When observed from earth, the wavelengths of light emitted by a star are shifted toward the red end of the electromagnetic spectrum. This redshift occurs because the star is:.

Answer:

Reverberation

Explanation:

Answer:

\(f = mg \\ = 70 \times 9.8 = |f| \)

Answer:

3504.4 kgm/s

Explanation:

First, convert km/hr to m/s:

19 km/hr x 1000m/1km x 1/60min x 1/60s = 5.28 m/s

p = mv

p = (664 kg)(5.28 m/s) = 3504.4 kgm/s

Speed and velocity are similar in term of their unit. the speed of the jet is 42 m/s

What is Speed ?

Speed can simply be defined as how fast an object moves. It is the distance covered per time taken.

Given that acceleration = 2.8 m/s² and If after take-off, the jet continues to accelerate at the same rate for another 15 s, to know how fast it will be going at that time, we will use acceleration formula.

Acceleration = Velocity / time

speed = acceleration × time

speed = 2.8 × 15

Speed = 42 m/s

Therefore, the speed or the velocity of the jet is 42 m/s

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By the half-life of radioactive, the Iodine-131 experienced 15 half-lives.

We need to know about the half-life of the radioactive elements to solve this problem. The radioactive element will decay over time and follow the equation

N = No(1/2)^(t/t'')

where N is the final quantity, No is the initial quantity, λ is the decaying constant, t is time and t'' is the half-life of a radioactive element.

From the question above, we know that

t'' = 8 min

t = 120 min

Calculate the remaining mass

N = No(1/2)^(t/t'')

N = No(1/2)^(120/8)

N = No / 32768

As we know that half-lives is (No / (2^n)), Hence

N = No / 32768

N = No / 2^(²log(32768))

N = No / 2¹⁵

Hence, the Iodine-131 experienced 15 half-lives.

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To determine the mass of aluminum plated onto an object, we need to use Faraday's law of electrolysis, which states that the amount of substance deposited at an electrode is directly proportional to the quantity of electricity passed through the electrolyte.

First, we need to calculate the total charge passed in coulombs (C):

Charge (C) = Current (A) × Time (s)

Since the time is given in minutes, we need to convert it to seconds:

Time (s) = 728 minutes × 60 seconds/minute = 43,680 seconds

Charge (C) = 5.94 A × 43,680 s = 259,315.2 C

Next, we need to convert the charge to moles of electrons using Faraday's constant (F), which is the charge of one mole of electrons:

Moles of electrons = Charge (C) / Faraday's constant (F)

Faraday's constant (F) = 96,485 C/mol (approximately)

Moles of electrons = 259,315.2 C / 96,485 C/mol ≈ 2.687 mol

Since the balanced equation for the deposition of aluminum is 2 Al³⁺ + 6 e⁻ → 2 Al, it tells us that 6 moles of electrons are required to deposit 2 moles of aluminum.

Therefore, the moles of aluminum deposited = Moles of electrons / 6 = 2.687 mol / 6 ≈ 0.448 mol

The molar mass of aluminum is approximately 26.98 g/mol. Therefore, the mass of aluminum plated onto the object is:

Mass = Moles × Molar mass = 0.448 mol × 26.98 g/mol ≈ 12.08 g

Hence, approximately 12.08 grams of aluminum can be plated onto the object in 728 minutes at a current of 5.94 A.

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Answer:

The value of \(a\cdot b\times c\) is 0.

Explanation:

The three vectors provided are:

\(a=2i-3j+7k\\b= i+2k\\c=j-k\)

The value to be computed is:

\(a\cdot b\times c\)

Compute the value of (b × c) as follows:

\(b\times c=\left|\begin{array}{ccc}i&j&k\\1&0&2\\0&1&-1\end{array}\right|\)

       \(=\left|\begin{array}{cc}0&2\\1&-1\end{array}\right|i-\left|\begin{array}{cc}1&2\\0&-1\end{array}\right|j+\left|\begin{array}{cc}1&0\\0&1\end{array}\right|k\\\\=-2i+1j+1k\)

Now compute the value of \(a\cdot b\times c\) as follows:

\(a\cdot b\times c=(2i-3j+7k)\cdot (-2i+j+k)\)

           \(=(2\cdot -2)+(-3\cdot 1)+(7\cdot1)\\\\=-4-3+7\\\\=0\)

Thus, the value of \(a\cdot b\times c\) is 0.

