a car moving on the road with the rain is falling vertically downward. why does the front windscreen get wet where is the back screen remains dry?

Answer:

For example, 1300 with a bar placed over the first 0 would have three significant figures (with the bar indicating that the number is precise to the nearest ten).

Explanation:

hope it helps :)

Answer:

Heavier than 2.2 pounds

Explanation:

Answer:

Science is

Explanation:

he intellectual and practical activity encompassing the systematic study of the structure and behaviour of the physical and natural world through observation and experiment.

The weight of the original mass on the spring was approximately 6.75 Newtons (N). The period of oscillation of a spring/mass system is determined by the mass and the spring constant.

Let's assume the original mass on the spring is represented by M and the corresponding weight is W.

Given that the original period is 5 seconds and the modified period is 3 seconds, we can set up the following equation using the formula for the period of an oscillating spring/mass system:

T = \(2\pi \sqrt{M/k}\), Where T is the period, M is the mass, and k is the spring constant.

For the original system with a period of 5 seconds, we have:

5 = \(2\pi \sqrt{M/k}\) ...(1)

When 12 N are removed from the spring, the modified system has a period of 3 seconds. This implies that the spring constant has changed, but the mass remains the same. Let's assume the new spring constant is k'.

3 = \(2\pi \sqrt{M/k'}\) ...(2)

Dividing equation (1) by equation (2), we can eliminate the mass M:

5/3 = \(\sqrt{k'/k}\)

Squaring both sides of the equation gives:

25/9 = k'/k.

Rearranging the equation gives:

k' = (25/9)k.

Since the spring constant is directly proportional to the weight of the mass, we can conclude that the weight of the original mass W is also reduced by a factor of (25/9).

Let's assume the weight of the original mass on the spring is W0. Thus, the weight of the modified mass is (W0 - 12 N).

Using the proportionality, we have: (W0 - 12 N) = (25/9)W0.

Simplifying the equation, we find: (9/9)W0 - (12 N) = (25/9)W0, (-16/9)W0 = 12 N.

Multiplying both sides by (-9/16) gives: W0 = (-9/16)(12 N), W0 = -6.75 N

Therefore, the weight of the original mass on the spring was approximately 6.75 Newtons (N).

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In order to properly configure the smart light bulbs, buyers should read the installation and configuration manual before connecting or using them to avoid possible damage due to improper handling.

The user manual is a tool that is usually included when we buy a product. This manual includes all the characteristics of the object that we have purchased and the recommendations and specifications for correct use and maintenance to prolong its useful life.

According to the above, what the common buyer should do to prevent their bulbs from being damaged and to be able to configure them properly is to read the user manual.

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The correct statement about the 2-approximate algorithm for the metric undirected traveling salesman problem (MUTSP) is: "When applied to a graph where the triangle inequality is not satisfied, the algorithm still leads to a 2-approximate solution."

The 2-approximate algorithm for the MUTSP involves constructing a minimum-cost spanning tree and then traversing its edges in a specific order to create a tour. However, this algorithm does not guarantee an optimal solution.

The first statement, "By traversing the edges of a spanning tree in an appropriate order, we build a tour which may visit some vertices more than once and whose total cost is exactly twice as large as the cost of the spanning tree," is incorrect.

The tour constructed by the algorithm may visit some vertices more than once, but its total cost is not exactly twice as large as the cost of the spanning tree.

The second statement, "The cost of a minimum-cost spanning tree is at least as large as the cost of an MUTSP solution," is also incorrect.

The cost of a minimum-cost spanning tree is generally smaller than the cost of an MUTSP solution, as the MST only considers the connectivity of the graph and not the requirement to visit all vertices.

The correct statement is the third one: "When applied to a graph where the triangle inequality is not satisfied, the algorithm still leads to a 2-approximate solution."

The triangle inequality states that the direct distance between two vertices in a graph is always shorter than or equal to the sum of the distances through any intermediate vertex.

Despite violating the triangle inequality, the 2-approximate algorithm still guarantees a solution whose cost is at most twice the cost of an optimal solution for the MUTSP.

