in milgram's follow-up studies, which of the following variant conditions was most effective at reducing the percentage of subjects who used the maximum voltage?

Explanation: Forest-Based Micro and Small Enterprises (FMSEs) are a major source of income for both rural and urban areas of most developing countries, and Nepal is no exception. Though not given enough priority compared to large-scale enterprises, FMSEs have contributed substantially to supporting the livelihoods of marginalized and low-income people by generating employment opportunities throughout the developing world. FMSEs provide employment to a large number of rural people in Nepal who have limited access to other off-farm employment opportunities, however, they are not growing well due to the ineffectiveness of enterprise promotion activities. Different enterprise promotion approaches are being implemented in Nepal by External Development Agencies (EDAs) in collaboration with local Community-Based Organizations (CBOs).In this review, we analyzed the effectiveness of existing FMSEs’ intervention approaches followed by EDAs, current status and constraints, and possible scopes of FMSEs in Nepal with reference to other developing countries. Five implementing approaches that are currently adopted by EDAs in Nepal have succeeded in the promotion of FMSEs to some extent. However, those approaches do still need to be improved in order to fully suit local contexts. In this respect, we suggest an Integrated FMSEs Development approach for Nepal and other developing countries coupled with some recommendations.

Answer:

If the resistance remains constant and the voltage doubles, the power will increase by a factor of 4 (option E)

The following data were obtained from the question:

P = V² / R

Resistance is constant, we have

V₁² / P₁ = V₂² / P₂

V² / P = (2V)² / P₂

V² / P = 4V² / P₂

Cross multiply

V² × P₂ = P × 4V²

Divide both side by V²

P₂ = P × 4V² / V²

P₂ = P × 4

From the above, we can conclude that the power will increase by a factor of 4 (option E)

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2. The primary reflection from the sand/mud interface will arrive first at the hydrophone. To determine which event arrives first, we need to calculate the two-way travel times (TWTT) for each event. The TWTT for the primary reflection from the sand/mud interface is:

TWTT = (2 × depth × sin (angle of incidence)) / velocity

TWTT = (2 × 509 × sin (9)) / 1478TWTT = 0.317 s

The TWTT for the double bounce off the seafloor is:TWTT = (2 × depth) / velocityTWTT = (2 × 509) / 1478TWTT = 0.689 s

Therefore, the primary reflection arrives first at the hydrophone. Here is a simple drawing of the two-event seismic trace:

3. To calculate the time it takes for energy to travel directly from the air gun to the first hydrophone, we need to determine the distance between them and divide it by the velocity of sound in seawater. Using the given values, we have:

Distance = depth + (thickness of sand/mud) + (thickness of mudstone)

Distance = 509 + 1003 + 373

Distance = 1885 m

Velocity of sound in seawater = 1478 m/s

Time = Distance / VelocityTime = 1885 / 1478Time = 1.276 s

Therefore, it takes 1.276 seconds for energy to travel directly from the air gun to the first hydrophone.

4. The maximum takeoff angle at which seismic energy can reflect from the sand/mud interface is called the critical angle. This angle can be calculated using Snell's law:

n1 × sin (angle of incidence) = n2 × sin (angle of refraction)

where n1 and n2 are the velocities of the two materials and the angle of refraction is 90 degrees (since seismic energy travels along a horizontal path once it reaches the interface).

For the sand/mud interface, the critical angle is:

n1 × sin (critical angle) = n2 × sin (90)n1 / n2 = cos (critical angle)critical angle = cos^-1 (n1 / n2)

Using the given values:

n1 = 2793 m/s (sandstone velocity)n2 = 2240 m/s (mudstone velocity)critical angle = cos^-1 (2793 / 2240)

critical angle = 35.9 degrees

Seismic energy cannot reflect from the sand/mud interface at angles greater than the critical angle. For larger angles, the energy will be transmitted into the mudstone.

