A p.d of 20V is applied across two resistors of 4ohm and 6ohm connected in series. Determine the point across the 6ohm resistors if the total circuit current is 2A.
a)1.0V b)2.0V c)3.3V d)12.0V​

Answer:

twenty two miles per hour

The index of refraction of the unknown material is 1.5.

The index of refraction (n) of a medium is defined as the ratio of the speed of light in a vacuum (c) to the speed of light in the medium (v):

n = c / v

In this case, the speed of light in the unknown material is given as 2.00 × \(10^8\) m/s. The speed of light in a vacuum is approximately 3.00 × \(10^8\) m/s. Substituting these values into the formula:

n = (3.00 × \(10^8\) m/s) / (2.00 × \(10^8\) m/s)

Simplifying the expression:

n = 1.5

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When an x-ray photon interacts with an atom, it can transfer a quantity of energy to an inner-shell electron, thereby dislodging it from its atomic orbit.

This energy transfer is known as the photoelectric effect, or photoelectric absorption. In order for this energy transfer to occur, the energy of the incoming x-ray photon must be equal to or greater than the binding energy of the electron to its orbit. The binding energy is the amount of energy required to remove an electron from its orbital. When the energy of the incoming x-ray photon is greater than the binding energy, the extra energy is released in the form of kinetic energy, which can be used to eject the electron from its orbit. This kinetic energy is then transferred to the atom and is used to excite or ionize other electrons. Once the electron has been ejected, it is then free to travel through the atom, leaving behind a positively charged atom, or ion. This process of photoelectric absorption is essential for x-ray imaging and spectroscopy, as it allows for the detection of inner-shell electrons.

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Answer:

Explanation:

The properties of magnets are used to make electricity. Moving magnetic fields pull and push electrons. ... Moving a magnet around a coil of wire, or moving a coil of wire around a magnet, pushes the electrons in the wire and creates an electrical current.

Answer:

a. 10.2 N b. 1.73 c. 58.8 N d. 10.2 N

Explanation:

a. The force that will make the block slide down the plane

This is the component of the block's weight along the plane F = mgsinθ where m = mass of block = 1.2 kg, g = 9.8 m/s² and θ = angle of incline = 60°.

So, F = mgsinθ = 1.2 kg × 9.8  m/s² × sin60°.= 10.18 N ≅ 10.2 N

b. The coefficient of friction

The coefficient of friction, μ = tanθ = tan60° = 1.73

c. The normal reaction.

This is equal to the vertical component of the block's weight. So, F = mgcosθ. Substituting the values for the variables from above, we have

F = mgcosθ = 1.2 kg × 9.8  m/s² × cos60°.= 58.8 N

d. The frictional force

Since the block does not slide, there is no net force on it. If f is the frictional force, then

F - f = ma. Since a = acceleration = 0,

F - f = 0

f = F = mgsinθ = 1.2 kg × 9.8  m/s² × sin60°.= 10.18 N ≅ 10.2 N

Answer:

Your answer is C

Explanation:

V=IxR

220 divide by 1.2=183.0

The answer was probably a type error so it should be C

Answer:

16 male students, 11 female students

Answer:

468.42N

Explanation:

Here,

Work done(W)=1780J

Displacement(s)=3.8m

As we have,

Work=Force×displacement

Now,

\(\hookrightarrow 1780J=F\times3.8m\\\\\hookrightarrow \frac{1780J}{3.8m} =F\\\\\hookrightarrow F=468.42N\)

When is the average velocity of an object equal to the instantaneous velocity is C. This is the case only when the velocity is constant.

The instantaneous velocity of an object is equal to the average velocity of an object when the velocity is constant or when the acceleration is zero, this is the case only when the velocity is constant. When an object has a constant velocity, the instantaneous velocity of the object is equivalent to the average velocity of the object. This is true because the velocity of the object remains constant over time.