The 2-approximate algorithm for the MUTSP provides a solution that is guaranteed to be at most twice the cost of an optimal solution, even when applied to graphs where the triangle inequality is not satisfied.

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The rms velocity of the gas change is \(2V_{rms}\).

The molecules are interacting with the container walls as well as one another as they move through it at varying speeds and directions. As a result, even at the same temperature, each particle's velocity and, consequently, kinetic energy, are constantly changing.

The value of the square root of the sum of the squares of the stacking velocity values divided by the number of values is the root-mean-square (RMS) velocity. The RMS velocity is the speed of a wave traveling along a certain ray path across subsurface strata with various interval velocities.

According to the formula

\(V_{rms}\) = √3RT/M

T"=4T

Then, \(V'_{rms}\)= √(3R×4T)/M

\(V'_{rms}\)=2√(3RT/M)

\(V'_{rms}\)=2V_rms

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Answer:

KE = 1/2 M V^2 = 1/2 * 25 * 10^2 = 1250 J

Check

M2 = 1/2 M1

V2 = V1 / 2

E2 = 1/2 * 1/4 E1 = E1 / 8 = 10000 / 8 = 1250 J

Answer:

B. 1,250 joules

Explanation:

In conclusion, to find the molar mass of the unknown gas, we use Graham's law of effusion and compare the effusion rates of the unknown gas and H2. By substituting the given values and solving the equation, we can determine the molar mass of the unknown gas.

To determine the molar mass of the unknown gas, we can use Graham's law of effusion. Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
Let's use the given information to solve the problem. The unknown gas took 11.1 minutes to effuse, while hydrogen gas (H2) took 2.42 minutes. Since the volumes are the same, we can set up the following ratio:
Rate of effusion of unknown gas / Rate of effusion of H2 = √(molar mass of H2 / molar mass of unknown gas)
Plugging in the given values, we get:
11.1 min / 2.42 min = √(molar mass of H2 / molar mass of unknown gas)
Simplifying, we find:
√(molar mass of unknown gas) = √(molar mass of H2) * (11.1 min / 2.42 min)
Squaring both sides, we have:
Molar mass of unknown gas = (molar mass of H2) * (11.1 min / 2.42 min)^2
Now, we need to know the molar mass of H2, which is 2 g/mol.

Plugging in this value and solving the equation, we find the molar mass of the unknown gas.
In conclusion, to find the molar mass of the unknown gas, we use Graham's law of effusion and compare the effusion rates of the unknown gas and H2. By substituting the given values and solving the equation, we can determine the molar mass of the unknown gas.

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Energy cannot be generated or destroyed; it can only be changed from one form of energy to another, according to the rule of conservation of energy. This implies that a system always possesses the same amount of energy, barring external energy addition.

According to the rule of conservation of energy, energy cannot be generated or destroyed. It can only be changed from one form to another, though. It is simply changed from a form that we can use to one that is less useful. Energy problem as a result of this

Energy efficiency is necessary to save expenses and extend the life of the resources.

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Answer:

Q1) C, D

Q2) A

Q3) E

Q4) C

Q5) B

Q6) C

Explanation:

I think these are correct, hopefully it helps!

The space station ISS orbits the Earth approximately every 90 minutes. That is, the space station he orbits the Earth about 16 times in 24 hours a day.

The International Space Station is the largest modular space station in low earth orbit. Five space agencies are involved in this project.

These include NASA in the United States, Roscosmos in Russia, JAXA in Japan, ESA in Europe, and his CSA in Canada. Each orbit takes 90-93 minutes, so there are about 16 orbits per day (24 hours). Depending on the altitude of the ISS, the exact number of orbits per day is usually less than 16 (typically 15.5-15.9 orbits per day). 

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Answer:

B. a large change in velocity over a short time interval

Explanation:

A "chjange in velocity" is acceleration.  Acceleration is in units of meters/sec/sec or meter/sec^2.

)meters/sec) is velocity.  Sec is the time.

