A train starts from rest and accelerates uniformly until it has traveled 5.6 km and acquired a velocity of What is the approximate acceleration of the train during this time?

Answer:

Explanation:

It's graph A because the pressure in the tire is increasing as the amount of air going into it increases. B says the pressure drops exponentially as air goes in, and C says that the pressure stays the same as air goes in. Pressure in a tire increases proportionally to the amount of air in it.

Hi there!

(a)

Recall that:
\(W = F \cdot d = Fdcos\theta\)

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

\(W =248(56)cos(30) = 12027.36 J\)

To the nearest multiple of ten:
\(W_A = \boxed{12030 J}\)

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
\(\boxed{W_g = 0 J}\)

(c)
Similarly, the normal force is perpendicular to the displacement, so:
\(\boxed{W_N = 0 J}\)

(d)

\(F_{f} =\mu_k N\)

In this instance, the normal force is equivalent to the downward force of gravity and the vertical component of the applied force.

\(N = F_g + F_A sin(30)\\\\N = mg + F_A sin(30)\\\\N = 56(9.8) + 248sin(30) = 672.8 N\)

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
\(W_f = -\mu_k Nd\\W_f = - (0.1)(672.8)(56) = -3767.68J\)

In multiples of ten:
\(\boxed{W_f = -3770 J}\)

(e)
Simply add up the above values of work to find the net work.

\(W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3767.68) = 8259.68 J\)

Nearest multiple of ten:
\(\boxed{W_{net} = 8260 J}}\)

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
\(F_{net} = F_{Ax} - F_f\)

\(W = F_{net} \cdot d = (F_{Ax} - F_f)\)

\(W = (F_Acos(30) - \mu_k N)d\\W = (248cos(30) - 0.1(672.8)) * 56 \\\\W = 8259.68 J\)

Nearest multiple of ten:
\(\boxed{W_{net} = 8260 J}\)

Answer:

60600N

Explanation:

lnput =effort=202 N

The speed of the sound wave is 660 meters per second.

To calculate the speed of the sound wave, we need to use the formula:
Speed = Frequency x Wavelength
Here, the frequency of the sound wave is given as 880Hz, and the wavelength is given as 0.75. To get the answer, we just need to plug these values into the formula and solve for the speed:
Speed = 880 x 0.75
Speed = 660 meters per second
It's important to note that the speed of sound depends on the medium through which it is traveling. In air, the speed of sound is approximately 343 meters per second at standard temperature and pressure. However, this value can change depending on factors such as temperature, humidity, and altitude.
Understanding the speed of sound is important in various fields, such as music, engineering, and physics. For example, in music, the speed of sound determines the pitch of a note, while in engineering, it can be used to design and optimize acoustic systems. In physics, it's used to study the properties of waves and to explain phenomena such as Doppler effect and sonic booms.

Therefore, the speed of the sound wave is 660 meters per second.

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Answer:

Using the VSEPR theory, the electron bond pairs and lone pairs on the center atom will help us predict the shape of a molecule. The shape of a molecule is determined by the location of the nuclei and its electrons. The electrons and the nuclei settle into positions that minimize repulsion and maximize attraction.Explanation:

Answer:

radiation

Explanation:

t = 0 and t = 2 are the values of t when the velocity of the particle is zero. However, since the problem states that t is in seconds and t can't be negative, the time at which the particle reaches its minimum velocity is t = 2 seconds.

To find the time at which the particle reaches its minimum velocity, we need to find the time at which the particle's velocity is zero.

Velocity is the derivative of position with respect to time, so we can find the velocity of the particle by taking the derivative of x with respect to t:

v(t) = dx/dt = (9t^2 - 18t) m/s

To find the time at which the velocity is zero, we set the velocity equal to zero and solve for t:

0 = 9t^2 - 18t

t(9t-18)=0

So, t = 0 and t = 2 are the values of t when the velocity of the particle is zero. However, since the problem states that t is in seconds and t can't be negative, the time at which the particle reaches its minimum velocity is t = 2 seconds.

