Use the equation in the sample problem on the previous page to find the area of a rectangle that is 48.5 cm long and 3.77 cm wide. Round your answer to the correct number of significant figures.

The ball hits the ground with a velocity of 19.6 m/s and it takes approximately 2.02 seconds for the ball to fall.

To determine the velocity the ball hits the ground at, we can use the equation v = gt, where g is the acceleration due to gravity, approximately 9.8 m/s². Since the ball falls for 2 seconds, the velocity is given by v = 9.8 m/s² * 2 s = 19.6 m/s. Therefore, the ball hits the ground with a velocity of 19.6 m/s.

To determine the time it takes for the ball to fall, we can use the equation h = (1/2)gt², where h is the height and t is the time. Rearranging the equation, we have t = √(2h/g). Substituting the given height of 20 m, the time is t = √(2 * 20 m / 9.8 m/s²) = √4.08 s ≈ 2.02 s. Therefore, the time for the ball to fall is approximately 2.02 seconds.

If the resistance of an electric circuit is 12 ohms and the voltage in the circuit is 60 V the current flowing through the circuit is 5A

Ohm's law states that the voltage across a conductor is directly proportional to the current flowing through it, provided all physical conditions and temperatures remain constant

resistance = 12 ohm

voltage = 60 V

using ohm's law

V = IR

I = V/R

 = 60 / 12

 = 5 A

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Approximately 0.0305 wine bottles can be purged with the argon in the canister.

Let's go through the calculations step by step:

Given:
- Pressure (\(P\)) = 1.26 atm
- Volume (\(V\)) = 0.75 L (converted from 750.0 ml)
- Temperature (\(T\)) = 283 K
- Ideal Gas Constant (\(R\)) = 0.0821 L·atm/(mol·K)

Now, use the ideal gas law to find the number of moles [tex](\(n\)):\[ n = \frac{PV}{RT} \][/tex]

[tex]\[ n = \frac{(1.26 \, \text{atm}) \times (0.75 \, \text{L})}{(0.0821 \, \text{L·atm/(mol·K)}) \times (283 \, \text{K})} \]\[ n = \frac{0.945}{23.2263} \]\[ n \approx 0.0408 \, \text{mol} \][/tex]

Now, let's assume that the argon canister contains pure argon, and we need to find out how many moles of argon are in a wine bottle. Argon's molar mass is approximately 39.95 g/mol.

[tex]\[ \text{Moles of argon in a wine bottle} = \frac{\text{Molar mass of argon} \times \text{Volume of the bottle}}{\text{Molar volume at STP}} \]\[ \text{Moles of argon in a wine bottle} = \frac{(39.95 \, \text{g/mol}) \times (0.75 \, \text{L})}{22.4 \, \text{L/mol}} \]\[ \text{Moles of argon in a wine bottle} \approx 1.34 \, \text{mol} \][/tex]

Finally, to find the number of wine bottles that can be purged with the argon in the canister:

[tex]\[ \text{Number of bottles} = \frac{\text{Total moles in canister}}{\text{Moles in one bottle}} \]\[ \text{Number of bottles} = \frac{0.0408 \, \text{mol}}{1.34 \, \text{mol/bottle}} \]\[ \text{Number of bottles} \approx 0.0305 \][/tex]

Therefore, approximately 0.0305 wine bottles can be purged with the argon in the canister.

An insulated lunch bag uses materials that slow down heat transfer.

Importance of Features in Insulated Containers

An insulated lunch bag keeps food warm by utilizing materials that slow down heat transfer, complying with the principles of thermodynamics. To understand which features are most important in minimizing the rate of a hot beverage cooling, we need to consider the mechanisms of heat transfer: conduction, convection, and radiation. A metal lining might conduct heat, leading to quicker heat loss, unless it's part of a vacuum layer such as in a thermos. Insulating cup material, on the other hand, is crucial as materials like polystyrene or foam are poor conductors of heat and therefore reduce heat loss through conduction. The importance of a lid can't be overstated because it traps hot air and reduces heat loss through convection.

The problem involves the principles of projectile motion. The ball thrown upward reaches its peak in 2.5 seconds. The ball dropped from the building cover 15 meters in about 1.75 seconds. Therefore, after the peak, the first ball will take an additional 0.75 second to descend to the point where both balls are at the same height, about 3.25 seconds from the start.

This physics problem involves applying the principles of projectile motion. We start by calculating the time it takes for the ball thrown upward to reach its peak. Since acceleration acting against the motion of the first ball is gravity (9.8 m/s²), using the formula final velocity = initial velocity - (acceleration* time), and since the final upward velocity at the top is 0, we calculate the time it takes the throw up ball to peak as 2.5 seconds (0 = 25m/s - 9.8m/s² * t).