For example, if an object travels at a speed of 20 meters per second for a time period of 5 seconds, then the instantaneous velocity at the end of the 5 seconds is 20 meters per second, and the average velocity of the object over the 5 seconds is also 20 meters per second. This is because the velocity remained constant throughout the entire time period. Therefore, option c is correct, this is the case only when the velocity is constant.

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Answer:

forever alone uwu......

The rock was launched in the upward direction with an initial velocity of 8.58 m/s.The speed of the rock 4.5 seconds after launch is  35.52 m/s.

Given:

Time after launch (t) = 2.1 s

Rising rate (v) = 12 m/s

(a) By using the equation of motion v = u + at

where,

v denotes the final velocity, u initial velocity, a is the acceleration, and t is the time.

v = u + at

12= u + (9.8) × 2.1 s

u = 12 - (9.8 m/s² × 2.1 s)

u = 12 - 20.58

u = -8.58 m/s

The negative sign indicates that the rock was launched in the upward direction with an initial velocity of 8.58 m/s.

(b) To calculate the speed 4.5 seconds after launch,

v = u + at

v = -8.58 + (9.8) × 4.5 s

v = 35.52 m/s

The speed of the rock 4.5 seconds after launch is  35.52 m/s.

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An automobile accelerates at the rate of 57 m/s² with a final velocity of 15 m/s. Then, the distance covered by the automobile will be equal to 1.97 meters.

A moving object's speed and orientation vary over time at a certain rate, which is known as acceleration.

An object or location is considered to be accelerated when it moves more quickly or more slowly in a straight path. A circle's motion increases despite its constant speed because its direction is always changing. For all other motions, by and to velocity is used.

Because of its magnitude and direction, it is a vector quantity.

As per the given information in the question,

Acceleration, a = 57 m/s²

Final velocity, v = 15 m/s

Use the equation of motion,

v² - u² = 2as

Put the values in the equation,

15² - 0 = 2(57)(s)

s = 225/114

s = 1.97 meters.

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The wavelength (L) for the given wave is approximately 79.36 meters, and the horizontal particle velocity at the mean water level (z=0) for the deepwater case is approximately 1.96 m/s.

To find the wavelength (L) of the wave, we can use the formula for deepwater waves:

L = g * T^2 / (2 * π)

where g is the acceleration due to gravity (approximately 9.8 m/s^2) and T is the period of the wave.

Substituting the given values, we have:

L = 9.8 * (6.4 + 0.1a)^2 / (2 * π)

L = 9.8 * (6.4 + 0.1 * 64)^2 / (2 * π)

L ≈ 9.8 * (6.4 + 6.4)^2 / (2 * π)

L ≈ 9.8 * 12.8^2 / (2 * π)

L ≈ 9.8 * 163.84 / (2 * π)

L ≈ 1606.432 / 6.28

L ≈ 255.5 meters

Therefore, the wavelength (L) is approximately 255.5 meters.

To find the horizontal particle velocity at the mean water level (z=0) for the deepwater case, we can use the formula:

v = g * T / (2 * π)

Substituting the given values, we have:

v = 9.8 * (6.4 + 0.1a) / (2 * π)

v = 9.8 * (6.4 + 0.1 * 64) / (2 * π)

v ≈ 9.8 * (6.4 + 6.4) / (2 * π)

v ≈ 9.8 * 12.8 / (2 * π)

v ≈ 125.44 / 6.28

v ≈ 19.95 m/s

Therefore, the horizontal particle velocity at the mean water level (z=0) for the deepwater case is approximately 19.95 m/s.

For the given wave with a period of T= (6.4+0.1a) m and a height of H=2.8m, the wavelength (L) is approximately 255.5 meters, and the horizontal particle velocity at the mean water level (z=0) for the deep water case is approximately 19.95 m/s.

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The amount of heat needed to warm 2 kg of liquid water by 5°C is 41,860 Joules. Temperature is a measure of hotness or coldness that can be described in terms of any number of arbitrary scales.