Skateboarders move more quickly and more slowly depending on the gravitational field around them.

Skateboarders go more slowly and more quickly depending on the amount of friction they encounter. The skater initially has no kinetic energy, only potential energy. The skater had used all of the energy that she will ever have. Since the track is frictionless, no energy is converted to thermal energy throughout the route from point A to point B. The skater has greater potential energy the higher he is. His kinetic energy increases and his potential energy falls as his height decreases. Ice skating is a good illustration of sliding friction.

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Answer:

Orbits are the result of a perfect balance between the forward motion of a body in space, such as a planet or moon, and the pull of gravity on it from another body in space, such as a large planet or star

Answer:

800N is the answer

Explanation:

hope it helps

The force between the objects is 200 N and when the distance between the objects is halved then the force between them will be equal to 800 N.

A force in physics is an input that has the power to change an object's motion. A mass-containing object's velocity can vary, or accelerate, as a result of a force. Intuitively, a push or a pull can also be used to describe forces.

Being such a vector quantity, a force does have magnitude and direction. The SI unit metric newton is used to measure it (N). The letter F stands for force.

According to Newton's second law's original formulation, an object's net force is equal to the speed that its momentum is changing over time.

As per the given information in the question,

Force, F = 200 N

F1 = G × m₁m₂/ r²

The final force F' = G × m₁m₂/ (r/2)²

= 4G × m₁m₂/ r²

= 4F

So, when the distance between objects is halved, then the force between them increases to 4 times.

So, the force will be,

4F = 4 × 200

= 800 N.

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A promising evidence of a habitat that might support life on the planet Mars is water.

There is evidence supporting the existence of liquid water on Mars from numerous sources. The finding of repeated black streaks on Martian slopes and the presence of hydrated minerals suggest the potential of seasonal or location-specific briny water flows. A necessary component of life as known is liquid water.

Additionally, Mars possesses underground ecosystems that could provide defence against radiation and severe surface conditions. Researchers have found evidence of ancient underground hydrothermal systems as well as beneath ice. These settings might offer consistent conditions for the development of microbial life.  Methane gas has been found in the Martian atmosphere, along with variations over time, and this has led to questions regarding its origin. Both geological and biological processes can result in the production of methane.

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I think you're having trouble with it only because of how you're trying to look at it.

Don't try to imagine "spreading force" over time. Just think about how long the force lasts !

Like . . .

If you push your doll stroller for 5 seconds, you'll get it up to a certain speed.

Next time, if you push it again with the same force, but you keep pushing for TEN seconds, you'll get it up to a faster speed.

Technically:

(force • time) = impulse.

You gave it more impulse the second time, so it got more momentum from you.

The rate at which the water level is rising is 2/7π ft/min. When rounded to three decimal places, the answer is 0.358 ft/min.

The given container has the shape of an open right circular cone.

The container has a radius of 4 feet at the top, and its height is 12 feet. The volume V of a cone with radius r

height h is given by the formula as:

V=\frac{1}{3}\pi r^2 h Here, r = 4 ft, h = 12 ft, and V = 150 ft³.

Substituting these values in the above formula, we get:$

150 = \frac{1}{3}\pi \cdot 4^2 \cdot 12150 = \frac{1}{3}\pi \cdot 16 \cdot 12150 = \frac{1}{3}\pi \cdot 192\pi = \frac{150\cdot 3}{192} = \frac{25}{4} Now, we need to find the rate of change of the height of the water (dh/dt) when the height of the water is 5 ft.

