A bridge 148.0 long is built of a metal alloy having a coefficient of expansion of 12.0 x 10-6/K. If it is built as a single, continuous structure, by how many centimeters will its length change between the coldest days (-29.0) and the hottest summer day (41.0)?

The frequency of the waves when a 0.5 kg mass is hanging on the end of a piece of fishing line is determined by the tension and length of the line.

The frequency of the waves on a piece of fishing line can be determined using the formula f = (1/2L)√(T/μ), where f is the frequency in hertz, L is the length of the line, T is the tension in newtons, and μ is the linear mass density of the line in kg/m.

In this case, the mass of the hanging object is 0.5 kg, so the tension of the line will be affected by the weight of the object as well as any other forces acting on it.

Assuming the line is under tension and the mass is held steady, the length of the line will determine the frequency of the waves. A shorter line will have a higher frequency, while a longer line will have a lower frequency.

Additionally, the tension of the line will affect the frequency - a higher tension will result in a higher frequency, while a lower tension will result in a lower frequency.

Therefore, the frequency of the waves when a 0.5 kg mass is hanging on the end of a piece of fishing line will depend on the tension and length of the line.

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As the charge, ( Q ) is equal and constant, the voltage drop across the capacitor is determined by the value of the capacitor only as V = Q ÷ C.

What is charge?

Charged material experiences a force when it is exposed to an electromagnetic field due to the physical characteristic of electric charge. You might have a positive or negative electric charge. Unlike charges attract one another while like charges repel one another.

As the charge, ( Q ) is equal and constant, the voltage drop across the capacitor is determined by the value of the capacitor only as V = Q ÷ C. A small capacitance value will result in a larger voltage while a large value of capacitance will result in a smaller voltage drop.

While the voltage drop across a resistor is proportional to the current and there is a current at the beginning, the voltage drop across a capacitor is related to its charge and is uncharged at the beginning. However, when charge on the capacitor begins to accumulate, some voltage is now lost across the capacitor.

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Answer:

mass (m) = 0.075 kg

acceleration (a) = 550000 m/\(s^{2}\)

F = ( 0.075 kg )( 550000 m/\(s^{2}\) ) = 41250 N

Explanation:

F = ma ; force is equal to mass times acceleration

mass (m) = 0.075 kg

acceleration (a) = 550000 m/\(s^{2}\)

F = ( 0.075 kg )( 550000 m/\(s^{2}\) ) = 41250 N.

do you get these options in the drop down?

Answer:

1,34 m/s

Explanation:

KE= 1/2 mv^2

massa = 3,6 × 10 ^4/ 10 ^3 kg = 36 kg

KE = 1/2 × 36 × v^2

32 × 2/36 = v^2

64/36 = V^2

√64/36 = V

V= 8/6

V = 1,34 m/s

Answer:

ocean-continent convergent boundaries

Examples of ocean-continent convergent boundaries are subduction of the Nazca Plate under South America (which has created the Andes Range) and subduction of the Juan de Fuca Plate under North America (creating the mountains Garibaldi, Baker, St.

The speed of traveling waves on the string is 48.6 m/s.

The fundamental frequency of a string that is bound at both ends and is under tension is calculated as follows:

f = v/2L

Where v is the velocity of waves on the string and L is the length of the string.

The third harmonic frequency of the standing wave can be expressed as f₃ = 3f₁. Therefore,864 = 3f₁.

Simplifying the above expression, we obtain f₁ = 864/3 = 288

Using the formula above, we can calculate the velocity v of the string as follows:

v = 2Lf₁

Substituting the values, we get:

v = 2(0.94 m)(288 Hz)

Evaluating the above expression gives us the velocity of the string as v = 48.6 m/s. Thus, the speed of traveling waves on the string is 48.6 m/s.

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Complete question is;

A 50.-ohm resistor, an unknown resistor R, a 120-volt source, and an ammeter are connected in a complete circuit. The ammeter reads 0.50 ampere.

