Which properties describe elements that are nonmetal gases? Select all that apply.

A)
composed of an extended structure

B)
composed of individual atoms or diatomic molecules

C)
high melting point

D)
low melting point

E)
strong attractions between particles

F)
weak attractions between particles

Given data:

* The mass on the frictionless table is,

* The mass hangs freely from the string is,

* The hanging mass falls a distance is,

Solution:

(1). The normal force of mass on the frictionless table is,

As the displacement of the mass m_1 on the frictionless table is in the hroizontal direction.

Thus, the work done by teh normal force is,

As both the normal force and displacement are perpendicuular to each other.

Thus, the work done by the nromal force on the mass m_1 is,

Thus, the work done by the normal force on m_1 is zero.

Explanation:

Speed or velocity (V) = 35 m/s

Kinetic energy (K. E) = 1500 Joule

mass (m) = ?

We know

K.E = 1/2 * m * v²

1500 = 1/2 * m * 35²

1500 * 2 = 1225m

m = 3000 / 1225

m = 2.45 kg

The mass of the object is 2.45 kg

Hope it will help :)

The California sea lions communicate  is  option C: 0.375 to 3 meters.

To calculate the approximate wavelengths of sound with which the California sea lions communicate, we can use the formula:

Wavelength = Speed of Sound / Frequency

Given:

Speed of Sound in Sea Water = 1,500 m/s

Frequency Range = 500 Hz to 4,000 Hz

For the lower frequency of 500 Hz:

Wavelength = 1,500 m/s / 500 Hz = 3 meters

For the higher frequency of 4,000 Hz:

Wavelength = 1,500 m/s / 4,000 Hz = 0.375 meters

Therefore, the approximate wavelengths of sound with which the California sea lions communicate are approximately 0.375 to 3 meters.

So, the correct answer is option C: 0.375 to 3 meters.

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The phase of a low-mass star's life that has a core with active helium fusion and a shell of hydrogen fusion is called the red giant phase. During the red giant phase, low-mass stars exhibit a core with active helium fusion and a shell of hydrogen fusion, allowing them to sustain their energy production and prolong their stellar evolution.

During the red giant phase, a low-mass star undergoes significant changes in its structure and energy generation processes. As the star exhausts the hydrogen fuel in its core, it starts to expand and cool, becoming a red giant. At this stage, the core contracts under its own gravity, becoming denser and hotter.

In the core of a red giant, the temperature and pressure conditions become suitable for helium fusion to occur. Helium nuclei combine to form heavier elements, releasing energy in the process. This active helium fusion in the core contributes to the star's energy production.

Simultaneously, a shell of hydrogen fusion forms around the core. The outer layers of the star still contain hydrogen, and as the core contracts and heats up, the hydrogen in the shell surrounding the core becomes hot enough to undergo fusion reactions. This shell of hydrogen fusion contributes to the overall energy output of the red giant.

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In the problem given, the vertical velocity of the rock when it hits the ground is approximately 21.53 m/s, and it hits the ground after 1.41 seconds.

To solve this problem, we need to break the initial velocity into its horizontal and vertical components.

The horizontal component of the velocity is given by:

v_x = v * cos(theta) = 12 m/s * cos(42°) ≈ 9.04 m/s

where theta is the angle above the horizontal.

The vertical component of the velocity is given by:

v_y = v * sin(theta) = 12 m/s * sin(42°) ≈ 7.93 m/s

We can use the vertical component of the velocity to find the time it takes for the rock to hit the ground. We can use the equation:

y = v_y * t + 1/2 * a * t^2

where y is the vertical displacement, a is the acceleration due to gravity (9.81 m/s^2), and t is the time.

Since the rock lands 9.5 m below the initial level, we have:

y = -9.5 m (taking downwards as negative)

Substituting the values, we get:

-9.5 m = 7.93 m/s * t + 1/2 * 9.81 m/s^2 * t^2

Simplifying the equation, we get:

4.905t^2 + 7.93t + 9.5 = 0

Using the quadratic formula, we get:

t = (-7.93 ± sqrt(7.93^2 - 4 * 4.905 * 9.5)) / (2 * 4.905)

t ≈ 1.41 s (ignoring the negative root)

So it takes approximately 1.41 seconds for the rock to hit the ground.

