"The White Shark" allows riders to start from rest on a tube and then slide down a 44 meter slide. It takes the rider 6.2 seconds to reach the bottom. What is the average acceleration of the ride?
A. 2.3 m/s^2 B. 3.4 m/s^2 C. 4.2 m/s^2 D. 5.0 m/s^2

I did (44 - 0)/6.2 but that's not any of the answers??

The form of mechanical weathering is exfoliation, which refers to the peeling off of the outer layers of a rock due to physical forces. Hydration, carbonation, and oxidation represent forms of chemical weathering.

The weathering process that exemplifies a form of mechanical weathering is option C, exfoliation. Mechanical weathering, also known as physical weathering, refers to the process where rock is broken down into smaller pieces by physical forces without any changes in its chemical composition.

Exfoliation is a form of mechanical weathering that occurs when the outer layers of rock peel off in layers due to differential heating and cooling, or freeze-thaw cycles. In contrast, options A (hydration), B (carbonation), and D (oxidation) all depict processes of chemical weathering, wherein the rock's mineral composition itself changes.

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Answer:

(a) Emerging adulthood.

Explanation:

Emerging adulthood is a period between teenagers' dependence on guardians and grown-ups' long haul duties in adoration and work, and during these years, rising grown-ups center around themselves as they build up the information, abilities, and self-understanding they will requirement for grown-up life.

So the correct option is (a)

For an object starting from rest and accelerating with a constant acceleration, the distance travelled is proportional to the square of the time. So, if an object travels 2.0 furlongs in the first 2.0 seconds, it will travel 8.0 furlongs in the first 4.0 seconds.

In the problem, we have an object accelerating from rest and we know that distance covered is proportional to the square of the time. In the first 2.0s, the object has covered a distance of 2.0 furlongs. Hence, if we double the time from 2.0s to 4.0s, since distance covered is proportional to the square of time, the object will cover 2*(2.0s)^2 = 8.0 furlongs in the first 4.0s. This relationship is based on the physics principle of displacement which defines that in constant acceleration, distance travelled is proportional to square of time.

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A driver sets out on a journey. for the first half of the distance she drives at the leisurely pace of 30 mi/h; during the second half she drives 60 mi/h, the driver's average speed on this trip is 40 mi/h.

To find the average speed for the entire trip, we can use the formula for average speed:

Average Speed = Total Distance ÷ Total Time.

In this case, the driver covers the first half of the distance at 30 mi/h and the second half at 60 mi/h.

Let's assume the total distance of the trip is [tex]\(D\)[/tex] miles. The first half of the distance is [tex]\(D/2\)[/tex] miles, and the second half of the distance is also [tex]\(D/2\)[/tex] miles.

Let's calculate the time taken for each half of the distance:

Time taken for the first half = Distance / Speed

= [tex]\((D/2) \, \text{miles} / (30 \, \text{mi/h})\)[/tex].

Time taken for the second half = Distance / Speed

= [tex]\((D/2) \, \text{miles} / (60 \, \text{mi/h})\).[/tex]

Total time for the trip = Time for the first half + Time for the second half.

Total time = [tex]\((D/2) / (30) + (D/2) / (60)\)[/tex].

Total time = [tex]\(\dfrac{D}{60} + \dfrac{D}{120}\)[/tex].

Total time = [tex]\(\dfrac{3D}{120} = \dfrac{D}{40}\)[/tex].

Now we can calculate the average speed:

Average Speed = Total Distance / Total Time.

Average Speed = [tex]\(D / \dfrac{D}{40}\)[/tex].

Average Speed = [tex]\(40\) mi/h[/tex].

Thus, the driver's average speed on this trip is 40 mi/h.

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The driver's average speed on this trip is 30 mi/h.

To find the average speed on this trip, we can use the formula: average speed = total distance / total time. Let's assume the total distance is D miles. According to the given information, the driver spends the first half of the distance at a speed of 30 mi/h and the second half at a speed of 60 mi/h. Therefore, the total time taken for the trip is D / (30 mi/h) + D / (60 mi/h) = D / (1/30 h) + D / (1/60 h) = 2D / (1/60 h) = 120D h. We can substitute this value into the average speed formula and simplify: average speed = D / (120D h) = 1/120 h. So, the driver's average speed on this trip is 1/120 h = 0.00833 h = 0.00833 x 60 min = 0.5 min/h = 0.5 x 60 = 30 mi/h. Therefore, the driver's average speed on this trip is 30 mi/h.

