Consider a steel tape measure with cross-sectional area, A = 0.0625 inches squared, and length L = 3, 600 inches at room temperature. How much error will occur if this tape measure is used on a hot day? Assume it is 130F and the coefficient of thermal expansion is α = 5× 10 −6 1/F. Does the error depend on the distance being measured?

The work done (W) in lifting the crate of bananas is equal to the force applied (F) times the distance moved (d). Therefore, W = F * d. Since we know the work done (W) and the distance moved (d), we can solve for the force applied (F).


W = 40 W

d = 5 m


F = W / d

F = 40 W / 5 m

F = 8 N


Therefore, the force applied is 8 N and the work done is 40 J.

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Answer:

Earth have a liquid water supply and Mars does not because of the difference in their magnetic fields. Therefore the option B is correct.

Explanation:

The atmosphere of Earth is shielded from solar winds and radiation by a potent magnetic field, which keeps the atmosphere from being torn away. This enables water to remain liquid on the surface of the planet.

Mars, in contrast, has a weaker magnetic field, which results in a thinner atmosphere that is unable to support liquid water. Mars is hence more colder and drier than Earth.

One of the main reasons that Mars has no life as we know it is because it lacks a magnetic field, whereas the magnetic field of Earth is crucial in fostering the conditions necessary for life to flourish.

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The average speed of the whole journey is 54 mph.

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False, all sounds are flucuations that hinders the preception of a signal. So by that standard, everything is a noise.

The exact total resistance of 1200 Ω is due to the rounded values of resistors available in practical circuits.

To determine the values of the resistors, we can use Ohm's Law:

Voltage (V) = Current (I) × Resistance (R)

Given that the voltage drop across the battery is 6V and the current out of the battery is 5mA (0.005A), we can calculate the total resistance:

Total Resistance (R_total) = Voltage (V) / Current (I)

R_total = 6V / 0.005A

R_total = 1200 Ω

Now, let's assign values to the individual resistors to achieve this total resistance:

R1 = 220 Ω

R2 = 470 Ω

R3 = 330 Ω

R4 = 680 Ω

R5 = 820 Ω

R6 = 350 Ω

With these values, the total resistance of the circuit would be:

R_total = R1 + (R2 || R3) + (R4 || R5) + R6

R_total = 220 Ω + (470 Ω || 330 Ω) + (680 Ω || 820 Ω) + 350 Ω

R_total ≈ 220 Ω + 214.8 Ω + 351.5 Ω + 350 Ω

R_total ≈ 1136.3 Ω

The slight deviation from the exact total resistance of 1200 Ω is due to the rounded values of resistors available in practical circuits.

Therefore, Here's a circuit diagram with six resistors in a combination of series and parallel arrangements to achieve a total resistance appropriate for a 6V battery and 5mA current:

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The tension in the strings are 31.47 and 19.25 N respectively.

Mass of the block, m = 3 kg

From the figure, consider the vertical components,

T₁ sin45° + T₂ sin30° = mg

(T₁/√2) + (T₂/2) = 3 x 9.8 = 29.4

Also, consider the horizontal components,

T₁ cos45° = T₂ cos30°

T₁/√2 = T₂ x√3/2

T₁ = T₂ x √3/2 x √2

So,

T₁ = 0.612T₂

Applying in the first equation,

(T₁/√2) + (T₂/2) = 29.4

(0.612T₂/1.414) + 0.5T₂ = 29.4

0.434 T₂ + 0.5 T₂ = 29.4

0.934 T₂ = 29.4

Therefore, the tension,

T₂ = 29.4/0.934

T₂ = 31.47 N

So, the tension,

T₁ = 0.612 T₂

T₁ = 0.612 x 31.47

T₁ = 19.25 N

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The electron number cannot be used instead because electrons can be removed from or added to an atom (option C)

The element of an atom is determined by its proton count, while the electron count can exhibit variability. Take, for instance, a sodium atom, which encompasses 11 protons and 11 electrons. However, it has the capacity to relinquish one electron, transforming into a sodium ion housing only 10 electrons.