Finally, we can use the time and the horizontal velocity to find the horizontal displacement of the rock using the equation:

x = v_x * t

x = 9.04 m/s * 1.41 s ≈ 12.77 m

Therefore, the rock lands approximately 12.77 m away from the base of the castle wall.

Since we know the time it takes to hit the ground, we can also find the vertical velocity of the rock when it hits the ground. We can use the equation:

v_y = v_y0 + a * t

where v_y0 is the initial vertical velocity (which we already calculated as 7.93 m/s), and t is the time it takes to hit the ground (which we just calculated as 1.41 s).

Substituting the values, we get:

v_y = 7.93 m/s + 9.81 m/s^2 * 1.41 s ≈ 21.53 m/s

Therefore, the vertical velocity of the rock when it hits the ground is approximately 21.53 m/s, and it hits the ground after 1.41 seconds.

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Answer:

ooops ignore this it wont let me leave this screen it does this weirdly s

Explanation:

Answer:

Newton.........

.....

Answer:

When you place an object in front of a mirror, you see the same object in the mirror. This image that appears to be behind the mirror is called the image. The object is the source of the incident rays, and the image is formed by the reflected rays. An image formed by reflection may be real or virtual.

The angular magnification (M) obtained with a thin lens used as a simple magnifier when the object is at the focal point is -1.

The angular magnification is given by the formula:

M = -f / (f - d)

Where:

M is the angular magnification,

f is the focal length of the lens,

d is the distance between the object and the lens.

In this case, since the object is at the focal point, the distance between the object and the lens (d) is equal to the focal length of the lens (f). Substituting the values, we have:

M = -f / (f - f)

M = -f / 0

M = -1

Therefore, when the object is at the focal point, the angular magnification obtained with the lens is -1.

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Therefore, the density of water would be 28g/48mL.

The density of an object can be calculated by dividing its mass by its volume.

In this case, we have a density bottle filled with water.

The mass of the empty dry density bottle is 25g, while the mass of the bottle filled with water is 53g.

The volume of water in the bottle is given as 48 cm³.

To find the density of water, we can subtract the mass of the empty bottle from the mass of the bottle filled with water to determine the mass of the water alone.

In this case, the mass of the water would be 53g - 25g = 28g.

Next, we divide the mass of the water by its volume to calculate the density.

The density of water is given by the equation density = mass/volume.

So, the density of water would be 28g/48cm³.

To simplify the units, we can convert cm³ to mL (milliliters), as they are equivalent.

Therefore, the density of water would be 28g/48mL.

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A. Controlled Variable.

I hope this helps!

A dedicated PV system circuit breaker, suitable for backfeed and positioned at the opposite end of the bus from the main circuit breaker, is a requirement of NEC 705.12(B).

A completed circular conduit through which electricity flows is known as a circuit in electronics. A simple circuit includes a current provider, conductors, and a load. The term "circuit" can broadly be used to describe any ongoing path via which electricity, information, or a signal might go.

A channel for the transfer of electric current is known as an electric circuit. Electric energy is transferred to components that transform it into various types of energy that can perform work, such as supplying power to light, appliances, and other devices, when electrical current flows across a circuit.

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(a) It can be stated that the 10-gram marble hits the ground first because it has a higher acceleration and travels faster.

(b) Both marbles will travel the same distance since their horizontal components of velocity and initial velocity with respect to the ground are the same.

According to the given statement, a 10-gram marble has a speed that is 5 times faster than that of a 100-gram marble. Both marbles roll off the table at the same time. The questions to answer are as follows:

(a) The acceleration due to gravity is constant at 9.8 m/s², according to Newton's Law of Gravitation.

According to the law of physics, heavier objects fall faster than lighter ones, but since the 10-gram marble has a speed that is five times faster than the 100-gram marble, it implies that the 10-gram marble covers more ground in less time than the 100-gram marble as acceleration is directly proportional to the force applied.

