of friction that keeps the person sitting on the edge of the rotating platform is 360 N.

 The answer is: [tex]360\text{ N}[/tex]

Answer:

Both forces have the same magnitude

Explanation:

1) Notation

[tex]F_{bus}[/tex]= force exterted by the bus to the firefly

[tex]F_{firefly}[/tex]= force exterded by the firefly to the bus

2) Analysis for the situation

The reason why is from the Third law of Newton that states: "For every action, there is an equal reaction force on the opposite direction to the original action force".

For this special case on math terms we have:

[tex]F_{bus}=F_{firefly}[/tex]

The fact that the firefly splatter on this case is because since have a smaller mass, it is less able to stand up to the larger acceleration from the interaction.

3) Conclusion

Based on this analysis, each force would have the same magnitude, so none force is greater than the other.

The force on the firefly and the force on the bus are equal in magnitude but opposite in direction, demonstrating Newton's third law of motion.

When a firefly collides with the windshield of a bus, we have a real-life illustration of Newton's third law of motion. This law states that for every action there's an equal and opposite reaction.

In the case of the firefly and the bus, both objects exert forces on each other that are equal in magnitude but opposite in direction.

The mess that is made when the firefly hits the bus might suggest to us that the force on the firefly is greater, but according to Newton's third law, the force on the firefly and the force on the bus are actually the same.

This symmetry in nature is fundamental to our understanding of physics and can be observed in many situations, like a professor walking across a room or a car accelerating down a road.

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Answer:

F = 4.47 10⁻⁶ N

Explanation:

The expression they give for the strength of the tide is

      F = 2 G m M a / r³

Where G has a value of 6.67 10⁻¹¹ N m² / kg² and M which is the mass of the Earth is worth 5.98 10²⁴ kg

They ask us to perform the calculation

      F = 2 6.67 10⁻¹¹ 135  5.98 10²⁴ 13 / (6.79 10⁶)³

      F = 4.47 10⁻⁶ N

This force is directed in the single line at the astronaut's mass centers and the space station

Answer:

The total distance is 16.9 m

Explanation:

We understand work in physics as certain force exerted through certain distance. To reach that point below the water, the work done by the diver must be equal to the work done by the water's force of resistance. Therefore, we determine both work expressions and we solve the equation for the diver distance, which is the total distance between the diving board and the stopping point underwater.

[tex]W_{d}: work done by diver\\W_{R}: work due the force of resistance\\W_{d} = W_{R}\\F_{d}\times d_{d}=F_R \times d_{R}\\d_{d}= \frac{F_R \times d_R}{F_d}= \frac{F_R \times d_R}{m_d \times g}= \frac {1598 N \times 5.2 m}{50.1 kg \times 9.81 m/s^2}\\d_{d}= 16. 91 m[/tex]

The mass of the aluminum added is calculated through the principle of conservation of energy, specifically thermal energy. By considering the heat lost by the aluminum and gained by the water, we can rearrange the equation for heat transfer and find that the mass of the aluminum is approximately 37.9 grams.

In this physics question, we're looking at a thermodynamic process involving a chunk of aluminum and water. Given the known values of their respective specific heats, the mass of water, and their final equilibrium temperature, we're aiming to find the mass of the aluminum.

We begin by understanding that in a closed system, the heat gained by one body is equal to the heat lost by another. In this case, the aluminum is losing heat, and the water is gaining it. The equation for heat transfer (Q = mcΔT), where m is mass, c is specific heat, and ΔT is change in temperature.

The heat gained by the water = mass of water * specific heat of water * change of temperature in water = 200g * 4.18J/g°C * (18.9°C - 15.5°C) = 2836.4J.

This is equal to the heat lost by the aluminum. Solving the analogous heat equation for the mass of the aluminum gives us the answer:

m = Q / (c * ΔT) = 2836.4J / (0.897J/g°C * (91.4°C - 18.9°C)) = 37.9g

So the mass of the aluminum is approximately 37.9 grams.