This occurs due to the relatively loose binding of electrons to the nucleus, enabling their removal through the influence of an electric field or alternative mechanisms.

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The final velocity of the red ball is 18 m/s.

The term momentum has to do with the product of the mass and the velocity of an object We know that the momentum is always conserved in accordance with the Newton third law. Also it is clear that the momentum before collision is equal to the total momentum after collision and we are going to apply this principle here.

Then;

Mass of the blue ball =  6 kg

Mass of the red ball =  1 kg

Initial velocity of the blue ball =  4 m/s

Initial velocity of the red ball = 0 m/s

Final velocity of the red ball = ??

Final velocity of the blue ball =  1 m/s

We now have;

(6 * 4) + (1 * 0) = (1 * v) + (6 * 1)

24 = v + 6

v = 24 - 6

v = 18 m/s

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Answer:

Explanation:

The order of mixtures according to their particle size from smallest to the largest is: solution< colloid< suspension. So, the first option is correct.

Solvent is defined as the substance in which the solute particles dissolve and forms a mixture. Solute is defined as the particles that get dissolved in the solvent to form the mixture.

Here,

The particles of the solutions, colloids and suspension varies according to their size and the type of mixture they comes under.

Colloid is a heterogenous mixture, that means it has a non-uniform composition of particles. In colloid, the particles have an intermediate size between 1 nm to 1000 nm which is the size range between that of solution and suspension.

Solution is a homogenous mixture in which the particles have a diameter of less than 1 nm. The size of the particles in solution is the smallest when compared to that of colloid and suspension.

Suspension is a homogenous mixture in which the size of the particles is larger than 10000 nm. The particles of suspension are the largest among the particles of colloid and solution.

Hence,

The order of mixtures according to their particle size from smallest to the largest is: solution< colloid< suspension.

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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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The angle of incidence for a complete polarization of incident ray on water is 53.1⁰.

The given parameters;

The angle of incidence for a complete polarization of incident ray on water is calculated by applying Brewster's law as shown below;

\(\mu = tan \ \theta\)

where;

The polarizing angle is calculated as;

\(\theta = tan^{-1}(1.333)\\\\\theta = 53.1^ 0\)

Thus, the angle of incidence for a complete polarization of incident ray on water is 53.1⁰.

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Answer:

7000 meters

Explanation:

obviously 99.99 is less than 7000

10 cubed is 10 × 10 × 10, 1000, × 4.5 is 4500, which is less than 7000

10 squared is 100, which × 9 is only 900, which is also less than 7000

have a good day

Answer:

C

Explanation:

I think it's C, because at that point, you are going fastest. Sorry if im wrong, hope this helps.

Answer:

In between and the middle one

Explanation:

Answer:

Explanation:

The work done by the first person in lifting the weight to a height H is given by:

W = F × d = 200 × H

The power output P of the first person is then given by:

P = W / T = (200 × H) / T

To lift the same weight to a height of 2H in time T/2, the second person would have to do twice the work done by the first person, since work done is equal to force times distance. Therefore, the work done by the second person is given by:

W = 2 × (200 × 2H) = 800H

The power output of the second person is then given by:

P' = W / (T/2) = (800H) / (T/2) = (800H × 2) / T = 1600H / T

Comparing the power output of the first person (P) and the second person (P'), we have:

P' / P = (1600H / T) / (200H / T) = 8

Therefore, the power output of the second person is 8 times the power output of the first person. So, the correct option is (3) 2P.

Answer:

The frequency of a string wave is 922.5 Hz

Explanation:

Given;

length of the string, L = 0.65 m

initial speed of the string wave at n = 5, v₁ = 320 m/s

initial frequency of the wave speed, f₁ = 1230 Hz

final speed of the string wave at n = 5, v₂ = 240.0 m/s

final  frequency of the wave speed, f₂ = ?

speed of wave is related to frequency of wave and wavelength by the following equation;

V = Fλ

If the wavelength is constant at n = 5, then

\(\lambda = \frac{V_1}{f_1} = \frac{V_2}{f_2}\\\\f_2 = \frac{f_1V_2}{V_1} \\\\f_2 =\frac{1230*240}{320} \\\\f_2 = 922.5 \ Hz\)

Therefore, the frequency of a string wave, if speed of 240.0 m/s is used is 922.5 Hz

Answer:

O Wavelengths and frequency can vary inversely to produce the same wave speed.

