Why must the difference between the water level in the eudiometer tube and the water level in the beaker be measured

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What mass of aluminum has a total nuclear charge of 2.9 C?

Aluminum has atomic number 13. Suppose the aluminum is all of the isotope with 14 neutrons.

Answer:

The  mass is   \(T_m  = 6.252 *10^{-5}\  g\)

Explanation:

From the question we are told that

   The  total nuclear charge is  \(q =  2.9 \  C\)

   The  atomic number is  \(u  =  13\)

    The number of neutron is  \(k  =  14\)

Generally the number of positive charge is mathematically represented as

      \(N  = \frac{q}{p}\)

here p is the charge on a single proton with value  \(p =  1.60*10^{-19} \  C\)

So      

     \(N = \frac{2.9}{1.60*10^{-19}}\)

=>    \(N = 1.813*10^{19} \ protons\)

Now since 1 atom contains 13 proton

The  number of atoms present is

        \(a =  \frac{1.813*10^{19}}{13}\)

       \(a =  1.395 *10^{18} \  atoms\)

Then the number of moles present is mathematically represented as

      \(n = \frac{a}{N_k}\)

Where  N_k is the Boltzmann constant with value  

       \(N_k  =  6.023*10^{23}\)

So

       \(n  =  \frac{1.395 *10^{18}}{ 6.023*10^{23}}\)

        \(n = 2.315 *10^{-6}\  moles\)

Generally one mole of aluminum is equal to 27 g

So

 The total  mass of aluminum  is

          \(T_m  = n *  27\)

=>        \(T_m  = 2.315 *10^{-6}  *  27\)

=>        \(T_m  = 6.252 *10^{-5}\  g\)

Answer:b

Explanation: If you look at the line on the graph, you can see that it is going downward, meaning it has a negative slope, and choice b is the only one that has a negative slope

Answer:

Key components of games are goals, rules, challenge, and interaction.

Explanation:

Hope I helped.

Answer:

the name of the games are goals, rules, challenges and interaction and mabe there can be more.

Explanation:

I hope this helped your day better this is the correct answer i got it correct on the test.

The vector described as "40 feet to the right" can also be expressed as a displacement vector. It represents the change in position of an object as it moves from one point to another.

Specifically, the vector denotes a horizontal displacement of 40 feet in the positive x-direction from the starting point.

Another way to describe the vector is to use a coordinate system. If we place the starting point at the origin (0,0) and define the x-axis to be the horizontal direction and the y-axis to be the vertical direction, then the vector can be represented as (40,0), where the first number corresponds to the x-coordinate and the second number corresponds to the y-coordinate. This notation emphasizes the fact that the vector has no vertical displacement, only a horizontal displacement of 40 feet.

Alternatively, we could describe the vector using magnitude and direction. The magnitude of the vector is simply its length, which in this case is 40 feet. The direction of the vector can be described as "to the right" or "in the positive x-direction." This notation is useful when comparing vectors with different magnitudes but the same direction or vice versa.

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The potential energy of the roller coaster is 176,400 J (joules).

The potential energy of an object is given by the formula PE = mgh, where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height or vertical position of the object.

In this case, the roller coaster is at a peak of 20m and has a mass of 900kg. The acceleration due to gravity, g, is approximately 9.8 \(m/s^2\).

Using the formula, we can calculate the potential energy:

PE = mgh

= (900 kg)(9.8 \(m/s^2\))(20 m)

= 176,400 J

Therefore, the potential energy of the roller coaster is 176,400 J (joules).

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The car is moving at approximately 12.5 meters per second.

The kinetic energy (KE) of an object can be calculated using the formula:

KE = 1/2 * m * \(v^2\)

where

KE = kinetic energy,

m =Mass of the object, and

v = velocity.

In this case, we are given the mass (m) of the car as 1600 kg and the kinetic energy (KE) as 125,000 J. To find the velocity .

