This question deals with air track experiment and collisions of the track cars. Was it necessary to have equal length intervals in the experiment to investigate properly the conservation of momentum? Explain. Please be as elaborate as you can.

Answer:

Explanation:

Since we are looking at speed over a time unit, we have an acceleration graph. The definition of acceleration mathematically is:

\(a=\frac{v_f-v_0}{t}\) (that's the change in velocity over the change in time). The slope of any line in this graph wil represent the acceleration. Slope is rise over run, or y over x. Therefore, if acceleration is velocity over time, then the y axis is the velocity axis and the x axis is the time axis. It makes perfect, beautiful sense!!

Answer:

7.8 N

Explanation:

Applying,

I = Ft................. Eqaution 1

Where I = Impulse on the egg test dummy, F = Force on the egg test dummy during the time interval, t = time interval

make F the subject of the equation

F = I/t.................. Equation 2

From the question,

Given: T = -0.39 N.s, t = 0.050 s

Substitute these vales into equation 2

F = -0.39/0.050

F = -7.8 N

Hence the force that act on the egg test dummy is 7.8 N

Answer:

Work, in physics, measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement.To express this concept mathematically, the work W is equal to the force f times the distance d, or W = fd.

We would apply the equation of continuity. Recall, it states that in case of flow of an incompressible fluid, the rate of flow at any time is constant at all points. The volume of fluid entering a section of flow in a period of time equals the volume of fluid that leaves at that same period of time. This means that

A1V1 = A2V2

where

A1 and A2 are the areas of the inlet and outlet

V1 and V2 are the velocities at inlet and outlet

AV is the volume flow rate in m^3/s

From the information given,

radius of inlet = 1 cm

Recall,

1 cm = 0.01 m

thus, radius of inlet = 0.01

Also,

Area of circle = pi x radius^2

Area of inlet = 3.14 x 0.01^2 = 0.000314

V1 = 3.84

A1V1 = 0.000314 x 3.84 = 0.00121

inner radius of outlet = 0.2 cm

Converting to meters, inner radius = 0.2 x 0.01 = 0.002

Area = 3.14 x 0.002^2 = 0.00001256

A2V2 = 0.00001256V2

Thus,

0.00121 = 0.00001256V2

V2 = 0.00121/0.00001256 = 96.33

The outlet velocity is 96.33 m/s

(a)The pressure at B is 0.1248 atm.

(b)The temperature at C is 727.1 K.

(c)The total work done by the gas in one cycle is -1979J

General calculation:

We can use the First Law of Thermodynamics to analyze the heat engine cycle:

ΔU = Q - W

where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system. For a complete cycle, ΔU = 0, so:

Q = W

We can also use the ideal gas law to relate the pressure, volume, and temperature of the gas:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the absolute temperature.

(a)How to find the pressure at B segment?

To find the pressure at B, we can use the fact that the segment AB is an isothermal expansion. This means that the temperature remains constant, so:

PV = nRT

PB = (nRT)/(2V) = (2.00 mol)(0.0821 L·atm/mol·K)(600 K)/(2V) = (0.0821 L·atm/mol)(600 K)/V

Since the pressure at A is 5.00 atm, we can use the fact that the temperature is constant to find the volume at A:

PV = nRT

VA = (nRT)/P = (2.00 mol)(0.0821 L·atm/mol·K)(600 K)/5.00 atm = 197.76 L

Since the volume at B is twice the volume at A, we have:

VB = 2VA = 395.52 L

Substituting into the expression for PB, we get:

PB = (0.0821 L·atm/mol)(600 K)/395.52 L = 0.1248 atm

Therefore, the pressure at B is 0.1248 atm.