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Answer:D

Explanation:

For Rod 1

length,diameter and Tension is L, d and T

for Rod 2 length,diameter and Tension is 2L, 2d and 2T

Change in Length is given by

[tex]\Delta =\frac{PL}{AE}[/tex]

where P=load

L=length

A=area of cross-section

E=young's Modulus

[tex]\Delta _1=\frac{TL}{\frac{\pi d^2}{4}E}[/tex]

since both are aluminium rod therefore E is common

[tex]\Delta _2=\frac{2T\cdot 2L}{\frac{\pi (2d)^2}{4}E}[/tex]

[tex]\Delta _2=\frac{TL}{\frac{\pi d^2}{4}E}=\Delta _1[/tex]

Thus extension in both the rods are same

The change in length of a rod under tension or compression depends on several variables. The change in length is directly proportional to the force and the original length and inversely proportional to the cross-sectional area. Therefore, if Rod #1 is under tension T and Rod #2 is under tension 2T, the change in length of Rod #2 will be double the change in length of Rod #1.

The change in length of a rod under tension or compression depends on several variables. These variables include the force applied (T or 2T), the original length of the rod (L or 2L), and the cross-sectional area of the rod (πd^2/4 or π(2d)^2/4). The change in length is directly proportional to the force and the original length and inversely proportional to the cross-sectional area. Therefore, if Rod #1 is under tension T and Rod #2 is under tension 2T, the change in length of Rod #2 will be double the change in length of Rod #1.

The projectile reaches a maximum height of 26.4 m.

The question is demanding us to find the;

i) Time of flight:

T = 2usinθ/g

u = initial velocity

θ = angle of projection

g = acceleration due to gravity

T = 2 × 30.0 × sin 50.0°/10

T = 4.6 s

ii) The range;

R = u^2sin2θ/g

R = (30.0)^2 × sin 2( sin 50.0°)/10

R = 88.6 m

ii) The maximum height;

H = u^2sin^2θ/2g

H =  (30.0)^2 × sin^2(50.0°)/2 × 10

H = 26.4 m

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Answer:

Option (E)

Explanation:

The Rock glaciers are a type of landform that is formed due to the downward motion of angular blocks under the influence of gravity and the continuous cycle of freeze and thaw method. It mostly occurs in the steep slopes where the soil particles and the rock materials are prone to landslide. Here, the materials are mixed and covered with ice. This also occurs in the region where the permafrost layer undergoes creeping.

This downward motion of ice erodes the soil particles and carries with them further, from the site of talus cone or moraines and extends outward.

Thus, the correct answer is option (E).

Answer:

one of it is e

Explanation:

Answer:

Explanation:

Due to changing magnetic field , there will be emf induced in the region . EMF induced will create electric field which will be circular in shape and will be uniform along its circular path

The magnitude of circular electric field can be calculated as follows

We should apply Faraday law of electro magnetic induction

e = - dФ / dt = - ∫ E dl

Here Ф = π r² B

π r² dB / dt = - ∫ E dl

π r² dB / dt = E x 2π r

E = - r / 2 x dB / dt

For a circular electric field having a particular radius , magnitude of field will be constant .

The direction of electric field will be known by lenz's law

In the given case , magnetic field is upwards and it is reducing , therefore electric field induced will be such as to prevent this change of flux.

So electric field will be anticlock-wise . Hence direction of acceleration will also be  anticlock-wise on proton at 1.5 cm from the centre.

When the mass of Object 1 is reduced to one-fourth, the mass of Object 2 is tripled, and the separating distance is halved, the new gravitational force between these objects will be three times the original force. Thus, the new gravitational force will be 216.0 units.

The question asks us to determine the new gravitational force between two objects given that the mass of Object 1 is reduced to one-fourth of its original mass (so M1 becomes 0.25M1), the mass of Object 2 is tripled (so M2 becomes 3M2), and the distance separating the objects is halved (so R becomes 0.5R). Since gravitational force is calculated using the equation F = G*(M1*M2)/(R^2), where G is the universal gravitational constant, we can plug our new values into this equation.

So, the new gravitational force F' = G*(0.25M1 * 3M2)/(0.5R)^2. This simplifies further to F' = G*(0.75M1M2)/(0.25R^2) = 3 *(G*M1M2/R^2) = 3 * original force. Therefore, if the original force was 72.0 units, the new gravitational force is 3 * 72.0 = 216.0 units.

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To find the new gravitational force between the objects, we can use Newton's Law of Universal Gravitation.

