This string is 5 meters long. What is the measurement of one wavelength? (Just * type the number, not the units)

pic attached below

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Answer:

Frequency = 25 / 10 = 2.5 Hz

Explanation:

If the wave oscillates up and down 25 times in 10 seconds, then the frequency can be calculated as follows:

Frequency = Number of oscillations / Time

In this case, the number of oscillations is 25, and the time is 10 seconds. Therefore, the frequency is:

Frequency = 25 / 10 = 2.5 Hz

So the frequency of the wave is 2.5 Hertz, which means it completes 2.5 cycles or oscillations in one second.

The just float in volume of the piece of wood is 0.008 cubic meters.

Let's denote the volume of the piece of wood as V_wood, and the volume of the lump of aluminum as V_aluminum.

We can start by using the information given to write two equations:

Equation 1: (density of wood) * V_wood + (density of aluminum) * V_aluminum = (mean density) * (V_wood + V_aluminum)

Substituting the given values, we get:

400 * V_wood + 270 * V_aluminum = 100 * (V_wood + V_aluminum)

Simplifying and rearranging, we get:

300 * V_wood = 70 * V_aluminum

Equation 2: The total mass of the arrangement is the sum of the masses of the wood and aluminum:

(density of wood) * V_wood + (density of aluminum) * V_aluminum = (total mass)

Substituting the given values, we get:

400 * V_wood + 270 * V_aluminum = 0.01

Now we can use Equation 1 to eliminate V_aluminum from Equation 2:

300 * V_wood = 70 * V_aluminum

V_aluminum = (300/70) * V_wood

Substituting this into Equation 2, we get:

400 * V_wood + 270 * [(300/70) * V_wood] = 0.01

Simplifying and solving for V_wood, we get:

V_wood = 0.008 m^3

Therefore, the just float in volume of the piece of wood is 0.008 cubic meters.

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The current that is supplied by the batteries is 0.01333 Amp.

An electric current is a stream of charged particles, such as electrons or ions, moving through an electrical conductor. It is measured as the net rate of flow of electric charge through a surface or into a control volume.

Eeq = E1 + E2 ( batteries in series )

=> Eeq = 3+4.5 = 7.5 V

Veq = V1 + V2 + V3 ( resistances in series )

=> V3 = 7.5 - 2 - 3.5 = 2 V

=> current = V/R = 2/150 = 0.01333 Amp

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During the earliest stages of the universe, the only things that existed were subatomic particles such as protons, neutrons, and electrons. These particles came together to form the first atoms, which were primarily hydrogen and helium.

The universe was also filled with a hot, dense plasma of particles and radiation, known as the cosmic microwave background radiation.

Hence, as the universe expanded and cooled, these atoms and radiation would play a key role in the formation of galaxies, stars, and the larger structures we see today.

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The small ball loses 2.37% of its kinetic energy during the elastic head-on collision with the stationary bigger ball.

[tex]\frac{Kf}{Ki} =[/tex] [tex]1- [\frac{1}{2}][ \frac{M}{m}] [\frac{2mv^{2} }{[M+m]^{2} }[/tex]

M = 12m

Kf/Ki = 1 - (1/2)(12m/m)[(2mv^2)/(13m)^2]

Kf/Ki = 165/169

(1 - Kf/Ki) x 100% = (1 - 165/169) x 100% = 2.37%

So, the small ball loses 2.37% of its kinetic energy during the elastic head-on collision with the stationary bigger ball.

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The Ceres is (a).The largest asteroid identified to date is correct option.

With a diameter of around 590 miles (940 km), Ceres is the biggest asteroid between Mars and Jupiter. It is regarded as a minor planet, like Pluto, and was found in 1801 by Giuseppe Piazzi. Ceres is made of rock and ice, and due to its low density, it is likely that it contains a substantial amount of water ice inside.

The NASA Dawn spacecraft visited Ceres in 2015 and orbited the dwarf planet for more than three years, gathering useful information and taking stunning pictures of it.

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Answer:

Sound travels faster in liquids than in gases because molecules are more tightly packed. In fresh water, sound waves travel at 1,482 meters per second (about 3,315 mph). That's well over 4 times faster than in air!

Explanation:

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If enough experimental data supports a hypothesis, it is considered a well-supported scientific theory, but it is not considered to be 100% true or proven. Scientific theories are always open to further investigation and revision based on new evidence. Therefore, option C ("Is proven 100% true") is incorrect.