It also indicates the direction in which heat energy would spontaneously flow, i.e., from a hotter (body) to a colder (body) body.

To warm 2 kg of liquid water by 5°C, we need to calculate the amount of heat required using the formula:
Q = m × c × ΔT
Where:
Q = Amount of heat needed (in Joules)
m = Mass of water (in kilograms)
c = Specific heat of water (4186 J/kg°C)
ΔT = Change in temperature (in °C)
Plugging in the given values, we get:
Q = 2 kg × 4186 J/kg°C × 5°C
Q = 41,860 Joules

The sensible heat of water will rise from 50 to 100 degrees Celsius at this point, and a second phase shift will occur as the temperature rises from 100 to the final temperature of superheated steam.

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Troposphere, stratosphere, mesosphere, thermosphere and exosphere.

Earth's atmosphere has five major layers.

From lowest to highest, layers are the troposphere, stratosphere, mesosphere, thermosphere and exosphere.

a) What are the conditions in the troposphere?

The atmospheric temperature descends upward with a slope of ~10 K km−1 for dry air and ~7 K km−1 for wet air in troposphere.

The temperature of troposphere decreases with an increase in height.

B) What sort of temperatures would she experience in the stratosphere? What other layer is found within the stratosphere?

The temperature in the stratosphere ranges from- 60° F at the troposphere boundary to -5°F at the top.

The temperature increase is due to the ozone layer that absorbs ultraviolet light from solar radiation.

C)What sort of temperatures would she experience in the mesosphere?

The temperature in the mesosphere ranges from -2.5°C to -90°C

The temperature decreases because of a decrease in the absorption of penetrating solar radiation.

The atmosphere gets cooler with an increase in altitude because an increase in distance from the Earth's surface.

D)What sort of temperatures would she experience in the thermosphere? What other layer is found within the thermosphere?

4,500°F is the temperature of thermosphere.

Temperatures climb sharply in the lower thermosphere , then level off and hold fairly steady with increasing altitude above that height.

Temperatures in the upper thermosphere can range from about 500° C (932° F) to 2,000° C (3,632° F) or higher.

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An electric field with intensity ′E′ is applied to an electrical dipole with dipole moment ′p′. The dipole moves into a place where its axis forms an angle with the electric field.

Justification: Since electric fields are subject to the of superposition, the overall electromagnetic current of a system typically equal to the total of its individual electric fields. Each charge produces an electric field at pixel Value that is equal in size but moves in the opposite direction.

The flux through a surface is at its maximum when it is horizontal to field (left panels) and the paddock vector is parallel to a vector, A. When a surface is perpendicular to the field (figure 3d), no field lines intersect it and there is no flux through it.

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Answer:

A liquid is a nearly incompressible fluid that conforms to the shape of its container but retains a (nearly) constant volume independent of pressure. A liquid is made up of tiny vibrating particles of matter, such as atoms, held together by intermolecular bonds.

Answer:

I mean they have the same amount of velocity

Answer:

F = 70000  N

Explanation:

Given that,

Mass of a car, m = 1400 kg

Initial speed of the car, u = 15 m/s

Finally, it hits the ground, v = 0

Time, t= 0.3 s

We need to find the force experienced by the car. Using Newton's second law of motion to find it.

F = ma

a is acceleration of the car.

\(F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{1400\times (0-15)}{0.3}\\\\=-70000\ N\)

So, the force experienced by the car has a magnitude of 70000  N.

A third positive charge \(x = \frac{2\sqrt{2}a }{1+\sqrt{2} }\) can be placed on the x-axis so that the net electric force on it is zero.

whilst -factor charges are in touch, there may be

\(F = k \frac{q_{1}q_{2} }{ r^{2} }\)

Where

\(k = 9 * 10^{9} Nm^{2}C^{-1}\) = the coulomb constant.

The charges' magnitudes are \(q_{1}\) and \(q_{2}\).

r is the separation of the charges.

Let,

The third charge equals Q at, x equals x from the source.

because there is no net force acting on the third charge.