The new gravitational force can be calculated using Newton's Law of Universal Gravitation, which states that the force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Let's denote the original mass of Object 1 as m1, the original mass of Object 2 as m2, and the original distance separating them as d. The original gravitational force can be represented as:

F1 = G * (m1 * m2) / d²

After the given changes, the new mass of Object 1 is 1/4 * m1, the new mass of Object 2 is 3 * m2, and the new distance is 1/2 * d. Plugging these values into the equation, we find the new gravitational force:

F2 = G * ((1/4 * m1) * (3 * m2)) / (1/2 * d)²

Simplifying this equation will give us the answer.

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Answer:

32 miles

Explanation:

Assuming that the speed of 60 mph is constant

Distance = Speed × Time

Time is 20 minutes

Converting to hours

[tex]\frac{20}{60}=\frac{1}{3}\ h[/tex]

[tex]Distance=60\times \frac{1}{3}\\\Rightarrow Distance=20\ mi[/tex]

The car covered 20 mi in 20 minutes.

Distance to Longwood would be [tex]52-20=32\ mi[/tex]

Distance to Longwood is 32 miles

Answer:

Trophosphere

Explanation:

The troposphere is the atmospheric layer closest to the planet and is characterized because contains the largest percentage of the mass of the total atmosphere.

On special characteristic is that in this layer the temperature and water vapor content decrease a lot respect to the altitude. Also on this layer the Water vapor is important in order to regulate the air temperature since on this zone we have absorption of the solar energy.

The troposphere contains almost all the water vapor in the atmosphere. And specially on the tropics we have an accumulation of the water vapour.

All weather phenomena occur within the troposphere. Tropos means "change" and Troposphere means "region of mixing".

Above this layer, we have the tropopause, ranges in height from 5 miles near the poles up to 11 miles above the equator. And the height depends of the seasons, with an special characteristic: the is highest height occurs in the summer and lowest height occurs in the winter.

The troposphere contains almost 75% of the mass of the entire atmosphere. The air on this layer is composed by 78% nitrogen, 21% oxygen and 1% is made of argon, water vapor, and carbon dioxide.

So for this reason this is the Region that contains the majority of molecules in the atmosphere.

Answer:

Explanation:

Actual weight, Wo = 5 N

Apparent weight, W = 4.5 N

density of water = 1 g/cm^3 = 1000 kg/m^3

density of gold, = 19.32 g/cm^3 = 19.32 x 1000 kg/m^3

Buoyant force = Actual weight - Apparent weight

Volume x density of water x g = 5 - 4.5

V x 1000 x 9.8 = 0.5

V = 5.1 x 10^-6 m^3

Weight of gold = Volume of gold x density of gold x gravity

W' = 5.1 x 10^-6 x 19.32 x 1000 x 9.8 = 0.966 N

As W' is less than W so, it is not pure gold.

Using Archimedes' Principle, we determined the density of the crown to be 10.00 g/cm³, which is significantly lower than the density of pure gold (19.32 g/cm³). Therefore, the crown is not made of pure gold.

This problem involves determining whether a crown is made of pure gold using Archimedes' Principle.

According to the given information,

Using Archimedes' Principle, we know that the buoyant force equals the weight of the water displaced:

Since the density of water (ρw) is 1.00 g/cm³, and using the relation 1N = 0.10197 kg in gravitational terms, the weight of the water displaced in kilograms is

The volume of water displaced, which is also the volume of the crown (V), is

Density of the crown = mass / volume.

The mass of the crown (m) is calculated from its weight in air,

Thus, the density of the crown = 509.6 g / 50.96 cm³ = 10.00 g/cm³.

Since the density of pure gold is 19.32 g/cm³ and the calculated density is 10.00 g/cm³, the crown is not made of pure gold.

Answer:

Q=1.42 C

Explanation:

Given that

[tex]I=I_oe^{\dfrac{-t}{\tau}}[/tex]

When t= 0 ,Io = 3 A

τ = Time constant = 0.5 s

[tex]I=3e^{-\dfrac{t}{\tau}}[/tex]

We know that

[tex]I=\dfrac{dQ}{dt}[/tex]

Q=Charge ,I =Current

Q = ∫I.dt

Given that

t= 0 to t= 3τ= 1.5 s

The charge Q

[tex]Q=\int_{0}^{1.5}3e^{-\dfrac{t}{0.5}}dt[/tex]

Q=1.42 C

Therefore charge flow conductor will be 1.42 C.

Answer: Part of Europe, part of Africa, part of Antarctic and the entire America.

Explanation: Western hemisphere or west hemisphere encompasses all regions located west of the longitude 0 °, or meridian of Greenwich.