Answer:

Mass, m = 2000

Explanation:

Given that,

Force acting on the truck, F = 2000 N

Acceleration of the truck, [tex]a=1\ m/s^2[/tex]

To find,

The mass of the car

Solution,

The second law of motion gives the relationship between the mass force and the acceleration. It is given by :

[tex]F=ma[/tex]

[tex]m=\dfrac{F}{a}[/tex]

[tex]m=\dfrac{2000\ N}{1\ m/s^2}[/tex]

m = 2000 kg

So, the mass of the car is 2000 kg.

The best answers to complete this sentence would be the following:

“sidereal day”

“upper median”

It keeps the time, in the two successive upper meridian of the sun during the crossing. Also it is when the star crosses the celestial meridian.

Explanation:

“sidereal day”

“upper median”

It keeps the time, in the two successive upper meridian of the sun during the crossing. Also it is when the star crosses the celestial meridian.

Hope this helps!      

The minimum length runway this aircraft can use is about 4.1 km

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

Let us now tackle the problem !

Let's look at the table in the attachment

Given:

u = 0 m/s

v = 79 m/s

a = 23 / 10 = 2.3 m/s²

Unknown:

d = ?

Solution:

Let's calculate the takeoff distance.

[tex]v^2 = u^2 + 2as[/tex]

[tex]79^2 = 0^2 + 2(2.3)s[/tex]

[tex]6241 = 4.6s[/tex]

[tex]s = 6241 \div 4.6[/tex]

[tex]s \approx 1400 ~ m[/tex]

The length of the runway must be three times the takeoff distance.

[tex]d = 3s[/tex]

[tex]d = 3(6241 \div 4.6)[/tex]

[tex]d \approx 4100 ~ m[/tex]

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate

To determine the minimum runway length for an aircraft, multiply the takeoff distance by three. If the takeoff distance is known, this simple multiplication will provide the minimum runway requirement.

The question asks for the minimum runway length required for an aircraft, given that the runway must be three times the takeoff distance for safety reasons in the event of an aborted takeoff. To determine this, if we know the takeoff distance required for the aircraft, we simply multiply that distance by three to find the minimum runway length the aircraft can use. For example, if an aircraft requires 1 kilometer (1,000 meters) to take off, the minimum runway length must be 3 kilometers (3,000 meters) for safety reasons.

Answer:

Force of Earth will be dominating and the object will fall towards Earth

Explanation:

As we know that

Mass of Earth

[tex]M_e = 5.98 \times 10^{24} kg[/tex]

Mass of Moon

[tex]M_m = 7.35 \times 10^{22} kg[/tex]

since we know that gravitational force depends on mass and the distance between two objects

so here if an object is placed midway between Moon and Earth then as we can see that mass of Earth is approx 100 times more than the mass of Moon

So here we can say that Force of Earth will be dominating and the object will fall towards Earth

Buoyant force due to the surrounding air on a man is 0.84 Newton

[tex]\texttt{ }[/tex]

The basic formula of pressure that needs to be recalled is:

Pressure = Force / Cross-sectional Area

or symbolized:

[tex]\large {\boxed {P = F \div A} }[/tex]

P = Pressure (Pa)

F = Force (N)

A = Cross-sectional Area (m²)

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

Density of Air = ρ_air = 1.20 kg/m³

Weight of the man = w = 700 N

Density of the man = ρ = 1000 kg/m³

Asked:

Buoyant Force = F = ?

Solution:

We will use Archimedes' principle to solve the problem as follows:

[tex]F = \rho_{air} g V[/tex]

[tex]F = \rho_{air} g \frac{m}{\rho}[/tex]

[tex]F = \rho_{air} g \frac{w}{g\rho}[/tex]

[tex]F = \rho_{air} \frac{w}{\rho}[/tex]

[tex]F = 1.20 \times \frac{700}{1000}[/tex]

[tex]F = 0.84 \texttt{ Newton}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Pressure

[tex]\texttt{ }[/tex]

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant , Liquid , Pressure

The buoyant force due to the surrounding air on a man weighing 700 N is approximately 0.84 N. The ratio of the buoyant force to his weight is approximately 0.00120, which shows the buoyancy effect in air is minimal compared to his weight.

