One acre is equal to 43560ft².How much is it in m²?​

Answer:

Asexual production they can be eukaryotes or prokaryotes

Explanation:

The strength of the resulting magnetic field at a distance of 61.1 cm from the wire is approximately 2.97 × 10⁻⁵ T.

To find the strength of the resulting magnetic field at a distance of 61.1 cm from the wire with a 27.5 A current flowing through it, we will use the following formula:

Magnetic field strength (B) = (μ₀ * I) / (2 * π * r)

where,
- B is the magnetic field strength
- μ₀ is the permeability of free space (4π × 10⁻⁷ T m/A)
- I is the current in the wire (27.5 A)
- r is the distance from the wire (61.1 cm, which is 0.611 m)


1. Convert the distance from cm to m: 61.1 cm = 0.611 m
2. Apply the formula: B = (4π × 10⁻⁷ T m/A * 27.5 A) / (2 * π * 0.611 m)

3. Simplify and solve for B:
  B = (4π × 10⁻⁷ T m/A * 27.5 A) / (2 * π * 0.611 m)
  B ≈ (1.21 × 10⁻⁶ T m * 27.5 A) / (1.222 m)
  B ≈ 2.97 × 10⁻⁵ T

At a distance of 61.1 cm from the wire, the resulting magnetic field has a strength of roughly 2.97 × 10⁻⁵ T.

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Answer:

The answer is water

Explanation:

it has a netural ph of 7.0

why is the only option milk?

The Ph for milk is about 6.5 to 6.7,  it varies becauses milks gets sour when it rots.

Answer:

In both scenarios the speed would increase

Answer:

What would happen to the speed of a sound wave if it moved from ocean water to air?

✔ It would slow down.

What would happen to the speed of a sound wave if it moved from a cooler room to a warmer one?

✔ It would speed up.

:)

The paper should be placed at the focal point of the lens to obtain a sharp image.

The focal point of a lens is the point where parallel rays of light converge or diverge after passing through the lens. Placing the paper at this point ensures that the incoming light rays from the object on the other side of the lens converge precisely onto the paper, resulting in a focused and sharp image.

The distance between the lens and the paper should be equal to the focal length of the lens. The unit of focal length is typically measured in millimeters (mm) or meters (m), depending on the scale of the lens. Therefore, the paper should be positioned at the focal point of the lens to achieve a sharp image, with the distance measured in units of length such as millimeters or meters.

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Answer:

erosion

Explanation:

because i just got it right on study island

1)the kinetic energy of 150 kg object that is moving at 7 m/s is 3675 J.

2)velocity of 60 kg object if its kinetic energy is 1080 J  will be  6 m/s.

3)the gravitational potential energy of a 10 kg object that is raised to a place of 3.0 m high is 294 J.

4) you must lift the 5.25 kg object to a height of 14.04 m if the gravitational potential energy is increased by 729.56 J

1. An item weighing 150 kg and travelling at 7 m/s has kinetic energy.

The equation for kinetic energy is KE= 1/2mv2, where m stands for mass and v for velocity.

When the supplied values are substituted into the formula, we obtain KE = 1/2mv2 = 1/2(150 kg)(7 m/s)² = 3675 J

Therefore, the 150 kilogram object's kinetic energy at 7 m/s equals 3675 J.

2. The speed of a 60 kilogram object with a 1080 J kinetic energy

Kinetic energy is defined as KE= 1/2mv2.

Inputting the values provided yields:

KE = 1/2mv2

1080J = 1/2(60 kg)v2

v2 = (1080 J x 2)/60 kg = 36 m²/s²

Consequently, v = (36 m²/s²) = 6 m/s.

3. A 10 kilogram item is elevated to a height of 3.0 m, and its gravitational potential energy

PE = mgh, where m denotes mass, g denotes gravitational acceleration, and h denotes height, is the formula for gravitational potential energy.

With the above numbers substituted, we obtain PE = mgh = (10 kg)(9.8 m/s²)(3.0 m) = 294 J.

Therefore, a 10 kilogram object elevated to a height of 3.0 m has a 294 J gravitational potential energy.

4. The height to which a 5.25 kg item must be lifted in the event that the gravitational potential energy is raised by 729.56 J

PE = mgh is the equation for gravitational potential energy.

With the provided numbers substituted, we obtain PE = mgh 729.56 J = (5.25 kg)(9.8 m/s²)

h ⇒ h = 14.04 m

Therefore, if the gravitational potential energy is raised by 729.56 J, you must lift the 5.25 kg item to a height of 14.04 m.

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Answer:

The light traveling in any one direction in a straight line is called a ray of light. A group of light rays given out from a source is called a beam of light.

Explanation:

Candle and electric bulbs. The rays coming out from the candle and electric bulb is divergent beam of light.A beam of light is made up of bundles of closely spaced parallel rays.

please give me brainiest

Answer:

The light going in any one course orderly is known as a beam of light. A light beam gathering from a source is known as light emission.

That's the definition of a beam of light.

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Answer:750

Explanation:

Answer:

I believe the answer is C

Explanation:

The mouse would run faster because it has mass.

Answer:

A

Explanation:

A-P-E-X

Answer:

be mature, it's the only way to be called independent, I will help my parents with doing chores, you will meet new friends

Explanation:

it's just my own understanding

Answer:

1eating junk food

2 I will be a good girl

The frequency of the given light wave is 4.3 × 10¹⁴ hertz.

The given information includes the speed of the light wave, which is 3.0 × 10⁸ meters/second, and the wavelength, which is 7.0 × 10⁻⁷ meters.The formula to calculate the frequency of the wave is given below:frequency = speed of light / wavelengthf = c / λwhere:f = frequencyc = speed of light = 3.0 × 10⁸ meters/secondλ = wavelength = 7.0 × 10⁻⁷ metersSubstitute these values in the above formula and solve for the frequency:f = c / λf = 3.0 × 10⁸ meters/second / 7.0 × 10⁻⁷ metersf = 4.3 × 10¹⁴ hertzTherefore, the frequency of the given light wave is 4.3 × 10¹⁴ hertz, which is the option B.

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The star's radius would be approximately 10 times larger than the radius of the sun.

This is because the luminosity of a star is proportional to its radius raised to the fourth power, and the surface temperature is related to the star's luminosity and radius. Using the Stefan-Boltzmann law, we can calculate that the star's radius is approximately 10 times larger than the sun's radius, assuming both stars have similar compositions. The star's radius is approximately 3.19 times larger than the sun's radius. This means that the star is roughly 10 times larger in volume and 1000 times more luminous (since luminosity is proportional to radius to the fourth power) than the sun.

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We will have the following:

First, we know that the centripetal acceleration is given by:

So, the centripetal acceleration is:

Formulas change from F to degree C : C = (F – 32) / 1.8

so we have (176-32) /1.8= 80 oC

Formulas change from F to degree K : K = 5 / 9 x (F – 32) + 273.15

so we have 5/9 x (176-32) + 273.15 = 353 oK

ok done. Thank to me :>

Answer:

an underground conduit for carrying off drainage water and waste matter

Explanation:

Answer:

an underground conduit for carrying off drainage water and waste matter.

Explanation:

Answer:

\(50\:\mathrm{m\: North}\\50\sqrt{3}\:\mathrm{m\: East}\)

Explanation:

We can create a 30-60-90 triangle. The distance she walked is then the hypotenuse of the triangle, and using 30-60-90 triangle rules, we have the following:

The North leg is opposite to the \(30^{\circ}\) angle. Therefore, if we call this distance \(y_N\), we have the following:

\(\sin 30^{\circ}=\frac{y_N}{100},\\\frac{1}{2}=\frac{y_N}{100},\\y_N=\fbox{$50\:\mathrm{m}$}\).

The East leg is opposite to the \(60^{\circ}\\\) angle. If we call this distance \(x_E\), we have:

\(\sin 60^{\circ}=\frac{x_E}{100},\\\frac{\sqrt{3}}{2}=\frac{x_E}{100},\\x_E=\fbox{$50\sqrt{3}\:\mathrm{m}$}\).

The KE of a 6 kg mass at 4 m/s will be 48 J. The kinetic energy is obtained, found as the product of mass and the square of the velocity.

The energy possessed by the body by the virtue of motion is known as the Kinetic energy. It is found as the product of mass and the square of the velocity.

The given data in the problem is;

The mass is, m=6 kg

The velocity is,v= 4 m/s

The kinetic energy is found as;

\(\rm KE = \frac{1}{2}mv^2 \\\\\ \rm KE = \frac{1}{2}6 (4) ^2\\\\\ KE= 48 \ J\)

Hence, the KE of a 6 kg mass at 4 m/s will be 48 J.

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Answer:

Gravity is a force.

Explanation:

Gravity is a very important force. Every object in space exerts a gravitational pull on every other, and so gravity influences the paths taken by everything traveling through space. It is the glue that holds together entire galaxies.

Answer:

an important or critical or essential or serious or vital or necessary force....

Explanation:

taking the direction of Bryan as positive that is 45° south of East as positive Displacement =  35. 8 km ( approx. )

Relative velocity of an object is given as it's velocity with respect to other body ( Or reference ) . Here other body is known as reference frame wrt to which the velocity is measured .

we know that  , Displacement = Avg Velocity x time

     Avg Velocity = 3.53m/s   ,

     time = 169 min= 169*60sec = 10140 seconds

  So   Displacement = 10140 x 3.53 m

          Displacement =35794.2 m \(\Rightarrow\) 35.7942 km (  as 1000m = 1km  )

So  ,  Displacement =  35. 8 km ( approx. )

           Direction  =  45° south of East  

 Relative velocity of swimmer with respect to air:-

              \(V_{sa }= V_s -V_a\)

where \(V_s\) = velocity of swimmer wrt to earth= 3.53m/s

            \(V_a\) = velocity of air wrt to earth = -2m/s

​​​​​  so     \(V_{sa}\)  = 3.53 -(-2) \(\Rightarrow\) 5.53m/s    wrt to air

          Direction = 45° South of East

( Relative velocity wrt to air x time taken)

Relative velocity wrt to air mass = \(V_{sa}\) = 5.53m/s

  Total time taken  = 169min =169 x 60s => 10140 sec

  So

    Net  Displacement  = 5.53 * 10140 => 56074.2m

     Net  Displacement  =  = 56.0742 km

  Net displacement wrt to air = 56.1 km ( approx.)

          Direction = 45° South of East.

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Answer:

The half-life of a radioactive isotope is the time such that the initial amount of the isotope is reduced to its half.

Thus, if we start with A grams of a given radioactive isotope, after a 1 half-life, we will have A/2 grams of the radioactive isotope.

In this case, we know that the sample has 110g of a radioactive isotope.

Then, after 1 half-life, we should have half of 110g, which is:

110g/2 = 55g

Then we should have 55 g of a radioactive isotope.

The answer that is closer to this result is option d (50 g), so that is the correct one.

The transfer function is Vc(s)/Vs(s) = (R + 1/(sC)) / (sL + R + 1/(sC)), the pole-zero map includes poles at -R/L and zeros at -1/(sC), the response to Vs(t) = u(t)V can be calculated using inverse Laplace transform techniques and the response to Vs(t) = o(t)V can also be determined using inverse Laplace transform techniques.

To find Vc(s)/Vs(s), we need to consider the given electrical system with components R, L, and C. By applying Kirchhoff's laws and solving for the output voltage Vc(s) and input voltage Vs(s) in the Laplace domain, we can derive the transfer function as (R + 1/(sC)) / (sL + R + 1/(sC)).

The pole-zero map provides insights into the stability and behavior of the system. In this case, the transfer function has poles at -R/L, indicating a time constant associated with the system's dynamics. The transfer function also has zeros at -1/(sC), which affect the frequency response characteristics.

To find the response to Vs(t) = u(t)V, where u(t) represents the unit step function, we can apply inverse Laplace transform techniques to the transfer function Vc(s)/Vs(s). This will yield the time-domain response of the system to a step input.

Similarly, to find the response to Vs(t) = o(t)V, where o(t) represents the unit impulse function, we can use inverse Laplace transform techniques on the transfer function Vc(s)/Vs(s). This will give us the time-domain response of the system to an impulse input.

By calculating the inverse Laplace transforms of the transfer functions in cases 3) and 4), we can obtain the time-domain responses of the electrical system to the respective inputs.

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Answer:

I think the answer is d; as the object is revolving around the earth in a circular motion due to centripetal force provided by gravity and if you "turned off" the gravity then it would tend to continue away from the planet but in the same direction.

Answer:

0.22 km/h

Explanation:

72 hours in 3 days

16/72 = 0.222222222

round to 0.22 km/h

For future reference, Distance/Time= Speed

Answer: Some elements like potassium, uranium, and thorium are naturally radioactive and give off gamma rays as they decay, but all elements can be excited by collisions with cosmic rays to produce gamma rays. The HEND and Neutron Spectrometers on GRS directly detect scattered neutrons, and the Gamma Sensor detects the gamma rays.

Explanation: searching

To calculate the efficiency and effort applied, we can use the formulas:

Efficiency = (Output / Input) × 100%

Effort = Input / MA

Given:

Load (Output) = 1000 N

MA = 2

VR = 3

First, let's calculate the Input (Effort) using the formula:

Effort = Input / MA

Effort = (Load / VR) / MA

Effort = (1000 N / 3) / 2

Effort = 166.67 N

Next, let's calculate the Output:

Output = Load / VR

Output = 1000 N / 3

Output = 333.33 N

Now, we can calculate the Efficiency using the formula:

Efficiency = (Output / Input) × 100%

Efficiency = (333.33 N / 166.67 N) × 100%

Efficiency = 2 × 100%

Efficiency = 200%

Therefore, the efficiency is 200% and the effort applied is 166.67 N.

Answer:

182 V

Explanation:

We do:V=I.R

V is energy so

I = 7

R = 7+7+3+3+6

V= 7.26

V=182

Because of the earth's large mass and solid structure, the skier moves but the earth does not.

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The magnitudes of T2 and T3 are approximately 89.71 N and 57.35 N, respectively, in order to maintain the equilibrium of the traffic light.

To solve for the magnitudes of T2 and T3, we will use the equations derived from the principle of equilibrium:

Horizontal forces:

T2 * cos(angle 2) - T3 * cos(angle 1) = 0

Vertical forces:

T2 * sin(angle 2) + T3 * sin(angle 1) - T1 = 0

Given:

angle 1 = 33 degrees

angle 2 = 57 degrees

T1 = 72 N

Let's substitute the known values into the equations:

For the horizontal forces equation:

T2 * cos(57°) - T3 * cos(33°) = 0

For the vertical forces equation:

T2 * sin(57°) + T3 * sin(33°) - 72 N = 0

Simplifying the equations:

0.5403T2 - 0.8387T3 = 0 (equation 1)

0.8480T2 + 0.5446T3 = 72 N (equation 2)

We have a system of two linear equations with two unknowns (T2 and T3). We can solve this system of equations using various methods such as substitution or elimination.

Using the substitution method, we solve equation 1 for T2:

T2 = (0.8387T3) / 0.5403

Substituting this value of T2 into equation 2:

(0.8387T3 / 0.5403) * 0.8480 + 0.5446T3 = 72 N

Simplifying the equation:

0.8387T3 * 0.8480 + 0.5446T3 = 72 N

0.7107T3 + 0.5446T3 = 72 N

1.2553T3 = 72 N

T3 = 72 N / 1.2553

T3 ≈ 57.35 N

Now, substituting this value of T3 back into equation 1:

0.5403T2 - 0.8387 * 57.35 = 0

0.5403T2 ≈ 48.42

T2 ≈ 89.71 N

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--The complete Question is, A traffic light is suspended by three cables. If angle 1 is 33 degrees, angle 2 is 57 degrees, and the magnitude of T1 is 72 N, what are the magnitudes of the other two cable tensions, T2 and T3, required to maintain the equilibrium of the traffic light? --