Substituting the  values , we have:

125,000 J = 1/2 * 1600 kg *\(v^2\)

Now, we can solve for v by rearranging the equation:

\(v^2\) = (2 * 125,000 J) / 1600 kg

\(v^2\) = 156.25 \(m^2/s^2\)

Taking the square root, we find:

v = √156.25\(m^2/s^2\)

v ≈ 12.5 m/s

Therefore, the car is moving at approximately 12.5 meters per second.

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Astronomers can use how long it takes an object's brightness to vary to estimate the physical size of the object through a method known as photometry. This method involves observing an object's brightness over time and analyzing the patterns of variation.

For example, consider a binary star system in which two stars orbit each other. As one star passes in front of the other, the combined brightness of the system will decrease. The duration of this decrease in brightness can be used to estimate the physical size of the stars, as the duration of the decrease is related to the size of the stars and the distance between them.

Similarly, if an asteroid or other small body passes in front of a star, the star's brightness will decrease for a short period of time. The duration of this decrease can be used to estimate the size of the asteroid, as the duration is related to the size of the asteroid and the distance between it and the observer.

In general, the size of an object can be estimated using photometry by comparing the observed variation in brightness to the expected variation based on the physical characteristics of the object. This can provide valuable information about the properties and behavior of celestial objects and can help astronomers to better understand the structure and evolution of the universe.

(a) The work done by the force applied by the tractor is 79,968.47 J.

(b) The work done by the frictional force on the tractor is 55,977.93 J.

(c) The total work done by  all the forces is 23,990.54 J.

The work done by the force applied by the tractor is calculated as follows;

W = Fd cosθ

W = (5000 x 20) x cos(36.9)

W = 79,968.47 J

W = Ffd cosθ

W = (3500 x 20) x cos(36.9)

W = 55,977.93 J

W(net) = work done by applied force  -  work done by friction force

W(net) = 79,968.47 J -  55,977.93 J

W(net) = 23,990.54 J

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Answer:

91 rpm

Explanation:

The angular velocity in rpm of the bowling ball is equal to 71.32 rpm.

Given the following data:

Radius = \(\frac{Diameter}{2} = \frac{0.22}{2} =0.11\;m\)

To determine the angular velocity in rpm of the bowling ball:

First of all, we would calculate the moment of inertia of the bowling ball by using the formula:

\(I = \frac{2}{5} mr^2\)

Where:

Substituting the given parameters into the formula, we have;

\(I = \frac{2}{5} \times 5.8 \times 0.11^2\\\\I=2.32 \times 0.0121\\\\I=0.0281\;Kgm^2\)

To find the angular velocity in rpm:

Mathematically, angular momentum is given by the formula:

\(L=I\omega\\\\\omega = \frac{L}{I} \\\\\omega = \frac{0.21}{0.0281}\\\\\omega = 7.47 \;rad/s\)

Converting rad/s to rpm:

\(\omega = 7.47 \times \frac{60}{2\pi} \\\\\omega = \frac{448.2}{6.284}\\\\\omega = 71.32 \;rpm\)

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Answer:

Potential and kinetic

Explanation:

If you are doing ck12 this is the answer

The simple conclusion I can make from the task given above is that all the objects in the universe possess constant acceleration under the influence of a free fall is actually due to gravity.

It has been practically proven that all objects or bodies; regardless of their masses have the same acceleration as long as they are under a free fall as a result of the Earth's gravitational force.

So therefore, it can now be deduced with understanding that all free falling objects have the same acceleration due to gravity.

Complete question:

What conclusion can you make on acceleration and free falling objects grade?

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To push a box up an inclined plane, the force required is smaller if you push parallel to the incline.

This is also referred to as a ramp and is a flat surface which is tilted at an angle.

Pushing parallel to the plane doesn't generate friction which means a smaller amount of force will be required.

