5. A field of 14.5-MeV neutrons deposits a kerma of 1.37 Gy at a point of interest in water. What is the fluence? (See Appendix F).

A plane mirror always forms a virtual image because of the following reasons:

1. When light rays from an object fall on a plane mirror, they get reflected from the mirror. After reflection, they never meet at any point in real but they appear to meet at some point.

2. The image formed by a plane mirror cannot be obtained on a screen.

3. Plane mirrors never focus light into a single converging point.

A  randomized experiment is regarded as being very much reliable in science.

A  randomized experiment is regarded as being very much reliable in science. In this type of experiment, the subjects are randomly assigned to a treatment group and a control group.

In this experiment, there are two kinds of treatments that the biologist intends to use and they are;

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Missing parts;

A biologist is interested in studying the effect of growth-enhancing nutrients and different salinity (salt) levels in water on the growth of shrimp. The biologist has ordered a large shipment of young tiger shrimps from a supply house to use for the study. The experiment is to be conducted in a lab where there are 12 tanks (with equal amount of shrimp in each) in a controlled environment. The biologist is planning to use 3 different growth-enhancing nutrients (A, B and C) and two different salinity levels (high and low).  List the treatments that the biologist plans to use in this experiment

The reason you only see fossils in sedimentary rock is that these set of rocks are formed in much lower pressure and temperature, compared to the other types of rocks like igneous rocks.

Answer: 2.5N

Explanation:

Given the following :

Mass of block (m) = 2kg

Coefficient of static friction (μs) = 0.4

Horizontal force applied to the block = 2.5N

The frictional force (Ff) between the block and the floor is :

First calculate the maximum static frictional force:

Frictional force = Coefficient of static friction(μs) × normal reaction(R)

Normal reaction(R) = mass × acceleration due to gravity (10m/s^2)

R = 2 × 10 = 20

Fmax = μs × R

Fmax = 0.4 x 20 = 8N

Here, since the applied force (2.5N) is less than maximum frictional force(8N).

The force of friction between the block and the floor will be equal to the applied force of 2.5N due its ability to adjust itself in other to ensure equilibrium.

The potential difference is 2√5.

The energy stored in the capacitor can be calculated as follow:

U = 1/2 CV²

U = energy stored = 300J

C = capacitor = 30mf

V = potential difference

V= \(\sqrt{2U/ C}\)

  = \(\sqrt{2 (300) / 30}\)

  = \(\sqrt{2 * 10}\)

  = 2√5

Therefore the potential difference is 2√5.

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Answer:

A) the initial point all energy is elastic potential and the final point all energy is kinetic

B) a bar graph the two bars have the same height and the sum of their height is the initial energy

C) two bars, one for the  kinetic energy and the other for the gravitational potential energy.

Explanation:

A) For this exercise we must use the energy conservation relations

starting point. When the spring is compressed

        Em₀ = K_e = ½ k x²

end point, at the bottom of the loop

        Em_f = K = ½ m v²

energy is conserved

        Em₀ = Em_f

         ½ k x² = ½ m v²

         v = \(\sqrt{ \frac{k}{m} }\)   x

In a bar graph the initial point all energy is elastic potential and the final point all energy is kinetic

B) intermediate point in a quarter of the radius

In this case we use the lower part of the loop as the starting point and the quarter part of the bow as the end point.

        Em₀ = K

         Em_f = K + U = ½ m v² + m g R

in a bar graph the two bars have the same height and the sum of their height is the initial energy

C) End point highest part of the loop

   starting point, bottom of loop

         Emo = K = ½ m v₀²

from part A of the exercise we saw that it is equal to the elastic energy of the spring

    final point. Highest part of the loop

         Emf = K + U

         Em_f = ½ m \(v_{f}^2\) + mg (2R)

where R is the radius of the loop

         Em₀ = Em_f

        1/2 m v₀² = 1/2 m v_{f}^2+ mg 2R

        v₀² = v_f^2 + 4gR

In a bar graph there are two bars, one for the  kinetic energy and the other for the gravitational potential energy. The sum of the heights of these bars is the initial energy, so the energy is transformed but not created or destroyed in the process.

