a 3.550 l sample of gas is cooled from 87.50°c to a temperature at which its volume is 2.150 l. what is this new temperature? assume no change in pressure of the gas.

Answer:

300m/s/s

Explanation:

a=f/m

f=150N

m=.5kg

150/.5=300

The smallest radius of an unbanked track around which a bicyclist can travel with a speed of 29km/h is 20.66m.

The maximum velocity with which we can travel on a unbanked road is given by,

V = √(urg)

Where you is the coefficient of friction,

R is the radius of the track,

g is the acceleration  due to gravity.

Here it is given that the bicyclist  travel with the speed of 29km/h and the coefficient of friction between the tyre and the track is 0.32.

Speed = 29km/h = 8.05 m/s.

So, putting the values, we get,

(8.05)² = r(9.8)(0.32)

r = 20.66m.

So, the smallest radius of an unbaked tract is 20.66 m.

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Answer Multiply the acceleration by time to obtain the velocity change: velocity change = 6.95 * 4 = 27.8 m/s . Since the initial velocity was zero, the final velocity is equal to the change of speed.

​  

v, start subscript, a, v, g, end subscript, equals, start fraction, delta, x, divided by, t, end fraction, equals, start fraction, minus, 4, start text, space, m, end text, divided by, 5, start text, space, s, end text, end fraction, equals, minus, 0, point, 8, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction

The minus sign indicates the average velocity is also toward the rear of the plane.

The average velocity of an object does not tell us anything about what happens to it between the starting point and ending point, however. For example, we cannot tell from average velocity whether the airplane passenger stops momentarily or backs up before he goes to the back of the plane. To get more details, we must consider smaller segments of the trip over smaller time intervals. For instance, in the figure below, we see that the total trip displacement,

Answer:

bdjdiosfjigjdigjtrht <3

Explanation:

Ca(OH)2 has a molar mass of 74.092 g/mol.

Calcium hydroxide, sometimes referred to as slaked lime, has the chemical formula Ca(OH)2. It is an inorganic material that when solid appears white and powdered.

The atomic masses of each element in the periodic table must fist be determined.

Ca = 40.078 (Calcium)

O(Ozone) = 15.999

1.008 H(Hydrogen)

The following equation can be made:

40.078 + (15.999 + 1.008) × 2

40.078 + (17.007) × 2

40.078 + 34.014

= 74.092 g/mol

The molar mass, or molecular mass, of Ca(OH)2 is 74.093 grammes. Th molecule consists of one calcium atom, two oxygen atoms, and two hydrogen atoms. Their molar masses are totaled in this The molar mass of Ca(OH)2 is 74.092 g/mol.

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From the given options electrical force is most sensitive to the distance.

In electrostatics, the electrical force between two charged particles is inversely proportional to the distance of separation between the two particles. By Increasing the separation distance between particles, the force of attraction or repulsion between the particles can be decreased. And by decreasing the separation distance between particles, the force of attraction or repulsion between the particles can be increased.

As the distance between two charge particles are of atomic level, or we can say very small, as compared to the gravitational force or other force, so we can say that electrical forces are extremely sensitive to distance. A small change in the distance between the charged particles can cause major changes in the force between them.

Hence the correct option is B.

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The theoretical efficiency is greater than that of the actual efficiency of the engine. This is because heat engine always produces some waste heat.

The Second Law of Thermodynamics states that a heat engine cannot be 100% efficient. In practice, a heat engine is only 100% efficient when it is operating at about 30-50% efficiency.

If we were to multiply this by 100, we would get the efficiency as a percent: 49%. This is the theoretical maximum efficiency. If we were to actually build an engine, it would be less efficient than the theoretical engine. The theoretical engine that can achieve this theoretical maximum efficiency is called the Carnot Engine.

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Therefore, the final velocity of the train when the nose leaves the station is 27.4 m/s. Therefore, the nose of the train is in the station for 4.32 seconds. Therefore, the velocity of the end of the train as it leaves the station is 20.5 m/s. Therefore, the end of the train leaves the station after 15.74 seconds.

