how to find freezing point 0.1 mole of sodium chloride in 100 ml of water (m.w.= 58.44 g/mol)

Answer:

1st

Explanation:

20 meters per second

Explanation:

look at the chart

Answer:

300m

Explanation:

It is very simple.

200m+100m= 300m

(c) Its water molecules were broken apart, and hydrogen was lost to space.

The evidence that the Venus once used to have water was the deuterium to hydrogen ratio in the atmosphere of Venus is about 150 times of earth.

Proofs are Venus must have started with oceans worth water.

The water is now escaped because when the UV rays from the sun as fallen upon the water the water molecules were broken and hydrogen went to the space.

And therefore, there is no water now present on the planet Venus.

scientist are still finding ways to discover life on the planet Venus and other planets.

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the force between the two wires if they are separated by 2.50 cm is 0.525 N.

Given that force F between two parallel wires carrying electric currents is inversely proportional to the distance d between the wires and that a force of 0.750 N exists between wires that are 1.75 cm apart and that we are supposed to find the force between them if they are separated by 2.50 cm.

Let the initial force be F₁ and the initial distance be d₁.

Therefore, we can write the relationship between force and distance as;

F₁d₁ = F₂d₂

Where

;F₁ = 0.750 N (initial force)

d₁ = 1.75 cm (initial distance)

F₂ = ? (force at new distance)

d₂ = 2.50 cm (new distance)

Let us find F₂;F₁d₁ = F₂d₂F₂ = F₁d₁/d₂

Now substitute the values we know;

F₂ = (0.750 N x 1.75 cm) / 2.50 cmF₂ = 0.525 N

Therefore, the force between the two wires if they are separated by 2.50 cm is 0.525 N.

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Two metric units that is used during the measurement activity to represent volume are m³ ( Cubic meter ) and L ( Litre )

Volume is the space occupied in a three dimensional space. The unit of volume is expressed in length cubed. Litre is used a measure of volume in terms of liquids. 1 litre = 1000 cm³.

Volume is used in different formulas such as:

Therefore, two metric units that is used to represent volume are m³ ( Cubic meter ) and L ( Litre )

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Answer:

the magnitude of the displacement after 5s is 137.31 m.

Explanation:

Given;

initial velocity of the projectile, u = 60 m/s

angle of projection, θ = 60°

time of motion, t = 5s

the vertical component of the velocity, \(u_y= u\ sin \theta = 60sin(60^0)\)

The magnitude of the displacement after 5s is calculated as;

\(h = u_yt -\frac{1}{2} gt^2\\\\h = 60sin (60^0)\times 5 - \frac{1}{2} (9.8)(5)^2\\\\h = 259.81-122.5\\\\h = 137.31 \ m\)

Therefore, the magnitude of the displacement after 5s is 137.31 m.

The vertical acceleration of the ball at its highest altitude is 0 m/s.

It is possible for an item to have both zero acceleration and non-zero velocity because acceleration is the change in velocity. Since there is no change in velocity while an object is moving at a constant speed, it experiences no acceleration.

The bullet reaches its highest point when its vertical velocity is zero. The vertical portion of the velocity vector will now point downward. The projectile's horizontal range, which is determined by the object's beginning velocity, is referred to as the projectile's range.

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Yes, Vincent is correct that 50 g of the sample would remain after 200 years since it has a half life of 50 g.

We know that the term radioactivity has to do with the fact that the nucleus is able to disintegrate and decay into daughter nuclei. The mass of the radioactive nucleus that can be obtained after a given time can be obtained by the use of the first order decay equation.

No we are told that Nickel-63 is a radioactive element that decays to copper-63. Nickel-63 has a half-life of approximately 100 years and Vincent predicts that about 50 grams of a 200-gram sample of nickel-63 remains after 200 years.

If the half life is 100 years then

- 100 g would remain after 100 years

- 50 g would remain after 200 years

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Gravitational fields are generated when two objects interact through the force of gravity.

Gravitational field is the region around an object in which its gravitational force is felt. It is a vector field that describes the strength and direction of the gravitational force at every point in space. The force of gravity is the result of the interaction between two objects, each of which has mass. The gravitational field is produced by any mass, and it is this field that causes objects to move towards each other in a process known as gravity.

This force depends on the mass of the objects, the distance between them, and the gravitational constant, which is a universal constant. Therefore, for a gravitational field to be generated between two objects, the objects must have mass, must be at a finite distance from each other, and the gravitational constant must be nonzero. It is also important to note that the objects must be in free-fall, meaning that they are not being held up by any other force, such as friction. If any of these conditions are not met, then there will not be a gravitational field between the two objects.

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Complete questions as follows-

For a gravitational field to be generated between two objects, what must be true?

No other forces can be presents

The two objects must be in contact

One object must be pure iron

Both objects must have mass

The radio antenna broadcasts a 1M Hz radio wave with 25 kW power.

1) The wave's intensity 30 km from the antenna is 8.842 * 10⁻⁶ W/m².

2) The electric field amplitude at this distance is 81.4 V/m.

The intensity of wave equation is given as

I = P/4πr²

where P is the power of radio wave (25 * 10³ W)

r is the distance from the source (30 * 10³ m)

I = (25 * 10³) / 4π (30 * 10³)² = 0.0000088419 W/m² = 8.842 * 10⁻⁶ W/m²

The relation between intensity of the wave and the electric field is given as

I = 0.5 * c * ε₀ * E²

where, c is the speed of light

ε₀ is the electric permitivity

E is the electric field

Now, let us make E as the subject,

E = √(2I/c ε₀) = √(2 * 8.8 * 10⁻⁶)/(3 * 10⁸ * 8.85* 10⁻¹²) = √(17.6 * 10⁴ / 26.55) = 0.814 * 10² = 81.4 V/m

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a. The probability that a randomly selected soda can has a volume less than 12 fluid ounces is approximately 0.3085

b. The probability that the sample mean volume is less than 12 fluid ounces, when a sample of size n = 32 is taken, is approximately 0.0023

c. The distribution of the sample mean becomes narrower and more concentrated around the population mean. Consequently, the probability of obtaining a sample mean less than 12 fluid ounces decreases because the sample mean is less likely to deviate significantly from the population mean.

a) Let X be the volume of a randomly selected soda can. We are given that the mean (μ) is 12.01 fluid ounces and the standard deviation (σ) is 0.02 fluid ounces.

We need to calculate P(X < 12). To do this, we standardize the variable using the z-score formula:

z = (X - μ) / σ

Substituting the given values, we have:

z = (12 - 12.01) / 0.02

= -0.5

Now, we can use a standard normal distribution table or calculator to find the probability associated with the z-score of -0.5. From the table, we find that the probability is approximately 0.3085.

b) When a sample of size n = 32 soda cans is drawn randomly from the population, the mean volume of the sample (denoted by X-bar) follows a normal distribution with the same mean (μ = 12.01 fluid ounces) but a smaller standard deviation (σ-bar) given by:

σ-bar = σ / sqrt(n)

Substituting the values, we have:

σ-bar = \(0.02 / \sqrt{(32)\)

= 0.02 / 5.6569

≈ 0.00354

Now, we need to calculate P(X-bar < 12). Again, we standardize the variable using the z-score formula:

z = (X-bar - μ) / σ-bar

Substituting the given values, we have:

z = (12 - 12.01) / 0.00354

≈ -2.8249

Using the standard normal distribution table or calculator, we find that the probability associated with the z-score of -2.8249 is approximately 0.0023.

c) As larger and larger sample sizes are taken, the probability that the sample mean volume is less than 12 fluid ounces tends to decrease. This is because as the sample size increases, the sample mean becomes a better estimate of the population mean. The larger the sample size, the more reliable and representative the sample mean is of the true mean. Hence, the sample mean is more likely to be closer to the population mean.

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Answer:

Bacteria

Explanation:

UV kills bacteria and its 8 letters

The ratio of the centripetal acceleration at the end of the blade to that which exists at a point located 6.16 m from the centre of the circle: 1.29

Centripetal acceleration is given by :

a = ω2R

where,

ω is the angular speed

R is the distance of the point from the centre

Let the point at the tip be A and that at distance 6.16 m from centre be B

Distance of point A from centre (RA) = 7.99 m

Distance of point B from centre (RB) = 6.16 m

Centripetal acceleration at A (aA) = ω2RA = 7.99ω2

Centripetal acceleration at B (aB) = ω2RB = 6.16ω2

The ratio (aA/aB) = 7.99/6.16 = 1.29

centripetal acceleration is defined as the motion characteristic of an object moving along a circular path. Any object moving in a circle whose acceleration vector points towards the centre of that circle is called centripetal acceleration. You must have seen various examples of medium acceleration in your daily life. When you drive a car around a circle, your car experiences an average acceleration, and the average acceleration is also observed by a satellite orbiting the Earth. Middle means towards the centre.

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Most helpful Answer~

There are no options~

Anyway If the ball is of bad quality it will get deflated/ or strike out.

' The must reasonable thing that could happen is that the ball will bounce'

*Smile* :)  

All we need to do in order to find out the direction is to take into account the sign of the charge. As they are both positive, the particles will repel each other, which can be seen on the following drawing:

Thus, particle q2 will try to push particle q1 to the left

The answer is: -

Answer: the motion of electric charges

Explanation: because there is relative motion between the charge and someone observing the charge

In the PET (Positron Emission Tomography) technique, detectors record the emission of Positrons from active substrates to form a tomographic image of the cross-sectional distribution of tissue concentration.

Detectors are employed in the PET (Positron Emission Tomography) method to capture the positron emission from active substrates. The subatomic particle known as a proton has the opposite charge to that of an electron but the same mass. They are released into the body by certain radioactive isotopes called positron-emitting radionuclides.

An immediate collision between an emitting positron and an electron is known as annihilation. The positron and electron collide during annihilation, which causes the conversion of their masses into gamma-ray energy. High-energy photons make up these gamma rays. The gamma rays released during annihilation events are intended to be detected and measured by the detectors in a PET scanner.

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The planetary body with the fastest orbit is Mercury, and the one with the slowest orbit is Neptune.

There is a general pattern between orbital velocity and orbital radius known as Kepler's second law of planetary motion. According to this law, a planet sweeps out equal areas in equal times as it orbits the Sun. This implies that planets closer to the Sun have smaller orbital radii and must travel faster to cover the same area in the same amount of time.

The relationship between orbital velocity and orbital radius can be expressed as v ∝ 1/r, where v represents the orbital velocity and r denotes the orbital radius. This relationship shows that as the orbital radius increases, the orbital velocity decreases. In other words, planets farther from the Sun have slower orbital velocities compared to those closer to the Sun.

This pattern is consistent with observations in our solar system. The inner planets, such as Mercury, have smaller orbital radii and faster orbital velocities, while the outer planets, like Neptune, have larger orbital radii and slower orbital velocities.

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Answer:

a) The distance covered by the moving object is approximately 18.85 km

b) The magnitude of the displacement of the object is 0 (no displacement)

Explanation:

From the drawing, we have;

The sale of the drawing is 1 unit = 1 km

The radius of the circle in the drawing along which the object moves = 3 units

Therefore;

The radius of the circle along which the object moves, r = 1 km/unit × 3 units = 3 km

a) The distance covered by the object that moves from point A to B to C to D and finally to A is equal to the circumference of the circle

Therefore;

The distance covered by the moving object, d = 2·π·r

By pugging the value for r = 3 km

d = 2 × π × 3 km = 6·π km ≈ 18.85 km

The distance covered by the moving object, d ≈ 18.85 km

b) The displacement = The object's change in position

The initial position of the object = Point A

The final position of the object = Point A

Therefore, the displacement of the object, A = No displacement = 0 (No difference between the initial and final location of the object

The force that the spring exerts is = 1.31 N

The spring force in the spring depends on the spring's displacement and the force constant of the particular spring. The below expressions show the force that the spring exerts,

\(F=kx\)

Here, we are given,

k= The spring constant of a spring = 8.75 N/m.

x= The displacement of the string is = 0.150 m

Now, we replace the known values in the above equation, we can find :

\(F=kx\)

Or, \(F= 8.75 \times 0.150\)

Or, \(F= 1.3125 N\)

Or, F≈ 1.31 N

Now, from the above calculation we can conclude that the force that the spring exerts is 1.31 N.

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Answer:

acceleration = 2m/s^2

distance= 100 meters

Explanation:

acceleration = (final velocity - initial velocity) / time

acceleration = (20 m/s - 0 m/s) / 10 s

-----------------------------------------------------------------------------------

distance = (initial velocity * time) + (0.5 * acceleration * time^2)

since the car starts from rest, the initial velocity is 0.

distance = 0.5 * 2 m/s^2 * (10 s)^2

In order to determine the probability that an electron can be detected in the middle one-third of a well region, we need to take into account the wave function and the boundary conditions.The wave function represents the probability density of finding the electron in a particular location within the well. The boundary conditions are determined by the geometry of the well, which can be rectangular, triangular, or other shapes.