The time taken by both marbles to reach the ground is given byt = √(2h/g)where h is the height from which the marbles were dropped, and g is the acceleration due to gravity.

The height from which the marbles were dropped is the same in both cases, so it can be stated that the 10-gram marble hits the ground first because it has a higher acceleration and travels faster.

(b) The range traveled by both marbles is determined by the horizontal component of their velocity. It's worth noting that the horizontal components of their velocities are identical since they were launched from the same height, so there's no advantage for either marble.

The range of a projectile is determined by the formula:

R = u²sin(2θ)/g where R is the range, u is the velocity of the object, θ is the angle of the initial velocity with the horizontal, and g is the acceleration due to gravity.

Since the angle of the initial velocity with the horizontal is the same for both marbles, and their initial velocity is also the same, it can be stated that both marbles travel the same distance.

Therefore, both marbles will travel the same distance since their horizontal components of velocity and initial velocity with respect to the ground are the same.

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Answer: Hot air is transferred through heat being radiated from a thermal burner.

The oven absorbs air from the environment and passes it through a thermal resistance (thermal burner) to heat it and then this hot air is the one that cooks the food.

Answer:

he is wronggg

Explanation:

Answer:

1.An object at rest has zero velocity - and (in the absence of an unbalanced force) will remain with a zero velocity. Such an object will not change its state of motion (i.e., velocity) unless acted upon by an unbalanced force.

3.Let speed of balls are v

1

and v

2

.

There is no external force acting, momentum will be conserved.

m

1

u

1

+m

2

u

2

=m

1

v

1

+m

2

v

2

⇒m×6−m×6=mv

1

+mv

2

⇒v

1

=−v

2

Coefficient , e=−

u

1

−u

2

v

1

−v

2

⇒

3

1

=−

6+6

v

1

−v

2

⇒v

1

=−2m/sWhen two objects with the same mass collide, Newton's laws tell us that they will accelerate the same amount but in opposite directions.

Collision Forces - 5/5

1. A) Motion will be either created, stopped or changed.

2. D) 5,208 N

3. B) the mass and acceleration of each person

4. C) balanced forces

5. B) 13.125 N to the right

Hope this helps ya out peeps!

Answer:

The expression written for the momenum of a body is dimensionally incorrect.

Explanation:

P = momentum, m = mass, E = Energy & t = time Now, we know: Dimensional formula of momentum (p) = [MLT-1] Dimensional formula of mass (m) = [M] Dimensional formula of Energy (E) = [ML2T-2] Dimensional formula of time (t) = [T] Here, Dimensional formula of LHS = [MLT-1] And, Dimensional formula of RHS = 2/M212T-21 [T] -=√2 M²LT 3]=[MLT/2] [2 is constant] 2[M] [MLT-21 [T] Since the dimensional formula of LHS # dimensional formula of RHS. [MILT-1] [MLT-3/2] hence the given relation is dimensionly incorrect

Answer: 5.96m/s

Explanation:

Given the following :

Mass of car (m) = 1500kg

Velocity (V) = 5.25m/s

Forward force of engine = 1250N

Diatance moved = 4.8m

Final Velocity =?

Final kinetic energy = Initial kinetic energy + work done by engine

Initial kinetic energy = 0.5 × mass × velocity^2

Initial kinetic energy = 0.5 × 1500 × 5.25^2

Initial kinetic energy = 20671.875 J

Work done by engine = Force × distance

Work done by engine = 1250 × 4.8 = 6000J

Final kinetic energy = (20671.875 + 6000) J

= 26671.875 J

From kinetic energy = 0.5mv^2

26671.875 = 1/2 × 1500 × v^2

53343.75 = 1500v^2

v^2 = 35.5625

v = sqrt(35.5625)

v = 5.96m/s

A)

The moon orbiting the Earth

It is crucial to ensure that the equipment is in compliance with safety regulations and standards to ensure the protection of individuals working with or near live parts. These measures help reduce the risk of electrical accidents and promote a safer working environment.