The formal regions located in the western hemisphere is Europe, Africa, Antarctic and America.

Final answer:

Western Europe can be divided into several smaller regions, including northern Europe, southern Europe, Central Europe, and the British Isles. The western hemisphere also includes various formal regions such as Central America, the Caribbean islands, and parts of South America.

Explanation:

Western Europe can be divided into several smaller regions, including northern Europe, southern Europe, Central Europe, and the British Isles. These regions are based on geographical location or cultural similarities. Additionally, the western hemisphere includes various formal regions such as Central America, the Caribbean islands, and parts of South America.

Answer:896 N

Explanation:

Given

mass of bullet [tex]m=8.7 gm[/tex]

Length of barrel [tex]d=105 cm=1.05 m[/tex]

final velocity [tex]v=465 m/s[/tex]

initial velocity [tex]u=0 m/s[/tex]

Work done by all the forces is equal to change in kinetic Energy

[tex]F\cdot d=\frac{1}{2}\times mv^2-0[/tex]

[tex]F_{avg}\times 1.05=\frac{1}{2}\times 8.7\times 10^{-3}\times 465^2[/tex]

[tex]F_{avg}=\frac{0.5\times 8.7\times 10^{-3}\times 465^2}{1.05}[/tex]

[tex]F_{avg}=895.789 Na\pprox 896 N[/tex]

To find the speed of Anissa after 2.0 seconds, we need to consider the forces acting on her due to gravity and friction. By calculating the force parallel to the slide using the coefficient of kinetic friction and the normal force, we can determine the acceleration of Anissa. Finally, using the equations of motion, we can find her speed after 2.0 seconds.

To find the speed of Anissa after 2.0 seconds, we need to consider the forces acting on her. The force of gravity pulling her down the slide can be divided into two components: a force parallel to the slide and a force perpendicular to the slide. The perpendicular force is canceled out by the normal force, leaving only the parallel force to consider. This force can be calculated using the coefficient of kinetic friction.

The force parallel to the slide is given by the equation F_parallel = μ_k * N, where F_parallel is the force parallel to the slide, μ_k is the coefficient of kinetic friction, and N is the normal force. The normal force can be calculated using the equation N = m * g * cos(θ), where m is the mass of the object, g is the acceleration due to gravity, and θ is the angle of the slide.

Once we have the force parallel to the slide, we can use Newton's second law to find the acceleration of the object: F_parallel = m * a. Finally, we can use the equation v = u + a * t to find the speed of Anissa after 2.0 seconds, where v is the final velocity, u is the initial velocity (which is 0 since she starts from rest), a is the acceleration, and t is the time.

Plugging in the given values: μ_k = 0.15, m = 55 kg, g = 9.8 m/s^2, θ = 25 degrees, and t = 2.0 s, we can calculate the normal force, force parallel to the slide, acceleration, and final velocity.

The final velocity will depend on the length of the slide and other factors that are not mentioned in the question.

Answer:

Paleoclimate proxies

Explanation:

Paleoclimate proxies are materials which are preserved and which can be analysed and correlated with parameters that relates to the environment and climate change. These materials are physical, biological or chemical and are preserved in paleoclimate archives

Scientists always seek clues from paleoclimate archives to understand how climate varied over the recent past few millions of years

The final volume of the gas is 238.9 mL

Explanation:

We can solve this problem by using Charle's law, which states that for a gas kept at constant pressure, the volume of the gas (V) is proportional to its absolute temperature (T):

[tex]\frac{V}{T}=const.[/tex]

Which can be also re-written as

[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}[/tex]

where

[tex]V_1, V_2[/tex] are the initial and final volumes of the gas

[tex]T_1, T_2[/tex] are the initial and final temperature of the gas

For the gas in the balloon in this problem, we have:

[tex]V_1 = 7.00\cdot 10^2 mL = 700 mL[/tex] is the initial volume

[tex]T_1=20.0^{\circ}C+273=293 K[/tex] is the initial absolute temperature

[tex]V_2[/tex] is the final volume

[tex]T_2 = 1.00\cdot 10^2 K = 100 K[/tex] is the final temperature

Solving for [tex]V_2[/tex],

[tex]V_2 = \frac{V_1 T_2}{T_1}=\frac{(700)(100)}{293}=238.9 mL[/tex]

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Answer:

120°

Explanation:

given,

mass of two objects are same , Let it be m

speed of both the objects = v

after inelastic collision the

mass are moving with velocity = v/2

let the angle between them be θ₁ and θ₂

writing the equation along y-axis

m v sinθ₁ - m v sinθ₂ = 0

m v sinθ₁ = m v sinθ₂

θ₁ = θ₂ = θ

writing the equation along x- axis

m v cos θ + m v cos θ = 2 m V

 v cos θ =  v/2

 [tex]cos\theta = \dfrac{1}{2}[/tex]

 [tex]\theta = cos^{-1}(\dfrac{1}{2})[/tex]

 [tex]\theta = 60^0[/tex]

angle between the two objects = 2 θ

                                                     = 2 x 60° = 120°

Answer:

Explanation:

They are infrared waves which mean they take the form of heat.

Answer: A

Explanation:

The speed of the mass when moving through the equilibrium point is 1.22 m/s.

But before that the angular frequency is

[tex]= \sqrt{\frac{80}{0.4} }[/tex]

= 14.14 rad/s

Now

The mass speed should be

[tex]= 14.14 \sqrt{(0.10)^2 - (0.05)^2}[/tex]

= 1.22 m/s

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The speed of a 0.40-kg mass moving through the equilibrium point in simple harmonic motion with a spring constant of 80 N/m and an initial displacement of 0.10 m is 2 m/s.

The speed of the mass when moving through the equilibrium point in a scenario of simple harmonic motion can be determined using the formula for the maximum speed of an object in simple harmonic motion. This formula is v = √(k/m) * A, where v is the speed, k is the spring constant, m is the mass, and A is the amplitude, which in this case is the initial displacement of the object from equilibrium.

So, to calculate the speed of this 0.40-kg mass with a spring constant of 80 N/m as it passes through equilibrium, we plug those values into the formula:

v = √[(80 N/m) / (0.40 kg)] * (0.10 m) = 2 m/s.

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Answer:

k = 86,646,076.92 J

Explanation:

We know that:

K = [tex]\frac{1}{2}IW^2[/tex]

where K is the kinetic energy, I is the moment of inertia and W is the angular velocity.

First we have to find the I with the equation:

I = [tex]\frac{1}{2}MR^2[/tex]

where M is the mass and R the radius of the cylinder.

so:

I = [tex]\frac{1}{2}(507 kg)(1.2m)^2[/tex]

I = 365.04 kg*m^2

Now we replace all the data in the first equation as:

K = [tex]\frac{1}{2}IW^2[/tex]

K = [tex]\frac{1}{2}(365.04)(689)^2[/tex]

k = 86,646,076.92 J

Answer:

True

Explanation:

Magma is known as melted rock deep within the Earth, normally coming from the melting of the upper mantle or crust.  Magma is formed by the partial melting of the mantle and crust and this can occur in different  ways. One way can be  called heat-transfer melting. Rising magma or rock will bring heat with it, and so can melt the surrounding mantle or crustal rock. For example, magmas generated in the mantle tend to be around 1200 degrees Celsius, whereas the more silicate minerals such as quartz and orthoclase feldspar (common in continental crustal rocks) begin to partially melt at around 650-850 degrees Celsius. Therefore, the crustal rock will begin to partially melt due to the introduction of heat from rising magma. A Another way of melting rock is known as decompression melting. During decompression melting, rock from within the mantle is brought to the surface adiabatically (no exchange of heat or energy with its surroundings) and so the lithostatic pressure decreases. This means that the parcel of rising rock crosses the solidus, and so at this point the thermal vibration of the molecules is no longer counteracted by the lithostatic pressure and the rock begins to partially melt.

Answer:

The answer is true

Explanation:

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The horse's velocity after completing a mile on an oval track and ending at the starting point is 0 mph. However, if considering speed, the horse runs at approximately 8.57 mph. The given options do not accurately represent this calculation.

The question pertains to determining the velocity of a horse after running a complete race on an oval track, where the start and end points are the same. In physics, velocity is a vector quantity, meaning it has both magnitude and direction. Since the horse starts and finishes at the same point, the displacement over the course of the race is zero. Consequently, the average velocity is also zero because velocity is displacement divided by time.