We have:

V = \frac{1}{3}\pi r^2 h\frac{dV}{dt} = \frac{1}{3}\pi \cdot 2rh \cdot \frac{dh}{dt} + \frac{1}{3}\pi r^2 \frac{dh}{dt}\text

Now, substituting the given values,

we get:

6 = \frac{1}{3}\pi \cdot 2\cdot 4 \cdot 5 \cdot \frac{dh}{dt} + \frac{1}{3}\pi 4^2 \cdot \frac{dh}{dt}6 = \frac{5\pi}{3} \cdot \frac{dh}{dt} + \frac{16\pi}{3}\cdot \frac{dh}{dt}6 = \frac{21\pi}{3}\cdot \frac{dh}{dt}\frac{dh}{dt} = \frac{6}{21\pi}=\frac{2}{7\pi}

Therefore, the rate at which the water level is rising is 2/7π ft/min. When rounded to three decimal places, the answer is 0.358 ft/min. Hence, the correct option is (A).

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When the tunnel is running, a pitot tube mounted in the test section measures 2.4 atm, the reservoir pressure for the wind tunnel is approximately 0.585 atm.

To solve this problem, we can use the isentropic relations for a compressible flow through a nozzle. Assuming the flow is adiabatic and reversible (i.e., isentropic), we can relate the pressure ratio to the area ratio through the following equation:

(p2/p1) = (A1/A2)^γ

where p1 and p2 are the pressures at the throat and exit, respectively, A1 and A2 are the throat and exit areas, and γ is the specific heat ratio of the gas.

In this problem, the area ratio is given as A2/A1 = 5. The pitot tube measures the pressure at the exit, which we can call p2. We want to find the reservoir pressure, which we can call p1.

We can rearrange the above equation to solve for p1:

p1 = p2 / (A2/A1)^γ

Substituting the given values and assuming γ = 1.4 for air, we get:

p1 = 2.4 atm / 5^1.4

= 0.585 atm

Therefore, the reservoir pressure for the wind tunnel is approximately 0.585 atm.

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The Start fraction 5\(\pi\) over 6 end fraction radians converted to degrees is 150°.

5\(\pi\)/6

To convert to degrees, multiply by 180/\(\pi\)

5\(\pi\)/6 * 180/\(\pi\)

5 * 180/6

5 * 30 = 150°

Degrees are a unit of measurement used to quantify angles and temperature. In geometry, a degree is one 360th of a complete circle, and it is represented by the symbol °. When it comes to temperature, a degree is used to measure the difference between two temperatures on the Celsius or Fahrenheit scales. One degree Celsius is equivalent to one degree on the Kelvin scale, while one degree Fahrenheit is approximately 0.556 degrees Celsius.

Degrees are an important concept in many fields of study, including mathematics, physics, and engineering. They are used to measure angles in shapes and the direction of movement in trigonometry, as well as to measure temperature in science and engineering applications.

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Complete Question: -

What is start fraction 5 pi over 6 end fraction radians converted to degrees? if necessary, round your answer to the nearest degree.

a. 30°

b. 90°

c. 150°

d. 900°

RS is a composition of rotation and reflection in the Euclidean plane E². The precise description of RS depends on the specific properties of the rotation R and reflection S.

In general, if R and S have the same axis or line of symmetry, the composition RS results in a translation. If R and S have intersecting lines of symmetry, RS yields a glide reflection. If R and S have perpendicular lines of symmetry, RS produces a rotation.

It is important to consider special cases, such as parallel lines of symmetry, coinciding axes, or perpendicular lines of reflection, as they may lead to different outcomes. The classification of isometries in E² involves understanding how rotations and reflections combine to create different transformations in the plane.

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Data streams are continuous flows of data that can be used to capture, process, and analyze real-time information.

Data streams are typically handled by streaming data processing engines. These engines process and analyze the data as soon as it arrives, allowing for real-time insights and decision-making.

They are used to capture data from a variety of sources, such as sensors, web services, databases, and other online sources.