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Answer:

24 cm/s

Explanation:

Applying

Pythagoras theorem,

a² = b²+c²............. Equation 1

Where a = resultant, b = vertical component, c = horizontal component

From the question,

Given: a = 26 cm/s, c = 10 cm/s

Substitute these values into equation 1

26² = b²+10²

676 = b²+100

b² = 676-100

b² = 576

b = √576

b = 24 cm/s

Answer:

35.2644

I suppose mew is refractive index

Explanation:

( sin i ) / (sin r) = refractive index

( sin 60) / (sin r) = 1.5

( sin 60) / 1.5 =sin r

r=35.844

sorry if I'm wrong

In the case of rain, the static friction is halved, meaning the new static friction coefficient is 0.5μs, while the kinetic friction is reduced to one-third, resulting in a new kinetic friction coefficient of (1/3)μk.

To determine the maximum velocity at which you can make a turn without slipping in the rain at the intersection, we need to consider the changes in friction.

Let's assume the normal static friction and normal kinetic friction are represented by μs and μk, respectively.

In the case of rain, the static friction is halved, meaning the new static friction coefficient is 0.5μs, while the kinetic friction is reduced to one-third, resulting in a new kinetic friction coefficient of (1/3)μk.

To avoid slipping during the turn, we need to ensure that the centripetal force required for the turn is less than or equal to the maximum frictional force available.

The centripetal force is given by the equation mv²/r, where m is the mass, v is the velocity, and r is the radius of the turn.

The maximum frictional force in the rain can be calculated as (0.5μs)mg, where g is the acceleration due to gravity.

Thus, to avoid slipping, we set the centripetal force equal to the maximum frictional force:

mv²/r = (0.5μs)mg

Simplifying the equation, we find:

v = √(0.5μsgr)

By plugging in the values for μs, g, and the radius of the turn, we can calculate the maximum velocity.

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Answer:

☆Cellular respiration is the process in which cells break down glucose, release the stored energy, and use it to make ATP. The process begins in the cytoplasm and is completed in a mitochondrion. Cellular respiration occurs in three stages: glycolysis, the Krebs cycle, and electron transport.

Explanation:

Each cell in the body undergoes cellular respiration in the cytoplasm and mitochondria. The three stages of  cellular respiration are  Krebs cycle, glycolysis, and the electron transport chain.

The Krebs cycle, glycolysis, and the electron transport chain, where oxidative phosphorylation takes place, are the three primary stages of cellular respiration. While glycolysis can take place in anaerobic environments, the TCA cycle and oxidative phosphorylation need oxygen to function.

In the cytoplasm, glucose is first broken down into pyruvate, a three-carbon compound. The pyruvate is subsequently transported into the mitochondrial matrix, where it undergoes a conversion process known as pyruvate oxidation. Pyruvate dehydrogenase does this by converting three-carbon pyruvate to two-carbon acetyl-CoA.

Finally, a proton gradient is produced by a series of redox processes fueled by high energy electrons called the electron transport chain, which pumps protons across the membrane. Together, they provide an electrochemical gradient.

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The runner who reaches their maximum speed more quickly has the larger maximum speed. In a 100-meter dash, reaching maximum speed more quickly generally indicates a higher level of acceleration, which is related to maximum speed.

The runner who reaches their maximum speed more slowly may have a longer time to build up speed, but once both runners have reached their maximum speed, they are both running at the same speed.

So, the runner who reached maximum speed more quickly will have had a higher maximum speed.

Speed is known as the rate of change of position of an object in any direction. Speed is calculated as the ratio of distance to the time in which the distance was covered.

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Answer:

H = 34.43 m

Explanation:

Given that,

Initial speed of the object, u = 30 m/s

The angle of projection, \(\theta=60^{\circ}\)

We need to find the maximum height reached by the object. Let it is H. Using the formula for maximum height reached by the projectile.

\(H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(30)^2\times \sin^2(60)}{2\times 9.8}\\\\H=34.43\ m\)

So, the maximum height reached by the object is 34.43 m.

I mean if it will be anything the asnwer would be 23^90'200

Answer:

Use the formula F=ma

Force is going to be 88, because the frictional force acting against it increases the force. So we need to add the frictional and applied force together.

f= 88

m= 20

a=?