A) Calculate the equivalent resistance of the circuit shown.

B) Determine the resistance of resistor R shown in the diagram.

Answer:

A) R_eq = 240 Ω

B) R = 190 Ω

Explanation:

A) To get the equivalent resistance, we will use the formula;

R = V/I

Where;

V is Voltage

I is current

R is equivalent resistance

From the question, V = 120 V and I = 0.5A

Thus;

R_eq = 120/0.5

R_eq = 240 Ω

B) From the image, we see that the resistors are connected in series.

Formula for resistors in series is;

R = R1 + R2 +..... Rn

Thus;

240 = 50 + R

R = 240 - 50

R = 190 Ω

A) The equivalent resistance of the circuit shown will be  240 Ω.

B) The resistance of resistor R will be 190 Ω

Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. Resistance is the enemy of the flow of current.

The given data in the problem is;

R is the resistance = 50.-ohm

v is the voltage = 120-volt source

I is the value of the current =0.50 ampere.

A) The equivalent resistance of the circuit shown will be  240 Ω.

According to ohm's law

\(\rm R= \frac{V}{I} \\\\ \rm R= \frac{120}{0.5} \\\\ \rm R=240 \ ohm\)

Hence the equivalent resistance of the circuit shown will be 240 Ω.

B) The resistance of resistor R will be 190 Ω

The given resistors are connected in the series;

R = R1 + R2 +..... Rn

240 = 50 + R

R = 240 - 50

R = 190 Ω

Hence the resistance of resistor R will be 190 Ω

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Answer:

A subscript

Explanation:

it tells you how many oxygen molecules there are in the equation

Answer: The pressure at the bottom : 19600 N/m²

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

\(a = \frac{f}{m} \\ \)

f is the force

m is the mass

From the question we have

\(a = \frac{35000}{1500} = \frac{350}{15} = \frac{70}{3} \\ = 23.33333...\)

We have the final answer as

Hope this helps you

Answer:

It reveals that light is a wave

Explanation:

Diffraction is the property of a wave in which there is a bending of the wave about the corners of an obstacle or aperture into the geometrical shadow of the obstacle or aperture.

This simply implies that a wave bends or spreads out when it passes through openings. Since the light diffracts through small slits and diffraction has been shown to occur in water waves and sound waves, this property of diffraction can only be characteristic of a wave and thus, this evidence reveals that light is a wave.

Answer:

It shows that light can travel in different wavelengths.

Explanation:

Light is similar to a wave in the way that it travels in a straight line, but when it is passing through the slits, it diffracts due to this difference. Therefore, light refracts in the same way as a wave would, and it is a mode of transferring energy. The light travels in a way similar to a wave

Answer:b

Explanation:

b

The mutual inductance between the coils is 6.25μH. the coefficient of coupling between the coils is approximately 0.447.

The mutual inductance between the coils can be determined using the formula:M = (Δφ_Y) / (N_X * ΔI_X)
Where M represents the mutual inductance, Δφ_Y is the change in flux in coil Y, N_X is the number of turns in coil X, and ΔI_X is the change in current in coil X.
Plugging in the values given, we have: M = (5mWb) / (200 * 4A)

M = 5mWb / 800A

M = 6.25μH. Therefore, the mutual inductance between the coils is 6.25μH.

(b) The coefficient of coupling (k) can be calculated using the formula:

k = M / √(L_X * L_Y)

Where k represents the coefficient of coupling, M is the mutual inductance, L_X is the self-inductance of coil X, and L_Y is the self-inductance of coil Y.
Substituting the given values: k = (6.25μH) / √((80mH) * (60mH))

k = 6.25μH / √(4.8mH^2)

k ≈ 0.447. Therefore, the coefficient of coupling between the coils is approximately 0.447.

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The work done in stretching the spring from its natural length to 10 inches beyond its natural length is 112 lb·in.