To guard live parts over 50V but less than 600V, equipment is permitted to be appropriately insulated. This means that the equipment needs to have insulation that can withstand the voltage and provide protection against electric shock. Insulation can be achieved through various means, such as using insulating materials like rubber or plastic.
For example, electrical wires are often coated with insulating material like PVC (polyvinyl chloride) to prevent contact with live parts and reduce the risk of electrical shock.

Similarly, electrical devices like switches, outlets, and circuit breakers are designed with insulation to prevent accidental contact with live electrical components.
It is crucial to ensure that the equipment is in compliance with safety regulations and standards to ensure the protection of individuals working with or near live parts. These measures help reduce the risk of electrical accidents and promote a safer working environment.

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nts have an uncertainty of 0.1 ohms each, we can calculate the following:

(a) Calculate the resistance R when R1 and R2 are wired in parallel:

Using the formula 1/R = 1/R1 + 1/R2, we can substitute the given values:

1/R = 1/7 + 1/10

(b) Calculate the percent uncertainty in R1:

Percent uncertainty in R1 = (Uncertainty in R1 / R1) * 100

Percent uncertainty in R1 = (0.1 ohms / 7 ohms) * 100

(c) Calculate the percent uncertainty in R2:

Percent uncertainty in R2 = (Uncertainty in R2 / R2) * 100

Percent uncertainty in R2 = (0.1 ohms / 10 ohms) * 100

(d) Calculate the percent uncertainty in R:

To calculate the percent uncertainty in R, we need to consider the uncertainties in R1 and R2:

Percent uncertainty in R = (Percent uncertainty in R1 + Percent uncertainty in R2)

You can substitute the given values into the equations to calculate the desired values.

Note: The uncertainty in R is calculated by combining the uncertainties in R1 and R2. Since the formula for parallel resistance is an addition of terms, the percent uncertainties in R1 and R2 can simply be added.

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The seasoned mini golfer must give the ball an initial speed of approximately 1.95 m/s to land directly in the hole on tricky hole number 5.

To land directly in the hole on tricky hole number 5 of mini golf, the seasoned golfer must launch the ball up a 41.5° ramp with a height of 0.540 m. The ball needs to travel a distance of 1.00 m to reach the hole. Assuming no slipping occurs and the ball maintains constant angular speed after launching, the golfer needs to give the ball an initial speed of approximately 1.95 m/s.

To determine the required initial speed (v1) of the ball, we can break down the problem into two parts: the ball's motion along the ramp and its motion through the air. Firstly, let's consider the motion along the ramp.

The ball moves up the ramp against gravity, and we can analyze its motion using the principles of projectile motion. The vertical component of the initial velocity (v1y) is given by v1y = v1 * sin(θ), where θ is the angle of the ramp. The ball must reach a height of 0.540 m, so using the equation for vertical displacement, we have:

h = (v1y^2) / (2 * g), where g is the acceleration due to gravity.

Solving for v1y, we get v1y = sqrt(2 * g * h). Substituting the given values, we find v1y ≈ 1.30 m/s.

Next, we consider the horizontal motion of the ball. The horizontal component of the initial velocity (v1x) is given by v1x = v1 * cos(θ). The ball needs to travel a horizontal distance of 1.00 m, so using the equation for horizontal displacement, we have:

d = v1x * t, where t is the time of flight.

Rearranging the equation to solve for t, we get t = d / v1x. Substituting the given values, we find t ≈ 0.517 s.

Now, considering the vertical motion, we know that the vertical velocity of the ball just before reaching the hole is zero. Using the equation for vertical velocity, we have:

v2y = v1y - g * t.

Substituting the values we found, we get v2y = 0. To land directly in the hole, the ball should have zero vertical velocity at the end. Therefore, we need to launch the ball with a vertical velocity of v1y ≈ 1.30 m/s.

Finally, to find the required initial speed (v1), we can use the Pythagorean theorem:

v1 = sqrt(v1x^2 + v1y^2).

Substituting the values we found, we get v1 ≈ 1.95 m/s.

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The ball's velocity decreases as it ascends, reaching a zero velocity peak at 11.025 meter's, the highest point. The speed of the ball remains constant during the toss.