Option A ("Is accepted as true until proven false") is also incorrect because scientists do not accept a hypothesis as true until it has been rigorously tested and supported by a large body of evidence. Even then, scientists recognize that any scientific theory is subject to revision or falsification if new data or evidence emerges that contradicts it.

Option B ("Becomes an Observational Law") is also incorrect because scientific laws are typically descriptive, rather than explanatory. They describe what happens in a given set of circumstances, but they do not explain why it happens. Hypotheses and theories, on the other hand, attempt to explain why certain phenomena occur, and they are supported by experimental evidence.

Therefore, none of the options are completely accurate, but the most appropriate answer is that the hypothesis becomes a well-supported scientific theory.

A wall less than or equal to 14.7 m in height is the highest the player can clear.

Use the kinematic equations of motion to solve the problem. First, we can find the time it takes for the ball to travel 99.1 m horizontally:

d = vt

t = d / v

t = 99.1 m / 36.2 m/s

t ≈ 2.74 s

Now, use the vertical motion equation to find the maximum height the ball reaches:

y = yo + vot + (1/2)at²

where:

yo = 0.996 m (initial height)

vo = v sinθ = 36.2 m/s x sin(30°) ≈ 18.1 m/s (initial vertical velocity)

a = -9.81 m/s² (acceleration due to gravity, pointing downward)

t = 2.74 s (time of flight)

y = 0.996 m + 18.1 m/s x 2.74 s + (1/2) x (-9.81 m/s²) x (2.74 s)²

y ≈ 14.7 m

Therefore, the tallest wall the player can clear is a wall with a height less than or equal to 14.7 m.

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The force of buoyancy acting on the sample is 2.65 N.

If the sample's supporting string is cut, it will float to the surface of the water.

a. The buoyant force acting on the sample can be calculated using the formula Fb = ρVg, where ρ is the density of water, V is the volume of the submerged part of the sample, and g is the acceleration due to gravity. Since the sample is immersed in water, its volume is equal to the volume of water it displaces, which can be calculated using the formula V = l × w × h, where l, w, and h are the dimensions of the submerged part of the sample. Since the sample is cubical, all sides have the same length of 3.0 cm. Thus, the submerged part of the sample has a volume of V = 3.0 cm × 3.0 cm × 3.0 cm = 27.0 cm³ = 0.027 m³. The density of water is ρ = 1000 kg/m³, and the acceleration due to gravity is g = 9.81 m/s². Therefore, the buoyant force acting on the sample is:

Fb = ρVg = 1000 kg/m³ × 0.027 m³ × 9.81 m/s² ≈ 2.65 N

b. The sample will float to the surface of the water if its supporting string is cut. This is because the buoyant force acting on the sample is greater than its weight, which means that there is a net upward force on the sample. This net force causes the sample to accelerate upward, and it will continue to accelerate until it reaches the surface of the water. At the surface, the upward force is balanced by the weight of the sample, and it will float on the water.

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Complete question:

2. Suppose the metal sample in Figure 10.2 is immersed in water, is cubical with side 3.0 cm, and has a mass of 54 g. a. Calculate the buoyant force acting on the sample. b. Describe the behavior of the sample if its supporting string is cut. Explain how you arrive at your answer. balance arm water metal sample water overflow can beaker Figure 10.2 MEASUREMENTS

The weight of someone on Earth with a mass of 75kg is 750N.

option B.

The weight of someone on Earth is calculated using Newton's second law of motion.

W = mg

where;

For the mass is 75kg and the gravitational field strength on Earth is 10N/kg.

W = 75 kg x 10 N/kg

W = 750 N

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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:

Therefore, the voltage required is 20 volts

Explanation:

To find the voltage that will send a current of 5 amperes through a bell circuit with a resistance of 4 ohms, we can plug in the values into the equation and solve for V: V=IR

V=(5A)(4Ω)

V=20V

The resultant x and y components are 53.23 and 41.51, respectively.

What are the components?

To determine the resultant x and y components of the vectors, we can use the following equations:

Rx = ΣFx = F1x + F2x + ...

Ry = ΣFy = F1y + F2y + ...

where F1x, F2x, ... are the x components of the vectors, F1y, F2y, ... are the y components of the vectors, and ΣFx and ΣFy are the total x and y components of the resultant vector, R.