5. When seismic energy is transmitted into the mudstone, it travels in all directions away from the source. However, the energy will be attenuated (reduced in amplitude) as it travels through the mudstone due to its relatively low velocity compared to the sandstone and seawater.

As a result, the mudstone acts as a barrier that blocks or reduces the energy that would otherwise be transmitted deeper into the subsurface.

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Answer:

The correct option is;

C) Equatorial coordinate system

Explanation:

The equatorial coordinate system is which has the most wide spread use in coordinate system for astronomy for mapping the location of celestial bodies such as stars by use of an imaginary projected celestial sphere or to rectangular coordinates with the Earth at the center.  Extending the Earth's axis onto the celestial sphere is essentially the projection of the Earths axis outwards to intersect the sphere at the celestial poles.

c, 'equatorial coordinate system'

its the correct answer on the test,,, goodluckk :))!!

a. The object having weight 700N in air, experiences buoyancy force of 200N in water.

b. The density of the object is 3.5g/cm3

a. Archimedes Principle:

When a body is partially or fully immersed in a fluid, its weight appears to decrease and apparent decrease in its weight is equal to weight of the fluid displaced by the body.

The object weight in air is 700N and it's weight in water is 500N.

Buoyancy force = Apparent loss of weight

= 700N - 500N

= 200N

b. Density of the object:

Buoyancy force = Vdg ( where v is volume of the object, d is the density of liquid and g is gravity of earth)

200 = Vdg

200 = V × 1000 × 10 (density of water= 1000kg/m3)

V = 2/100

m/d = 2/100 (V = m/d ,where d is density of solid)

(mg = 700N given in question)

(m = 70)

d = 70× 100/2

d = 3500kg/m3

= 3.5g/cm3

hence density of the object is 3.5g/cm3

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Therefore, the correct answer is c. 20 N/kg

To solve this problem, we need to consider the relationship between weight, mass, and gravitational field strength.

Weight is the force experienced by an object due to gravity, and it is given by the formula:

Weight = mass * gravitational field strength

Let's assume the gravitational field strength on Earth is g, and the weight of the astronaut on Earth is W.

According to the problem, the weight of the astronaut at the given location is 25 percent of what he would weigh on Earth. Mathematically, this can be expressed as:

Weight at location = 0.25 * Weight on Earth

Using the formula for weight, we can rewrite this as:

mass * gravitational field strength at location = 0.25 * (mass * gravitational field strength on Earth)

The mass of the astronaut cancels out from both sides of the equation, and we are left with:

gravitational field strength at location = 0.25 * gravitational field strength on Earth

Now, we know that the weight of an 80 kg astronaut on Earth is equal to the force of gravity acting on him, which is given by:

Weight on Earth = mass * gravitational field strength on Earth

W = 80 kg * g

We can substitute this value into the equation for the gravitational field strength at the location:

gravitational field strength at location = 0.25 * (80 kg * g)

gravitational field strength at location = 20 kg * g

So, the magnitude of the gravitational field at the location where the astronaut weighs 25 percent of what he would weigh on Earth is 20 N/kg.

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The resistance of the combination of resistors given is; 1.5 ohms

Since 3 of the resistors are in series and each of the resistor has a resistance of 2 ohms, then;

Total resistance; R = 2 + 2 + 2

R = 6 ohms

Now, formula for sum of resistance in parallel is;

1/R = 1/R1 + 1/R2 + 1/R3 ....

Since we have four of the resistor combination above in parallel, then;

1/R = 1/6 + 1/6 + 1/6 + 1/6

1/R = 4/6

R = 6/4

R = 1.5 ohms

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Answer:

LOL i belive its 200 because i did this exact same thing yesterday for homework and got it right bcs i got it lol, yw btw! hahaha! good luck also, subscribe to my channel!

Explanation:

Answer:

it would say ask and see the tutor

Answer:

You type out your question and it should ask you where you want to post it. You choices are tutor or community. Choose community.