The ball continues to move upward in the same direction until it reaches highest point. As the ball ascends higher, its velocity will slow down. The following is true for the ball's velocity and acceleration when it reaches its highest point. Its acceleration is not zero, despite its zero velocity. Option B is the appropriate response, thus. Any object's rate of change in velocity with respect to time is referred to as acceleration.

Maximum height reached = H = \(V^{2} /2g\)

Maximum height reached = \((14.7)^{2}/(2*9.8)\)

                                          = 11.025 meters

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The given statement is that the temperature in degrees Fahrenheit in times square during a day in August can be predicted by the function t(x).

Hence, it can be stated that the function t(x) provides a prediction for the temperature in degrees Fahrenheit in Times Square for a given value of x. Here, x represents the input variable.

It is important to note that the input variable must be provided in the correct units and format as required by the function t(x).The content loaded refers to the data that has been previously collected and is available for analysis. In the context of predicting temperature using function t(x),

the content loaded can refer to temperature data collected in Times Square during August on previous days or years. This data can be analyzed and used to develop the function t(x) that predicts temperature for a given input variable x.

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Answer:

x = 5[km]

Explanation:

We must convert the time from minutes to hours.

\(t=30[min]*\frac{1h}{60min}= 0.5[h]\\\)

We know that speed is defined as the relationship between space and time.

\(v=x/t\)

where:

x = space [m]

t = time = 0.5 [h]

v = velocity [m/s]

Now replacing:

\(x = 10[\frac{km}{h} ]*0.5[h]\\x=5[km]\)

The total momentum after collision is 143.466 kgm/s.

The momentum of a body is the product of the mass and the velocity of the body.

Assuming the right direction as positive and left direction as negative:

Momentum to the right = 46.3 * 5.24 = +242.612 kgm/s

Momentum to the left = 55.4 * -1.79 = -99.166 kgm/s

Total momentum before collision =  +242.612 kgm/s - 99.166 kgm/s

Total momentum before collision = 143.466 kgm/s

Thus, total momentum after collision is 143.466 kgm/s.

In conclusion, momentum is conserved in an isolated system of colliding bodies.

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Answer:

0.6 ☻︎☻︎☻︎☻︎☻︎☻︎☻︎☻︎☻︎☻︎☻︎

Given: The position of a particle moving along the x-axis is x(t)=sin(2t)−cos(3t) for time t≥0, and we have to find the acceleration of the particle when t=π.Solution:In order to find the acceleration of the particle, we need to take the derivative of x(t) twice.

Derivative of x(t):x'(t) = 2cos(2t) + 3sin(3t) [using chain rule]Second Derivative of x(t):x''(t) = -4sin(2t) + 9cos(3t) [using chain rule]When

t = π, we get: x'(π)

= 2cos(2π) + 3sin(3π)

= 2(1) + 3(0) = 2x''(π)

= -4sin(2π) + 9cos(3π)

= -4(0) + 9(-1)

= -9

Thus, the acceleration of the particle when t = π is -9.We have found that the acceleration of the particle when t = π is -9. The given equation for position of a particle moving along the x-axis is

x(t)=sin(2t)−cos(3t)

for time t≥0. The question asks for the acceleration of the particle when t=π. The acceleration can be calculated by taking the derivative of the given function of position. The derivative of x(t) is x'(t) = 2cos(2t) + 3sin(3t). We can find the acceleration of the particle by taking the second derivative of

x(t), x''(t) = -4sin(2t) + 9cos(3t).

Now, we can find the acceleration of the particle when t=π by plugging π into the first and second derivative equations.

x'(π) = 2cos(2π) + 3sin(3π)

= 2(1) + 3(0) = 2. x''(π)

= -4sin(2π) + 9cos(3π)

= -4(0) + 9(-1)

= -9.

Thus, the acceleration of the particle when t = π is -9.

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The optimal values of \(q_1\) and \(q_2\) are determined by these equations, which are functions of Michael's income and the prices of the goods.