In this scenario however, we are looking for speed, which is a scalar quantity and only considers magnitude, not direction. Speed is total distance divided by time. The horse runs a one mile track in 7 minutes. To find the speed, first convert 7 minutes to hours:

7 minutes = 7/60 hours = 0.1167 hours

Then divide the distance (1 mile) by the time in hours to find the speed:

Speed = 1 mile / 0.1167 hours = approximately 8.57 mph

Since none of the options matches this calculation, it seems there might be a mistake in the provided answer choices. If we had to choose from the given options and ignore the direction of the horse's travel, B. 7 mph would be the closest to the calculated value, albeit still not accurate.

Answer:

309 m

Explanation:

time apart = 0.8 s

speed of sound in air = 343 m/s

speed of sound in concrete = 3000 m/s

lets assume it time it takes to travel trough concrete = T

the time it takes to travel through air = T + 0.8 since they are 0.8 s apart

the distance traveled by both waves is the same, so we can equate the distance for both waves

distance  = speed x time

now equating the two distance together we have

3000 x T  = 343 x (T + 0.8)

3000T = 343T + 274.4

3000T - 343T = 274.4

T = 0.103 s

recall that distance = 3000 x T = 3000 x 0.103 = 309 m

Answer:

P-value of the test statistic is 2.499

Explanation:

given data:

size sample is 24

sample mean is 414 gm

standard deviation is 15

Null hypotheis is

[tex]H_0: \mu = 420 gm[/tex]

[tex]H_1 \mu< 420[/tex]

level of significance is 0.01

from test statics

[tex]t = \frac{\hat x - \mu}{\frac{s}{\sqrt{n}}}[/tex]

degree od freedom is  =  n -1

df = 24 -1 = 13

[tex]t = \frac{414 - 420}{\frac{15}{\sqrt{24}}}[/tex]

t = -1.959

from t table critical value of t at 0.1 significane level and 23 degree of freedom  is 2.499

Answer:

h = 7.54 m

t = 1.24 s

Explanation:

1.Let g = 9.81 m/s2 is the gravitational acceleration. Since the formula for potential energy is:

[tex]E_p = mgh[/tex]

where m = 0.25 is the ball mass and h is the height. We can solve for h

[tex]h = \frac{E_p}{mg} = \frac{18.5}{0.25*9.81} = 7.54m[/tex]

2. The time it take for the ball to reach a distance of 7.54m with a gravitational acceleration of 9.81m/s2:

[tex]h = \frac{gt^2}{2}[/tex]

[tex]t^2 = \frac{2h}{g} = \frac{2*7.54}{9.81} = 1.54[/tex]

[tex]t = \sqrt{1.54} = 1.24 s[/tex]

Answer:Misses the Target

Explanation:

Given

Distance between window and Professor is h=11 m

she dropped the balloon 2 sec earlier

time taken by balloon to cover the 11  m

using

[tex]h=ut+\frac{at^2}{2}[/tex]

where h=height

u=initial velocity

t=time

a=acceleration

[tex]11=0+\frac{9.8t^2}{2}[/tex]

[tex]t=\sqrt{\frac{2\times 11}{10}}[/tex]

[tex]t=\sqrt{2.2}[/tex]

[tex]t=1.483 s[/tex]

Therefore she misses the Professor by 0.51 s

Final answer:

The water balloon will hit the ground before the professor is directly underneath the window since it takes approximately 1.48 seconds to fall 11 meters, which means the student is unlikely to hit the professor.

Explanation:

The student's question involves calculating the time it takes for a water balloon dropped from a height to hit a target. This is a problem in projectile motion and free fall within the context of physics, specifically under the topic of mechanics. As the professor is 2 seconds away from the point directly below the student's window, which is 11 meters above the sidewalk, we can determine if the balloon will hit the target by calculating how long it takes for the balloon to reach the ground.

Using the equation of motion for free fall s = 1/2gt^2, where g is the acceleration due to gravity (approximated to 10 m/s^2), and s is the distance (11 meters), we can solve for t (time): s = 1/2gt^2 implies t = sqrt(2s/g). Substituting the given values, we get t = sqrt(2*11/10), which gives t = sqrt(2.2), or approximately t = 1.48 seconds. Because the professor is 2 seconds away from being underneath the window, the water balloon will hit the ground before the professor reaches the point below the student's window, meaning the student's attempt to hit the professor with the balloon will likely fail unless the professor's speed changes.

Answer:

Explanation:

This means that the ratio of the speed of the incident beam of light in a vacuum to the speed of light in the glass is 1.5

This means that the speed of light in vacuum is 1.5 times the speed of light in flint glass.