Explanation:

I just did it and it worked. Good luck

Answer:

14,400

Explanation:

I guessed and I think it's right Soo

The correct unit used to measure the electric potential difference is Volts.

Joules is used to measure energy.

Newtons is used to measure forces.

Centimeters is used to measure distance.

Therefore the correct option is the first one.

Answer:

(I). The Schwarzschild radius is \(2.94\times10^{8}\ km\)

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is \(1.1\times10^{-7}\ km\)

(IV). The Schwarzschild radius is \(7.4\times10^{-29}\ km\)

Explanation:

Given that,

Mass of black hole \(m= 1\times10^{8} M_{sun}\)

(I). We need to calculate the Schwarzschild radius

Using formula of radius

\(R_{g}=\dfrac{2MG}{c^2}\)

Where, G = gravitational constant

M = mass

c = speed of light

Put the value into the formula

\(R_{g}=\dfrac{2\times6.67\times10^{-11}\times1\times10^{8}\times1.989\times10^{30}}{(3\times10^{8})^2}\)

\(R_{g}=2.94\times10^{8}\ km\)

(II). Mass of block hole \(m= 6 M_{sun}\)

We need to calculate the Schwarzschild radius

Using formula of radius

\(R_{g}=\dfrac{2MG}{c^2}\)

Put the value into the formula

\(R_{g}=\dfrac{2\times6.67\times10^{-11}\times6\times1.989\times10^{30}}{(3\times10^{8})^2}\)

\(R_{g}=17.7\ km\)

(III). Mass of block hole m= mass of moon

We need to calculate the Schwarzschild radius

Using formula of radius

\(R_{g}=\dfrac{2MG}{c^2}\)

Put the value into the formula

\(R_{g}=\dfrac{2\times6.67\times10^{-11}\times7.35\times10^{22}}{(3\times10^{8})^2}\)

\(R_{g}=1.1\times10^{-7}\ km\)

(IV). Mass = 50 kg

We need to calculate the Schwarzschild radius

Using formula of radius

\(R_{g}=\dfrac{2MG}{c^2}\)

Put the value into the formula

\(R_{g}=\dfrac{2\times6.67\times10^{-11}\times50}{(3\times10^{8})^2}\)

\(R_{g}=7.4\times10^{-29}\ km\)

Hence, (I). The Schwarzschild radius is \(2.94\times10^{8}\ km\)

(II). The Schwarzschild radius is 17.7 km.

(III). The Schwarzschild radius is \(1.1\times10^{-7}\ km\)

(IV). The Schwarzschild radius is \(7.4\times10^{-29}\ km\)

The energy conservation allows to find the Schwarschild radius for several bodies of different masses are:

  1)  Black hole quasar is:  r = 2.9 10⁸ km

  2) Blsck hole supernove is:  r = 17.7 km

  3)  Mini black hole is:   r = 1.1 10⁻⁷ km

  4) Human body is:  r=  7 10⁻²⁹ km

The schwarschild radius is defined as the distance from a black hole center at radius  which the escape velocity is equal to the light speed, in some cases it is also called the event horizon.

Let's use Newton's second law where force is the universal law of attraction and acceleration is centripetal.

        F = ma

        F = \(G \frac{Mm}{r^2}\)        

Where F is the force, M the mass of the black hole, m the handle of the body, r the radius and v the speed of the body.