First, we need to find the x and y components of each vector:

Vector 1: magnitude = 26.5, angle = 56°

F1x = 26.5cos(56) = 14.28

F1y = 26.5sin(56) = 21.44

Vector 2: magnitude = 44, angle = 28°

F2x = 44cos(28) = 38.95

F2y = 44sin(28) = 20.07

Now we can add the x and y components of the vectors to find the total x and y components of the resultant vector:

ΣFx = F1x + F2x = 14.28 + 38.95 = 53.23

ΣFy = F1y + F2y = 21.44 + 20.07 = 41.51

Therefore, the resultant x and y components are 53.23 and 41.51, respectively.

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Answer:

a) To solve for the speed of the car, we can use the conservation of energy equation:

(1/2)mv^2 = mgh - (1/2)kx^2

where m is the mass of the car, v is the speed of the car, g is the acceleration due to gravity, h is the height of the car above the starting point, k is the spring constant, and x is the distance the spring is compressed.

Plugging in the given values, we get:

(1/2)(0.250 kg)v^2 = (0.250 kg)(9.81 m/s^2)(0.110 m) - (1/2)(450.0 N/m)(0.0600 m)^2

Solving for v, we get:

v = 1.23 m/s

Therefore, the speed of the car when it reaches 0.110 m above the starting point is 1.23 m/s.

b) To estimate the power expended by a child raising a 0.25-kg toy 0.75 m in 2.0 s, we can use the work-energy theorem:

W = ΔE = mgh

where W is the work done, m is the mass of the toy, g is the acceleration due to gravity, h is the height the toy is raised, and ΔE is the change in energy.

Plugging in the given values, we get:

W = (0.25 kg)(9.81 m/s^2)(0.75 m) = 1.84 J

Next, we can use the definition of power:

P = W/t

where P is power, W is work, and t is time.

Plugging in the given values, we get:

P = 1.84 J / 2.0 s = 0.92 W

Finally, we can convert watts to horsepower:

1 hp = 746 W

0.92 W = 0.92 / 746 hp = 0.0012 hp

Therefore, the power expended by a child raising a 0.25-kg toy 0.75 m in 2.0 s is approximately 0.0012 hp.

The gauge pressure at point 1 = 145.2 kPa.

The equation states that the sum of the pressure, kinetic energy, and potential energy per unit volume of a fluid is constant along a streamline, in the absence of external forces like friction or viscosity.

The principle of continuity of mass flow rate and Bernoulli's equation, which connects pressure, velocity, and height of a fluid in a flow, can be used to solve this issue.

With the provided data, we can first determine the soft drink's mass flow rate:

mass flow rate =[tex](220 cans/min) x (0.355 L/can) x (1 kg/L) x (1 min/60 s)[/tex] = [tex]1.235 kg/s[/tex]

Then, we may compare the soft drink's velocities at positions 1 and 2 using the principle of continuity:

[tex]A1 v1 = A2 v2[/tex]

where A1 and A2 are the pipe's cross-sectional areas at points 1 and 2, respectively, and v1 and v2 are the soft drink's velocities at those same locations. When we solve for v1, we get:

[tex]v1 = (A2/A1) v2 = (8.00 cm^2)/(2.00 cm^2) v2 = 4 v2[/tex]

Now, we can link the pressures, velocities, and heights of the soft drink at points 1 and 2 using the Bernoulli's equation:

P1 + (1/2)ρv1²+ ρgh1 = P2 + (1/2)ρv2²+ ρgh2

where P1 and P2 are the gauge pressures at points 1 and 2, respectively; ρ is the soft drink's density; g is gravity's acceleration; and h1 and h2 are the heights of the respective points 1 and 2 above a reference level.

As the soft drink contains primarily water, we may calculate its density using the formula: = 1000 kg/m3. Moreover, we can decide to set the reference level at point 2, making h2 = 0.When the supplied values and the velocity relation are substituted, we obtain:

[tex]P1 + (1/2)(1000 kg/m^3)(16 v2^2) + (1000 kg/m^3)(9.81 m/s^2)(1.35 m)[/tex] =[tex]152 kPa + (1/2)(1000 kg/m^3)(v2^2)[/tex]

By condensing and figuring out P1, we get at:

[tex]P1 = 152 kPa + (1/2)(1000 kg/m^3)(15 v2^2) - (1000 kg/m^3)(9.81 m/s^2)(1.35 m)[/tex] = [tex]152 kPa + 6.75 v2^2 - 13.33 kPa[/tex]

We must now locate v2. We can link the velocity and the cross-sectional area at point 2 using the mass flow rate relation and the density of the soft drink:

[tex]A2 v2[/tex] = (mass flow rate)/ρ =[tex]1.235/(1000 kg/m^3)[/tex]= [tex]0.001235 m^3/s[/tex]

Upon solving for v2, we obtain:

v2 =[tex]0.001235 m^3/s / (8.00 cm^2 / 10000 cm^2)[/tex]= [tex]0.1544 m/s[/tex]

Last but not least, we can insert this value into the P1 expression to obtain:

[tex]P1[/tex] =[tex]152 kPa + 6.75 (0.1544 m/s)^2 - 13.33 kPa[/tex] = [tex]145.2 kPa[/tex]

As a result, point 1's gauge pressure is 145.2 kPa.