The work done in stretching a spring is given by the formula:

\(\[ W = \frac{1}{2} k (x_f^2 - x_i^2) \]\)

In this case, the spring is stretched 4 inches beyond its natural length, so the initial displacement is 4 inches. The force required to hold the spring at this displacement is 16 lb. We can use Hooke's Law to find the spring constant:

\(\[ k = \frac{F}{x_i} = \frac{16 \, \text{lb}}{4 \, \text{in}} = 4 \, \text{lb/in} \]\)

Now, we can calculate the work done in stretching the spring to 10 inches beyond its natural length:

\(\[ W = \frac{1}{2} (4 \, \text{lb/in}) \left( (10 \, \text{in})^2 - (4 \, \text{in})^2 \right) = 112 \, \text{lb·in} \]\)

Therefore, the work done in stretching the spring is 112 lb·in.

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The required force need to accelerate the rocket is 120000N while the force need to balance the weight is 117600 N and the total force is 237,600 Newtons.

Force is defined as an object which is in a state of motion then its rate of change of moment is called force.

Here,
We can use the equations of motion to solve this problem.

First, we can find the acceleration of the rocket

a = (v - u) / t

where v is the final velocity, u is the initial velocity (which is 0 in this case), and t is the time taken.

a = (400 m/s - 0) / 40 s

a = 10 m/s²

Next, we can use Newton's second law, F = ma,

F = ma

F = 12,000 kg x 10 m/s²

F = 120,000 N

So the force needed to accelerate the rocket is 120,000 Newtons.

The weight of the rocket is given by,

weight = m x g

weight = 12,000 kg x 9.8 m/s²

weight = 117,600 N

To balance the weight, a force of 117,600 N must be exerted on the rocket in the opposite direction.

total force = force to accelerate + force to balance weight

total force = 120,000 N + 117,600 N

total force = 237,600 N

So the engines must provide a total force of 237,600 Newtons.

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Answer:9.81 m/s2.

Explanation:

If a 2 kg object is falling at 3 m/s, at what rate is gravity working on the object ? The object is falling at 3 m/s. Gravity is working on the object at a rate of 9.81 m/s2.

The minimum thickness of the oil slick at that spot is 313 nm.

The refractive index of the oil is 1.3,

Therefore, the interface between the two rays will give two stage change or zero stage change.

The position for the constructive interface :

\(2 n t=m \lambda\)

Where, n = refractive index of oil

          m = order

So, for the minimum thickness (i.e. m = 1)

\(t_{\min }=\frac{\lambda}{2 n}\)

\(t_{\min }=\frac{750}{2 \times 1.2}\)

\(t_{\min }=313 \mathrm{~nm}\)

What is The refractive index?

A primal physical property of a substance, it is usually used to identify a particular substance, confirm its chastity, or scale its concentration is called refractive index.

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this question is centered around newtons second law f=ma

First you need to find the acceleration

\( acceleration = \frac{8 - 0}{22} \)

that gives you 0.36 recurring(on both numbers) m/s²

mass= force ÷ acceleration

\( mass = \frac{31}{0.36recurring} \)

mass= 85.25 kg

if you want to check

f=ma

85.25 × 0.36 recurring (remember its on both numbers not just 6) = 31 N

Answer:

Explanation:

Givens

F = 31 N

vi = 0

vf = 8 m/s

t = 22 seconds

m = ?

Formula

F = m ( vf - vi)/t

Solution

31 = m (8 - 0) / 22             Multiply both sides by 22

31 * 22 = 8 * m

8m = 682                          Divide by 8

m = 682 / 8

m = 85.25

Therefore, the center of mass of the system is approximately located at x ≈ 3.28 m.

The center of mass of a distribution of mass in space (sometimes referred to as the balance point) is the unique point at any given time where the weighted relative position of the distributed mass sums to zero.

For x-axis:

xcom=m1x1+m2x2+m3x3+...m1+m2+m3+...