Next, we calculate the time it takes for a ball to fall 15m from rest (assume the drop from the building begins from rest). We use the formula h = 0.5*g*t² which we rearrange to solve for time, yielding t=√(2h/g). Using this formula with h = 15m, we get t ≈ 1.75 seconds.

Therefore, the ball thrown upwards must start to descend after reaching peak so that both balls meet at the same height. Subtracting the two times, we get 2.5 seconds - 1.75 seconds = 0.75 second. Ball 1 will take 0.75 seconds to descend so that both balls reach the same height at the same instance. So, after approximately 3.25 seconds from the start of projectile motion, the two balls will be at the same height.

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The direction of the current in a wire is influenced by the direction of the magnetic field and the movement of the wire in relation to this field. The thickness of the wire affects current amount but not direction. The type of magnet and the ammeter's needle do not determine the current's direction.

The direction of the current moving through a wire is affected by various factors, but let's discuss the pertinent ones here:

Factors such as the type of magnet and the direction of the needle on an ammeter do not affect the direction of the current. An ammeter is a tool used to measure current, not influence it. As for the type of magnet, it's the strength and orientation of the magnetic field it produces that matter, not its inherent type. The relationship between current and magnetic fields is a fundamental concept in electromagnetism, and understanding this relationship is crucial as it forms the basis of electrical motors and generators.

Friction is a force that exists between two surfaces in contact. Some of the applications of friction in our everyday lives include;

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Answer:

D.  the density of the surrounding air and the density of water.

The buoyancy depends on D. the density of the surrounding air and the density of water.

be aware of how the buoyant pressure most effective depends on the density of the fluid ρ in which the object is submerged, the acceleration due to gravity g, and the quantity of the displaced fluid V f V_f Vf​V, start subscript, f, quit subscript. fairly the buoyant force doesn't rely upon the general depth of the item submerged.

the tendency of a frame to glide or to upward push whilst submerged in a fluid trying out an item's buoyancy.  chemistry: the electricity of a fluid to exert upward pressure on a frame positioned in it the buoyancy of water also : the upward force exerted.

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If the acceleration of an object is negative, the object must be slowing down, which is true, as acceleration is the rate at which an object changes its velocity, if the acceleration is negative, it means that the velocity is decreasing over time.

Acceleration is a vector quantity that describes the rate at which an object's velocity changes and is the derivative of velocity with respect to time, so if the acceleration is negative, it means that the velocity is decreasing with respect to time. and can happen if the object is moving in the opposite direction from its initial velocity. This can be described by taking an example: if a car is moving forward with a velocity of 30 meters per second and its acceleration is -5 meters per second squared, it means that its velocity is decreasing at a rate of 5 meters per second.

Hence, if the acceleration of an object is negative, the object must be slowing down, which is true.

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First we calculate the force.

F = 8 *3 * 6 i + 13 j = 144 i + 13 j

F = sqrt (144^2 + 13^2) = 144.59 N

Then calculate the displacement d.

d = sqrt (3^2 + 6^2) = 6.71 m

Work is the produce of force and displacement, therefore:

W = F d = 144.59 N * 6.71 m

W = 970 J

The work done by the force on the particle which moves first along the y-axis from the origin and then parallel to the x-axis is 970 J.

Work done is the force applied on a body to move it over a distance. Work done can be calculated with the following formula.

[tex]W=F\times d[/tex]

Here (F) is the magnitude of force and (d) is the distance traveled.

The particle moves first along the y axis from the origin to the point (0,6m) and then parallel to the x axis until it reaches (3m,6m).

The particle is subject to a force,

[tex]\vec F=cxy\hat i+d\hat j[/tex]

Here,  c = 8.0 N/m2 and d = 13 N. The final point of the displacement of the particle is (3 m, 6 m).

Put the values of (x,y), c and d to find the value of force.

[tex]\vec F=(8)(3)(6)\hat i+(13)\hat j\\\vec F=144\hat i+13\hat j[/tex]

Solving this equation as,

[tex]F=\sqrt{(144)^2+(13)^2}\\F=144.59\rm\; N[/tex]

The initial position of the particle is at the origin (0, 0) and the final position of the particle is (3 m, 6 m). Thus, the displacement is,

[tex]d=\sqrt{(3-0)^2+(6-0)^2}\\d=6.71\rm\; m[/tex]

The work done will be,

[tex]W=144.59\times(6.71)\\W=970\rm\; J[/tex]

Hence, the work done by the force on the particle which moves first along the y-axis from the origin and then parallel to the x-axis is 970 J.