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Answer:

Explanation:

Here we go !!!!

I hope this information helps!

Answer:

When unpolarized light is reflected off a non-metallic surface at a specific angle of incidence, the reflected light becomes polarized. This angle of incidence is known as Brewster's angle and can be calculated using the formula:

tan θp = n

where θp is Brewster's angle and n is the refractive index of the medium the light is passing into.

In this case, the light is passing from air (which has a refractive index of approximately 1) into water with a refractive index of 1.3. Therefore, we can plug in the values into the formula and solve for θp:

tan θp = 1.3

θp = tan^-1(1.3)

Using a calculator, we get θp = approximately 53.1 degrees.

Therefore, light reflected from water will be completely polarized at an angle of incidence of approximately 53.1 degrees.

The pressure of the system, that the large piston produces a force of 37775 pounds is 3007.56 pound / in²

The following data were obtained from the question:

Pressure is defined as force per unit area as shown by the following formula

Pressure  = Force / Area

Inputting the value of the force and area, we have

Pressure of system = 37775 / 12.56

Pressure of system = 3007.56 pound / in²

Thus, from the calculation made above, we can conclude that the pressure of the system is 3007.56 pound / in²

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Answer:

100

Explanation:

F = m * a

Given that the mass (m) of the toy car is 20 grams (or 0.02 kilograms, since 1 kilogram = 1000 grams) and the acceleration (a) is 5 m/s^2, we can plug these values into the formula to calculate the force (F):

F = 0.02 kg * 5 m/s^2 = 0.1 N

So, the amount of force required to accelerate the 20 gram toy car at 5 m/s^2 is 0.1 N, which is equivalent to 100 N when rounded to the nearest whole number. Therefore, the correct answer is 100 N.

Answer:

The first set of errors, which include friction and imperfect elasticity, are systematic errors because they arise from consistent factors that affect the measurements in a predictable way. These errors will be present in every trial of the experiment and will cause a consistent deviation from the true value.

The second set of errors, which include small amounts of friction and the initial velocity of glider 2, are also systematic errors because they arise from consistent factors that affect the measurements in a predictable way. These errors will also be present in every trial of the experiment and will cause a consistent deviation from the true value.

Explanation:

Answer:

Volcanic mountains can be formed by a convergent boundary.

Answer:

To solve for the amount of stretch in the nylon rope, we can use the equation for the elongation (stretch) of a rope under tension:

ΔL = FL / AE

where ΔL is the change in length of the rope, F is the force on the rope, L is the original length of the rope, A is the cross-sectional area of the rope, and E is the Young's modulus of the rope.

First, we need to find the force on the rope. This is equal to the weight of you and the rope:

F = mg = (75.0 kg + mass of rope)g

Next, we need to find the cross-sectional area of the rope. The diameter of the rope is given as 0.600 cm, so the radius is 0.300 cm = 0.00300 m. Therefore, the cross-sectional area is:

A = πr^2 = 2.83 × 10^-6 m^2

Now we can plug in the given values and solve for ΔL:

ΔL = FL / AE = [(75.0 kg + mass of rope)g](23.0 m) / [(2.83 × 10^-6 m^2)(5.00 × 10^9 N/m^2)]

We can simplify this expression by using the fact that the mass of the rope is much smaller than your mass, so we can assume that the force on the rope is equal to your weight:

ΔL = (mg)(23.0 m) / (A E) = (75.0 kg)(9.81 m/s^2)(23.0 m) / [(2.83 × 10^-6 m^2)(5.00 × 10^9 N/m^2)]

Solving for ΔL, we get:

ΔL = 0.068 cm

Therefore, the nylon rope stretches by approximately 0.068 cm when you hang 23.0 m below a rock outcropping.

The moment of the weight of the 1.50 kg mass about the CG of the 5.00 kg object will be considered as part of the object.