The energy of the gravitational field is

        F = \(- \frac{dU}{dr }\)  

         U = \(-G \frac{Mm}{r}\)  

Let's use conservation of energy

        Em₀ = K + U = ½ m v² -  \(G \frac{Mm}{r}\)  

In infinity the energy

        Em_f = 0

energy is conserved

       Em₀ = Em_f  

       ½ m v² - \(G \frac{Mm }{r}\)  = 0

       r = \(\frac{2GM}{v^2}\)

From the definition of the Schwarschild radius this speed is equal to the light speed

        v = c

        r = \(\frac{2GM}{c^2 }\)  

They ask to calculate the radius for several cases of different mass, claculate the constant value

      V = \(\frac{2 \ 6.67 \ 10^{-11} }{(3 \ 10^8) ^2 }\)  

      V =  1.482 10⁻²⁷

1) A black hole of mass M = 1 10⁸ \(M_{sum}\)

The tabulated mass of the sun is \(M_{sum}\) = 1.989 10³⁰ kg

Let's  substitute

        r =  1.482 10⁻²⁷   1 10⁸ 1.989 10³⁰

        r = 2.94 10⁸ km

With two significant figures

        r = 2.9 10⁸ km

2) A black hole of mass M = 6 \(M_{sum}\)

        r =  1.482 10⁻²⁷ 6 1.989 10-30

        r = 17.7 km

3) a mini black hole with the mass of the moon

    Tabulated mass of the moon M = 7.35 10²² kg

       r =  1.482 10⁻²⁷  7.35 10²²    

       r = 1.1 10⁻⁷ km

4) A person of M = 50 kg

    r =  1.482 10⁻²⁷ 50

    r=  7 10-29 km

In conclusion using the conservation of energy we can find the Schwarschild radius for several bodies of different masses are:

  1)  Black hole quasar is:  r = 2.9 10⁸ km

  2) Blsck hole supernove is:  r = 17.7 km

  3)  Mini black hole is:   r = 1.1 10⁻⁷ km

  4) Human body is:  r=  7 10⁻²⁹ km

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Answer:

141m

Explanation:

Answer:

t= 2.0s

Explanation:

Khan Academy

Answer: It will push back with an equal amount of force.

The relatively stable scale degree that often resolves directly to the tonic is the fifth degree. The fifth degree is also known as the dominant scale degree, and it is one of the most essential elements of Western tonal music.

When a composer or songwriter wants to create a sense of resolution in a piece of music, they will often use the dominant chord, which is built on the fifth degree of the scale, to lead back to the tonic chord, which is built on the first degree of the scale. This is known as a V-I cadence, and it is one of the most common and effective ways to create a sense of closure in Western tonal music.In addition to its function as a dominant chord, the fifth scale degree is also used in a variety of other ways in Western music. For example, it is often used as a passing tone, connecting two other scale degrees in a melodic line. It can also be used as a pedal tone, where it is repeated over and over while other parts of the music change around it.Overall, the fifth scale degree is a crucial element of Western tonal music, and its stable and predictable nature makes it an excellent tool for composers and songwriters to use when they want to create a sense of resolution or closure in their music.

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Answer:

Explanation:

The only word I can think of is confirm.

I choose confirm because they have already used empirical evidence (evidence that is primarily found by the senses. We hear something, we see something, We smell something -- although this is rather rare. Probably we would hear or see something first). We've used logical thinking, and we've likely even performed experiments.

So we've done all the ground work. All of these things are needed to confirm a scientific thought or premise that has been proposed. All of them aim at confirming what  we proposed.

Answer:

a = 6.4 [m/s²]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration. In this way, we have the following equation.

∑F = m*a

where:

∑F = sum of forces (horizontal force = 40 [N] and friction force = 8 [N])

m = mass of the block = 5 [kg]

a = acceleration [m/s²]

Now replacing:

\(40-8=5*a\\32 = 5*a\\a=6.4[m/s^{2} ]\)

Note: the sign of the friction force is negative since this force is acting against the movement of the block.

The maximum current in the electric oscillator is approximately 0.0127 A or 12.7 mA.

To answer your question about an electric oscillator made with a 0.10 μF capacitor and a 1.0 mH inductor, initially charged to 4.0 V, we'll calculate the resonant frequency and maximum current in the circuit.