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Streets and highways implement speed limits, not velocity limits, as speed is concerned with only how fast an object is moving, while velocity includes both speed and direction. Imposing velocity limits is impractical as the direction of a moving vehicle continuously varies. Speed limits are enforced to regulate the speed of vehicles, thereby ensuring road safety.

Streets and highways have speed limits instead of velocity limits because speed concerns only the magnitude of movement while velocity includes both speed and direction. Speed is a scalar quantity, meaning it involves only magnitude. On the other hand, velocity is a vector quantity, taking into account both the speed of an object and its direction of motion.

For instance, when a vehicle runs through an intersection without regard to the speed limit, a traffic signal can make the vehicle slow down. This highlights the need for enforcing speed limits.

The final velocity of a vehicle depends on both the magnitude of acceleration and the distance it covers. A vehicle going twice its speed doesn't stop exactly at twice the distance but goes much further before stopping, as exemplified by reduced speed zones near schools. This concept reinforces the necessity of speed limits to ensure public safety and reduce road accidents.

In conclusion, while speed refers to the rate at which an object moves, velocity refers to both the rate and direction of an object's movement. It's impractical to impose velocity limits because direction varies continuously for a car in motion, hence, speed limits are established to regulate the rapidness of vehicles, ensuring road safety.

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A bag of sugar that weighs 1.50 lb (or 6.68 N) on Earth would weigh 1.11 N on the moon because the moon's gravity is only one-sixth as strong as Earth's gravity.

To find the weight of the bag of sugar on the moon, we first need to convert its weight on earth from pounds to newtons. As 1.0 lb is equivalent to 4.45 N, a 1.50-lb bag of sugar would weigh 6.68 N on Earth. On the moon, where the free-fall acceleration is one-sixth of what it is on Earth, this weight would be reduced by a factor of 6, giving a weight of approximately 1.11 N.

So, a 1.50-lb bag of sugar which weighs 6.68 N on Earth would weigh 1.11 N on the moon.

To understand why this is the case, it's necessary to understand that weight varies depending on location because it depends on the force of gravity. On the moon, gravity is only one-sixth as strong as it on Earth, which is why objects weigh less there.

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Final answer:

Scenario C correctly describes resonance, where a pipe open at both ends has a length equal to half the wavelength of the sound from the tuning fork, allowing for a standing wave and resonant sound.

Explanation:

The scenario that accurately describes a condition in which resonance can occur is scenario C: A pipe's length is equal to 1 ⁄ 2 of the wavelength of the sound waves produced by a tuning fork vibrating over one end of the pipe that's open at both ends. In this case, the half-wavelength of the sound matches the length of the pipe, allowing a standing wave to be set up with antinodes at both ends and a node in the middle. This results in constructive interference and a resonant amplification of the sound.

Answer

Friction between the wheels and the ground, normal force, and friction force in the brakes.

Explanation

A truck moves on the surface simply because the driving force is greater than frictional force. The is always the normal force acting perpendicularly on the ground. The frictional force will be applying in two areas that is in the brakes and between the road and the wheels.

So, the forces exerted on the object are; friction and the normal force.

Final answer:

The pram starts moving when the applied force reaches 75 N due to static friction, and then it can be kept moving with a force of 37.5 N due to kinetic friction. The coefficients of static and kinetic friction are 0.75 and 0.375, respectively.

Explanation:

When the pram's brakes are locked, it requires a certain amount of force to overcome the static friction and get it moving. According to the question, the pram starts moving when the applied force is three-quarters of the normal force, which is 100 N.

Therefore, the force required to start moving the pram is (3/4) × 100 N = 75 N. Static friction is what prevents the pram from starting to move until this force is applied.

Once the pram starts moving, the force required to keep it moving drops to half of the force required to start moving it, which is (1/2) × 75 N = 37.5 N, indicating the kinetic friction between the pram and the surface is now in effect.

The coefficient of static friction (μs) is calculated using the formula μs = (force of static friction)/(normal force) = 75 N / 100 N = 0.75. Similarly, the coefficient of kinetic friction (μk) = (force of kinetic friction)/(normal force) = 37.5 N / 100 N = 0.375.

The wavelength of light emitted when an electron in a hydrogen atom transitions from n = 5 to n = 2 is 434 nm, which falls within the blue spectrum.

When the electron in a hydrogen atom moves from n = 5 to n = 2, light with a specific wavelength is emitted. This transition falls under the Balmer series, which includes the transitions of electrons from higher energy levels to n = 2. The specific wavelengths of the Balmer series for n = 3 to n = 2 is 656 nm (red), for n = 4 to n = 2 is 486 nm (green), for n = 5 to n = 2 is 434 nm (blue), and for n = 6 to n = 2 is 410 nm (violet). Therefore, the wavelength of light emitted when an electron transitions from n = 5 to n = 2 is 434 nm.