The primary indication of an object's position and speed is its velocity. It is defined as the distance traveled by an item in one unit of time. The displacement of an item in unit time is defined as velocity. The directional speed of an item in motion as an indicator of its rate of change in position as perceived from a certain frame of reference and measured by a specific standard of time (e.g., 60 km/h northbound) is known as velocity. Simply put, velocity is the rate at which something travels in a certain direction. For example, the speed of a car driving north on a highway or the speed of a rocket after launch.

Here,

(a) To find the final velocity of the train when the nose leaves the station, we can use the equation:

v² = u² + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled. We can split the distance traveled into two parts: the distance covered while accelerating, and the length of the station.

For the distance covered while accelerating, the initial velocity is 22.0 m/s, the acceleration is -0.150 m/s² (opposite to the motion of the train), and the distance is the length of the station minus the initial distance covered by the train:

s₁ = (210.0 m - 0 m) - (130.0 m + 22.0 m/s × t) = 80.0 m - 22.0 m/s × t

where t is the time taken to cover the distance.

For the distance covered in the station, the initial velocity is the final velocity from the acceleration phase, and the acceleration is 0 (since the train is no longer accelerating):

s₂= 130.0 m

The total distance traveled is the sum of s₁ and s₂:

s = s₁+ s₂ = 80.0 m - 22.0 m/s × t + 130.0 m

Substituting the given values into the first equation, we get:

v² = (22.0 m/s)² + 2 × (-0.150 m/s²) × [80.0 m - 22.0 m/s × t + 130.0 m]

Simplifying, we get:

v² = 22.0² - 0.3t + 2.0 × 130.0

v² = 484 - 0.3t + 260

v² = 744 - 0.3t

Taking the square root of both sides, we get:

v = √(744 - 0.3t)

At the moment when the nose of the train leaves the station, the distance traveled is 210.0 m. Therefore, we can set s = 210.0 m and solve for t:

210.0 = 80.0 m - 22.0 m/s × t + 130.0 m

Simplifying, we get:

t = 4.32 s

Substituting this value into the expression for v, we get:

v = √(744 - 0.3 × 4.32)

v = 27.4 m/s

Therefore, the final velocity of the train when the nose leaves the station is 27.4 m/s.

(b) The time that the nose of the train spends in the station is the time it takes to travel the length of the station at a constant speed, which is the time between the end of the acceleration phase and the moment when the nose leaves the station. From part (a), we know that the acceleration phase lasts for:

t₁= (v - u) / a = (27.4 m/s - 22.0 m/s) / (-0.150 m/s²) = 36.67 s

Therefore, the total time it takes for the nose to leave the station is:

t₂ = t₁ + 4.32 s = 41.0 s

The time that the nose of the train spends in the station is:

t_station = t₂- t₁ = 4.32 s

Therefore, the nose of the train is in the station for 4.32 seconds.

(c) The velocity of the end of the train as it leaves the station can be found using the same kinematic equation:

v² = u²+ 2as

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.

In this case, the initial velocity is 22.0 m/s, the acceleration is -0.150 m/s², and the distance is the length of the station:

s = 210.0 m

Substituting these values, we get:

v² = (22.0 m/s)² + 2(-0.150 m/s²)(210.0 m)

v² = 484.0 m²/s² - 63.0 m²/s²

v² = 421.0 m²/s²

Taking the square root of both sides, we get:

v = 20.5 m/s

Therefore, the velocity of the end of the train as it leaves the station is 20.5 m/s.

(d) To find the time when the end of the train leaves the station, we can use the same kinematic equation used in part (b):

s = ut + (1/2)at²

where s is the displacement of the end of the train, which is 210.0 m + 130.0 m = 340.0 m, u is the initial velocity of the train, and a is the acceleration of the train.

Substituting the given values, we get:

340.0 m = (22.0 m/s)t + (1/2)(-0.150 m/s²)t²

Simplifying and solving for t, we get:

t = 15.74 s

Therefore, the end of the train leaves the station after 15.74 seconds.