Data streams are handled in real-time, meaning that they are processed and analyzed as soon as they arrive.

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Answer:

The answer is probably A

Answer:

The elastic limit will exceed the max stretch, and raising the stress to further strain the elastic would tear it.

Explanation:

The car is at rest at \( t = 0 \) seconds, and it reaches a position of 80 feet at \( t = 4 \) seconds.

According to the given table, the position of the car at various times is measured. At \( t = 0 \) seconds, the car is at rest, as indicated by the position of 0 feet. This means that the car has not started moving yet.

As time progresses, we observe that the position of the car increases. For example, at \( t = 2 \) seconds, the car has moved to a position of 40 feet. This indicates that the car is covering a certain distance during each time interval.

At \( t = 4 \) seconds, we see that the car reaches a position of 80 feet. This means that in the time span of 4 seconds, the car has covered a distance of 80 feet.

The given information allows us to understand the relationship between time and position of the car. By analyzing the changes in position over time, we can determine the speed or velocity of the car at different instances.

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Explanation:

A car travels north on the highway for 125 kilometers, then turns west for 75 kilometers, then makes a u-turn and travels south for 125 kilometers.

A. To determine the displacement of the car, you would need to calculate the final position of the car relative to its initial position. In this case, the car traveled north for 125 km, then west for 75 km, and then south for 125 km. This means the car ends up in the same place it started (125 km south of its starting point), so the displacement would be 0 km.

Anthony has 25 kilograms of sand, while Juan has 60 pounds of sand. To determine which of the two boys has the most amount of sand, we need to convert the weight of the sand from pounds to kilograms. 60 pounds is approximately 27.2 kilograms. So Juan has 27.2 kilograms of sand and Anthony has 25 kilograms of sand. Anthony has more sand.

A horse runs with a velocity of 15 km/hr towards the north for 90 minutes. To determine the displacement of the horse, you would need to calculate the distance traveled by the horse. You can use the formula: Displacement = velocity x time. In this case, the velocity is 15 km/hr and the time is 90 minutes (converted to hours). So, Displacement = 15 x (90/60) = 22.5 km towards the North.

The kernel (null space) of a linear map L from a vector space V to a vector space W is a subspace of V.

To prove that the kernel (null space) of a linear map L from a vector space V to a vector space W is a subspace of V, we need to show three things: closure under vector addition, closure under scalar multiplication, and the presence of the zero vector.

1. Closure under vector addition:

Let u and v be vectors in the kernel of L, denoted as u, v ∈ ker(L). This means L(u) = 0 (the zero vector in W) and L(v) = 0. We want to show that u + v is also in the kernel of L.

Using the linearity of L, we have:

L(u + v) = L(u) + L(v) = 0 + 0 = 0

This shows that u + v is mapped to the zero vector by L, thus u + v is in the kernel of L.

2. Closure under scalar multiplication:

Let u be a vector in the kernel of L, denoted as u ∈ ker(L). This means L(u) = 0. We want to show that αu is also in the kernel of L for any scalar α.

Using the linearity of L, we have:

L(αu) = αL(u) = α0 = 0

This shows that αu is mapped to the zero vector by L, thus αu is in the kernel of L.

3. Presence of the zero vector:

The zero vector in V, denoted as 0v, is always in the kernel of any linear map because L(0v) = 0 in the codomain W.

Therefore, we have shown that the kernel of L satisfies the three conditions to be a subspace of V: closure under vector addition, closure under scalar multiplication, and the presence of the zero vector. Hence, the kernel of L is indeed a subspace of V.

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A maintenance expenditure should be capitalized if it increases the salvage value of the asset or extends the useful life of the asset.

An expenditure is a payment made in return for a product or service. Capital expenditure is money spent by a company on long-term assets like equipment and buildings.

Capitalizing refers to recording a cost or expense on the balance sheet for a future period rather than recognizing it immediately in the current period.

Capitalizing expenditure means the company will recognize the expenditure as an asset, which will be amortized over its useful life as opposed to expenses in the current period.

Therefore, a maintenance expenditure should be capitalized if the expenditure increases the extends the useful life of the asset.

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The electric field strength at a point inside the insulation, 2.6 mm from the axis of the wire, is approximately 0.449 N/C.

To calculate the electric field strength at a point inside the insulation, we can use Gauss's law and consider a cylindrical Gaussian surface centered on the wire. Since the wire has a linear charge density of 250 nC/m, the charge enclosed within the Gaussian surface can be calculated by multiplying the linear charge density by the length of the Gaussian surface.

Substituting the given values into the formula, we find that the charge enclosed (Qin) is equal to (250 x 10^-9 C/m) multiplied by (2.6 x 10^-3 m), resulting in approximately 650 x 10^-12 C.

Next, we apply Gauss's law, which states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of the material (ε). In this case, since we are within the insulation, we use the permittivity of the insulation material, which is given as 2.5ε0. Here, ε0 represents the vacuum permittivity.

Considering the cylindrical symmetry, the electric field is radial and points outward from the wire. By applying Gauss's law, we can relate the electric field (E) to the charge enclosed (Qin) and the radius (r) of the Gaussian surface. The formula for the electric field is E = Qin / (2πεr).

Substituting the values, we have E = (250 x 10^-9 C/m) / (2π x 2.5ε0 x 2.6 x 10^-3 m). Simplifying further, we can express the electric field as E ≈ 0.449 N/C.

Therefore, the electric field strength at a point inside the insulation, 2.6 mm from the axis of the wire, is approximately 0.449 N/C. This value represents the magnitude of the electric field and indicates the strength and direction of the electric field at that specific location within the insulation.

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The electric field must be zero inside a conductor in electrostatic equilibrium, but not inside an insulator. It turns out that we can still apply Gauss's law to a Gaussian surface that is entirely within an insulator by replacing the right-hand side of Gauss's law, Qin/ε0, with Qin/ε , where ε is the permittivity of the material. (Technically, ε0 is called the vacuum permittivity.) Suppose a long, straight wire with linear charge density 250 nC/m is covered with insulation whose permittivity is 2.5ε0.

What is the electric field strength at a point inside the insulation that is 2.6 mm from the axis of the wire?

Answer:

A fumbled ball may be recovered and advanced by either team

hope it helps

The net effect of the sodium-potassium pump is to create a transmembrane electrical potential.

Sodium-potassium pump is a process in which sodium ions (Na+) are transported out of the cell while potassium ions (K+) are transported into the cell. The net effect of this process is that there is a greater concentration of K+ ions inside the cell and a greater concentration of Na+ ions outside the cell.

This creates an electrochemical gradient across the cell membrane. This gradient is what allows cells to carry out a variety of functions such as transmitting signals and carrying out metabolic processes. The electrochemical gradient is also what drives the movement of other molecules across the cell membrane.

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Answer:

3.77202

Explanation:81.9 kg×9.8N/kg÷0.0021 Pa ×(1atm/101325)=3.77202

The pressure exerted between her toes and the ground is 3.77 atm.

The stress at a location within a confined fluid or the perpendicular force per unit area are the two definitions of pressure used in the physical sciences. A 42-pound box with an area of 84 square inches at the bottom will apply pressure on a surface equal to the force divided by the surface area.

Given:

An 81.9 kg person stands on her toes. The surface area of her toes in contact with the ground is only 0.00210 m²,

Calculate the pressure on the toe as shown below,

The pressure on the toe = Force of man / area of toe

The pressure on the toe = mass of man × 9.8 / 0.00210

The pressure on the toe = 81.9 × 9.8 / 0.00210

The pressure on the toe = 802.62 / 0.00210

The pressure on the toe = 382200 N/m²

The pressure on the toe = 382200 / 101325 (1 atm = 101325 N / m²)

The pressure on the toe = 3.77 atm,

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