find the mass of a football player who's momentum is 630 kg•m/s backward and. is traveling at 7 m/s.
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1.14m far the crate was moved along the floor when a total of 80.5 j of work is done on the crate if the coefficient of kinetic friction is 0.100.

Given the mass of crate (m) = 30kg

The horizontal force applied on crate (f) = 100N

The coefficient of friction between the crate and floor is (µ) =  0.100

The total work done on the crate (W) =  80.5J

The distance the crate was moved along the floor = d

We know that the work done is equal to the force exerted in moving an object to a certain distance such that W = Fd

The force of friction acting on the object Fr = µ*N where N is the normal force = mg such that g = 9.8m/s^2

Then Fr = 0.100 * 30 * 9.8 = 29.4N

The net force exerted on the crate (F) = f - Fr = 100 - 29.4 = 70.6N

distance travelled(d) = W/F = 80.5/70.6 = 1.14m

Hence the distance the crate moved is 1.14m

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1) The missing properties are \(\nu = 0.001093\,\frac{m^{3}}{kg}\), \(u = 639.54\,\frac{kJ}{kg}\) and \(h = 640.59\,\frac{kJ}{kg}\). (Subcooled liquid)

2) The missing properties are \(T = 263.94\,^{\circ}C\), \(\nu = 0.0163\,\frac{m^{3}}{kg}\), \(u = 1719\,\frac{kJ}{kg}\) and \(x = 0.394\). (Liquid-gas mix)

3) The missing properties are \(T = 422.825\,^{\circ}C\), \(\nu =0.0603\,\frac{m^{3}}{kg}\) and \(u = 3251.7\,\frac{kJ}{kg}\). (Superheated steam)

4) The missing properties are \(P = 0.36873\,kPa\), \(u = -337.836\,\frac{kJ}{kg}\), \(h = -337.458\,\frac{kJ}{kg}\) and \(x = 2.988\times 10^{-3}\). (Solid-gas mix)

In this question we must determine the phase associated to the given pressure (\(p\)), in kilopascals, temperature (\(T\)), in degrees Celsius, specific volume (\(\nu\)), in cubic meters per kilogram, internal energy (\(u\)), in kilojoules per kilogram, enthalpy (\(h\)), in kilojoules per kilogram, or quality (\(x\)), no unit.

By steam tables we know that saturation temperature associated with pressure is 151.83 °C, which means that water is a subcooled liquid and quality does not apply herein.

The missing properties are summarized below:

\(\nu = 0.001093\,\frac{m^{3}}{kg}\), \(u = 639.54\,\frac{kJ}{kg}\), \(h = 640.59\,\frac{kJ}{kg}\)

The missing properties are \(\nu = 0.001093\,\frac{m^{3}}{kg}\), \(u = 639.54\,\frac{kJ}{kg}\) and \(h = 640.59\,\frac{kJ}{kg}\). (Subcooled liquid) \(\blacksquare\)

By steam tables we know that water is a mix of gaseous and liquid phases. Hence, the missing properties are summarized below:

\(T = 263.94\,^{\circ}C\), \(\nu = 0.0163\,\frac{m^{3}}{kg}\), \(u = 1719\,\frac{kJ}{kg}\), \(x = 0.394\)

The missing properties are \(T = 263.94\,^{\circ}C\), \(\nu = 0.0163\,\frac{m^{3}}{kg}\), \(u = 1719\,\frac{kJ}{kg}\) and \(x = 0.394\). (Liquid-gas mix) \(\blacksquare\)

By steam tables we know that water is a superheated steam. Hence, missing properties are summarized below:

\(T = 422.825\,^{\circ}C\), \(\nu =0.0603\,\frac{m^{3}}{kg}\), \(u = 3251.7\,\frac{kJ}{kg}\)

The missing properties are \(T = 422.825\,^{\circ}C\), \(\nu =0.0603\,\frac{m^{3}}{kg}\) and \(u = 3251.7\,\frac{kJ}{kg}\). (Superheated steam) \(\blacksquare\)

By ice tables we know that properties correspond to a mix of solid and gaseous phases. Thus, missing properties are summarized below:

\(P = 0.36873\,kPa\), \(u = -337.836\,\frac{kJ}{kg}\), \(h = -337.458\,\frac{kJ}{kg}\), \(x = 2.988\times 10^{-3}\)

The missing properties are \(P = 0.36873\,kPa\), \(u = -337.836\,\frac{kJ}{kg}\), \(h = -337.458\,\frac{kJ}{kg}\) and \(x = 2.988\times 10^{-3}\). (Solid-gas mix) \(\blacksquare\)

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Answer:

A. Capitalist economy

Explanation:

The United States is a capitalist economy. Hope this helps, thank you !!

The slope of the velocity time graph of an object moving with constant acceleration is constant

It will take approximately 3 seconds for the center of mass of Object X to reach point J near the bottom of the incline

The reason why the above time value is correct is given as follows:

Known parameters:

Initial velocity of the objects, u = 0

The graph  in the question is a straight line graph with data points

(0, 0), (0.5, 1.0), (1.0, 2), (3.0, 6), and (3.5, 7)

Given that the slope of the velocity-time graph is constant, we have that the acceleration is constant and is given as follows;

\(a = \dfrac{\Delta v}{\Delta t } = \dfrac{v_2 - v_1}{t_2 - t_1}\)

Therefore;

\(a = \dfrac{6 - 2}{3.0 - 1.0} = 2\)

The acceleration, a ≈ 2 m/s²

The distance from the center of mass of the Object X to the point J near the bottom = 9 m

The equation for distance travelled is given as follows;

\(s = u\cdot t + \dfrac{1}{2} \cdot a \cdot t^2\)

Which gives;

\(9 = 0\times t + \dfrac{1}{2} \times 2 \times t^2 = t^2\)

t = √9 = 3

The time it will take the center of mass of Object X to reach point J near the bottom of the incline is t = 3 seconds

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Printer relies on a photoconductive process is Laser Printer. Option C is the answer.

A laser printer is a type of printer that uses a photoconductive process to produce high-quality text and graphics. The printer contains a cylindrical drum coated with a photoconductive material that is charged with static electricity. As the drum rotates, a laser beam is directed at the surface, discharging the static electricity in certain areas to create a pattern of charged and uncharged regions.

Toner, a fine powder, is attracted to the charged areas of the drum, then transferred to paper and fused by heat. This process produces sharp, precise images with high resolution, making laser printers popular for office and home use. Option C is the answer

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Answer

A spherical bob creates more control for the simple pendulum experiment. An irregular bob like a large piece of paper for instance, will create too much air resistance for a basic classical experiment to yield predictable results within the academic lab.

When a neutral object (such as a lamp) is brought near a charged object (such as a Van de Graaff generator), electrons can be transferred between the two objects. This can result in the neutral object becoming either positively or negatively charged.

In the case where both ends of the lamp are equidistant from the charged Van de Graaff generator, the charges induced in the lamp will be equal and opposite, resulting in a net neutral charge overall. This means that there is no potential difference between the two ends of the lamp, and therefore no current flows through the circuit, and the lamp does not glow.

However, when one end of the lamp is closer to the dome than the other end, the charge induced in the lamp on that end will be greater than on the other end. This creates a potential difference between the two ends of the lamp, and a current can flow through the circuit, causing the lamp to glow.

In summary, the closer proximity of one end of the lamp to the charged Van de Graaff generator creates an imbalance in the induced charges on the lamp, leading to a potential difference and the establishment of a current in the circuit.

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The final velocity of the two freight cars after the collision will be 1.1 m/s.

To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum of an isolated system remains constant. In this case, the two freight cars stick together after the collision, so we can consider them as a single system.

The initial momentum of the system is:

pinitial = m₁ × v₁+ m₂ × v₂

where m₁ and m₂ are the masses of the two freight cars, and v₁ and v₂ are their initial velocities. Substituting the given values, we get:

pinitial = 30 000 kg × 2.2 m/s + 30 000 kg × 0 m/s

= 66 000 kg.m/s

After the collision, the two freight cars stick together and move with a common final velocity, which we can calculate using the conservation of momentum principle as:

pfinal = (m₁ + m₂) × vfinal

where vfinal is the final velocity of the combined system. Since the total momentum is conserved, we have:

pinitial = pfinal

Substituting the given values and solving for vfinal, we get:

66 000 kg.m/s = (30 000 kg + 30 000 kg) × vfinal

vfinal = 66 000 kg.m/s / 60 000 kg

= 1.1 m/s

Therefore, the final velocity of the two freight cars after the collision is 1.1 m/s.

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Metals are opaque to electromagnetic radiation due to their ability to absorb electromagnetic radiation over a wide range of frequencies, particularly in the ultraviolet and visible regions.

Metals are opaque to electromagnetic radiation because they have the ability to absorb electromagnetic radiation over a wide range of frequencies, particularly in the ultraviolet and visible regions. A metal is a good conductor of electricity, which means that the electrons in its atoms are free to move around the metal's structure and form a "sea" of electrons, which is what causes the metal's characteristic luster.

Metallic bonding occurs when the metal atoms share their outermost electrons with each other, forming a crystal lattice. The delocalized electrons move freely throughout the metal lattice, and they are responsible for a metal's electrical conductivity and thermal conductivity.

Metals' ability to absorb electromagnetic radiation is linked to the behavior of their electrons. Electrons in a metal's lattice can be excited by absorbing energy from electromagnetic radiation, and the energy is subsequently lost when the electrons relax back to their original energy states.

Metals are opaque because their electrons absorb so much energy from the electromagnetic radiation that the energy is completely absorbed by the metal before it has a chance to pass through it. Thus, metals appear opaque to most electromagnetic radiation.

This happens when the electrons in a metal's structure absorb photons (particles of light). They take the energy from the photons and use it to move around the metal's structure, which causes the metal to heat up. If the metal absorbs enough photons, it can become hot enough to emit light of its own.

In summary, metals are opaque to electromagnetic radiation due to their ability to absorb electromagnetic radiation over a wide range of frequencies, particularly in the ultraviolet and visible regions.

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Answer:

(a). The decay constant is \(1.55\times10^{-5}\ s^{-1}\)

The half life is 11.3 hr.

(b). The value of N₀ is \(2.38\times10^{11}\ nuclei\)

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity \(R_{0}=10\ mCi\)

Time \(t_{1}=4\ hours\)

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

\(R=R_{0}e^{-\lambda t}\)

\(\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})\)

Put the value into the formula

\(\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})\)

\(\lambda=0.0000154\ s^{-1}\)

\(\lambda=1.55\times10^{-5}\ s^{-1}\)

We need to calculate the half life

Using formula of half life

\(T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}\)

Put the value into the formula

\(T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}\)

\(T_{\dfrac{1}{2}}=44.719\times10^{3}\ s\)

\(T_{\dfrac{1}{2}}=11.3\ hr\)

(b). We need to calculate the value of N₀

Using formula of \(N_{0}\)

\(N_{0}=\dfrac{3.70\times10^{6}}{\lambda}\)

Put the value into the formula

\(N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}\)

\(N_{0}=2.38\times10^{11}\ nuclei\)

(c). We need to calculate the sample's activity

Using formula of activity

\(R=R_{0}e^{-\lambda\times t}\)

Put the value intyo the formula

\(R=10e^{-(1.55\times10^{-5}\times30\times3600)}\)

\(R=1.87\ mCi\)

Hence, (a). The decay constant is \(1.55\times10^{-5}\ s^{-1}\)

The half life is 11.3 hr.

(b). The value of N₀ is \(2.38\times10^{11}\ nuclei\)

(c). The sample's activity is 1.87 mCi.

According to the information we can infer that the travel angle is called a "drag" angle.

In welding terminology, the travel angle refers to the angle between the axis of the electrode and the workpiece surface. When the top of the electrode leads the welding end (meaning the electrode is inclined downward at an angle) and the welding arc is pointing back toward the weld bead, it is known as a "drag" angle.

In this configuration, the electrode is dragged along the surface of the workpiece, creating a trailing arc direction. This technique is commonly used in certain welding processes, such as shielded metal arc welding (SMAW) or stick welding, to control the heat input and the direction of the molten metal flow during the welding operation.

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The correct answer is(b). There is a physical change from the solid state to a gas state where energy between the molecules increases.

During sublimation, carbon dioxide transitions directly from the solid state (dry ice) to the gas state without passing through the liquid phase. In this process, individual molecules of carbon dioxide gain energy from the surroundings, leading to an increase in their kinetic energy and molecular movement.

As a result, the carbon dioxide molecules separate from each other and form a gas. This phase change is considered a physical change rather than a chemical change since the chemical composition of carbon dioxide remains the same throughout the process.

Therefore, the sublimation of carbon dioxide results in a physical change where the molecules transition from the solid state to the gas state, gaining energy and increasing their molecular movement.

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A superheated cloud of gas and dust swirling around a massive object (usually a black hole or neutron star) is called an accretion disk. The correct option is c.

Accretion disk is a heavy object surrounded by flow of gas, plasma, dust, or particles.  The materials surrounding the heavy object orbiting in the gravitational field of the object that loses energy and angular momentum as it slowly spirals inward.

This cloud can reach incredible speeds and temperatures as gravity causes it to build up more and more friction, eventually releasing x rays.

Thus, accretion disk is a superheated cloud of gas and dust swirling around a massive object.

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Answer:

KE=800,000

Explanation:

The formula for kinetic energy is KE=1/2mv^2 or Kinetic Energy= 0.5*mass*velocity^2

so 1000 is the mass and 40 is the velocity

KE=0.5*1000*40^2

KE=0.5*1,000*1,600

KE=800,000 Joules

Answer:The first spacecraft which did not merely fly by a jovian (or giant) planet, but actually went into orbit around it for an extended period of time was option a, Galileo. The Galileo spacecraft was launched in 1989 and orbited Jupiter for almost eight years, from 1995 to 2003.

Explanation:

An alternative way to explain Hubble's Law without relying on the standard Big Bang idea of a universe with no center and no edge is through the concept of a "tired light" hypothesis.

According to this hypothesis, the redshift observed in distant galaxies is not due to their motion away from us in an expanding universe, but rather the result of photons gradually losing energy as they travel through space.

In the tired light scenario, the wavelength of light is thought to stretch over time due to interactions with matter or other unknown processes. As a result, the light from distant galaxies appears redshifted, mimicking the effect of cosmic expansion. This explanation implies that the universe may have a finite size and a center, challenging the notion of an infinite and uniformly expanding universe.

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The kinetic energy of the electron just as it reaches the positive plate of the parallel plate capacitor is \(1.01 \times 10^{-16}\) Joules.

The electric field between the plates of a parallel plate capacitor can be calculated using the formula:

\(\[E = \frac{V}{d}\]\)

where E is the electric field, V is the voltage, and d is the separation between the plates. Rearranging the formula to solve for V, we get:

\(\[V = E \times d\]\)

Substituting the given values, we have:

\(\[V = (2.4 \times 10^6 \, \text{V/m}) \times (1.9 \times 10^{-2} \, \text{m}) = 4.56 \times 10^4 \, \text{V}\]\)

The potential difference V represents the change in electric potential energy of the electron as it moves from the negative plate to the positive plate. This change in potential energy is converted into kinetic energy as the electron accelerates. The equation for the kinetic energy of the electron can be expressed as:

\(\[K.E. = q \times V\]\)

where q is the charge of the electron. The charge of an electron is \(\(1.6 \times 10^{-19}\)\) Coulombs. Substituting the values, we get:

\(\[K.E. = (1.6 \times 10^{-19} \, \text{C}) \times (4.56 \times 10^4 \, \text{V}) = 1.01 \times 10^{-16} \, \text{Joules}\]\)

Therefore, the kinetic energy of the electron just as it reaches the positive plate is \(\(1.01 \times 10^{-16}\)\) Joules.

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Answer:

By raising her center of mass just at the beginning of the arc, the skater gains energy and thereby increases her speed. Pumping on a skateboard in this way is closely related to pumping on a swing.

Explanation:

I hope this helps you!

(ps: can i plz have the brainlyest answer?)

Answer:

They provide structure and support.

Explanation:

A. Its kinetic energy increases occurs as the potential energy of an object decreases.

The mass and height of an item affect how much gravitational potential energy it has. More gravitational potential energy is held by an item the heavier it is and the higher it is above the surface of the earth. As weight and height grow, gravitational potential energy rises.

The mass of the item has a significant role in determining the gravitational potential energy. The mass and potential energy are intimately related. As a result, an item with higher mass has more potential energy.

A falling object loses potential energy while gaining kinetic energy. The difference between the increase in kinetic energy and the reduction in potential energy is exact. Work is a further crucial idea.

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Answer:

Since the waves must carry a great deal of visual as well as audio information, each channel requires a larger range of frequencies than simple radio transmission. TV channels utilize frequencies in the range of 54 to 88 MHz and 174 to 222 MHz. (The entire FM radio band lies between channels 88 MHz and 174 MHz.)

Answer:

Since the waves must carry a great deal of visual as well as audio information, each channel requires a larger range of frequencies than simple radio transmission. TV channels utilize frequencies in the range of 54 to 88 MHz and 174 to 222 MHz. (The entire FM radio band lies between channels 88 MHz and 174 MHz.)

Explanation:

Ok

Answer:

\(velocity = \frac{distance}{time} \\ velocity = \frac{10}{12.20} \\ velocity = \frac{50}{61} \\ multiply \: by \: 3.6u \: can \: convert \: to \: km \: per \: hour \\ velocity = \frac{50}{61} \times 3.6 \\ v = \frac{180}{61} \\ v = 2.95 \frac{km}{h} \\ thank \: you\)

Answer:

Jsdiwkwdmxxj we

Explanation:

hcndjnjwidjxnwnjwi

Generally, the coefficient of static friction should be at least 0.7 for a car traveling at this speed, but this value could be higher depending on the specific conditions.

The coefficient of static friction, or μs, is the ratio of the maximum static force that can be applied without causing slipping to the normal force between two surfaces. The coefficient is affected by the surfaces in contact and the material they are made of.

The tires of a car also have an effect on the coefficient of static friction. Tires with a greater tread depth can provide better grip on the road and therefore have a higher coefficient of static friction than tires with a shallow tread. Tire pressure also plays a role, as low pressure can reduce the coefficient of static friction.

The weight of the vehicle is also an important factor. Heavier vehicles tend to have a higher coefficient of static friction, as the greater mass increases the normal force between the car and the surface.

In conclusion, the coefficient of static friction for a car traveling at 115 km/h around a turn depends on several factors, including the road surface, the vehicle's tires, and the vehicle's weight.

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It is 1.93 metres away from the post.

for vertical motion

\(S= u t + \frac{1}{2} g t^{2}\)

S = 0 +  \(\frac{1}{2} g t^{2}\)

1.7 = \(\frac{1}{2} 9.8 t^{2}\)

t = 0.589 sec

using law of  conservation of momentum, on the horizontal motion

\(m_{bullet} \ v_{bullet i} = m_{bullet} \ v_{bullet f} + m_{tin} \ v_{tin}\)

\(0.0020 = 0.0020 \ 3 \ 720 + 0.11 \ v_{tin}\)

\(v_{tin}\) = 3.273 m/s

S = \(v_{tin} \ t\)

   = 3.273 m/s * 0.589 s

   = 1.93 m

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It is important to note that the debate between scientific realism and antirealism is ongoing and complex, with various nuances and perspectives within each position. Different philosophers of science and scientists may hold different views on the nature of scientific terms and their relationship to reality.

According to a scientific realist perspective, scientific terms for unobservable phenomena such as "atom" and "black hole" are seen as referring to entities that truly exist in reality. Scientific realists believe that scientific theories and concepts accurately capture aspects of the world, including unobservable entities and phenomena. They argue that scientific theories provide the best explanation of the natural world and aim to describe the underlying structure and mechanisms of reality.

On the other hand, scientific antirealists hold a different view. They argue that scientific terms that refer to unobservable phenomena do not necessarily correspond to something that exists independently in reality. Antirealists often emphasize the instrumentalist view of science, which suggests that scientific theories are simply tools or frameworks that help us organize and predict observable phenomena, without making claims about the ultimate nature of reality.

Antirealists may argue that scientific theories are subject to revision and change over time as new evidence emerges, suggesting that the terms used to describe unobservable phenomena are not fixed and may not have a one-to-one correspondence with actual entities in reality. They may also highlight the role of social and cultural factors in shaping scientific knowledge, suggesting that scientific terms are influenced by human conventions and interpretations.

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A spherical object orbiting the sun that has objects of similar mass orbiting nearby or crossing its path is a dwarf planet.

A celestial entity that circles the sun and has enough mass to produce a spherical shape but has not completely purified its orbit of other trash or smaller particles is referred to as a dwarf planet. Dwarf planets, in contrast to planets, are said to circle other celestial bodies like asteroids and comets. Five dwarf planets in our solar system have been designated by the International Astronomical Union (IAU) : Ceres, Pluto, Haumea, Makemake, and Eris. These objects are categorized as a different category of solar system bodies because they are thought to be distinct from planets.

While dwarf planets have certain characteristics with planets, they lack the gravitational dominance necessary to be considered genuine planets. They are sufficiently massive to cause their self-gravity to draw them in a nearly circular form, but they have not yet swept other debris from their orbit.

Hence, a spherical object orbiting the sun that has objects of similar mass orbiting nearby or crossing its path is a dwarf planet.

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A fish appears closer to the surface of the water than it really is when observed from a riverbank. Refraction is the bending of light as it passes from one medium to another with a different optical density. In the case of the fish appearing closer to the surface of the water, this is an example of refraction because the light rays coming from the fish underwater undergo bending at the air-water interface.

When light passes from water (a denser medium) to air (a less dense medium), it changes its direction due to the difference in optical density. This bending of light causes the fish to appear higher or closer to the surface of the water than its actual position.

The other options mentioned in the question are not examples of refraction:

A parabolic mirror in a headlight focusing light into a beam is an example of reflection and focusing of light using a curved surface.

Light bending slightly around corners is an example of diffraction, not refraction.

In a mirror, when you lift your right arm, the left arm of your image is raised. This is an example of lateral inversion due to reflection, not refraction.

Therefore, the correct example of refraction from the given options is when a fish appears closer to the surface of the water than it really is when observed from a riverbank.

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