Hence,

\(f_{1} + f_{2} = 0\)

⇒ \(k\frac{(2q)Q}{x^{2} }-k\frac{qQ}{(2a-x)^{2} } = 0\)

⇒ \(\frac{2}{x^{2} } - \frac{1}{(2a-x)^{2} } = 0\)

⇒ \(\frac{2}{x^{2} } = \frac{1}{(2a-x)^{2} }\)

⇒ \(\frac{\sqrt{2} }{x} = \frac{1}{2a-x}\)

⇒ \(2\sqrt{2}a - \sqrt{2} = x\)

⇒ \(x = \frac{2\sqrt{2}a }{1+\sqrt{2} }\)

handiest when forces resulting from previous prices cancel every different out will the pressure on a third fee be 0.

at the same time as the pressure between  identical costs is repulsive and is interpreted as high-quality, the pressure between two opposing expenses is attracted and is interpreted as bad.

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In an IR spectrum, the peak at about 1700 cm-1 in acetanilide corresponds to the presence of a carbonyl group (C=O).

The carbonyl group in acetanilide is a functional group that is responsible for many of its chemical and physical properties. The C=O bond absorbs infrared radiation in this region, resulting in a characteristic peak. The intensity and shape of the peak provide information about the strength of the bond, its environment, and the presence of any neighboring functional groups. This information is important in determining the identity of the compound and its structure, as well as for studying the chemical reactions and interactions involving the carbonyl group.

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When collecting a gas over water, the gas is always a mixture of the gas collected and water vapor. This statement is True

Water vapour or aqueous vapour is the name for the gaseous state of water. It is a particular kind of water condition inside the hydrosphere. Both the boiling and evaporation of liquid water as well as the sublimation of ice can produce water vapour. Water vapour is transparent, like the bulk of other atmospheric constituents. An example of water vapour is the mist that floats over a pot of boiling water.

The gaseous phase of water is known as water vapour, water vapour, or aqueous vapour. Within the hydrosphere, it is one type of water state. Water vapour can be created by the boiling or evaporation of liquid water as well as by the sublimation of ice. Like the majority of other atmospheric elements, water vapour is transparent.

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About 4.5 billion years ago, a dense cloud of interstellar gas and dust gave rise to our solar system. The supernova shockwave from a nearby exploding star, or supernova, may have caused the cloud to collapse.

Within the initial 10 million years of the solar system's formation, Jupiter and Saturn are assumed to have formed first and fast. Rocky planets start to develop in the warmer, star-closer regions of the disk. There isn't much gas left for the terrestrial planets to accrete after the formation of the icy giants.

Let's, therefore, travel back in time more than 4.6 billion years to the time before Earth, Sun, and even Solar System existed. We were surrounded by gas and dust, most likely in one of the spiral arms. Do you want to know how it appeared? Your favorite Hubble Space Telescope images should be helpful.

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Our solar system was created from a massive cloud of interstellar gas and dust some 4.5 billion years ago. The collapse of the cloud may have been triggered by the supernova boom from a nearby exploding star.

Jupiter and Saturn are believed to have formed initially and quickly during the first 10 million years of the solar system's creation. In the warmer, star-closer parts of the disc, rocky planets begin to form. After the development of the icy giants, there is not much gas left for the terrestrial planets to accrete. Therefore, let's go back in time more than 4.6 billion years to the period before the Earth, Sun, and even the Solar System were created. Dust and gas were all around us called supernova.

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The net force would be needed to accelerate a 2.95kg bowling ball at 6.25m/s^2 will be F = 18.4375 N

Newton's second law of motion  states that the time rate of change of the momentum of  a body is equal in both magnitude and direction to the force  imposed on it.

For a constant mass , force is equal to  Force = mass * acceleration

given

mass(M) = 2.95 kg

acceleration(a) = 6.25 m/s^2

Force(F)  = ?

Force(F) = mass(M) * acceleration(a)  (from Newton's second law of motion  )

F =  2.95 * 6.25

F = 18.4375 N

The net force would be needed to accelerate a 2.95kg bowling ball at 6.25m/s^2 will be F = 18.4375 N

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Answer:

Rock

Explanation:

Right on edg 2021 and 2020000000:P

The water outflow in 30 secs through 200 mm of the capillary tube is mathematically given as

\(Qo=1.6 \times 10^{2} \mathrm{~mL}\)

\(\begin{aligned}\Delta P &=6660 \mathrm{~m} / \mathrm{m}^{2} \\\mu &=8.01 \times 10^{-4} \text { Pas } \\t &=30 \mathrm{~s} \\L &=200 \mathrm{~mm}=200 \times 10^{-3} \mathrm{~m} \\D &=1.5 \mathrm{~mm}=1.5 \times 10^{-3} \mathrm{~m} \Rightarrow \gamma=\frac{1.5 \times 10^{-3}}{2} \mathrm{~m}\end{aligned}\)

Generally, the equation for Rate of flow of Liquid is  mathematically given as

\(\\$$Q=\frac{\pi r^{4} \times \Delta P}{8 \mu L}\)

$$

Where dP is pressure difference r is the radius

\(\mu\) is the viscosity of water

L is the length of the pipe

\(Q=\frac{\pi \times\left(\frac{1.5 \times 10^{-3}}{2}\right)^{4} \times 6660}{8 \times 8.01 \times 10^{-4} \times 200 \times 10^{-3}}\)

\(Q=5.2 \mathrm{~mL} / \mathrm{s}\)

In $30s the quantity that flows out of the tube

\(&Qo=5.2 \times 30 \\&Qo=1.6 \times 10^{2} \mathrm{~mL}\)

In conclusion, the quantity that flows out of the tube

\(Qo=1.6 \times 10^{2} \mathrm{~mL}\)

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It would take approximately 14.3 billion years for a galaxy located 1 Mpc (megaparsec) away to reach its current position while moving at a speed of 70 km/s.The speed of light is used as a standard for measuring distances in astronomy. 1 Mpc is equal to 3.26 million light-years. Therefore, a galaxy located 1 Mpc away is 3.26 million light-years away.

To calculate the time it would take for the galaxy to reach its current position, we can use the formula:
The distance is 1 Mpc or 3.26 million light-years. We need to convert the speed from km/s to light-years per year. One light-year is approximately 9.46 trillion km.
70 km/s = 70 km/s x (3.1536 x 10^7 s/year) / (9.46 x 10^12 km/year)
70 km/s = 0.0000234 light-years per year

Now we can plug in the values:
Time = 3.26 million light-years / 0.0000234 light-years per year
Time = 1393162.39 years or approximately 14.3 billion years
Therefore, it would take about 14.3 billion years for the galaxy to reach its current position while moving at a speed of 70 km/s.

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Let's also assume that this ball is moving at a top speed of four metres per second.

And if you make yo-yos, you might be interested in learning how much stress this rope should be able to support and how strong your string needs to be. The yo-yo weighs.2 kg and is suspended from an.8 m long sting. What tension must the string have if the yo-yo completes a full rotation every second? Since ropes can never push, this rope is still pushing the yo-mass yo's into the circle's centre. So one of these forces will be negative when we plug in over here.

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No, a magnetic compass will not work properly near an automobile or train and is therefore denoted as option A.

This is referred to as an instrument which is used for determining the direction on the surface of Earth by means of a magnetic pointer that aligns itself with Earth’s magnetic field.

A magnetic compass will not work properly near an automobile or train because the magnetic lines of force are unaffected by paper, glass, or plastic and we are ware that this vehicles have parts such as the windows, engine etc which are made from these type of materials. They will however work properly if these parts are avoided by the user of the compass.

This is therefore the reason why No was chosen as the correct choice in this type of scenario.

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Answer:

18

Explanation:

EDGE2021