Which subatomic particles each have a positive electrical charge?
a. protons
b. neutrons
c. electrons

The rock has 19600 J of gravitational potential energy relative to the base of the cliff.

The gravitational potential energy [tex]\( E_p \)[/tex] of an object relative to a reference point (in this case, the base of the cliff) can be calculated using the formula:

[tex]\[ E_p = mgh \][/tex]

Where:

m is the mass of the object (20 kg in this case)

g is the acceleration due to gravity (approximately [tex]\( 9.8 \, \text{m/s}^2 \) on Earth)[/tex]

h is the height of the object relative to the reference point (100 m in this case)

Substituting the values:

[tex]\[ E_p = (20 \, \text{kg}) \times (9.8 \, \text{m/s}^2) \times (100 \, \text{m}) \]\[ E_p = 20 \times 9.8 \times 100 \, \text{J} \]\[ E_p = 19600 \, \text{J} \][/tex]

So, the gravitational potential energy that the rock possesses relative to the base of the cliff is [tex]\( 19600 \, \text{J} \).[/tex]

Answer:

Velocities

Explanation:

Given that, two cars are each traveling at 72 km/h one car is traveling northeast, and the other is traveling south. Since, both objects are moving with same speeds but the direction of both cars is opposite. In this case, both cars will have different velocities. Velocity of an object is vector quantity i.e. it will have same magnitude but different velocity.

Hence, two cars have different velocities.

To find the radius that maximizes air velocity during coughing, we differentiate the given velocity equation, set it to zero, and solve for the radius. The maximum air velocity occurs when the radius r is two-thirds of the normal radius R. Therefore, the radius that maximizes air velocity is 2R / 3.

First, let's rewrite the function for clarity:

[tex]\[ v(r) = k(R - r)r^2 \][/tex]

To find the maximum value, we need to take the derivative of [tex]\( v(r) \)[/tex] with respect to  r , set it equal to zero, and solve for  r .

[tex]\[ \frac{dv}{dr} = k \frac{d}{dr}[(R - r)r^2] \][/tex]

Using the product rule:

[tex]\[ \frac{dv}{dr} = k \left[ (R - r) \cdot \frac{d}{dr}(r^2) + r^2 \cdot \frac{d}{dr}(R - r) \right] \][/tex]

[tex]\[ \frac{dv}{dr} = k \left[ (R - r) \cdot 2r + r^2 \cdot (-1) \right] \][/tex]

[tex]\[ \frac{dv}{dr} = k \left[ 2r(R - r) - r^2 \right] \][/tex]

[tex]\[ \frac{dv}{dr} = k \left[ 2rR - 2r^2 - r^2 \right] \][/tex]

[tex]\[ \frac{dv}{dr} = k \left[ 2rR - 3r^2 \right] \][/tex]

[tex]\[ 0 = k \left[ 2rR - 3r^2 \right] \][/tex]

Since  k  is a constant and not equal to zero, we can divide both sides by  k :

[tex]\[ 0 = 2rR - 3r^2 \][/tex]

Factor out of the r :

[tex]\[ r(2R - 3r) = 0 \][/tex]

So, the solutions are:

[tex]\[ r = 0 \][/tex]

[tex]\[ 2R - 3r = 0 \][/tex]

Solve for r :

[tex]\[ 2R = 3r \][/tex]

[tex]\[ r = \frac{2R}{3} \][/tex]

The solution [tex]\( r = 0 \)[/tex] is not physically meaningful in this context since it would imply the trachea is completely closed. Thus, the radius that produces the maximum air velocity is:

[tex]\[ r = \boxed{\frac{2R}{3}} \][/tex]

Answer:

"At 40 mph, your response time for steering is ½ of a second and you will travel 29 feet during that time." The statement is true.

Explanation:

Speed, s = 40 mph

Converting mph to m/s :

1 mph = 0.44704 m/s

40 mph = 17.8816 m/s

Time taken, t = 1/2 seconds

Distance covered, d = speed × time

d = 17.8816 m/s × (1/2 s)

d = 8.9408 meters

Now converting meters to feet :

1 meter = 3.28084 foot

So, 8.9408 meters = 29.4 feets

or d = 29 feets

Hence, the given statement is true.

A 1-ton rock, assuming a roughly spherical shape and a density of rock three times that of water, would be approximately 2.82 feet in diameter, based on weight-volume relationships and the formula for the volume of a sphere.

To calculate the size of a 1-ton rock, we must use density. Given that the density of rock equals three times the volume of water, we will use the approximation that rock weighs about 168 pounds per cubic foot (62lb/cubic foot * 3).

A ton, in American usage, is 2000 pounds. So, a cubic foot of rock weighs around 168 pounds. Therefore, a 1-ton rock would be about 2000/168 ≈ 11.9 cubic feet in volume.

To convert volume (in cubic feet) to diameter (assuming a spherical rock), we use the formula for the volume of a sphere, V= 4/3πr³. In this case, we can rearrange the formula to solve for diameter: D= [(6V)/π]^(1/3).

For our 1-ton rock, the diameter, D = [(6*11.9)/π]^(1/3) ≈ 2.82 feet, which is less than 3 feet but more than 1 foot. So, a 1-ton rock would be around 2.82 feet in diameter.

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To calculate the atomic mass of the metal, we can use the ideal gas law. Given the pressure, volume, and temperature, we can determine the number of moles of hydrogen gas. By dividing the mass of hydrogen by the number of moles, we can calculate the atomic mass of the metal.

To calculate the atomic mass of the metal, we need to use the ideal gas law. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to convert the given pressure from Torr to atm by dividing it by 760. So, the pressure becomes 0.995789 atm.

Next, we convert the volume from mL to L by dividing it by 1000. So, the volume becomes 0.229 L.

Now, we can use the ideal gas law to calculate the number of moles of hydrogen. Rearranging the equation, we get n = (PV) / (RT).

Plugging in the values, we have n = (0.995789 atm * 0.229 L) / (0.08205 L atm /(K mol) * (25 + 273.15)K).

Simplifying the equation gives us n = 0.01012 mol.

Since hydrogen gas has a molar mass of 2.02 g/mol, the atomic mass of the metal can be calculated by dividing the mass of hydrogen by the number of moles. So, the atomic mass of the metal is (2.02 g/mol) / (0.01012 mol) = 199.60 g/mol.

Final answer:

The speed of the tire at the top of the 5m hill, calculated using conservation of energy principles and ignoring any work done by friction, is approximately 17.15 m/s.

Explanation:

To solve this problem, we can use the conservation of energy principle, which states that if no external work is done on the system (like work by friction), the total mechanical energy remains constant. This means that the potential energy lost by the tire as it rolls down from the higher hill will be converted into kinetic energy.

The potential energy at the top of the 20m hill is given by PE = mgh, where m is mass, g is acceleration due to gravity (9.8 m/s2), and h is the height of the hill. At the 20m hill, PE = 10kg × 9.8 m/s2 × 20m. When the tire reaches the top of the next hill, its potential energy will be PE = 10kg × 9.8 m/s2 × 5m.

We can then equate the initial potential energy minus the final potential energy to the kinetic energy at the top of the 5m hill: KE = ½ mv2, and solve for the speed v.

Conservation of energy: mgh1 - mgh2 = ½ mv2

Calculation:

PE at 20m: (10 × 9.8 × 20) J = 1960 J

PE at 5m: (10 × 9.8 × 5) J = 490 J

Kinetic energy at 5m hill: 1960 J - 490 J = 1470 J

1470 J = ½ × 10kg × v2

v2 = (1470 J × 2) / 10kg

v2 = 294 m2/s2

v = √294 m2/s2

v ≈ 17.15 m/s

Therefore, the speed of the tire at the top of the 5m hill is approximately 17.15 m/s.

The portion of a longitudinal wave where the atoms are closest to one another is called compression. A rarefaction is an area in a longitudinal wave in which the atoms are the furthest distance from one another. Compression refers to the process of compressing a media, and rarefaction refers to the process of spreading a form of media out.

Sound is a mechanical disruption from an equilibrium position that travels through an elastic medium of material. It is also possible to define sound solely subjectively, as that which is regarded by the ear, but this definition lacks clarity and is overly constrictive because it is useful to talk about sounds that are manufactured by devices other than the human ear, such as dog whistles and sonar machinery, which cannot do hear by human ear.

The features of sound waves must be examined first in any study of sound. Transverse and longitudinal waves are indeed the two fundamental forms of waves, and they vary by the direction in which they move.

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The question pertains to the concept of work done by a force over a displacement, emphasized by the calculation methods for constant and variable forces and illustrated through the area under a force vs. displacement graph.

The force component along the displacement varying with the magnitude of the displacement refers to the physical concept of work done by a force along a certain displacement. The work done is calculated by integrating the force component in the direction of displacement over the path taken. When the force component (F cos θ) is constant, the work done is simply the product of this force component and the displacement (d), represented as W = Fd cos θ. However, when the force varies along the displacement, the calculation involves dividing the area under the force vs. displacement graph into strips, calculating the work done for each strip as (F cos θ)i(ave) di, and summing these values to find the total work done. This method highlights that the total work done is equivalent to the area under the curve in a force versus displacement graph, which is a core principle in physics for understanding work and energy.

(a), The force component remains relatively constant

(b), There's a discernible increase in the force component

(c), The force component exhibits a steeper incline

The graph depicts how the force component changes concerning displacement magnitude across three intervals: (a) 0 to 1.0 m, (b) 1.0 to 2.0 m, and (c) 2.0 to 4.0 m.

In segment (a), the force component remains relatively constant, suggesting a consistent force acting within this range.

Transitioning to segment (b), there's a discernible increase in the force component, indicating a proportional rise in force with displacement.

However, in segment (c), the force component exhibits a steeper incline, suggesting a nonlinear relationship where the force increases more rapidly concerning displacement magnitude.

Such variation implies complex interactions between the force and displacement, possibly influenced by factors like material properties, external forces, or system dynamics.

Analyzing these intervals aids in understanding the system's behavior and optimizing its performance within different displacement ranges.

Answer: when shot scrapes off the top of the turf

Explanation: Divot is a term used in golf, where "Divot" is the piece of turf that is "cuted" out of the ground when the player tries to hit the ball.

Usually, the Iron or the Wedge are the ones that cause the divots, and while it may seem like a poor technique, it is actually pretty a common thing to see. At the point that in some cases, the divot itself is analyzed to see the technique of the player.

The correct option is the third one: "when shot scrapes off the top of the turf"

Answer:

move at constant velocity.

Explanation:

Newton's first law (also known as law of inertia) states that:

"when the net force acting on an object is zero, the object will keep its state of rest or if it is moving, it will continue moving at constant velocity".

In the case of the probe, friction in deep space is negligible, therefore when the engine is shut down, there are no more forces acting on the probe: the net force therefore will be zero, so the probe will move at constant velocity.

a = F/m 

a' = F/3m 

a'/a = 1/3

Based on this, the correct answer would be option A. 

A 4000kg truck is parked on a 15.0∘ slope, the friction force on the truck is approximately 10,293 N.

We must take into account the truck's weight and the slope's angle in order to calculate the friction force on a truck that is parked on a hill.

The following formula can be used to determine the truck's weight:

Weight = mass × gravitational acceleration

Weight = 4000 kg × 9.8 m/s²

= 39,200 N

Parallel Component = Weight × sin(angle)

The angle of the slope is given as 15.0 degrees. Converting this to radians, we get:

Angle in radians = 15.0 degrees × (π/180)

≈ 0.2618 radians

Now we can calculate the parallel component:

Parallel Component = 39,200 N × sin(0.2618)

≈ 10,293 N

Therefore, the friction force on the truck is approximately 10,293 N.

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Answer:

move at a constant velocity

Explanation:

When the friction is present the engine helps move the probe by constantly doing work against the friction. But when the friction is absent then there is no need for the engine to work all the time. According to newton's first law, no object can change its state of rest or uniform motion with constant velocity without an external force. When the engine is shut off, the probe will continue to move at a constant speed due to inertia.

The answer is,

D. Y’s molecules experience stronger London dispersion forces than X’s molecules.

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Answer: DC flows in one direction, and AC repeatedly switches direction

Explanation:

DC stands for direct current.

AC stands for alternating current.

When current flows only in single direction, it is known as direct current. When current changes direction i.e. it alternates direction, it is known as alternating current.

There are both AC generators and DC generators.

AC generators supply power to home appliances and small motors. DC generators are used to power large electric motors.

AC flows in one direction, and DC repeatedly switches direction.

AC flows in one direction, and DC repeatedly switches direction. This is incorrect. AC, or alternating current, periodically changes direction, while DC, or direct current, flows in one direction only. Examples of AC include household electrical outlets and power generated by generators, while DC is commonly used in batteries and electronic devices.

DC flows in one direction, and AC repeatedly switches direction. This is the correct answer. As mentioned earlier, DC flows in one direction, while AC repeatedly switches direction.

Therefore, the best comparison between AC and DC is that DC flows in one direction, and AC repeatedly switches direction.

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By using Newton's second law of motion, we can determine that the new acceleration of the sled when the net force is tripled and the mass is doubled is 3 m/s².

To calculate the new acceleration, we will use Newton's second law of motion which is F = m * a, where F is the net force, m is the mass of the object, and a is the acceleration. Initially, consider the force as F = m * a. After the changes, the new force and mass become F' = 3F = 2m * a', where F' is the new force, m' is the new mass, and a' is the new acceleration.

So, now you have the equation 3F = 2m * a'. Substitute the initial force F (m * a) into the equation and you get 3 * m * a = 2m * a'. Now you can solve for the new acceleration a', a' = (3/2) * a. Therefore, the new acceleration of the sled is 1.5 times the original, in this case 1.5 * 2 m/s² = 3 m/s². So the new acceleration of the sled is 3 m/s².

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It is an element

that his the answer :)