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The engine is providing a forward force of 259,800 N to keep the train moving at a constant velocity.

What is forward force?

Forward force is the force applied in the direction of motion, which causes an object to move or maintain a constant velocity.

Constant velocity refers to the motion of an object moving in a straight line at a steady speed, without changing its direction. The velocity of the object is constant when its speed and direction remain unchanged over time, regardless of whether the object is moving or stationary.

Since the train is moving at a constant velocity, the net force acting on the train is zero. Therefore, the forward force provided by the engine must be equal in magnitude and opposite in direction to the force of air resistance and surface friction.

So, the magnitude of the forward force provided by the engine is:

Magnitude of forward force = Magnitude of force of air resistance and surface friction

Magnitude of forward force = 259,800 N

Therefore, the engine is providing a forward force of 259,800 N to keep the train moving at a constant velocity.

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Answer: 18

Explanation:

i think

Answer: 30

Explanation: ck-12 momentum and impulse

The distance between the two buses is 48km.

Let us assume that the velocity of the first bus is 36 km/h and that of the second bus is 108 km/h.

To find the distance between them, we will consider that both buses move in opposite directions.

Distance travelled by bus 1 in 20 minutes = 36 km/h × 20 min ÷ 60 min/h = 12 km

Distance travelled by bus 2 in 20 minutes = 108 km/h × 20 min ÷ 60 min/h = 36 km

Total distance between them = distance travelled by bus 1 + distance travelled by bus 2= 12 km + 36 km= 48 km

Therefore, the distance between them in 20 minutes is 48 km.

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Answer:

C eq = .96uF

Explanation:

Answer: frequency

Explanation:

amplitude is the max height at which the wave reaches
wavelength distance b/w two waves

the speed at which the wave is oscillating

frequency is no. of oscillations of  a wave per unit length

Answer:

360 degrees is one full rotation starting at zero

it take 2 s  to the flag to rotate once in a full circle  with 3 rad/s of angular velocity.

Angular velocity is "rate of change of angular displacement with respect to time". i.e. ω= dθ/dt. it is also defined as angular displacement over time. i.e. ω = angular displacement/Time.

Angular velocity shows how much angle can be covered in unit time. It's SI unit rad/s.

Angular acceleration is rate of change of Angular velocity with respect to time.

i.e. α = dω/dt if an object changes its angular velocity in short time, we can say that it has greater angular acceleration. It is expressed in rad/s².

In this problem we have to calculate time, but the angular velocity id not given.

Consider the angular velocity is 3 rad/s.

Given,

Angular displacement θ = 2π

angular velocity ω = 3 rad/s

Time t = ?

Time = θ/ω

Time = 2π/3 rad/s = 2 s

Hence the answer is 2s.

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The maximum grade (in percent) is 14%.

Suppose angle be theta.

Power = force * velocity

35*746 = [1300*9.8* sin theta + 410]* (43*1000/3600)

sin theta = (35*746/(43*1000/3600) - 410)/(1300*9.8) = 0.139

grade = tan ( arcsin 0.139 )

= 0.140

= 14%

Drag is a mechanical pressure. it is generated via the interplay and speaks to a stable body with a fluid (liquid or fuel). It is not generated via a force area, within the feel of a gravitational field, or an electromagnetic field, wherein one item can have an effect on some other item without being in physical touch.

For instance drag on a ship moving in water or drag on an aircraft shifting in the air. therefore drag pressure is the resistance pressure resulting from the motion of a body thru a fluid like water or air. This drag force acts opposite to the path of the oncoming float velocity.

Drag is generated via the distinction in speed between the strong item and the fluid. There needs to be motion among the object and the fluid. If there's no movement, there's no drag. It makes no difference whether or not the item moves via a static fluid or whether or not the fluid movements beyond a static stable item.

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The maximum grade (in percent) is 14%.

Suppose angle be theta.