F=ma

88=20 x a

88/20 = a

a= 4.4 m/s^2

Considering the four electric charges forming a square with magnitude of charge of 2.0 nC on each. :

A) Magnitude of the force on the 5.5 nC charge in the middle of the figure = 3.48 * 10^-4 N

B) Direction of the force on the 5.5 nC charge in the middle of the figure = Leftward ( negative x -axis )

Using the given data :

size of square = 4 cm

magnitudes of four charges = 2.0 nC

a) magnitude of the force on the center charge

Electric force between two point charges = \(F = \frac{1}{4\pi *E_{o} } \frac{q1q2}{r^2}\) ----- ( 1 )

where ; \(\frac{1}{4\pi E_{o} } = 9 * 10^9 Nm^2/C^2\)

step 1 ; find r ( distance between charges )

r² = ( 2 )² + ( 2 )² = 8 cm²  

back to equation 1

F = 9 * 10⁹ * \(\frac{2 * 10^{-9} * (5.5 * 10^{-9}) }{8*10^{-4} }\) =  1.23 * 10^-4  N

∴ magnitude of the force on the center charge ( Fnet )= 4F cos 45°

        = 4 * ( 1.23 * 10^-4 ) * \(\frac{1}{\sqrt{2} }\)  = 3.48 * 10^-4 N

b) The direction of the force at the center is along the negative x-axis ( leftward )

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Answer:

 T² = (π² / 2G M_{s})    (R₁ + R₂)³

Explanation:

Let's use Newton's second law where the force is the universal gravitational force

          F = m a

the acceleration is centripetal

         a = v² / r

the out of gravitational universal is

           F = G \(M_{s}\) m / r²

we substitute

         G M_{s} m / r² = m v² / r

          G M_{s} / r = v²

the planet's speed in orbit is

          v = d / T

          v = 2π r / T

we substitute

           G M_{s}/ r = 4π² r² / T²

            T² = 4π² / G M_{s} r³

In the case of an elliptical orbit the distance r is the length of the semi-major axis, see attached for the nomenclature

            T² = 4π² / G M_{s} a³

indicates that the minimum distance is R₁ = a -c and the maximum distance is R₂ = a + c, let's add these two expressions

              R₁ + R₂ = 2 a

we substitute in the equation of the period

             T² = 4π²/ G M_{s}      (R₁ + R₂)³/2³

             T² = (π² / 2G M_{s})    (R₁ + R₂)³

Answer:

It's either B or D, I'm not positive which it is

Explanation:

Answer:

The answer is B.

Explanation:

The two resistors are in the same square space, while the others are not.

In a case whereby two springs are attached to two hooks and Spring A has a greater force constant than spring B. Equal weights are suspended from both then Spring B will have more extension than spring A.

Force constant or spring constant  serves as the measure of the stiffness of a spring which can be explained as the force per unit deformation of the spring.

In the case above, we can see that since Equal weights are suspended from both, then there will be more extension at the side of  Spring B  since A has a greater force constant.

Therefore, option B is correct.

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To determine the acceleration of each block, we need to apply the laws of motion and analyze the forces acting on each block.

the only force acting on it is the applied force P. The force of friction acting on Block A is equal to µN, where N is the normal force. Since Block B is resting on top of Block A, the normal force acting on Block A is equal to the weight of Block B plus the weight of Block A, which is (2m)g. Therefore, the force of friction acting on Block A is 0.25(2m)g.

F = ma, we can write the equation of motion for Block A as P - 0.25(2m)g = (2m + m)a. Solving for a, we get a = (P - 0.25(2m)g)/(3m).

the force acting on it is the tension in the cable CD, which is equal to the weight of Block B plus the weight of Block A, plus the force of friction acting between Block A and Block B. Using the same method as above, we can calculate the force of friction as 0.25mg. Therefore, the equation of motion for Block B can be written as T - 0.25mg = ma, where T is the tension in the cable.

their accelerations must be the same. Therefore, we can set the two equations equal to each other and solve for the tension T, which is equal to 0.75mg + P.

we can use the tension T to determine the acceleration of Blocks C and D, which are attached to Block B by the cables. Since the tension in the cables is the same throughout, the acceleration of Blocks C and D is also equal to a.