Explanation:

Analyzing the given data table, it is clear that the wave speed remains constant in every case. Since the wave speed does not change and has the same value in every instance. Therefore, the wavelength and frequency must vary to make the speed constant.

Now the data also indicates that whenever wavelength increases, the frequency decreases. While the frequency increases whenever the wavelength decreases. Hence, it can be concluded that the wavelength and the frequency are inversely proportional to each other.

So, the correct option is:

O Wavelengths and frequency can vary inversely to produce the same wave speed.

Answer:

i actually dont know im so sorry but i will keep trying to find the answerr

Answer:

When the lines (representing peaks) pass over each other, constructive interference occurs. To put it another way, when two waves are in phase, they constructively interfere. When two waves are entirely out of phase (a peak lies in the middle of two waves), destructive interference ensues.

Explanation:

Answer: 1 How many grams of KCl will dissolve in 1 liter of H2O at 50 °C? 5. 58.0 g of K2Cr2O7 is added to 100 g H2O at. 0 °C. With constant stirring, to what temp-.                                                                                                                                     2  34 °C? 4. How many grams of KCl will dissolve in 1 liter of H2O at 50 °C? 5. 58.0 g of ... A saturated solution of KClO3 was made with 300 g of H2O at. 34 °C.

Explanation:

Answer:The correct answer should be C. is filled with medication used to administer a fixed amount of medication per inhalation through the mouth.

Explanation:

Answer:

c. is filled with medication used to administer a fixed amount of medication per inhalation through the mouth.

The minimum horizontal velocity required for an object to orbit the earth just above the surface is 5.04 x 10¹⁰ m/s.

The orbital speed of Earth or a natural satellite is the speed at which it orbits around the center of the system.

The minimum horizontal velocity required for an object to orbit the earth just above the surface is calculated as follows;

V(min) = √GMR

where;

Substitute the given parameters and solve for the minimum orbital speed of the earth.

V(min) = √(6.67 x 10⁻¹¹ x 5.97 x 10²⁴ x 6.38 x 10⁶)

V(min) = 5.04 x 10¹⁰ m/s

Thus, the minimum horizontal velocity required for an object to orbit the earth just above the surface is 5.04 x 10¹⁰ m/s.

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Answer: Brainliest?

Explanation:

Isobars and isotherms are both types of contour lines used to represent data on weather maps, specifically for atmospheric pressure and temperature, respectively.

The similarities between isobars and isotherms are:

Both are contour lines that connect points of equal value on a map.

Both are used to depict weather patterns and conditions.

Both help to identify areas of high and low values.

The differences between isobars and isotherms are:

Isobars connect points of equal atmospheric pressure, whereas isotherms connect points of equal temperature.

Isobars are measured in units of pressure such as millibars, while isotherms are measured in units of temperature such as degrees Celsius or Fahrenheit.

Isobars are typically used to show pressure patterns associated with wind, while isotherms are used to show temperature patterns.

Isobars are often used to forecast weather conditions, including the movement and intensity of storm systems. Isotherms are used to identify areas of warm and cold air masses, which can affect local weather patterns.

In summary, both isobars and isotherms are useful tools for understanding weather patterns, but they represent different types of data and are used for different purposes.

Isobars and isotherms are both concepts used in meteorology and climatology to represent important variables that help to describe atmospheric conditions. While they share some similarities, they also have several key differences.

Isobars refer to lines of equal pressure, meaning they connect points on a map or graph where the atmospheric pressure is the same. Isobars are drawn on weather maps to indicate areas of high and low pressure, and to show the general movement of air masses. When isobars are closely spaced, it indicates a steep pressure gradient, which can result in strong winds.