The given problem describes Michael's utility maximization problem with a constant elasticity of substitution (CES) utility function. The objective is to find the optimal values of \(q_1\) and \(q_2\) in terms of Michael's income (Y) and the prices of the two goods (\(p_1\) and \(p_2\)).

1. Substitute the income constraint into Michael's utility function:

\(U(q_1, q_2) = (q_1^\rho + q_2^\rho)^(1/\rho)\)

  s.t. \(Y = p_1q_1 + p_2q_2\)

  We can rewrite Michael's budget constraint as \(q_2 = (Y - p_1q_1)/p_2\). Substituting this expression into his utility function, we have:

 \(U(q_1, p_2, Y) = (q_1^\rho + [p_2(Y - p_1q_1)/p_2]^\rho)^{(1/\rho)\)

  By making this substitution, we have converted the constrained maximization problem with two control variables (\(q_1\) and \(q_2\)) into an unconstrained problem with one control variable \((q_1)\).

2. Use the standard unconstrained maximization approach to determine the optimal value for \(q_1\). To obtain the first-order condition, we differentiate the utility function with respect to \(q_1\) and set it equal to zero:

\(\delta U / \delta q_1 = \rho(q_1^{(\rho-1)} + \rho[p_2(Y - p_1q_1)/p_2]^{(\rho-1)}(-p_1/p_2)) = 0\)

Simplifying and solving for \(q_1\):

\(\rho q_1^{(\rho-1)} - \rho(p_1/p_2)[p_2(Y - p_1q_1)/p_2]^{(\rho-1)} = 0\)

\(\rho q_1^{(\rho-1)} - \rho(p_1/p_2)[Y - p_1q_1]^{(\rho-1)} = 0\)

\(\rho q_1^{(\rho-1)} = \rho(p_1/p_2)[Y - p_1q_1]^{(\rho-1)}\)

\(q_1^{(\rho-1)} = (p_1/p_2)[Y - p_1q_1]^{(\rho-1)\)

\(q_1^{(\rho-1)} = (p_1/p_2)^{(1-\rho)}[Y - p_1q_1]^{(\rho-1)}\)

\(q_1^{(\rho-1)} = (p_1/p_2)^{(1-\rho)}(Y - p_1q_1)^{(\rho-1)}\)

\(q_1^{(\rho-1)} = (p_1/p_2)^{(1-\rho)}(Y^{(\rho-1)} - (\rho-1)p_1q_1(Y - p_1q_1)^{(\rho-2)})\)

  This equation represents Michael's optimal \(q_1\) as a function of his income (Y) and the prices (\(p_1\) and \(p_2\)).

3. Similarly, we can derive a similar expression for his optimal \(q_2\):

\(q_2^{(\rho-1)} = (p_2/p_1)^(1-\rho)(Y^{(\rho-1)} - (\rho-1)p_2q_2(Y - p_1q_2)^{(\rho-2)})\)

  This equation represents Michael's optimal \(q_2\) as a function of his income (Y) and the prices (\(p_1\) and \(p_2\)).

Therefore, these equations, which depend on Michael's income and the prices of the commodities, determine the ideal values of \(q_1\) and \(q_2\).

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9) A vector describe as a quantity that has both magnitude and direction. 10) We use vectors to describe in physics in physical quantities, including displacement, velocity, acceleration, force, momentum, and torque.

The vector is typically represented by an arrow, with the length of the arrow representing the magnitude of the vector and the direction of the arrow representing the direction of the vector. In addition to the magnitude and direction, vectors also have a starting point and an end point. Vectors can be added together to obtain a resultant vector that represents the sum of the individual vectors. Vectors can also be subtracted, multiplied by a scalar, or divided by a scalar.

For example, displacement is a vector that describes the distance and direction between two points. Velocity is a vector that describes the speed and direction of motion and acceleration is a vector that describes the rate at which velocity changes. Force is a vector that describes the push or pull on an object and momentum is a vector that describes the motion of an object and its resistance to change. Torque is a vector that describes the rotational motion of an object. So therefore vectors are an important tool for physicists because they allow them to describe physical quantities in a way that is both concise and precise.

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