Power = force * velocity

35*746 = [1300*9.8* sin theta + 410]* (43*1000/3600)

sin theta = (35*746/(43*1000/3600) - 410)/(1300*9.8) = 0.139

grade = tan ( arcsin 0.139 )

= 0.140

= 14%

Drag is a mechanical pressure. it is generated via the interplay and speaks to a stable body with a fluid (liquid or fuel). It is not generated via a force area, within the feel of a gravitational field, or an electromagnetic field, wherein one item can have an effect on some other item without being in physical touch.

For instance drag on a ship moving in water or drag on an aircraft shifting in the air. Therefore drag pressure is the resistance pressure resulting from the motion of a body thru a fluid like water or air. This drag force acts opposite to the path of the oncoming float velocity.

Drag is generated via the distinction in speed between the strong item and the fluid. There needs to be a move between the object and the fluid. If there's no movement, there's no drag. It makes no difference whether or not the item moves via a static fluid or whether or not the fluid moves beyond a static stable item.

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Answer:

A great review helps your employees identify growth opportunities and potential areas of improvement without damaging employee-manager relations, but writing a ...

Explanation:

This question involves the concepts of current, power, and charge.

a) The power of taser is "2.75 watt".

b) The magnitude of charge flow per second "2.5 x 10⁻³ C/s".

c) Electrons flow per second is "1.56 x 10¹⁶ electrons/s".

The power of the taser can be given by the following formula:

P = IV

where,

Therefore,

P = (2.5 x 10⁻³ A)(1100 V)

P = 2.75 Watt

The charge flow per second is equal to the current passing through taser.

\(\frac{q}{t}=I\)

where,

Therefore,

\(\frac{q}{t}=2.5\ x\ 10^{-3}\ C/s\)

The electron flow per second is given by the following formula:

\(I = \frac{q}{t} = \frac{ne}{t}\\\\\frac{n}{t}=\frac{q}{te}\)

where,

Therefore,

\(\frac{n}{t}=\frac{2.5\ x\ 10^{-3}\ C/s}{1.6\ x\ 10^{-19}\ C}\\\\\frac{n}{t}=1.56\ x\ 10^{16}\ electrons/s\)

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Neutral atoms contain the same amount of protons and electrons, protons are electrostatic attraction, stellar spectra are absorbance spectra, and spectra may show the chemical makeup of stars.

Information about a star's temperature, chemical make-up, and inherent brightness may be found in its spectrum. A series of photographs taken at various wavelengths of the split in the star's light make up spectrograms produced by slit spectrographs.

The transitions of electrons inside atoms or ions form spectral lines, which are typically dark absorbing lines but may also be dazzling emission lines in some things.

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Answer:

1200 watts at a standard 120v draws 10 amps

The power generated or output power of the turbine is determined as 42.96 W.

The power generated or output power of the turbine is calculated as follows;

eff = 0utput power/1nput power x 100%

13 = 0utput power/(0.62 x 533)

0.13 = 0utput power/330.46

0utput power = 0.13 x 330.46

0utput power = 42.96 W

Thus, the power generated or output power of the turbine is determined as 42.96 W.

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Since the JWST stops at L2, the centripetal force required to keep it in orbit is zero. Therefore, the velocity at this point is also zero.

To determine the total gravitational force acting on the JWST at Lagrange Point 2 (L2), we need to consider the gravitational forces from both the Sun and the Earth.

The gravitational force between two objects can be calculated using Newton's law of universal gravitation:

F = G * (m1 * m2) / r^2

where F is the gravitational force, G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2), m1 and m2 are the masses of the two objects, and r is the distance between their centers of mass.

For the JWST, the mass is 6500 kg, and the distance from the center of the Earth to L2 is 1.50 million km, which is equivalent to 1.5 × 10^9 meters.

The mass of the Sun is approximately 1.989 × 10^30 kg, and the distance from the center of the Sun to L2 is also 1.50 million km.