On the other hand, isotherms refer to lines of equal temperature, meaning they connect points on a map or graph where the temperature is the same. Isotherms are often drawn on weather maps to show the boundaries between warmer and cooler air masses, and to indicate areas where temperature changes rapidly.

One similarity between isobars and isotherms is that they are both used to describe atmospheric conditions in terms of spatial variation. They are also both used to infer information about the movement of air masses and the development of weather patterns.

However, there are also some key differences between isobars and isotherms. The most obvious difference is that isobars represent pressure while isotherms represent temperature. Additionally, while isobars are generally oriented parallel to each other and indicate the direction of winds, isotherms are typically oriented perpendicular to isobars and indicate the location of temperature gradients. Finally, while isobars are more commonly used to describe weather conditions associated with areas of high and low pressure, isotherms are often used to identify the location of fronts and other weather boundaries.

In summary, isobars and isotherms are similar in that they both describe atmospheric conditions in terms of spatial variation, and can be used to infer information about the movement of air masses and the development of weather patterns. However, isobars represent pressure and are oriented parallel to each other, while isotherms represent temperature and are oriented perpendicular to isobars.

Answer:

pascal

Explanation:

Answer:

It should be PCI

Explanation:

My head

Answer:

F_total = -65.5 N

Explanation:

Coulomb's law is

           F = \(k \frac{q_1 q_2}{r^2}\)k q1q2 / r2

In this exercise the force on the charge q3 is asked, for this we use that the force is a vector

          F_total = F₁₃ + F₂₃

let's look for every force

          F₁₃ = \(k \frac{q_1 q_3}{r_{13}^2}\)

it is indicated that charge 1 is equal to q₁ = -11.5 nC = -11.5 10⁻⁹ C, located at x₁ = -1.675 mm = -1.675 10⁻³ m and charge q₃ = 48.0 nC = 48.0 10⁻⁹ C located at x₃ this position is not written in the exercise, suppose the position x₃ = -0.5 mm = 0.5 10⁻³ m

the distance is

          r₁₃ = \(\sqrt{(x_3 - x_1 )^2}\)

          r₁₃ = \(\sqrt{(0.5-1.675)^2} \ 10^{-3}\)

          r₁₃ = 1.175 10⁻³ m

we calculate

          F₁₃ = \(\frac{9 \ 10^{9} 11.5\ 10^{-9} 48.0\ 10^{-9} }{(1.175 10^{-3})^2 }\)

          F₁₃ = 3.598 N

as the charge q₁ is negative and the charge q₃ is positive the force is attractive directed to the right

we look for F₂₃, where q₂ = 40 nC = 40.0 10⁻⁹ C located at x₂ = 0

             F₂₃ = \(k \frac{q_2q_3}{r_{23}^2}\)

            r₂₃ = \(\sqrt{(x_3-x_2)^2}\)

            r₂₃ = \(\sqrt{(0.5 -0)} \ 10^{-3}\)RA (-0.5 0) 2 103

            r₂₃ = 0.5 10⁻³ m

             F₂₃ = \(\frac{9 \ 10^{9}\ 40\ 10^{-9} \ 48.0\ 10^{-9} x}{(0.5 \ 10^{-3})^2 }\)

             F₂₃ = 6.912 10¹ N

             F₂₃ = 69.12 N

as the two charges are of the same sign, the force is repulsive, therefore it is directed to the left

the total force is

            F = total = 3.598 - 69.12

            F_total = -65.5 N

the negative sign indicates that the force is to the left

In a DC generator, the generated electromotive force (emf) is directly proportional to the rotational speed of the generator's armature and the strength of the magnetic field within the generator.

This relationship is described by the equation for the generated emf in a DC generator:

Emf = Φ * N * A * Z / 60

Where:

Emf is the generated electromotive force (in volts),

Φ is the magnetic flux density (in Weber/meter^2\(meter^2\) or Tesla),

N is the number of turns in the armature winding,

A is the effective area of the armature coil (in square meters),

Z is the total number of armature conductors, and

60 is a constant representing the conversion from seconds to minutes.