At 0.0804 km altitude the 1% of the mass of the atmosphere lies above and 99% of the mass lies below. Assumptions made are the global mean surface pressure of 1000 hPa is a representative value and the scale height is assumed to be constant throughout the entire atmosphere.

First, we need to calculate the pressure at the desired percentiles (1% and 99%) relative to the surface pressure.

For 1% of the mass lying above, we consider the pressure to be 1% of the surface pressure:

1% of 1000 hPa = 0.01 × 1000 hPa

= 10 hPa.

For 99% of the mass lying below, we consider the pressure to be 99% of the surface pressure:

99% of 1000 hPa = 0.99 × 1000 hPa

= 990 hPa.

Next, we use the exponential relationship between pressure and altitude:

P = P0 × exp(-z/H),

where P is the pressure at a given altitude, P0 is the surface pressure, z is the altitude, and H is the scale height.

To find the altitude z at which the pressure is equal to 10 hPa (1% of the surface pressure), we rearrange the equation:

10 hPa = 1000 hPa × exp(-z/H).

Taking the natural logarithm (ln) of both sides, we have:

ln(10 hPa / 1000 hPa) = -z / H.

ln(0.01) = -z / 8 km.

z = -8 km × ln(0.01).

Evaluating the expression:

z = -8 km × (-4.605)

= 36.84 km.

Therefore, 1% of the mass of the atmosphere lies above an altitude of approximately 36.84 km.

Similarly, to find the altitude z at which the pressure is equal to 990 hPa (99% of the surface pressure), we follow the same procedure:

990 hPa = 1000 hPa × exp(-z/H).

ln(990 hPa / 1000 hPa) = -z / H.

ln(0.99) = -z / 8 km.

z = -8 km × ln(0.99).

Evaluating the expression:

z = -8 km × (-0.01005)

= 0.0804 km.

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Answer:

By watching in the door lock

Answer:

.-. you can flip a switch and look at the bottom or the top of the door and see if there’s a light of there isn’t then flip another one.

also the other chat that we were talking on got deleted

Answer:

kenaoaccff da di rumah aja deh biar aman aja nih orang bikin emosi mudah marah sama aja deh kalo gue juga bingung nih anak gaul banget

Answer:

b

Explanation:

The force of friction is the force that opposes the direction in which a body is moving. It is the force that the earth applies to every object that is close to or rests on its surface. It is a force of touch. Furthermore, it is a force that does not make touch.Thus, option B is correct.

The amount of kinetic energy here is lower than in the case of a frictionless incline because gravitational potential energy is transformed to heat energy and kinetic energy in different proportions.

The force of friction is at work when two surfaces come into contact. It always moves counter to the flow of motion.

Because a portion of the Earth-block system's gravitational potential energy was converted into nonmechanical energy at the top of the slope, the experimental substance did not completely minimize the frictional force.

Therefore, some energy is consumed through friction.

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The maximum differential height of the slide that meets the condition that the children don't travel faster than 5 m/s when they reach the bottom is 4.8 meters. This is calculated using the formula

When a child slides down the slide, the potential energy at the top is converted to kinetic energy at the bottom. When the child reaches the bottom, the kinetic energy is equal to the potential energy at the top. Therefore, the potential energy can be calculated using the formula:PE = mgh, where m is the mass of the child, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the slide. The kinetic energy can be calculated using the formula:KE = 1/2mv², where v is the velocity of the child at the bottom of the slide. Since the slide is frictionless, all the energy at the top is converted to kinetic energy at the bottom, so PE = KE. Therefore, mgh = 1/2mv², and solving for h, we get:h = v²/2g. We want to find the maximum differential height that meets the condition that the children don't travel faster than 5 m/s when they reach the bottom. Therefore, we want to find the maximum value of h that satisfies the condition:v = 5 m/sh = v²/2g = 5²/(2×9.8) = 1.2755 m ≈ 1.28 mThe maximum differential height is the height of the slide minus the height at the bottom of the slide.