Therefore, the total gravitational force acting on the JWST at L2 is the sum of the gravitational forces from the Sun and the Earth.

F_total = F_Sun + F_Earth

F_Sun = G * (m_JWST * m_Sun) / r_Sun^2

F_Earth = G * (m_JWST * m_Earth) / r_Earth^2

Substituting the known values, we can calculate the gravitational forces.

Now, to verify that the total gravitational force is equal to the centripetal force required to keep the JWST in orbit, we need to compare it to the centripetal force.

The centripetal force required to keep an object in circular motion is given by:

F_c = (m_JWST * v^2) / r

where F_c is the centripetal force, m_JWST is the mass of the JWST, v is the velocity, and r is the radius of the orbit.

In this case, the JWST is effectively orbiting the Sun, so we can use the distance from the Sun to L2 as the radius of the orbit.

We get the following when we set the gravitational force equal to the centripetal force:

F_total = F_c

Finally, we can calculate the velocity of the JWST at the point where it stops after being released by the Ariane 5 rocket.

Since the JWST stops at L2, the centripetal force required to keep it in orbit is zero. Therefore, the velocity at this point is also zero.

In summary, we need to calculate the total gravitational force acting on the JWST at L2 by summing the gravitational forces from the Sun and the Earth. This total gravitational force should be equal to the centripetal force required to keep the JWST in orbit. At the point where the JWST is released by the Ariane 5 rocket, it reaches L2 and stops, so its velocity at this point is zero.

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The applied force (P) required to achieve a speed of 10 m/s in 5 seconds, considering a coefficient of kinetic friction of 0.2, is 198 N.

To analyze the situation, we can break it down into several components;

Determine the acceleration of the crate;

Using the formula v = u + at, where v is the final velocity, u is the initial velocity (which is 0 in this case), and t is the time taken, we can solve for acceleration (a);

10 m/s = 0 + a × 5 s

a = 10 m/s / 5 s = 2 m/s²

Calculate the force of kinetic friction;

The force of kinetic friction can be calculated using the formula kinetic friction = μk × N, where μk is the coefficient of kinetic friction and N is the normal force. The normal force is equal to the weight of the crate, which can be calculated as N = m × g, where m will be the mass of the crate and g is the acceleration due to gravity (approximately 9.8 m/s²).

N = m × g = 50 kg × 9.8 m/s² = 490 N

kinetic friction = μk × N = 0.2 × 490 N = 98 N

Determine the applied force;

Since the crate is accelerating, there must be a net force acting on it. The net force is the difference between the applied force (P) and the force of kinetic friction;

Net force = P - kinetic friction

Calculate the net force;

The net force can be determined using Newton's second law, which states that the net force is equal to the mass of the object multiplied by its acceleration;

Net force = m × a = 50 kg × 2 m/s² = 100 N

Determine the applied force (P);

Substituting the values into the equation from step 3, we can solve for the applied force;

Net force = P - kinetic friction

100 N = P - 98 N

P = 100 N + 98 N = 198 N

Therefore, the applied forcerequired to achieve a speed of 10 m/s in 5 seconds, considering a coefficient of kinetic friction of 0.2, is 198 N.

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The minimum force of friction necessary to stop the car just before hitting the deer is 3,000N.

Friction is a force between two surfaces that opposes motion when they rub against each other. It is the force that allows us to walk without slipping, and it is also the force that makes it difficult to move a heavy object. Friction is a type of non-conservative force, which means that it takes energy to overcome it, and that energy is not stored or conserved.

The force of friction necessary to stop the car can be determined using the equation F = mv²/2d. In this equation, F is the force of friction, m is the mass of the car (2,000kg), v is the initial velocity of the car (30m/s), and d is the distance between the car and the deer (300m). Plugging in the values, we get F = 2,000 ×(30)² / (2 × 300) = 3,000N. Therefore, the minimum force of friction necessary to stop the car just before hitting the deer is 3,000N.