From this equation, we can see that the generated emf is directly proportional to the magnetic flux density (Φ) and the product of the number of turns (N), effective area (A), and the total number of armature conductors (Z). This means that increasing any of these factors will result in a higher generated emf.

The magnetic flux density (Φ) can be increased by using stronger permanent magnets or increasing the strength of the field windings in the generator.

The number of turns (N) and the effective area (A) are design parameters and can be optimized for a specific generator. Increasing the number of turns or the effective area will result in a higher generated emf.

Similarly, the total number of armature conductors (Z) can be increased to enhance the generated emf.

By controlling and optimizing these factors, the generated emf in a DC generator can be increased, resulting in higher electrical output. However, it is important to note that there are practical limits to these factors based on the design and construction of the generator.

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The letter that represents the various descriptions is as follows:

5.) Angle of incidence: B

6.) Object: D

7.) Plain mirror:A

8.) Reflection: E

9.) Angle of reflection: C

10.) Normal: F

The law of reflection in a plain mirror states that the angle of reflection is equal to the angle of incidence.

The diagram explains the law of reflection on a plain mirror because it shows the reflection of the object D as E after creating an angle of incidence B on the plane mirror surface A which is equal to the angle of reflection C.

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Answer:

Explanation:

Let the equal charges be Q .

Force between charges = k Q² / R where R is distance between charges , k = 9 x 10⁹.

Putting the charges

Force F = 9 x 10⁹ x Q² /( .72 x 10⁻²)²

= 17.36 x 10¹³ Q² N

This force equals weight so

17.36 x 10¹³ Q² = 15.7 x 10⁻³

Q² = .90 x 10⁻¹⁶

Q = .95 x 10⁻⁸

= 9.5 x 10⁻⁹

= 9.5 nC.

The magnitude of the electric field along the positive z axis is kqz/√(z^2+a^2)^3/2 , the angular frequency of these oscillations is √(kqq0/ma^3)

1. Given that a uniformly charged ring in the xy plane, centered at the origin.

Radius of the ring(r) = a

positive charge distributed evenly along its circumference = q

k= 1/4piepsion

The electric field can be described as, E = kq/r^2

The force exerted on an electric charge by a charge is F = q0E

Any vector can be projected along the z = A(z)=Acosθ

We know that ∑F =ma

Acceleration along z axis = a =d^2z/dt^2

The equation of motion for a simple harmonic oscillation along z is d^2z/dt^2 = -ω^2z

Here,  ω is the cosine function in a right-angled triangle is cosθ = z/√(z^2+a^2)

dE(z)=dEcosθ we know that E = kq/z^2+a^2

To calculate the magnitude of the electric field caused by complete rings, integrate dE(z)=dEcosθ

E(z) = Ecosθ = kq/(z^2+a^2) x  z/√(z^2+a^2)

E(z) = kqz/(z^2+a^2)^3/2

2. We know that F= q0E = kqq0z/(z^2+a^2)^3/2

Given The ball will oscillate along the z axis between z=d and z=-d in simple harmonic motion. Let z =d

F = kqq0d/(d^2+a^2)^3/2 = kqq0d/a^3

Force at any point along z = - kqq0z/a^3

We know F = ma =  

- kqq0z/a^3 = md^2z/dt^2

d^2z/dt^2= - kqq0z/ma^3

This equation of motion can be compared with the equation of motion for simple harmonic motions to find the angular frequency.

ω =√(kqq0/ma^3)

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Answer:

v’= 279.66 m / s

Explanation:

We work this exercise using the conservation of the moment. For this we define the system formed by the two blocks, therefore the forces during the collision are internal of the action and reaction type.

Initial instant. Before the crash

        p₀ = m v₀ + 0

Final moment. After the crash

        p_f = m v + M v ’

how the tidal wave is preserved

       p₀ = p_f

       m v₀ = m v + M v ’

       v = \(\frac{m v_o - Mv'}{m}\)

let's calculate

       v ’= \(\frac{0.00467 \ 619 - 0.072 \ 22}{0.004676}\)

       v ’= \(\frac{2.89- 1.584}{ 0.00467}\)

       v ’= 279.66 m / s