For y-axis:

ycom=m1y1+m2y2+m3y3+...m1+m2+m3+...

xcom

and ycom

together will give the coordinates for the Centre of Mass of a system. This is the point to which a force may be applied to cause a linear acceleration without an angular acceleration.

To find the center of mass of the system, we need to consider the masses and their respective positions. The center of mass (x_com) can be calculated using the formula:

x_com = (m1 * x1 + m2 * x2 + m3 * x3) / (m1 + m2 + m3)

where m1, m2, and m3 are the masses, and x1, x2, and x3 are the corresponding positions.

Given:

Mass 1 (m1) = 12 kg, positioned at x1 = -3 m (3 m to the left)

Mass 2 (m2) = 15 kg, positioned at x2 = 2 m (2 m to the right)

Mass 3 (m3) = 20 kg, positioned at x3 = 8 m (8 m to the right)

Substituting these values into the formula, we have:

x_com = (12 kg * (-3 m) + 15 kg * (2 m) + 20 kg * (8 m)) / (12 kg + 15 kg + 20 kg)

x_com = (-36 kgm + 30 kgm + 160 kg*m) / 47 kg

x_com = 154 kg*m / 47 kg

x_com ≈ 3.28 m

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(a) The change in internal energy of the gas is 452 J, and (b) the energy that must be added to the gas by heat along the indirect path is 1752 J.

(a) The change in internal energy of a gas is given by the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:

ΔU = Q - W

where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system.

Along the diagonal path from I to F, no work is done by the system because the volume of the gas does not change. Therefore, W = 0.

The energy added to the gas by heat is Q = 452 J.

So, ΔU = Q - W = 452 J - 0 J = 452 J.

Therefore, the change in internal energy of the gas is 452 J.

(b) For the indirect path IAF, the change in internal energy of the gas is the same as in part (a), which is ΔU = 452 J.

Along the indirect path, the gas undergoes two processes:

From I to A, the volume of the gas is kept constant (isochoric process), so no work is done by the system. The heat added to the system is:

Q_1 = ΔU = 452 J

From A to F, the gas expands and does work on the surroundings. The work done by the gas is given by the area under the PV curve between points A and F. This area can be divided into a rectangular part and a triangular part:

The rectangular part represents the work done by the gas when it expands from A to C at constant pressure. The pressure is given by the value of P at point A, which is 4.5 × 10^5 Pa. The volume change is given by the difference in volume between A and C, which is 0.0025 m^3 - 0.0010 m^3 = 0.0015 m^3. Therefore, the work done by the gas during this process is:

W_1 = PΔV = (4.5 × 10^5 Pa)(0.0015 m^3) = 675 J

The triangular part represents the work done by the gas when it expands from C to F along the path CF. The pressure varies linearly with volume along this path, so the average pressure can be calculated as (P_C + P_F)/2. The volume change is given by the difference in volume between C and F, which is 0.0050 m^3 - 0.0025 m^3 = 0.0025 m^3. Therefore, the work done by the gas during this process is:

W_2 = (P_C + P_F)/2 × ΔV = (2.5 × 10^5 Pa)(0.0025 m^3) = 625 J

The total work done by the gas along the indirect path is:

W = W_1 + W_2 = 675 J + 625 J = 1300 J

Now, the energy that must be added to the gas by heat along the indirect path is:

Q = ΔU + W = 452 J + 1300 J = 1752 J

So, the energy that must be added to the gas by heat along the indirect path is 1752 J.

Therefore, The gas's internal energy changes by 452 J and heat must add 1752 J to the gas to make up the difference along the indirect path.

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The forces acting on the 0.100 kg ball as the child twirls it in a horizontal circle along the floor are tension and centripetal force.

Tension is the force exerted by the string on the ball, pulling it towards the center of the circle. This force is a result of the child holding the string and causing the ball to move in a circular path. The tension force keeps the ball from flying off in a straight line according to Newton's first law of motion.