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Isothermal means constant temperature.  On a P-V diagram, this appears as a curve.

Constant volume of course appears as a vertical line.

Work done BY an ideal gas at constant pressure is W = PΔV.

Work done BY an ideal gas at constant temperature is:

W = nRT ln(Vf / Vi)

W = nRT ln(Pi / Pf)

Heat added to an ideal gas at constant volume is Q = (Cᵥ/R) VΔP.

For a monotomic gas, Cᵥ = 3R/2.

For a diatomic gas, Cᵥ = 5R/2.

Change in internal energy equals heat added to the gas minus the work done BY the gas.

ΔE = Q − W

For an ideal gas, if the temperature is constant, ΔE = 0.

At state 1, the pressure is 94.0 kPa and the volume is 4.0 L.  From ideal gas law:

PV = nRT

(94.0 kPa) (4.0 L) = nRT

nRT = 376 J

At state 2, the volume triples to 12.0 L.  The work done is:

W = nRT ln(Vf / Vi)

W = (376 J) ln(12 / 4)

W = 413 J

The process is isothermal, so ΔE = 0 J.  Therefore, Q = 413 J.

The new pressure is:

P₁ V₁ = P₂ V₂

(94.0 kPa) (4.0 L) = P₂ (12.0 L)

P₂ = 31.3 kPa

At state 3, the volume is constant at 12.0 L, and the pressure rises back to 94.0 kPa.  Since there's no change in volume, W = 0 J.  The heat added is:

Q = (Cᵥ/R) VΔP

Q = (3/2) (12.0 L) (94.0 kPa − 31.3 kPa)

Q = 1130 J

So ΔE = Q − W = 1130 J.

From ideal gas law:

PV = nRT

(94.0 kPa) (12.0 L) = nRT

nRT = 1128 J

At state 4, the volume returns to 4.0 L.  The work done is:

W = nRT ln(Vf / Vi)

W = (1128 J) ln(4 / 12)

W = -1240 J

The process is isothermal, so ΔE = 0 J.  Therefore, Q = -1240 J.

The new pressure is:

P₃ V₃ = P₄ V₄

(94.0 kPa) (12.0 L) = P₄ (3.0 L)

P₄ = 376 kPa

Finally back to state 1, the volume is constant at 4.0 L, and the pressure drops back to 94.0 kPa.  Again, since there's no change in volume, W = 0 J.  The heat added is:

Q = (Cᵥ/R) VΔP

Q = (3/2) (4.0 L) (94.0 kPa − 376 kPa)

Q = -1690 J

So ΔE = Q − W = -1690 J.

For the entire cycle:

Q = 413 J + 1130 J − 1240 J − 1690 J = -1390 J

W = 413 J + 0 J − 1240 J + 0 J = -827 J

ΔE = 0 J + 1130 J + 0 J − 1690 J = -560 J

Answer:

Explanation:

An object following a circular path can be covering the same distance along the circle's circumference with every passing minute, giving it a constant speed. Since a change in either speed OR direction means a change in velocity, the object's velocity is not constant.

velocity is a vector so therefore direction affects it being constant.

An object's acceleration is the rate its velocity (speed and direction) changes. Therefore, an object can accelerate even if its speed is constant - if its direction changes. If an object's velocity is constant, however, its acceleration will be zero.  

Velocity is a vector so therefore direction affects it being constant.

When an item is moving, its velocity is the rate at which its direction is changing as seen from a certain point of view and as measured by a specific unit of time

An object following a circular path can be covering the same distance along the circle's circumference with every passing minute, giving it a constant speed. Since a change in either speed OR direction means a change in velocity, the object's velocity is not constant. An object's acceleration is the rate its velocity (speed and direction) changes. Therefore, an object can accelerate even if its speed is constant - if its direction changes. If an object's velocity is constant, however, its acceleration will be zero.  

Velocity is a vector so therefore direction affects it being constant.

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The image is located 15 cm behind the lens, and its height is half the height of the object.