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The stars that can be classified in spectral class A are the Red giants with an absolute magnitude of 0 and Blue giants with a temperature of 10,000 K

Stars can be classified based on their luminosity and temperature.

From the given graph, the various classification of stars include the following;

At the given spectral class A from the graph, the following stars can be found;

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Answer:

When waves overlap in-phase (crest meets crest or trough meets trough) the waves energy is additive and the amplitude increases.

Explanation:

When waves overlap out-of-phase (crest meets trough) the waves cancel and the amplitude (energy) decreases. When two interfering waves cancel each other out.

Answer:

   i = 1.09 10⁴ A

Explanation:

For this exercise we will look for the resistance of the wire

          R = ρ L / a

the area of ​​the wire is

          a =ππ r² = π πd² / 4

we substitute

         R = ρ L 4 / π d²

          R = 1.72 10⁻⁸   20  4/π 8 10⁻⁴

         R = 5.47 10⁻⁴ Ω

to calculate the current we use ohm's law

           V = R i

            i = V / R

            i = 6 / 5.47 10⁻⁴

            i = 1.09 10⁴ A

When a wave is too steep to support itself, the wave front collapses therefore creating a break.

This is defined as the propagation of disturbance from one place to another in an organized manner.

In situations where the wave is  too steep to support itself there is a break in the wavefront which  advances up the shoreline.

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The chi-square test assumes that the variables being analyzed are measured at a nominal or ordinal level of measurement.

In statistics, the level of measurement refers to the nature and properties of the data being collected. There are four levels of measurement: nominal, ordinal, interval, and ratio. Nominal and ordinal levels are considered categorical, while interval and ratio levels are considered numerical. The chi-square test is specifically designed for analyzing categorical data, where the observations can be classified into distinct categories or groups. It is used to determine whether there is a significant association or relationship between two categorical variables.

The test calculates the difference between the observed frequencies and the expected frequencies under the assumption of independence between the variables. It compares the observed and expected frequencies using a chi-square statistic and determines the p-value to assess the statistical significance of the association. Therefore, the chi-square test assumes that the variables being analyzed are measured at a nominal or ordinal level because it deals with categorical data and evaluates the relationship between different categories or groups.

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Answer:

h= 0.45 m

Explanation:

PE= 1/2 mv^2             KE= mgh

v= 3m/s

vf= 0 m/s

h=?

PE= 1/2(1kg)(3m/s)^2

PE= 4.5 J

4.5 J/ 1kg(9.8 m/s^2)

h=0.45 m

The height of the ball above the ground when it stops is 0.46 m.

The initial kinetic energy of the ball is equal to the final potential energy of the ball.

⇒ Formula:

⇒ Where:

⇒ make h the subject of the equation

From the question,

⇒ Given:

⇒ Substitute these values into equation 2

Hence, the height of the ball above the ground when it stops is 0.46 m.

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To answer the questions posed in Figure 21, we need to apply the principles of electric circuits and the properties of the devices used to measure current and voltage.

Figure 21 shows a circuit diagram with a 12 V battery connected to a resistor. To answer the questions posed, we need to consider the properties of the components in the circuit.
a. The ammeter is a device that measures the current flowing in the circuit, which is the flow of electric charge. The ammeter reading would depend on the resistance of the resistor, which is not provided in the question. However, assuming that the resistor has a resistance of 10 ohms, the ammeter reading would be 1.2 A (using Ohm's Law, I = V/R).
b. The voltmeter is a device that measures the potential difference between two points in the circuit. In this case, the voltmeter reading would be 12 V since the battery provides a constant voltage.
c. Power is the rate at which energy is transferred or transformed. The power delivered to the resistor can be calculated using the formula P = VI (where V is the voltage and I is the current). Using the values from part (a), the power delivered to the resistor would be 14.4 W (P = 12 V x 1.2 A).
d. Energy is the ability to do work. The energy delivered to the resistor per hour is the power delivered multiplied by the time in hours. Using the values from part (c) and assuming a time of 1 hour, the energy delivered to the resistor would be 14.4 Wh.

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