Centripetal force is the net force that acts on the ball towards the center of the circle, causing it to move in a circular path. In this case, the tension in the string provides the centripetal force needed to keep the ball moving in a circle. Centripetal force is given by the formula F_c = m*v^2/r, where m is the mass of the ball (0.100 kg), v is its speed (2.0 m/s), and r is the radius of the circle (1.0 m). The centripetal force in this scenario is the same as the tension force in the string since there are no other forces acting horizontally on the ball.

In summary, the tension force in the string and the centripetal force keep the 0.100 kg ball moving in a horizontal circle with a speed of 2.0 m/s. These forces are necessary for maintaining the circular motion and preventing the ball from flying off in a straight path.

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Answer:

The true statement about the maximum speed of each car will be that the power of each car directly proportional to the speed of each car.

Explanation:

Given that,

Power of first car = 100 watt

Power of second car = 150 watt

Power of third car = 200 watt

We know that,

Power :

The power is directly proportional to the cube of speed.

In mathematically,

\(P\propto v^3\)

Where, P = power

v = speed

If the power is increased then the speed will be increase.

We need to find the maximum speed of car

According to question,

The power of car is maximum so the speed of car will be maximum.

For first car,

The power of first car is 100 watt.

So, the speed of first car will be 1000000 m/s.

For second car,

The power of second car is 150 watt.

So, the speed of second car will be 3375000 m/s.

For third car,

The power of third car is 200 watt.

So, the speed of first car will be 8000000 m/s.

Hence, The true statement about the maximum speed of each car will be that the power of each car directly proportional to the cube of speed of each car.

Answer:

If the car is initially travelling at u m/s, then the stopping distance d m travelled by ... the speed of the car at the instant the brakes are applied. ... Common usage will force us to depart from this later in the notes. ... The history of these equations is not absolutely clear, but we do have some ... Newton (1642–1727) and Leibniz.

Explanation:

hope this helped

The apparent depth of a fish floating motionless 7.10 cm under the surface of the water in a tank with a mirrored bottom can be determined using the concept of refraction. The index of refraction of water is given as 1.33.

Part A: The apparent depth of the fish when viewed at normal incidence to the water can be calculated using the formula for apparent depth: \(\[d_{\text{apparent}} = \frac{d_{\text{actual}}}{\text{refractive index}}.\]\)Substituting the given values, we have \(\[d_{\text{apparent}} = \frac{7.10}{1.33} = 5.34\] cm\). Therefore, the apparent depth of the fish is 5.34 cm.

Part B: When the fish is viewed through the mirrored bottom of the tank, we consider both the refraction of light at the air-water interface and the reflection from the mirror. The apparent depth of the reflection can be calculated using the same formula as in Part A, as the reflected light undergoes refraction at the air-water interface. Therefore, the apparent depth of the reflection of the fish in the bottom of the tank is also 5.34 cm.

In summary, the apparent depth of the fish floating motionless 7.10 cm under the surface of the water when viewed directly or through the mirrored bottom of the tank is 5.34 cm.

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The part of the microscope that the objectives are attached to is called the (C) nosepiece.

The nosepiece is a rotating mechanism located below the microscope's body tube. It holds the objectives, which are the lenses responsible for magnifying the specimen. The nosepiece typically has multiple positions, allowing the user to switch between different objective lenses for varying levels of magnification.

This convenient feature eliminates the need to manually remove and replace objectives when changing magnification. By rotating the nosepiece, different objectives can be brought into position above the specimen. This allows for quick and efficient adjustments in magnification without disrupting the viewing process.

Hence, the nosepiece plays a critical role in the microscope's functionality by providing a convenient way to switch between objectives and adjust the magnification level.

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Here is the complete question:

What is the name of the part of the microscope that the objectives are attached to? (Choose the best answer)

A. Ocular

B. Stage

C. Nosepiece

D. Arm