The center of gravity of the additional 1.50 kg mass should be located 1.60 cm to the right of the current CG of the 5.00 kg object.

What is the moment ?

We can use the principle of moments to determine the location of the center of gravity of the additional 1.50 kg mass that needs to be glued on.

The principle of moments states that the sum of the moments of all the forces acting on an object is equal to the moment of the resultant force about any point. In this case, we can take moments about the current center of gravity (CG) of the 5.00 kg object to determine the location of the center of gravity of the additional 1.50 kg mass.

The moment of a force about a point is given by the product of the magnitude of the force and the perpendicular distance from the point to the line of action of the force. Since the gravitational force acting on the masses is the only force acting on the system, we can take moments of the weight of each mass about the CG of the 5.00 kg object.

The moment of the weight of the 5.00 kg object about its CG is zero, since the CG is the point about which the object is in rotational equilibrium. Therefore, we only need to consider the moment of the weight of the 1.50 kg mass about the CG of the 5.00 kg object.

The distance between the CG of the 5.00 kg object and the desired new CG is 1.60 cm to the left, so we need to find the distance between the current CG and the new CG. Let x be the distance from the current CG to the new CG.

The moment of the weight of the 1.50 kg mass about the current CG is:

(1.50 kg)(9.81 m/s²)(x) = 14.72x Nm

The moment of the weight of the 1.50 kg mass about the new CG is:

(1.50 kg)(9.81 m/s²)(1.60 cm + x) = 14.72x + 23.54 Ncm

Setting these two moments equal, we have:

14.72x Nm = 14.72x + 23.54 Ncm

Converting the units, we get:

0.1472x m = 0.2354 m

x = 1.60 cm (to two significant figures)

Therefore, the center of gravity of the additional 1.50 kg mass should be located 1.60 cm to the right of the current CG of the 5.00 kg object.

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The period (in second) of the 0.95 Kg mass, given that it has an amplitude of 0.21 m and an angular velocity of 9.5 rad/s is 0.66 second

From the question given above, the following data were obtained:

The period of the mass can be obtained as shown below:

ω = 2π/ T

9.5 = (2 × 3.14) / T

9.5 = 6.28 / T

Cross multiply

9.5 × T = 6.28

Divide both sides by 9.5

T = 6.28 / 9.5

T = 0.66 second

Thus, from the above calculation, we can conclude that the period of the mass is 0.66 second

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The resulting angular speed of the flat, uniform circular disk is 0.237 rad/s.

Angular momentum is a fundamental concept in physics that describes the rotational motion of an object around an axis. It is defined as the product of the moment of inertia of an object and its angular velocity with respect to a chosen axis.

We can use conservation of angular momentum to solve this problem. The initial angular momentum of the disk is zero because it is stationary. The final angular momentum of the system (disk + person) is:

L = Iω

where I is the moment of inertia of the disk and person about the axis of rotation, and ω is the resulting angular speed of the disk.

The moment of inertia of the disk about its axis is:

I_disk = (1/2)mr²

where the disk's radius is r and its mass is m.  Substituting the given values, we get:

I_disk = (1/2)(150 kg)(5.44 m)² = 2226.24 kg·m²

The moment of inertia of the person about the axis can be approximated as:

I_person = mr²

where r is the distance from the axis to the person. Substituting the given values, we get:

I_person = (47.0 kg)(1.54 m)² = 109.64 kg·m²

The total moment of inertia of the system is:

I = I_disk + I_person = 2226.24 kg·m² + 109.64 kg·m² = 2335.88 kg·m²

The final angular momentum of the system is:

L = Iω

where ω is the resulting angular speed of the disk. Substituting the given values, we get:

(2335.88 kg·m²)ω = (197.64 kg·m²/s)(2.80 m/s)

Solving for ω, we get:

ω = (197.64 kg·m²/s)(2.80 m/s) / (2335.88 kg·m²) = 0.237 rad/s.

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The speed of sound in air can be calculated using the following formula:speed of sound = distance / timeIn this case, we know that the distance between the person and the wall is 40 meters, and the time it takes for the echo to be heard after the original shot is 0.24 seconds. However, we need to account for the fact that sound travels to the wall and back, so the total distance the sound wave covers is twice the distance to the wall, or 2 x 40 = 80 meters.Thus, using the formula above, we can calculate the speed of sound:speed of sound = distance / time = 80 m / 0.24 s ≈ 333.33 m/sTherefore, the speed of sound in air is approximately 333.33 meters per second.