To find the temperature at C, we can use the fact that the segment BC is an adiabatic expansion. This means that no heat is added or removed from the system, so:

\(PV^\gamma\)= constant

where γ is the ratio of specific heats (for a monoatomic gas, γ = 5/3). We can use the fact that the volume at C is equal to the volume at A to find the pressure at C:

\(PAV^\gamma = PCV^\gamma\)

PC =  \(PA(V/A)^\gamma\) = 5.00 atm\((1/2)^(^5^/^3^)\) = 1.556 atm

Since the segment BC is adiabatic, the temperature changes but no heat is added or removed from the system. Using the ideal gas law, we can relate the pressure, volume, and temperature:

PV = nRT

TC = (PCVC)/(nR) = (1.556 atm)(197.76 L)/(2.00 mol)(0.0821 L·atm/mol·K) = 727.1 K

Therefore, the temperature at C is 727.1 K.

The total work done by the gas in one cycle is the sum of the work done in each segment of the cycle:

W = WAB + WBC + WCD + WDA

For segment AB, the work done is:

WAB = -QAB = -∫PdV = -nRT∫(1/V)dV = -nRT ln(VB/VA) = -(2.00 mol)(0.0821 L·atm/mol·K)(600 K) ln(2) = -602 J

For segment BC, the work done is:

WBC = -QBC = -∫PdV = -nγRT∫(1/V)dV = -nγRT

We know that VB = 2VA and VC = 2VD, so we can express the ratio VB/VC in terms of VA/VD:

VB/VC = (2VA)/(2VD) = VA/VD

Substituting into the expression for WBC, we get:

WBC = -nγRT ln(VA/VD)

For segment CD, the work done is:

WCD = -QCD + PCDΔV = -nCpΔT + PCDΔV

where Cp is the specific heat at constant pressure, ΔT is the change in temperature, and ΔV is the change in volume. We know that the segment CD is isobaric, so ΔV = VB - VA = (2VA) - VA = VA. We can also use the ideal gas law to relate the pressure, volume, and temperature:

Substituting into the expression for WCD, we get:

WCD = -nCpΔT + (nRT/VD)VA = -nCp(TC - TD) + (nRT/VD)VA

For segment DA, the work done is:

WDA = -QDA + ΔU = -nCvΔT

where Cv is the specific heat at constant volume. We know that the segment DA is isovolumetric, so ΔV = 0. Using the First Law of Thermodynamics, we know that ΔU = 0 for a complete cycle, so:

QDA = -WDA = nCvΔT

Substituting into the expression for WDA, we get:

WDA = -nCvΔT

Adding up the work done in each segment, we get:

W = WAB + WBC + WCD + WDA

= -(2.00 mol)(0.0821 L·atm/mol·K)(600 K) ln(2)- (2.00 mol)(5/3)(0.0821 L·atm/mol·K)(727.1 K) ln(VA/VD)- (2.00 mol)(Cp)(TC - TD) + (2.00 mol)(0.0821 L·atm/mol·K)(600 K) ln(2)- (2.00 mol)(Cv)(TC - TA)

We know that Cp and Cv for a monoatomic gas are related by Cp = Cv + R, so we can express Cp in terms of Cv:

Cp = Cv + R = (3/2)R + R = (5/2)R

Substituting and simplifying, we get:

W = (2.00 mol)(0.0821 L·atm/mol·K)(600 K) ln(2)- (2.00 mol)(5/3)(0.0821 L·atm/mol·K)(727.1 K) ln(VA/VD)- (2.00 mol)(5/2)(0.0821 L·atm/mol·K)(727.1 K)+ (2.00 mol)(5/2)(0.0821 L·atm/mol·K)(600 K)

W = -966.2 J - 4957 J - 7476 J + 5154 J

    = -1979 J

Therefore, the total work done by the gas in one cycle is -1979 J

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A measure of how quickly a conductor moves across a field. The voltage induced increased as the conductor moved through the field more quickly. The induced voltage and speed are directly inversely proportional.

Current can only flow when a circuit is closed or complete, hence, unless the circuit is completed current will not flow

What is Electro-Magnetic Induction?

This a phenomenon whereby a wound wire or a soft core of conductor made to cut across magnetic line of force and an E.M.F is generated, this E.M.F forces current to flow if the circuit is completed.

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We noticed that our cart was moving down the track at a continuous acceleration. This suggests that the cart was under a continuous force and that the constant force was being generated by the mass of the cart.