Therefore, the maximum differential height is:4 m - 1.28 m = 2.72 mThe maximum differential height of the slide that meets the condition that the children don't travel faster than 5 m/s when they reach the bottom is 2.72 meters.

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Answer:

go girl

Explanation:

i'm feom public uni too

...

Explanation:

.................

Answer:

Solution given:

t=\(2π \sqrt{\frac{m}{g}}\)

where

t=time and it's dimension is [T¹]

m=mass and its dimension is [M¹]

dimension of constant term is nothing.

t=\(2π \sqrt{\frac{m}{g}}\)

squaring on both side

we get

t²=4π²*\(\frac{m}{k}\)

since 4π² has no dimension

Dimension of t²=[T²]

Dimension of m=[M¹]

Dimension of k=?

By using

The principle of homogeneity of dimension we get

Dimension of t²=dimension of \(\frac{m}{k}\)

[T²]=\(\frac{[M¹]}{K}\)

doing crisscrossed multiplication

K=\(\frac{[M¹]}{[T²]}\)

dimension of k is\( [M¹T^{-2}] \)or \([M¹L^0T^{-2}]\)

since length is absent

Answer:

Density =mass/volume 20/10=2

The density of a substance that has a mass of \(20 g\) and a volume of \(10 mL\) is \(2 g/cm^{3}\).

Density of substance is defined as its mass per unit volume. Density gives information about how tightly the substance is packed together. The formula for density is

\(d =\frac{m}{V}\)

Where,

\(d =\) density

\(m =\) mass

\(V =\) volume

Unit of density is grams per cubic centimetre.

One of the most common uses of density is in how different materials interact when mixed together.

Specific Gravity is the ratio of the substance's density to the density of water. An object having specific gravity less than one will float in water, while a specific gravity greater than one will sink.

Given

\(m = 20g\)

\(V = 10 mL= 10 cm^{3}\)

The density of substance is

\(d =\frac{m}{V}\\d =\frac{20}{10}\\d = 2 g/cm^{3}\)

Hence, The density of a substance that has a mass of \(20 g\) and a volume of \(10 mL\) is \(2 g/cm^{3}\).

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1. They have evolved their leaves into spikes for minimum water loss through transpiration.

2. They have a waxy layer for minimum water loss.

3. They have thick walls for minimum water loss.

4. They can take water from atmosphere.

5. They change the photo energy from Sun into an intermediate stage and store it, so that they can make food even in night.

Answer:

v = v +at = 22 + (1.6 6.8 s  ms) = 33 ms

Explanation:

By following these steps, you can accurately determine the temperature of the refrigerant at the compressor exit, which is crucial for understanding the performance and efficiency of the refrigeration system. To determine the temperature of the refrigerant at the compressor exit, you need to follow these steps:

1. Identify the refrigerant being used in the system, as different refrigerants have different properties and behaviors under varying conditions.

2. Obtain the pressure at the compressor exit. This can usually be found by consulting the system's specifications or measuring the pressure using a pressure gauge.

3. Consult the pressure-temperature (P-T) chart for the specific refrigerant being used. This chart provides the relationship between the pressure and temperature of the refrigerant at different points in the system.

4. Locate the exit pressure value on the P-T chart, and find the corresponding temperature. This temperature value will be the temperature of the refrigerant at the compressor exit.

By following these steps, you can accurately determine the temperature of the refrigerant at the compressor exit, which is crucial for understanding the performance and efficiency of the refrigeration system.

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Answer:

Kinetic energy is 2886.5Joule

The graph in the question that represents an object with constant acceleration is : Graph A

Acceleration is the change in velocity with time. For graph A the acceleration remained constant as the velocity of the object did not change with the change in time. As the time changes from 4 to 9 seconds the velocity of the object remained at 40 before decelerating to 0.  This shows that the acceleration of the object was constant.

Hence we can conclude that The graph that shows an object with constant acceleration is : Graph A

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The correct classification of each substance is as follows:

A conductor is a substance that can transmit electricity, heat, light or sound while an insulator is a substance that does not transmit heat, sound or electricity.

Conductors are usually metals while insulators are usually non-metals. Examples of conductors include;

Examples of insulators include;

Therefore, the correct classification of each substance is as follows:

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Answer:

Explanation:

E2020

(A) 300 J of labour are needed to stop the hoop. The hoop has a kinetic energy of \((1/2)mv2 = 300 J.\)  (B) The hoop will roll up the slope for 2.55 m before coming to a stop and rolling back down.

(A) The hoop has a kinetic energy of\((1/2)mv2 = 300 J\). The needed labour is also 300 J since an equivalent amount of work must be done to stop the hoop.

(b) When the hoop ascends the slope, its initial kinetic energy is transformed into potential energy. All of the kinetic energy has been transformed into potential energy at the highest point, making the potential energy there also 300 J. m represents the hoop's mass, and h is the height attained, therefore mgh = 300 J. We arrive at h = 3.06 m after solving for h. The distance climbed up the incline is then calculated using the formula \(d = h/sin(),\)  where is the incline's angle. With the values substituted, we obtain d = 2.55 m.

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Answer:

For any collision occurring in an isolated system, momentum is conserved. The total amount of momentum of the collection of objects in the system is the same before the collision as after the collision.

Explanation:

Hope this helps

Answer:

all of the above

Explanation:

The density of the object will be 60 kg/m³.

Density of any object is a derived quantity as it is obtained from mass and volume. But, it is not a derived unit. So, it is not a derived unit of measure.

Density of a substance is a physical intensive property. It can not be considered as an chemical property. We can measure the density of a  substance without changing of its chemical composition or identities.

Density of a substance gives the idea whether the substance will sink or float. If the density of a substance is more as compared to the liquid, then it will sink. Otherwise it will float.

Density of an object is given by mass/volume.

Here,

mass = 30 kg

Volume = 0.5 m³

Density = 30/0.5 = 60 kg/m³

Therefore the density of an object having mass of 30 kg and volume of 0.5 m³ is given as 60 kg/m³.

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Potential Energy - (Measured in Joules) - Potential energy is the energy that is stored in an object due to its position relative to some zero position.

Mass - (Measured in Kilograms) - Mass is the amount of matter in a body, regardless of its volume or the forces acting on it.

Acceleration Due to Gravity - (Measured in Meter/Square Second) - Acceleration due to gravity is the acceleration gained by an object due to the gravitational force.

Height - (Measured in Meters) - Height is the distance between the lowest and highest points of a standing person/shape/object.

Data:

m = 140 kg

h = 10 m

g = 9.8 m/s²

Ep = ?

The formula to calculate potential energy:

Ep = m * g * h

We clear, but it is not necessary, since we want to calculate the potential energy.

The reason why it is given in Joules is because N * m, are units of Joules, and N = Kg * m/s^2. It is the ratio of the Joules.

The potential energy of the body of mass m = 140 kilograms and located at height h = 10 meters is equal to 13720 Joules.

The point between the two charges where the electric field equal to zero is calculated to be 1.55 m.

The charge q₁ is given as 3 nc = 3 × 10⁻⁹ c

It is located at x₁ = 0.

The charge q₂ is given as 7.5 nc = 7.5 × 10⁻⁹ c

It is located at x₂ = 4 m.

Let us suppose that, electric field is zero at the point x, it is in between 0 and 4 m.

We know the formula for electric field as,

E = k q /r²

Let us equate the electric field at both the charges.

k q₁/x² = k q₂/(x₂ - x)²

q₁ (x₂ - x)² = q₂ x²

3 × 10⁻⁹× (4 - x)² = 7.5 × 10⁻⁹ × x²

3 × ( 16 + x² - 8x) = 7.5 ×x²

48 + 3x² - 24x = 7.5x²

4.5 x² + 24x - 48 = 0

The two values of x are,

x = 1.55, -6.883

The distance cannot be negative. So, the electric field at x = 1.55 m is said to be zero.

Thus, the required value of x is 1.55 m.

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