A metal sphere with a mass of 80.0g rolls along frictionless surface at 20.0m/s and strikes a stationary sphere having a mass of 200.0g. The First sphere stops completely. At what speed does the second sphere move away from the point of impact
...?

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.

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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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. The change in kinetic energy is –345.075 J

B. The amount of work done is –345.075 J

C. The cart travelled a distance of 34.51 m

ΔKE = ½m(v² – u²)

ΔKE = ½ × 15 × (3.2² – 7.5²)

ΔKE = 7.5 × –46.01

ΔKE = –345.075 J

The workdone in this case is equal to the change in energy of the cart.

Wd = Fd

Divide both side by F

d = Wd / F

d = –345.075 / –10

d = 34.51 m

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If the sun shrank in size but its mass remained the same, the Earth's orbit around the Sun would remain unchanged. This is due to the fact that it's the Sun's mass, not its size, that mainly determines the strength of its gravitational pull on Earth.

If the sun were to shrink in size but maintain the same mass, the gravitational pull it exerts on the Earth would remain the same. This is based on Newton's law of universal gravitation, which states that two bodies attract each other with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers. Therefore, if the sun's mass is unchanged, the Earth's orbit would remain unchanged as well because it is the sun's mass (not its size) that principally determines the strength of its gravitational pull on the Earth.

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The difference in heat, q, between the two processes cannot be determined without additional information, such as the specific heat capacities or the number of moles of gas. Since the final temperature is the same, the difference in work done relates to the difference in heat q due to the first law of thermodynamics, but exact values require further data.

To determine the difference between q for the two-step process and q for the one-step process, we need to apply the principles of thermodynamics. Since the gas is ideal, we can use the formula q = nCΔT, where n is the number of moles, C is the molar heat capacity, and ΔT is the change in temperature. However, since the final temperature is the same for both processes, ΔT will be the same, implying that the heat exchange, q, is simply dependent on the pathway taken by the process.

In the two-step process, the work done is the sum of the work in each step. According to the formula W = -PΔV (work done by the gas is negative when compressed), the external pressure multiplied by the change in volume gives us the work. Since the work done by the gas is different in the two-step and one-step processes, the heat q will also differ according to the first law of thermodynamics, which states that the change in internal energy is equal to the heat added to the system plus the work done on the system (ΔU = q + W).

Given that the final internal energy is the same in both cases because the final temperature is the same, the difference in work done between the processes will equal the difference in heat exchanged. However, without specific heat capacities or the amount of substance (moles), we cannot calculate the exact difference in q.

It is an element

that his the answer :)

The SI unit of power is the watt (W), which is equivalent to 1 joule per second (1 J/s). Power indicates the rate at which work is performed or energy converted, and it can also be expressed in horsepower (hp), where 1 hp equals 746 W.

The SI unit of power is the watt (W). Power, in physics, represents the rate at which work is performed or energy is converted. The definition of a watt is 1 joule per second (1 W = 1 J/s), meaning that when one watt of power is expended, one joule of energy is used every second. Additionally, power can be expressed in horsepower (hp), where 1 hp is equivalent to 746 W. This unit is commonly used to indicate the power output of engines and motors.

Electric power is specifically measured in watts as well, and electric energy is measured in joules. However, for practical purposes such as billing, energy companies use the kilowatt-hour (kWh), which measures the total energy used over a period of time.

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.

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:

False

Explanation: Lightening is not an example of matter in solid or liquid state. It is a phenomenon of spark of electricity in the atmosphere. Lightening can also be seen in volcanic eruptions,forest fires, thunderstorms, hurricanes etc. In the initial stage of development of lightening air acts as an insulator between positive and negative charge but when charge develops enough it break the wall of insulator air and sparking occurs.

The energy transformations chart that take place when energy from the wind is used to generate electricity is  best completed by kinetic energy transformed to mechanical energy.

Wind power, also known as wind energy, is the use of wind turbines to generate electricity. Wind energy is a popular, sustainable, renewable energy source that has a much lower environmental impact than burning fossil fuels.

Wind power has historically been used in sails, windmills, and wind pumps, but it is now primarily used to generate electricity. Wind farms are made up of many individual wind turbines that are linked to an electric power transmission network.

Hence, kinetic energy of wind transforms mechanical energy of the turbines in wind power stations.

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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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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.

In kinematics, motion equations are derived using the time interval, Δt, between events. Before the concept of special theory of relativity came to be, everyone thought that the time interval, Δt, is the same for every observer, whether the observer is at rest or in motion. This is called absolute time.

However, according to Einstein, time is not absolute; that is, the time interval, Δt, depends upon the velocity of the observer

Answer:

The height of the cliff is found to be 176.4 m

Explanation:

Since, the body is being dropped from a certain height and reaches the ground in some time. Thus, we can apply the equations of motion (modified for vertical motion) in this case, due to constant accelerated motion. We have the following data:

Acceleration due to gravity = g = 9.8 m/s²

Time to reach ground = t = 6 sec

Initial Velocity of the Rock = Vi = 0 m/s (Because, the rock will be at rest, initially)

Height of cliff = H = ?

Now, applying second equation of motion (modified for vertical motion), to the rock, between the top of the cliff and ground, we get:

H = Vi t + (1/2)gt²

Using values:

H = (0 m/s)(6 sec) + (1/2)(9.8 m/s²)(6 sec)²

H = 176.4 m

The light intensity at 2.0 m from the bulb would be I0/4. The rms magnetic field value at 2.0 m from the bulb would be B0/2. The average energy density of the waves at 2.0 m from the bulb would be u0/4.

1. The light intensity follows the inverse square law, which means that the intensity decreases as the distance squared increases. So at 2.0 m from the bulb, the light intensity would be I0/4.

2. The rms magnetic field value is related to the light intensity through the equation B0 = sqrt((2u0cI0)/(ε0c^2)), where c is the speed of light and ε0 is the vacuum permittivity. Therefore, at 2.0 m from the bulb, the rms magnetic field value would be B0/sqrt(4) = B0/2.

3. The average energy density of the waves is equal to the energy per unit volume. It can be calculated using the formula u0 = ε0E0^2/2, where E0 is the rms electric field value. At 2.0 m from the bulb, the average energy density of the waves would be u0/4.

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At 2.0 m from the 60-W light bulb, the light intensity is one-fourth of the intensity at 1.0 m, the RMS magnetic field value is half of the initial value, and the average energy density also becomes one-fourth of the initial value.

To solve the problem involving a 60-W light bulb radiating electromagnetic waves:

a = F/m 

a' = F/3m 

a'/a = 1/3

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