It is not possible for a thing to "have in it" or "contain" a force. One object experiences a force from another. The idea of a force includes both inanimate items and living things. A force can lower or modify the velocity of a massed item. Two simple ways to illustrate force are to push or pull. Given that it has both a magnitude and a direction, a force is a vector quantity.

Both direct force and distant action two different types of forces exist. It's clear that you utilize force consistently. Forces are just push and pull. You exert force on an object when you push or pull against it.

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Answer:

They first spend it on necessities.

They adjust pH so that acidification does not affect the organisms living in the ocean.

The ocean generates 50 percent of the oxygen we need, absorbs 25 percent of all carbon dioxide emissions and captures 90 percent of the excess heat generated by these emissions.

It is not just ‘the lungs of the planet’ but also its largest ‘carbon sink’ – a vital buffer against the impacts of climate change. The ocean is central to reducing global greenhouse gas emissions and stabilizing the Earth’s climate.

However, increasing greenhouse gas emissions have affected the health of the ocean  – warming and acidifying seawater – causing detrimental changes to life under water and on land, and reducing the ocean’s ability to absorb carbon dioxide and safeguard life on the planet.

Therefore,  They adjust pH so that acidification does not affect the organisms living in the ocean.

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Astronaut holds a rock 100 m above the surface of Planet X. The acceleration due to gravity on Planet X is 1.5 m/s².

Given information: Astronaut holds a rock 100 m above the surface of Planet X.The rock is thrown upward with a speed of 15 m/s.The rock reaches the ground 10 s after it is thrown.

The atmosphere of Planet X has a negligible effect on the rock when it is in free fall.

To find: acceleration due to gravitySolution: When the rock is thrown upward, its initial velocity, u = +15 m/s (upward velocity is taken as positive)The final velocity, v = 0 (at the maximum height, the velocity becomes zero). The distance traveled by the rock, s = 100 m. Total time taken by the rock to return to the ground, t = 10 s

Using the kinematic equation,v = u + gtv = u + gt0 = +15 - g x 10 where g is the acceleration due to gravityg = 15/10= 1.5 m/s²Therefore, the acceleration due to gravity on Planet X is 1.5 m/s².

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The force diagram of the player has been shown in the image attached here.

The force diagram of a player sliding to a stop depends on the specific situation and factors such as the player's mass, velocity, and the surface on which they are sliding. However, we can make some generalizations based on common assumptions and considerations.

We know that the force of kinetic friction that is sliding the player to a stop would tend to have a higher magnitude as shown by the force diagram.

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To find the Voltage, ( V ) [ V = I x R ] V (volts) = I (amps) x R (Ω)

To find the Current, ( I ) [ I = V ÷ R ] I (amps) = V (volts) ÷ R (Ω)

To find the Resistance, ( R ) [ R = V ÷ I ] R (Ω) = V (volts) ÷ I (amps)

To find the Power (P) [ P = V x I ] P (watts) = V (volts) x I (amps)

This is a defective question. It was WRITTEN by someone who is unclear on the concepts.  DON'T try and answer it.

It's trying to get us to use Newton's second law ... F = m • a.

But that only tells us how much force must act ON THE CAR in order to accelerate it. (45 kg) • (4 m/s^2) = 180 newtons.

This is NOT the force exerted BY the car when it hits something. THAT force depends on its speed WHEN it hits, AND how long it takes for the wreckage to actually come to rest, AND how hard or soft the wall is.

DON'T try to answer this question. Your answer will be wrong, you won't understand why, and the teacher you try to argue with probably won't either.

============================================

More explanation:

Think about jumping off of a ladder in your back yard.  Several times.

Your mass is the same every time.  Your acceleration is the same every time . . . 9.8 m/s² down, the acceleration of Earth gravity, every time.

BUT ...

-- I'll bet you would rather land on wood than on concrete. The force of landing would be less.

-- I'll bet you would rather land on dirt than on wood. The force of landing would be less.

-- I'll bet you would rather land on grass than on dirt. The force of landing would be less.

-- I'll bet you would rather land on a pile of blankets than on dirt. The force of landing would be less.

-- I'll bet you would rather land on a trampoline than on a pile of blankets. The force of landing would be less.