Velocity/Time:  A large change on the top (Velocity), divided by a small change in the bottom (sec) means the largest value for accceleration.

question: Please help!!!

If a bottle is being squeezed with a force of 10 Newtons and the area of the bottle is 15

squared inches. What is the pressure??

Answer:

1025.64 N/m²

Explanation:

Pressure: This can be defined as the ratio of force to area. The si unit of pressure is N/m².

From the question,

P = F/A........................ Equation 1

Where F = Force, A = Area.

Given: F = 10 Newtons, A = 15 Squared Inches = (15×0.00065) = 0.00975 m²

Substitute these values into equation 1

P = 10/0.00975

P = 1025.64 N/m²

Hence the pressure of the bottle is 1025.64 N/m²

Answer:

0.66 psi

Explanation:

Answer:

1 deg C = 5/9 deg F         (Ex. 212 F - 32) * 5/9 = 100

So deg F = 9/5 deg C = 9/5 * 2.9 = 5.2 deg F

Answer:current

Explanation:water flows down a river. Current flows down a wire (in the drude model, at least)

Answer:

electricity

Explanation:

Answer:

We can solve this problem by using the concept of relative motion. Let's assume that the first boy is running in the clockwise direction and the second boy is chasing him in the counterclockwise direction.

Since the second boy always remains on the radius connecting the center of the circle and the first boy, the distance between them is always equal to the radius of the circle, which is 28 m.

Let's denote the distance covered by the first boy as S1 and the distance covered by the second boy as S2. We know that the first boy is running with a constant speed of 4 m/s, so we can write:

S1 = u*t1

where t1 is the time taken by the first boy to complete the chase.

The second boy is moving with a constant velocity of 4 m/s towards the first boy, so we can write:

S2 = V*t2

where t2 is the time taken by the second boy to catch up with the first boy.

Since the second boy is always moving on the radius connecting the center of the circle and the first boy, the distance covered by him is equal to the distance on the circumference of the circle covered by the first boy, minus the distance covered by the first boy along the radius. We can write:

S2 = S1 - 2*pi*R

where pi is the mathematical constant pi (approximately equal to 3.14).

Substituting the values of S1 and S2, we get:

u*t1 = V*t2 + 2*pi*R

Since the time of chase is (10 + x) sec, we can also write:

t1 + t2 = 10 + x

We have two equations and two unknowns (t1 and t2), so we can solve for them. First, we can solve for t2:

t2 = (u*t1 - 2*pi*R) / V

Substituting this in the second equation, we get:

t1 + (u*t1 - 2*pi*R) / V = 10 + x

Simplifying this equation, we get:

t1*(1 + u/V) = 10 + x + 2*pi*R/V

Finally, we can solve for t1:

t1 = (10 + x + 2*pi*R/V) / (1 + u/V)

Substituting the given values of R, u, and V, we get:

t1 = (10 + x + 56*pi) / 20

Simplifying this expression, we get:

t1 = 2.8*pi + 0.5*x + 2.8

Therefore, the time taken by the first boy to complete the chase is 2.8*pi + 0.5*x + 2.8 seconds.

Explanation:

this gives me nightmare

The option with the smallest mass is e. 0.020 kg of Br2.Option E

To determine which option has the smallest mass, we need to compare the masses of each given quantity.

a. 10.0 mol of F2:

To find the mass, we can use the molar mass of F2, which is 38.0 g/mol. Therefore, the mass of 10.0 mol of F2 is:

10.0 mol * 38.0 g/mol = 380 g

b. 5.50 x 10^24 atoms of 12O:

To find the mass, we need to know the molar mass of 12O. However, the given molar mass is for F2, not for 12O. Therefore, we cannot determine the mass of this option.

c. 3.50 x 10^24 molecules of 12O:

Similarly, without the molar mass of 12O, we cannot determine the mass of this option.

d. 255 g of Cl2:

Since the molar mass of Cl2 is 70.9 g/mol, the number of moles in 255 g can be calculated as:

255 g / 70.9 g/mol = 3.59 mol

e. 0.020 kg of Br2:

The molar mass of Br2 is 159.8 g/mol. To convert 0.020 kg to grams, we multiply by 1000:

0.020 kg * 1000 g/kg = 20 g

Now we can determine the number of moles:

20 g / 159.8 g/mol ≈ 0.125 mol

Comparing the number of moles, we find that option d (255 g of Cl2) has the largest number of moles, indicating a larger mass compared to the other options. Among the remaining options, option e (0.020 kg of Br2) has the smallest mass. Option e

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The dewpoint when the air temperature is 28°C and the relative humidity is 47% is 15.9°C.

Temperature is a physical property of matter that quantitatively expresses hot and cold. It is measured with a thermometer, which may work through the bulk behavior of a thermometric material, detection of thermal radiation, or particle kinetic energy. Temperature is one of the principal parameters of thermodynamics. Temperature is also important in all fields of natural science, including physics, geology, chemistry, atmospheric sciences, medicine and biology.

The dewpoint is the temperature at which the air has reached its saturation point. At this point, water vapor will begin to condense into liquid droplets, forming dew.

To calculate the dewpoint when the air temperature is 28°C and the relative humidity is 47%, we use the formula:

Dewpoint (°C) = (Air Temperature (°C) * Relative Humidity) / 100 + (112 - Relative Humidity) / (x + 0.444)

Where x = (17.27 * Air Temperature (°C)) / ( Air Temperature (°C) + 237.3 ).

In this example, x = (17.27 * 28) / (28 + 237.3) = 0.39.

Plugging the values into the formula gives us:

Dewpoint (°C) = (28 * 47) / 100 + (112 - 47) / (0.39 + 0.444) = 15.9°C.

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1. In Joe's rigid box, the final pressure inside will remain the same as the initial atmospheric pressure, which is 100 kPa. The rigid box does not allow for any volume change, so the pressure remains constant throughout the heating process.

2. To determine the heat transfer required for each process, we can use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat transfer (Q) into the system minus the work (W) done by the system.

ΔU = Q - W

For Joe's process in the rigid box, since the volume remains constant, there is no work done (W = 0). Therefore, the heat transfer required (Q) can be calculated as:

Q = ΔU

For Bob's weighted piston-cylinder device, the volume can change, and work is involved in moving the piston against the external pressure. The work done can be calculated using the equation:

W = PΔV

Where P is the pressure and ΔV is the change in volume.

The heat transfer required (Q) for Bob's process can be calculated as:

Q = ΔU + W

To determine which process requires less heat transfer, we need to compare the values of Q for Joe's and Bob's processes.

3. To calculate the heating power (Watts) required for each process, we need to know the time required for the entire process to be completed. Let's assume the entire process must be completed in one minute (60 seconds).

The heating power (P) can be calculated using the equation:

P = Q / t

Where Q is the heat transfer and t is the time taken.

By calculating the heat transfer (Q) for each process and dividing it by 60 seconds, we can determine the heating power required for Joe's and Bob's processes.

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Answer:

i = 1.2 [amp]

Explanation:

In order to find the current in the circuit, we must perform a sum of voltages equal to zero, around the mesh.

∑V = 0

Let's take the direction of the current as shown in the drawing Clockwise.

\(-3+(i*2)+(i*1)+(i*2)+9=0\\6+(5*i)=0\\5*i = - 6\\i = - 1.2[amp]\)

Note: The negative sign means that the current moves in a counterclockwise direction.

The answer is 65.3 inches tall

Explanation:

To know the heigh of Sammy in inches it is necessary to convert the 5 feet to inches and add this number to 5.3 inches as the statement mentions "Sammy is 5 feet and 5.3 inches tall". Additionally, it is known each foot is equal to 12 inches ( 1 foot = 12 inches). According to this, the simplest method to convert feet to inches is to multiply the feet given by 12. The process is shown below:

1 foot  = 12 inches

feet to inches =  number of feet x 12

5 feet x 12 = 60 inches

This means 5 feet is equal to 60 inches. Now, 60 inches + 5.3 inches = 65.3 inches (total height of Sammy in inches)

The net force in the horizontal (x) direction is 97.54 N.

The net force in the horizontal (x) direction is calculated by applying the following formula.

Fx (net) = F1 cos (θ₁) + F2 cos(θ₂)

where;

Fx (net) = 49 N cos (27)  +  79 N cos (47)

Fx (net) = 97.54 N

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The speed at which your image appears to be moving away from you is 0.20 m/s.

When you back away from a plane mirror, the apparent motion of your image is determined by the concept of relative motion. The speed at which your image appears to move away from you is equal to twice the speed at which you are moving towards or away from the mirror.

In this case, since you are backing away from the mirror at a speed of 0.10 m/s, your image will appear to be moving away from you at a speed of 0.20 m/s.

This is because the light rays reflecting off your body and reaching the mirror have to travel the distance you move plus the distance your image appears to move, resulting in the doubling of apparent speed.

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When creating a design that minimizes the force upon an object during a collision, there are certain factors that need to be considered. These factors include the materials used, the shape of the object, and the speed of the collision. To minimize the force upon an object during a collision, the following tips can be followed:

The materials used in the design of an object can greatly affect the force experienced during a collision. Materials that are elastic in nature, such as rubber or foam, can absorb the force of the collision and reduce the impact on the object. Materials that are rigid, such as metal or wood, will transfer more force to the object and can cause more damage.

The shape of an object can also affect the force experienced during a collision. Objects that are designed to absorb impact, such as bumpers on a car, are often curved or have a crumple zone. These shapes are designed to distribute the force of the collision over a larger area, reducing the impact on any one point.

The speed of the collision can also affect the force experienced by an object. A slower collision will result in less force being transferred to the object, while a faster collision will result in more force being transferred. It is important to design objects with the appropriate speed of collision in mind.

This may mean adding additional safety features, such as airbags or seat belts, to minimize the force experienced by the object. In conclusion, a design that minimizes the force upon an object during a collision can be achieved by using elastic materials, designing the object to absorb impact, and taking into account the speed of the collision. These simple tips can be used in a school project to create a design that minimizes the force upon an object during a collision.

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Answer:

I think when the balloon is heated up the gas will expand making the balloon bigger. When things are heated up the molecules will get excited and expand. When they balloon is cooled then the it will retract.

Hope this helps! :)

When high-key lighting is used, characters and figures are clearly lit with bright images. High-key lighting is a lighting technique characterized by a predominance of light tones and minimal shadows.

It involves using an abundance of light sources or high-intensity lighting to evenly illuminate the scene, resulting in a well-lit and cheerful ambiance.

High-key lighting is commonly employed in genres such as comedies, romantic films, and musicals, where a bright and upbeat atmosphere is desired.

By reducing the contrast between light and shadow, high-key lighting creates a sense of openness, positivity, and a visually pleasing aesthetic, allowing the audience to focus on the characters and their expressions without distractions caused by dark or dramatic lighting.

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The thigh segment is 43% from the hip to the vertex. We need to identify the moment created by the thigh segment about the vertical (y) axis. The value would be closest to Option (a) 0.53.

This value can be calculated as follows:

Moment = W x d

Where, W = Weightd = Perpendicular distance between the vertical axis and the center of mass of the thigh segmentWe need to calculate the value of d using the given data.

Location of the center of mass of the thigh segment = 43% of the distance from the hip to the vertex

= 0.43 x 39.6 = 17.028 cm

Perpendicular distance between the vertical axis and the center of mass of the thigh segment = 4.1 - 3.9 = 0.2 cm

Therefore, the moment created by the thigh segment about the vertical (y) axis would be:

Moment = W x d= W x 0.2= (17.028 / 100) x 9.81 x 0.2= 0.335 x 0.2= 0.067 Nm ≈ 0.53 (closest to)

Therefore, option (a) 0.53 is the correct answer.

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