1. Calculate the resonant frequency using the formula f = \(1 / (2 * \pi  * \sqrt{(LC)} )\), where L is the inductance and C is the capacitance.

f =\(1 / (2 * \pi * \sqrt{(1.0 * 10^-^3 H * 0.10 * 10^-^6 F)} )\)
f ≈ 1591.55 Hz

The resonant frequency of the electric oscillator is approximately 1591.55 Hz.

2. Calculate the maximum current using the formula I_max = \(V / \sqrt{ (L/C)}\), where V is the initial voltage.

I_max = \(4.0 V / \sqrt{(1.0 * 10^-^3 H / 0.10 * 10^-^6 F)}\)
I_max ≈ 0.0127 A

The maximum current in the electric oscillator is approximately 0.0127 A or 12.7 mA.

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The light travels through water and gets reflected off on the glass surface into the water. There had been a 180° phase change between the incident and the reflected wave. This is called Total internal reflection (TIR).

In total internal reflection, in physics, a ray of light in a medium such as water or glass is completely reflected back into the medium from the surrounding surfaces. This phenomenon occurs when the angle of incidence is greater than a certain critical angle called the critical angle.

TIR only occurs when both of the following two conditions are met

Thus, the phases which include the TIR are the incident and the reflected phase and the incident light hits the surface while the reflected light reflects back.

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A body of groundwater that is porous, permeable, and has the water table as its upper surface is called an aquifer.

An aquifer is a body of groundwater that is porous, permeable and has the water table as its upper surface. It is an underground layer of water-bearing permeable rock or unconsolidated materials (gravel, sand, silt, or clay) from which groundwater can be extracted using a well or by pumping.The term “aquifer” comes from two Latin words: aqua, which means “water,” and ferre, which means “to carry.” There are two types of aquifers: confined and unconfined. Confined aquifers are those that are separated from the land surface by an overlying layer of low-permeability material, while unconfined aquifers are not.Aquifers are a vital resource for human activities, particularly in areas where surface water is scarce. Groundwater from aquifers is commonly used for drinking, irrigation, and industry. It is essential to manage and conserve aquifers to ensure their continued availability and sustainability.

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Answer:

h= 0.45 m

Explanation:

PE= 1/2 mv^2             KE= mgh

v= 3m/s

vf= 0 m/s

h=?

PE= 1/2(1kg)(3m/s)^2

PE= 4.5 J

4.5 J/ 1kg(9.8 m/s^2)

h=0.45 m

The height of the ball above the ground when it stops is 0.46 m.

The initial kinetic energy of the ball is equal to the final potential energy of the ball.

⇒ Formula:

⇒ Where:

⇒ make h the subject of the equation

From the question,

⇒ Given:

⇒ Substitute these values into equation 2

Hence, the height of the ball above the ground when it stops is 0.46 m.

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Answer:

they store information

Explanation:

Information are encoded on the magnetic part of a hard drive and the strip of a credit card.

Answer:

Tysm

Explanation:

Answer:

23: Acceleration 24:1m/s^2

Explanation:

\(a = (v_f - v_i) /delta(t)\\a=(16-10)/6\\a=6/6\\a=1m/s^2\)

2269 × 10-³ m can be written in scientific notation as follows: 2.269 × 10⁰.

Scientific notation is a method of writing, or of displaying real numbers as a decimal number between 1 and 10 followed by an integer power of 10.

It is an alternative format of such a decimal number immediately followed by E and an integer.

According to this question, 2269 × 10-³ metres is to be converted to scientific notation. We do this by shifting the decimal place backwards three times to obtain the following;

2.269 × 10⁰

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Deposition occurs when the agents (wind or water) of erosion lay down sediment. Deposition changes the shape of the land. Erosion, weathering, and deposition are at work everywhere on Earth. Gravity pulls everything toward the center of Earth causing rock and other materials to move downhill

Answer:

Bowling ball

Explanation:

The bowling ball's mass will make it roll down the incline faster than the lighter volleyball.

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