To calculate the buoyant force due to the surrounding air on a man weighing 700 N (equivalent to a mass of approximately 71.4 kg assuming g = 9.81 m/s2), we can use Archimedes' principle. This principle states that the buoyant force on an object immersed in a fluid is equal to the weight of the fluid displaced by the object. Since the man's average density is the same as that of water (about 1000 kg/m3), his volume V can be calculated using the formula:

V = mass / density = 71.4 kg / 1000 kg/m3 = 0.0714 m3.

The buoyant force (Fb) in air can then be calculated with the density of air (1.20 kg/m3):

Fb = density of air × volume × g = 1.20 kg/m3 × 0.0714 m3 × 9.81 m/s2 ≈ 0.841 N.

Therefore, the buoyant force is approximately 0.84 N.

To find the ratio of the buoyant force to the man's weight, we divide the buoyant force by the weight:

Ratio = Fb / weight = 0.841 N / 700 N ≈ 0.00120.

So, the ratio of the buoyant force to the man's weight is approximately 0.00120, which implies the effect of buoyancy in air is quite small compared to the weight of the man.

Final answer:

A vehicle traveling at 55 miles per hour covers approximately 80.67 feet in one second, calculated by converting miles per hour to feet per second.

Explanation:

To find out how many feet a vehicle travels in one second at a speed of 55 miles per hour, we can perform a unit conversion from hours to seconds. First, we need to determine how many feet are in a mile and then convert miles per hour to feet per second.

There are 5,280 feet in a mile, therefore:

Since there are 3,600 seconds in an hour, we divide the total feet per hour by the number of seconds in an hour to find the distance in feet per second.

So, a vehicle traveling at 55 miles per hour travels approximately 80.67 feet in one second.

One complete orbit of the earth around the sun is 365 and ¼ days. Because of this, the earth has to completely orbit around the sun in respect to the stars too no the sun only and so the earth spins 366.26 times every rotation. The ratio of the earth's orbital period about the sun to the earth's period of rotation about its own axis will then be 1 is to 366.26.

Final answer:

The ratio of the Earth's orbital period about the sun to its period of rotation about its axis is 365.26:1.

Explanation:

The ratio of the earth's orbital period about the sun to the earth's period of rotation about its own axis is 365.26 days to 1 day.

This means it takes approximately 365.26 days for the Earth to orbit the Sun once, while it takes 1 day for the Earth to complete one full rotation about its axis.

Therefore, the ratio can be simplified to 365.26:1.

Compared to energy-flow, which enters ecosystems as sunlight and leaves as heat, the flow of matter is continually recycled and conserved, obeying the law of conservation of mass.

Compared to energy-flow in ecosystems, the flow of matter is conserved and recycled. While energy enters an ecosystem, typically in the form of sunlight, and is eventually dissipated as heat, matter circulates within the ecosystem through various biotic and abiotic processes. The law of conservation of mass supports the notion that matter is neither created nor destroyed, but rather continuously reused and transformed. Substances like water, carbon, and nitrogen undergo recycling through ecosystems; essential for life, these elements are integral components of the food web, influencing the distribution and abundance of organisms.

The binding energy of electrons to the magnesium surface can be calculated using the equation KE = hf - BE, where KE is the kinetic energy of the ejected electrons, hf is the energy of the incident photons, and BE is the binding energy. Given that the maximum velocity of the ejected electrons is 3.45 × 10^5 m/s, we can calculate the kinetic energy using the equation KE = (1/2)mv^2. Using the given wavelength of the photons (310 nm), we can calculate the energy of the photons using the equation E = hc/λ, where h is Planck's constant (6.63 × 10^-34 J.s), c is the speed of light (3.00 × 10^8 m/s), and λ is the wavelength in meters. By rearranging the equation to solve for the binding energy, we find that the binding energy is equal to the energy of the incident photons minus the kinetic energy of the ejected electron.