-- I'll bet you would rather jump from a short ladder than from a tall one.  Your speed would be less when you landed, and the force of landing would be less.

==> Your mass is the SAME every time, and your acceleration is the SAME every time.  But the force when you hit is DIFFERENT every time.

The mass and acceleration of the car DON'T tell us the force of the hit when the car hits a wall.  

Answer:

• riding on a Ferris wheel whose entrance and exit are the same

• walking around the block, starting from and ending at the same house

• running exactly one lap around a racetrack

Explanation:

Displacement simply means the.change in position of an object. In a situation whereby the initial and final position are thesame, the displacement will be zero.

The statements that describe a situation with a displacement of zero include:

• riding on a Ferris wheel whose entrance and exit are the same

• walking around the block, starting from and ending at the same house

• running exactly one lap around a racetrack

Answer:

• riding on a Ferris wheel whose entrance and exit are the same

• walking around the block, starting from and ending at the same house

• running exactly one lap around a racetrack

Explanation:

B,C,E

Answer:

4.7m

Explanation:

Given parameters:

Mass of the book  = 1kg

Gravitational potential energy  = 46J

Unknown:

Height of the shelf  = ?

Solution:

The potential energy is due to the position of a body above the ground.

        Gravitational potential energy  = mgh

m is the mass,

g is the acceleration due gravity  = 9.8m/s²

h is the height which is unknown

                       46  = 1 x 9.8 x h

                       h  = 4.7m

A 1 kg box falls two metres from a shelf and lands on the ground after 0.021 seconds. The box impacted the ground with a force of around 524.7 N.

The box impacted the ground with a force of around 524.7 N.

Explanation: We can determine the force using the equation F = m * (v/t), where m is the box's mass, v is the velocity change (which is the ultimate velocity because the box starts at rest and comes to a halt), and t is the time it takes for the box to stop.

Given that the box falls 2 metres and its terminal velocity is 0 m/s, v = 2 m/s.

524.7 N is equal to F = 1.1 kg * (2 m/s / 0.021 s).

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Answer:

D. It focuses on the core muscles

Answer:

A. It is a total body exercise

Explanation:

In cardio kickboxing as described by my school is "

The first benefit is the one which we are focusing on because all the other answers are correct but A. Total body exercise is the best answer because it covers all of them as an compendious answer

Also I took the test and got this correct

Answer:The disturbance created by a source of sound in the medium travels through the medium and not the particles of the medium

Explanation:i hope this is right

Answer:

451.13 J/kg.°C

Explanation:

Applying,

Q = cm(t₂-t₁)............... Equation 1

Where Q = Heat, c = specific heat capacity of iron, m = mass of iron, t₂= Final temperature, t₁ = initial temperature.

Make c the subject of the equation

c = Q/m(t₂-t₁).............. Equation 2

From the question,

Given: Q = 1500 J, m = 133 g = 0.113 kg, t₁ = 20 °C, t₂ = 45 °C

Substitute these values into equation 2

c = 1500/[0.133(45-20)]

c = 1500/(0.133×25)

c = 1500/3.325

c = 451.13 J/kg.°C

Answer:

R = V^2 sin 2 θ / g        range formula

R is a maximum for θ = 45 and R = V^2  g

g = V^2 / R = 3^2 / 15 = .6 m/s^2

The given statement "A driver does not need to allow as much distance when following a motorcycle as when following a car." is FALSE.

When following a motorcycle, a driver should maintain the same safe following distance as when following a car. This distance provides adequate time to react in case the motorcycle stops suddenly or encounters an obstacle in the road. In general, drivers should follow the "3-second rule" when determining the safe following distance.

Motorcycles are smaller and lighter than cars, making them more vulnerable to road hazards such as potholes, debris, or uneven surfaces. Additionally, motorcycles can stop more quickly than cars, so maintaining a safe following distance is crucial to avoid a potential collision.

Motorcyclists may also need to make sudden maneuvers or adjust their position in the lane to avoid obstacles, maintain stability, or optimize visibility. Drivers should be aware of these factors and give motorcycles ample space to navigate safely.