Assuming that the object is located on the principal axis of the lens, we can use the thin lens equation to find the image distance and image height:

\(1/f = 1/do + 1/di\)

where:

f is the focal length of the lens

How far away is the object? (distance between the object and the lens)

di is the image distance (distance between the image and the lens)

In this case, f = 10 cm and do = 30 cm. Substituting these values into the equation, we get:

\(1/10 = 1/30 + 1/di\)

Simplifying this equation, we get:

\(1/di = 1/10 - 1/30 = 1/15\)

Therefore, \(di = 15 cm.\)

Now, we can calculate the image height using the magnification equation:

\(m = -di/do\)

where m is the magnification (ratio of the image height to the object height). The negative sign indicates that the image is inverted (upside-down) relative to the object.

Substituting the values of di and do, we get:

\(m = -15/30 = -0.5\)

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Answer:

closed

Explanation:

A closed system allows energy (usually heat) to be exchanged but not matter.

Your welcome

The correct statement which describes the relationship between earth and the sun is that earth orbits the sun in an ellipse, with the sun at one focus.

An orbit in celestial mechanics is the curved path taken by an object, such as the path taken by a planet around with a star, a celestial body around a planet, or a manufactured satellite around an object or location in space, such as a planet, satellite, meteorite, or Lagrange point.

Ordinarily, the physics term "orbit" implies a trajectory that repeats itself with a regular basis, though it can also denote a non-repeating trajectory.

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The displacement of the train after 2.23 seconds is 25.4 m.

The resultant velocity of the train is calculated as follows;

R² = vi² + vf² - 2vivf cos(θ)

where;

R² = 8.81² + 9.66² - 2(8.81 x 9.66) cos(76)

R² = 129.75

R = √129.75

R = 11.39 m/s

The displacement is calculated as follows;

Δx = vt

Δx = 11.39 m/s x 2.23 s

Δx = 25.4 m

Thus, the displacement of the train after 2.23 seconds is 25.4 m.

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To determine whether the batter will hit a home run, we need to analyze the ball's trajectory and determine if it will clear the 11.34m high Green Monster wall.

Let's break down the problem into steps:

Step 1: Calculate the time of flight (t) for the ball's horizontal motion.

We can use the horizontal distance and the average exit velocity to find the time it takes for the ball to reach the Green Monster wall. The horizontal distance (range) can be determined using the formula:

range = horizontal velocity * time

In this case, the range is given as 94.5m, and the average exit velocity is 41.0m/s. Let's solve for time:

94.5m = (41.0m/s) * t

Simplifying the equation, we have:

t = 94.5m / 41.0m/s

t ≈ 2.31s

Step 2: Find the initial vertical velocity (Viy) at the peak of the trajectory.

Since the ball brushes against the top of the Green Monster, we can assume it reaches its peak at half of the total time of flight (t/2). The vertical motion is influenced by gravity, so the equation to determine the initial vertical velocity is:

Viy = (displacement) / (time)

In this case, the displacement is half the height of the Green Monster, which is 11.34m/2 = 5.67m. The time is half of the total time of flight:

Viy = (5.67m) / (t/2)

Viy = (5.67m) / (2.31s/2)

Viy ≈ 2.46m/s

Step 3: Calculate the horizontal velocity (Vix).

Since the horizontal motion is unaffected by gravity, the horizontal velocity remains constant throughout the ball's trajectory. We can use the horizontal distance and time of flight calculated earlier to find the horizontal velocity:

Vix = (horizontal distance) / (time)

Vix = 94.5m / 2.31s

Vix ≈ 40.95m/s

Step 4: Determine the total initial velocity (Vi) using the Pythagorean theorem.

The total initial velocity of the ball can be calculated using the horizontal and vertical velocities:

Vi = √(Vix^2 + Viy^2)

Vi = √((40.95m/s)^2 + (2.46m/s)^2)

Vi ≈ √(1676.9025m^2/s^2 + 6.0516m^2/s^2)

Vi ≈ √(1682.9541m^2/s^2)

Vi ≈ 41.02m/s

Now we have found the total initial velocity of the ball, which is approximately 41.02m/s.

To determine whether it's a home run, we need to consider the ball's trajectory and the height of the Green Monster. Since the height of the wall is 11.34m and the ball's vertical velocity is 2.46m/s, the ball will not clear the Green Monster and will not result in a home run.

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Answer:

120.5

Explanation:

blc unverce form is this

Answer:

yes

Explanation:

Answer:

No, it cannot. The car needs the friction of the surface to drive because the car pushes the surface backwards, and the surfaces makes a reaction force pushing the car forward, and that works because of the friction. In a frictionless surface the tires would rotate in the same place

Answer:

PE = 137.2931 J

Explanation:

PE = 137.2931 J

Answer:

The answer is A.

Explanation:

the maximum bending stress in the beam is 24 kip/in².

Bending stress is the usual stress that an item withstands when it is exposed to external load at any cross-section. The ratio of the bending moment to the section modulus is another way to define the bending stress.

Given that, Base of the rectangular cross-section area,

Height of the rectangular cross-section area, h=6in=0.5ft

Refer attached free-body diagram,

Applying moment equilibrium at point A

Σ\(M_A\) = 0

−14 (4.5) + R (4.5+4.5+1.5)= 0

10.5 R =63

\(R_A\) = 6 kip

Resolve the force along vertical direction in the above diagram

\(P = R_B + R_A\)

\(14 = R_B +6\\R_B = 8 \ kip\)

From the diagram, it is clearly understood that maximum bending will occur at mid of the beam as compared to the ends, so the maximum bending moment can be calculated as

M = 4.5* \(R_A\)

M = 4.5 * 8

M = 36 kip

From the theory of bending, the maximum bending stress in the beam can be calculated as

σ = M*Y/I

Where I = bh³/12

σ = 3461.53 kip/ft²

σ = 24 kip/in²

Thus, the maximum bending stress in the beam is 24 kip/in².

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Complete question:

The amount of energy needed to power a 0.20 kW bulb for one minute would be just sufficient to lift a 2.5 kg object through a vertical distance of approximately 29.03 meters.

To calculate the energy required to lift a 2.5 kg object through a vertical distance, we need to consider the gravitational potential energy formula:

Potential energy (PE) = mass (m) × gravity (g) × height (h)

Where:

m = 2.5 kg (mass of the object)

g = 9.8 m/s² (acceleration due to gravity on Earth)

h = ? (height)

First, let's find the height (h) by rearranging the formula:

h = PE / (m × g)

Now, let's calculate the potential energy (PE) needed to lift the object. We are given that the power of the bulb is 0.20 kW, and we want to find the energy required for one minute. To convert kilowatts (kW) to joules (J), we multiply by the conversion factor of 3,600 (60 seconds × 60 minutes):

Energy (E) = power (P) × time (t)

E = 0.20 kW × 1 min × 3,600 J/kW

Now, we can substitute the values into the equation to find the height:

h = (0.20 kW × 1 min × 3,600 J/kW) / (2.5 kg × 9.8 m/s²)

Calculating the expression on the right side:

h ≈ 0.20 × 1 × 3,600 / (2.5 × 9.8) ≈ 29.03 meters (rounded to two decimal places)

Therefore, the amount of energy needed to power a 0.20 kW bulb for one minute would be just sufficient to lift a 2.5 kg object through a vertical distance of approximately 29.03 meters.

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The specific latent heat of fusion of ice obtained from this experiment is approximately 583.33 J/g.

To determine the specific latent heat of fusion of ice using the given experiment, we need to consider the energy transferred during the process.First, we need to calculate the energy lost by the water to cool down from 25 °C to 0 °C. The energy lost is given by:

Q1 = m1 * c * ΔT1

Where:

m1 = mass of water = 100 g

c = specific heat capacity of water = 4.2 J/g °C

ΔT1 = change in temperature = (0 °C - 25 °C) = -25 °C

Q1 = 100 g * 4.2 J/g °C * (-25 °C) = -10,500 J

Next, we calculate the energy released by the water to freeze and cool the remaining ice. The energy released is given by:

Q2 = m2 * Lf

Where:

m2 = mass of ice = 18 g

Lf = specific latent heat of fusion of ice (to be determined)

Q2 = 18 g * Lf

Since energy is conserved in the system, the energy lost by the water (Q1) is equal to the energy released by the water (Q2):

-10,500 J = 18 g * Lf

Solving for Lf:

Lf = -10,500 J / 18 g = -583.33 J/g

The negative sign indicates that energy is being released during the process of freezing.

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The magnitude of horizontal force, is 33.27 N.

Length of the uniform bar, L = 24 m

Mass of the uniform bar, m = 5 kg

Angle between the cord and the bar, θ₁ = 90°

Angle between the cord and the wall, θ₂ = 50°

For the horizontal equilibrium,

Hₓ = T sinθ₂

Hₓ = T x sin 50

Hₓ = 0.766 T

For the vertical equilibrium,

Hy + T cosθ₂ = mg

Hy + T cos 50 = 5 x 9.8

Hy + 0.642 T = 49 N

For rotational equilibrium,

mg (L/2) = T x L

So,

T = mg/2 = 5 x 9.8/2

T = 24.5 N

Therefore, the magnitude of horizontal force,

Hy = 49 - (0.642 x 24.5)

Hy = 49 - 15.73

Hy = 33.27 N

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Answer:

P=2*\(10^{5}\)Pa

Explanation:

P=P1+P2

P1=\(\frac{F}{A}\)

Putting the values

P1=\(\frac{400}{40*10^{-4} }\)

P1=1*\(10^{5}\)

It is the pressure exerted by 1 feet

Pressure exerted by 2nd feet is also equal to the pressure exerted by 1st feet then

P2=1*\(10^{5}\)

P=P1+P2

P=(1*\(10^{5}\))+(1*\(10^{5}\))

P=2*\(10^{5}\) Pa

Answer:

The product of (wavelength) x (frequency) is always the same number ... the wave's speed.  

So if the wavelength is somehow reduced to  1/4  its original length, the frequency is immediately multiplied by 4 .  That's the only way their product can remain the same.

Explanation:

Answer:

Bottom choice

Explanation:

Average speed is      total distance / total time

Answer:

No because it could run out of gas quicker or get into a reck.

Explanation:

78% of the houses in Elizabeth's town were for sale last summer.

To find the percentage of houses that were for sale last summer in Elizabeth's town, we need to divide the number of houses for sale by the total number of houses and then multiply by 100.

The total number of houses in Elizabeth's town is given as 9,000. Last summer, 7,020 houses were for sale. To calculate the percentage, we divide 7,020 by 9,000:

Percentage = (7,020 / 9,000) * 100

Simplifying the calculation:

Percentage = 0.780 * 100

Percentage = 78.0%

To explain further, we take the number of houses for sale (7,020) and divide it by the total number of houses (9,000) to get the ratio of houses for sale to the total. Multiplying this ratio by 100 gives us the percentage. In this case, 78.0% indicates that a significant portion of the houses in Elizabeth's town were available for sale during the summer.

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The mass rate of change of the space probe is approximately 28.49 kg/s .

To solve this problem, we can use the generalized form of Newton's Second Law, which states that the force acting on an object is equal to its mass times its acceleration:

F = ma

In this case, the force acting on the space probe is the thrust force generated by the rocket motor, which is equal to the rate of change of momentum of the ejected fuel:

F = (Δ m /Δt) * v

where;

Since the space probe is landing on the planet, the net force acting on it should be equal to the force of gravity pulling it down minus the upward thrust force generated by the rocket motor. So we can write:

F_net = m * g - (Δ m /Δt) * v

Plugging in the values and solving for delta m / delta t, we get:

2.00 m/s² = (175,000 kg * 8.25 m/s²) - (Δ m / Δt) * 35,000 m/s

Δ m / Δt = (175,000 kg * 8.25 m/s² - 2.00 m/s² * 35,000 m/s) / 35,000 m/s

Δm / Δt ≈ 28.49 kg/s

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