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The acceleration of block A is 17.55 m/s^2

Calculation for the given question is as following :-

Since block B is being held in place by cable CD, it is not moving and therefore has no acceleration. Thus, we only need to consider the motion of block A.

To solve the problem, we can use Newton's second law, which states that the net force on an object is equal to its mass times its acceleration:

F_net = ma

where F_net is the net force acting on the object, m is the mass of the object, and a is its acceleration.

First, we need to calculate the net force acting on block A. There are two forces acting on block A: the applied force P and the force of friction f. The force of friction is given by:

f = mu*N

where mu is the coefficient of kinetic friction and N is the normal force acting on the block. The normal force is equal in magnitude to the weight of the block, which is:

N = mg

where g is the acceleration due to gravity. Thus, the force of friction is:

f = mu*mg

Substituting the given values, we get:

f = 0.2559.81 = 12.27 N

The net force acting on block A is therefore:

F_net = P - f

Substituting the given value of P and the calculated value of f, we get:

F_net = 100 - 12.27 = 87.73 N

Finally, we can calculate the acceleration of block A using Newton's second law:

a = F_net/m

Substituting the given value of m and the calculated value of F_net, we get:

a = 87.73/5 = 17.55 m/s^2

Therefore, the acceleration of block A is 17.55 m/s^2.

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In an inclided plane the perpendicular component of the weight to the plane is always given by:

Plugging the values given in the problem we have:

Therefore, the perpendicular component of the weight is 30.029 N

The subject depicted in the attached image is Astronomy and Astrophysics.

Astronomy is the scientific study of celestial objects such as stars, planets, galaxies, and other phenomena that exist outside of Earth's atmosphere.

Astronomers use a variety of methods to observe and study these objects, including telescopes, spacecraft, and computer simulations.

Astronomy is a broad field that includes many different sub-disciplines, such as astrophysics, planetary science, and cosmology.

Astronomers study the physical properties and behavior of celestial objects, such as their composition, temperature, motion, and evolution.

They also seek to understand the structure and history of the universe as a whole.

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Answer: 19.80 cm

Explanation:

Given

focal length \(f=6.14\ cm\)   (as focal length is positive, it is converging lens)

Image distance \(v=1.5f\)

\(v=1.5\times 6.14\\v=9.21\ cm\)

using lens formula

\(\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\)

insert values

\(\dfrac{1}{u}=\dfrac{1}{6.14}-\dfrac{1}{8.9}\\\\\dfrac{1}{u}=0.1628-0.1123\\\\\dfrac{1}{u}=0.0505\\\\u=19.80\ cm\)

Thus, the distance of the object is 19.80 cm

Given:

Now,

The image distance will be:

→ \(v = 1.5 f\)

     \(= 1.5\times 6.14\)

     \(= 9.21 \ cm\)

By using the lens formula, we get

→ \(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)

By putting the values, we get

→ \(\frac{1}{u} = \frac{1}{6.14} - \frac{1}{8.9}\)

   \(\frac{1}{u} = 0.1628-0.1123\)

   \(\frac{1}{u} = 0.0505\)

   \(u = 19.80 \ cm\)

Thus the answer above is correct.    

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The average speed of the bus is 8 kilometers per hour (kph).

Distance traveled = 6 blocks east + 8 blocks north = (6 x 100m) + (8 x 100m) = 1000m

Time taken = 15 minutes

Hence the speed can be computed as

Speed = Distance/Time = 1000m/15min = 8 kph

The average speed of the bus is 8 kilometers per hour (kph), after traveling 6 blocks east and 8 blocks north, with each block being 100m long and the journey taking 15 minutes.

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