The binding energy of electrons to the magnesium surface can be calculated using the equation KE = hf - BE, where KE is the kinetic energy of the ejected electrons, hf is the energy of the incident photons, and BE is the binding energy. Given that the maximum velocity of the ejected electrons is 3.45 × 10^5 m/s, we can calculate the kinetic energy using the equation KE = (1/2)mv^2.

Using the given wavelength of the photons (310 nm), we can calculate the energy of the photons using the equation E = hc/λ, where h is Planck's constant (6.63 × 10^-34 J.s), c is the speed of light (3.00 × 10^8 m/s), and λ is the wavelength in meters.

By rearranging the equation to solve for the binding energy, we find that the binding energy is equal to the energy of the incident photons minus the kinetic energy of the ejected electrons.

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The binding energy of electrons to the magnesium surface is calculated to be approximately 0.62 eV.

To determine the binding energy of electrons ejected from a magnesium plate by photons with a wavelength of 310 nm, we use the photoelectric effect equation:

Calculate the energy of the photon (E(photon)):

Given:

Convert the photon energy from joules to electron volts (eV):

Calculate the kinetic energy (Ke) of the ejected electrons:

Convert the kinetic energy into electron volts (eV):

Calculate the binding energy (Eb):

Therefore, the binding energy of electrons to the magnesium surface is approximately 0.62 eV.

The direction of the induced magnetic field caused by this current will be the same direction as the external field.

As the external magnetic field decreases, an induced current flows in the coil. The direction of the induced magnetic field would be pointing to the screen.

The flux through the coil is said to decrease. In order to counter this change, the coil would generate or produce a magnetic field that is induced that would be pointing in the same direction as the external field that is flowing which is into the screen.

This is according to Lenz's law or the right-hand rule. It states that an induced current in a circuit that is due to the change or motion in the magnetic field should be directed opposing the change in the flux.

hence the direction of the induced magnetic field caused by this current will be the same direction as the external field.

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Answer:

C. 1,314,718 J

Explanation:

The heat energy needed to raise the temperature of the coal is given by:

[tex]Q=m C_s \Delta T[/tex]

where:

m = 5 kg is the mass of the coal

[tex]C_s = 1314 J/kg ^{\circ}C[/tex] is the specific heat of coal

[tex]\Delta T= 220^{\circ}C-20^{\circ}C=200^{\circ}C[/tex] is the increase in temperature

Substituting into the formula, we find

[tex]Q=(5 kg)(1314 J/kg ^{\circ}C )(200^{\circ}C)=1,314,000 J[/tex]

So, the closest option is

C. 1,314,718 J

The information, most important when passing near a lighthouse is the

Water's depth or depth of the water.

A lighthouse is simply a structure that emits a bright light that provides navigators with a constant or intermittent signal.

In conclusion, When passing near a lighthouse we consider looking at the water's depth around the area

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The force of impact is the same as driving your vehicle off a 10.0 story structure.

Given the following data:

Conversion:

In this exercise, you're required to compare the force of impact with an equivalent height. Thus, we would use the following formula to calculate the height:

[tex]H = \frac{V^2}{2g}[/tex][tex]H = \frac{V^2}{2g}[/tex]

Where:

Substituting the parameters into the formula, we have;

H = \frac{26.82^2}{2(9.8)}

H = 36.70 meters.

Assuming a distance of 3.6 meters:

Height = \frac{36.70}{3.6}

Height = 10.0 meters.

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Answer:

a

Explanation:

Part A: Make a free-body diagram of this ball just before it hits the wall.

The only force acting on the ball is the pull of the Earth, this is its weight; so the diagram is a vertical vector downwards.

Part B: Make a free-body diagram of this ball just after it has bounced free of the wall.

Again, the only force acting on the ball is the pull of the Earth, its weight, and the free-body diagram is identical to that of the part A.

Part C: Make a free-body diagram of this ball while it is in contact with the wall.

The total gauge pressure at the bottom of the cylinder would simply be the sum of the pressure exerted by water and pressure exerted by the oil.

The formula for calculating pressure in a column is:

P = ρ g h

Where,

P = gauge pressure

ρ = density of the liquid

g = gravitational acceleration

h = height of liquid

Adding the two pressures will give the total:

P total = (ρ g h)_water + (ρ g h)_oil

P total = (1000 kg / m^3) (9.8 m / s^2) (0.30 m) + (900 kg / m^3) (9.8 m / s^2) (0.4 - 0.30 m)

P total = 2940 Pa + 882 Pa

P total = 3,822 Pa

Answer:

 The total gauge pressure at the bottom is 3,822 Pa.

The total gauge pressure at the bottom of cylinder due to the oil and the water is  [tex]\boxed{3822\,{\text{Pa}}}[/tex].

Further Explanation:

Given:

The water in the cylinder is up to the height of 30 cm .

The total height of the liquid column in the cylinder is 40 cm .

The density of oil is  [tex]900\,{{{\text{kg}}}\mathord{\left/{\vphantom {{{\text{kg}}} {{{\text{m}}^{\text{3}}}}}}\right.\kern-\nulldelimiterspace} {{{\text{m}}^{\text{3}}}}}[/tex].

Concept:

The gauge pressure is the amount of pressure exerted by the liquid column on the surface below it.

The gauge pressure due to the height of the liquid pressure is given by:

[tex]P =\rho gh[/tex]

Here, [tex]P[/tex] is the gauge pressure, [tex]\rho[/tex] is the density of the liquid, g is the acceleration due to gravity and [tex]h[/tex] is the height of the liquid column.

The height of the oil present in the cylinder is:

[tex]\begin{aligned}{h_{oil}}&={h_{total}} - {h_{water}}\\&= 40 - 30\,{\text{cm}}\\&=10\,{\text{cm}}\\&\approx {\text{0}}{\text{.1}}\,{\text{m}}\\\end{aligned}[/tex]

The total gauge pressure at the bottom of the cylinder will be:

[tex]{P_{total}} = {\left({\rho gh}\right)_{water}} + {\left( {\rho gh}\right)_{oil}}[/tex]

Substitute the values in the above expression.

[tex]\begin{aligned}{P_{total}}&=\left({1000 \times 9.8 \times 0.30} \right) + \left( {900 \times 9.\times 0.1}\right)\\&=2940 + 882\\&=3822\,{\text{Pa}}\\\end{aligned}[/tex]

Thus, the total gauge pressure at the bottom of cylinder due to the oil and the water is [tex]\boxed{3822\,{\text{Pa}}}[/tex]

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Answer details:

Grade: High School

Subject: Physics

Chapter: Pressure

Keywords: Gauge pressure, water, oil, bottom of cylinder, 30cm of water, density of oil, 900kg/m^3, poured, floats on top, total liquid depth.

Since the system itself is giving off heat, this is a reduction in the internal energy.

heat = - 25,000 J

Since work is being done on the system, therefore it is an additional energy to the system. Work is given as:

work = - P dV

work = - 1.50 atm (6 L – 12 L)

work = 9 L atm

Since it is given that 1 L atm is equivalent to 101.3 J, therefore the total energy added is:

energy due to work = 9 L atm (101.3 J / 1 L atm)

energy due to work = 911.7 J

Therefore the total change in internal energy is the sum of heat and energy due to work:

Change in internal energy = - 25,000 J + 911.7 J

Change in internal energy = - 24,088.3 J

Therefore, approximately 24.1 kJ of energy is lost by the system in the total process.

Answer:

-24.1 kJ

The correct option is d. which is the change in internal energy is -24.9 kj.

As work is being done on the system, meaning additional energy is provided to the system. Therefore,

[tex]\begin{aligned}W&=- \int\limits {P} \, dv\\&=-P\ dv\\\\&=-P\ (V_f-V_i)\\&=- (1.50) (6-12)\\&= 9\ l\cdot atm\end{aligned}[/tex]

Also,

[tex]1\ l\cdot atm=101.3\ j\\9\ l\cdot atm=101.3\times 9\\9\ l\cdot atm=911.7\ j[/tex]

[tex]\Delta U = Q - W\Delta\\\\where,\\ U = change\ in\ internal\ energy\\Q = heat\ added\\W = work\ done\ by\ the\ system[/tex]

Substituting the values,

[tex]\begin{aligned}\\\Delta U&= Q - W\Delta\\&= -25000+911.7\\&=-24,088.3\ j\\&= -24.0883\ kj\end{aligned}[/tex]

Hence, the change in internal energy is -24.0883 kj.

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