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Answer:

satellite B

Explanation:

A .F= G (mM)/r²

B .F= G (2mM)/r² = 2G (Mm)r²

C .F= G (3mM)/(2r)² = ¾G (mM)/r²

D .F= G (4mM)/(2r)² = G (mM)/r²

To achieve an acceleration of 9.8 m/s², you should swing the 3.00 kg stone at a speed of approximately 2.71 m/s.

The centripetal acceleration formula, a = (v² / r), relates the acceleration (a) to the velocity (v) and the radius (r) of the circular path. Rearranging the formula to solve for v gives v = sqrt(a * r). In this case, the acceleration is given as 9.8 m/s² and the radius is half the length of the rope, which is 0.75 m. Substituting these values into the formula, we get v = sqrt(9.8 m/s² * 0.75 m) ≈ 2.71 m/s. Therefore, to achieve an acceleration of 9.8 m/s², you should swing the stone at a speed of approximately 2.71 m/s.

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Answer:

7 volts

Explanation:

Applying,

E = IR+Ir................ Equation 1

Where E = electromotive force of the cell, I = current. R = External resistance, r = internal resistance.

From the question,

Given: I = 2 amperes, R = 3 ohm, r = 0.5 ohm

Substitute these values into equation 1

E = (2×3)+(2×0.5)

E = 6+1

E = 7 volts

Answer:

There are several reasons why a hunter might choose to use a shotgun with slugs instead of a rifle.

Cost. Shotguns are typically less expensive than rifles.

Accuracy. Slugs are more accurate than buckshot at longer ranges.

Versatility. Shotguns can be used for a variety of hunting applications, including waterfowl, upland game, and big game.

Recoil. Shotguns have less recoil than rifles, making them easier to shoot for extended periods of time.

Penetration. Slugs can penetrate thicker materials than buckshot, making them a better choice for hunting large game.

Of course, there are also some disadvantages to using a shotgun with slugs.

Range. Slugs are not as effective at long ranges as rifle bullets.

Shotgun spread. Shotguns with slugs do not have a spread like shotguns with buckshot. This can make it more difficult to hit a target at close range.

Shotgun choke. The choke on a shotgun can affect the accuracy of slugs. A rifled choke is the best choice for shooting slugs.

Ultimately, the decision of whether to use a shotgun with slugs or a rifle depends on the individual hunter's needs and preferences.

Explanation:

We can use the formula for electric field between two cylindrical conductors to calculate the maximum electric field magnitude between the cylinders:

E = (V ln(b/a))/d

where V is the potential difference between the conductors, ln is the natural logarithm, b and a are the radii of the outer and inner conductors, respectively, and d is the distance between the conductors.

Given:

V = 600 V

a = 20 mm = 0.02 m

b = 80 mm = 0.08 m

The distance between the conductors is the difference in their radii:

d = b - a = 0.08 m - 0.02 m = 0.06 m

The electric constant, k, is also needed:

k = 8.98755 × 10^9 N·m^2/C^2

Substituting these values into the formula, we get:

E = (V ln(b/a))/d

E = (600 V ln(0.08/0.02))/0.06

E = 3.5983 × 10^8 V/m or approximately 3.60 × 10^8 V/m

Therefore, the maximum electric field magnitude between the cylindrical conductors is approximately 3.60 × 10^8 V/m.

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Answer:

The answer is 'more' as more mass can exert more pressure

Answer:

8.5 Ω

Explanation:

La resistencia de un material es directamente proporcional a su longitud e inversamente proporcional al área de la sección transversal.

La fórmula de la resistencia (R) viene dada por:

R = ρL/A

Donde ρ es la resistividad del material, L es la longitud del material y A es el área de la sección transversal del material.

Dado que:

L = 1 km = 1000 m, A = 2 mm² = 2 * 10⁻⁶ m², ρ (cobre) = 1.7 * 10⁻⁸ Ωm

Sustituyendo da:

R = 1,7 * 10⁻⁸ * 1000/2 * 10⁻⁶

R = 8.5 Ω