Furthermore, drivers should be extra cautious in adverse weather conditions or on wet roads, as motorcycles are more susceptible to losing traction, which can result in a skid or fall. Increasing the following distance in these situations can help ensure the safety of both the motorcyclist and the driver.

In summary, it is false to claim that a driver does not need to allow as much distance when following a motorcycle as when following a car. A safe following distance is crucial for preventing accidents and ensuring the well-being of all road users.

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Answer:

Little to no change in temperatures between seasons.

Reason: because i got it right on the test, and i smart

Answer:

Little to no change in temperatures between seasons.

Explanation:

I can confirm this is right

Answer:

B) t = 1.83 [s]

A) y = 16.51 [m]

Explanation:

To solve this problem we must use the following equation of kinematics.

\(v_{f} =v_{o} -g*t\)

where:

Vf = final velocity = 0

Vo = initial velocity = 18 [m/s]

g = gravity acceleration = 9.81 [m/s²]

t = time [s]

Note: the negative sign in the above equation means that the acceleration of gravity is acting in the opposite direction to the motion.

A) The maximum height is reached when the final velocity of the ball is zero.

0 = 18 - (9.81*t)

9.81*t = 18

t = 18/9.81

t = 1.83 [s], we found the answer for B.

Now using the following equation.

\(y = y_{o} + v_{o}*t - 0.5*g*t^{2}\\\)

where:

y = elevation [m]

Yo = initial elevation = 0

y = 18*(1.83) - 0.5*9.81*(1.83)²

y = 16.51 [m]

Use the following kinematic equation:

where,

v: final speed = 25m/s

vo: initial speed = 0m/s

t: time interval = 30.0s

a: acceleration = ?

replace the previous values of the parameters into the expression for a and simplify:

The acceleration of th train is approximately 0.83m/s^2

Answer:

Wala ko ka sabot ana oy sorry

To find the speed of the car when it has traveled 50 meters, we need to use the Work-Energy Theorem and equate the net work done to the change in kinetic energy.

The net work done (W_net) is given by integrating the force (f(x)) over the displacement (x₁ to x₂):

In this case, the force acting on the car is given by f(x) = 5.7x² + 1.5x.

The change in kinetic energy (∆KE) is given by:

Since the car starts from rest (v₁ = 0), the initial kinetic energy (KE₁) is 0.

Using the formula for kinetic energy KE = 1/2 mv², we can express the final kinetic energy (KE₂) in terms of the car's mass (m) and its final velocity (v₂):

According to the Work-Energy Theorem, W_net = ∆KE. Therefore, we can write:

Substituting the given force expression into the integral:

Now we can solve this equation to find the velocity v₂. However, the problem statement does not provide the values of x₁ and x₂, which are necessary to evaluate the integral and determine the velocity. Without those values, we cannot proceed with the calculation.

If you have the values of x₁ and x₂, please provide them, and I'll be happy to assist you further in finding the speed of the car when it has traveled 50 meters.

Kinetic energy or energy of motion is the energy possessed by an object due to its motion. The kinetic energy of an object is defined as the work required to move an object with a certain mass from rest to a certain speed.

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The weight of an 18-kg object on Earth is about 176.58 N, and on the Moon, it is approximately 29.16 N.

The weight of an object depends on the gravitational force acting on it. The gravitational force is a function of the mass of the object and the gravitational acceleration of the celestial body it is located on.

On Earth, the gravitational acceleration is approximately 9.81 m/s², whereas on the Moon it is about 1.62 m/s², which means that the gravitational force on the Moon is approximately one-sixth of the force on Earth.

Therefore, to calculate the weight of an 18-kg object on the Moon and on Earth, we can use the formula:

Weight = mass x gravitational acceleration

On Earth:

Weight = 18 kg x 9.81 m/s² ≈ 176.58 N

On the Moon:

Weight = 18 kg x 1.62 m/s² ≈ 29.16 N

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Answer:

yes

Explanation:

because you will be hydrated

Answer:

Its good because you can survive without food but you cant survive without water. Extra water can replace empty, sugary calories many people drink with meals.

Explanation: