gas is compressed by a piston in a cylinder from an initial pressure of 1.0 bar and initial volume of 0.4m3 , to a final pressure of 1.4 bar. During the compression process, the product of the gas pressure P and volume V remains constant. Determine the work done on the gas. If during the compression the gas loses 8.1 kJ of heat to the surroundings, determine the change in internal energy of the gas.​

Answer:

Approximately \(-15.37m/s^2\)

Explanation:

The acceleration can be found using the formula:

\(x_{f} = x_{i} + v_{i} (t)+\frac{1}{2} (a)(t^{2} )\)

The work is as shown:

\(45m = 0 + 39 m/s (3.3s)+\frac{1}{2} a(3.3s)^2\)

\(45m=128.7m + 5.445s^2 a\)

\(-83.7m=5.445s^2a\)

\(a = -15.37190083m/s^2\)

a is about -15.37\(m/s^2\)

The magnitude of the force is directly related to the amount of work done; the greater the force, the greater the amount of work done on the particle.

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Answer:

45 degrees

Explanation:

The textbooks say that the maximum range for projectile motion (with no air resistance) is 45 degrees.

the dig would -5 from where he first started.

C. 10,200 N is how much energy was lost due to air resistance while she was bouncing

The energy lost due to air resistance while the bungee jumper was bouncing can be calculated by finding the total mechanical energy of the system at the beginning of the jump and comparing it to the total mechanical energy at the end of the jump.

At the beginning of the jump, the total mechanical energy is given by:

Ei = mgh

where m is the mass of the bungee jumper, g is the acceleration due to gravity, and h is the height of the bridge. Therefore, at the beginning of the jump:

50 x 30 x 10 - 1/2 x 30^2 x 10

= 15000 - 4500

= 10,200 N

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The magnitude and direction of the resultant force (F=10N+5N+4N+6N) when four forces of magnitude 10N, 5N, 4N, and 6N operate on an object in the directions of north, west, east, and south, respectively, is 25N.

In terms of physics, magnitude is just "distance or quantity." It displays an object's size, direction, or motion in absolute or relative terms. It is employed to indicate the size or scope of something. The word "magnitude" in physics frequently denotes a size or quantity.. Something's magnitude is determined by its size. For example, a car moves faster than a bike with terms of speed. The speed difference between the car and the bike in this case is greater. It reveals the absolute or relative size or direction that an object moves when in motion.

Speed is defined as the magnitude of velocity, and the magnitude of the slope of the displacement-time graph provides the speed. Since the slope gets steeper as we move from A to B, Speed is increased as a result between points A and B.

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The car has traveled a distance of approximately 55.44 m.

In 7.0 s, a car accelerates uniformly from rest to a velocity at which its wheels are turning at 6.0 rev/s. If the tires of the car have a diameter of 42 cm, and they rolled on the ground without slipping, the car will travel a distance of approximately 50.9 meters in those 7.0 seconds.

To calculate the distance, we must first determine the car's final velocity. The angular velocity of the wheels is given by 6.0 rev/s. Since the diameter of the tires is 42 cm, the circumference is:πd = π(0.42 m) = 1.32 m. The velocity of the car can be calculated by multiplying the circumference by the angular velocity: v = 6.0 rev/s × 1.32 m/rev = 7.92 m/s.

Now that we know the final velocity of the car, we can use the formula:d = (vf + vi)t/2where vi = 0 m/s (since the car is initially at rest), vf = 7.92 m/s, and t = 7.0 sd = (7.92 m/s + 0 m/s)(7.0 s)/2 = 27.72 m. The car traveled approximately 27.72 m in the first half of the trip (from rest to the final velocity), and 27.72 m in the second half of the trip (from the final velocity to a complete stop).

Therefore, the total distance traveled by the car in those 7.0 s is approximately 27.72 m + 27.72 m = 55.44 m. However, this is the distance that the wheels have rolled, not the distance that the car has traveled. Since the wheels are not slipping, the distance that the car has traveled is equal to the distance that the wheels have rolled.

So, the car has traveled a distance of approximately 55.44 m. Rounding to the appropriate significant figures, the distance is approximately 50.9 meters.

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It would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

1. The decay rate of a radioactive isotope is proportional to the number of radioactive atoms present in the sample at any given time.

2. The decay rate can be expressed as a function of time using the formula: R(t) = R₀ * \(e^{(-\lambda t\)), where R(t) is the decay rate at time t, R₀ is the initial decay rate, λ is the decay constant, and e is the base of the natural logarithm.

3. The half-life of a radioactive isotope is the time it takes for half of the radioactive atoms in a sample to decay. In this case, the half-life is given as 210 days.

4. Using the half-life, we can find the decay constant (λ) using the formula: λ = ln(2) / T₁/₂, where ln(2) is the natural logarithm of 2 and T₁/₂ is the half-life.

5. Substituting the given half-life into the formula, we have: λ = ln(2) / 210.

6. Now, we need to find the time it takes for the decay rate to fall to 0.58 of its initial rate. Let's call this time "t".

7. Using the formula for the decay rate, we can write: 0.58 * R₀ = R₀ * e^(-λt).

8. Simplifying the equation, we get: 0.58 = \(e^{(-\lambda t\)).

9. Taking the natural logarithm of both sides, we have: ln(0.58) = -λt.

10. Substituting the value of λ from step 5, we get: ln(0.58) = -(ln(2) / 210) * t.

11. Solving for t, we have: t = (ln(0.58) * 210) / ln(2).

12. Evaluating the expression, we find: t ≈ 546.

13. Therefore, it would take approximately 546 days for the decay rate of the sample of this radioactive isotope to fall to 0.58 of its initial rate.

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Answer:

0.705 kg less

Explanation:

Altitude at sea level = 0 ft

Altitude after climbing = 30,000 ft = 30,000 x 0.3048 = 9,144 m

Weight = W = mg

Change in weight = ΔW = (M-luggage) × (g-sea level - g-altitude)

g at sea level:

g1 = (Gravitational constant × Mass of the Earth) / (Radius of the Earth)²

g at altitude of 30,000 ft:

g2 = (Gravitational constant × Mass of the Earth) / (Radius of the Earth + Altitude)²

Gravitational constant = 6.674 × 10^-11 m^3 kg^-1 s^-2

g1 = (6.674 × 10^-11 × 5.98 × 10^24) / (6.37 × 10^6)^2

g1 ≈ 9.8358 m/s^2

g2 = (6.674 × 10^-11 × 5.98 × 10^24) / (6.37 × 10^6 + 9,144)^2

g2 ≈ 9.8076 m/s^2

ΔW = (M-luggage) × (g1 - g2)

= 25 kg × (9.8358 - 9.8076)

≈ 0.705 kg

Therefore, your luggage would weigh approximately 0.705 kg less when you climb to 30,000 ft compared to its weight at the airport.

Can't guarantee this is right, but I checked the numbers a few times and this is the best I can do!

At room temperature,300 K, will a miscibility limit emerge. At the temperature less than, the miscibility will fall below 1%.

Temperature directs the hotness or coldness of a body. In clear terms, it is the method of finding the kinetic energy of particles within an entity. Faster the motion of particles, more the temperature.

Hence, at room temperature,300 K, will a miscibility limit emerge. At the temperature less than, the miscibility will fall below 1%.

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Answer:

False

Explanation:

A contact force is any force that requires contact to occur. Contact forces are ubiquitous and are responsible for most visible interactions between macroscopic collections of matter. Pushing a car up a hill or kicking a ball across a room are some of the everyday examples where contact forces are at work. 

Answer:

false

Explanation:

The height of the tower if the distance AB is 50m is 22.36m.

What is distance?

Distance is a measurement of how far apart two objects or points are, either numerically or occasionally qualitatively. Distance can refer to a physical length in physics or to an estimate based on other factors in everyday language (e.g. "two counties over"). The term is also frequently used metaphorically to denote a measurement of the degree of separation between two common pieces (such as statistical distance between probabilistic or edit distance between strings of text). This is because spatial cognition is a rich source of implicatures in human thought. The concept of a metric space is used in mathematics to formalise the majority of these notions of distance, both literal and figurative.

Sol-Here in right ∆AOB

AB= 50m (as per given question)

Let height of the tower is =OD = A meters

in ∆AOD , tan theta= DO/AO

Tan 38= A/∆O

∆O= A /tan 38= a/0.78

Also in ∆ BOD

tan 29 = DO/BO

BO = a/tan 29°

=9/0.55

So a^2 = 500

=>a=10√5=22.36 approx.

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The tension at point AE is 28.8 lb, and the reactions at point D are RD = 64.6 lb and RE = 145.4 lb.

the tension at point AE First, we need to calculate the weight of the CD bar, which is given by WC = Weight of CD bar = 5 × 32lbWC = 160 lb

Now we can find the total weight supported by the system as follows: W = Weight of AE bar + Weight of CD bar + Weight of AD bar W = 40 + 160 + 10 = 210 lb

As the weight is distributed evenly, the vertical forces at D and E should be equal: RD + RE = 210 lb

Next, we will determine the moments around point D.

This will help us find the tension at point AE.∑MD = 0(-32 × 2) + (40 × 4) + (160 × 7) + (10 × 5) + AE × 10 = 0Solving for AE,AE = 28.8 lb

Determine the reactions at point D

Now we can solve for the reactions at point D.

∑Fy = 0RD + RE - 210 = 0RD + RE = 210 lb∑MD = 0(-32 × 2) + (40 × 4) + (160 × 7) + (10 × 5) + AE × 10 = 0Solving for RD,RD = 64.6 lb

Now that we have RD, we can solve for RE:RE = 210 - RDRE = 145.4 lb

Therefore, the tension at point AE is 28.8 lb, and the reactions at point D are RD = 64.6 lb and RE = 145.4 lb.

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A controlled procedure that tests the effect of one variable on another is known as experiment.

An experiment is a set of actions and observations, performed in the context of solving a particular problem or question, to support or falsify a hypothesis or research concerning phenomena.

An experiment generally tests how one variable is affected by another.

Thus, a controlled procedure that tests the effect of one variable on another is known as experiment.

The correct answer is option B. "experiment"

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Answer:

The experiment.

We are given:

Mass of the Paint bucket (with paint) = 7000 grams

Mass of the paint bucket (without paint) = 500 grams

Volume of Paint in the Bucket = 5000 cm³

Mass of Paint in the Bucket:

To get the mass of the paint in the bucket, we will subtract the mass of the bucket from the mass of the paint bucket (with paint)

Mass of Paint = Mass of Paint bucket (with paint) - Mass of the paint Bucket (without paint)

Mass of Paint = 7000 - 500

Mass of Paint = 6500 grams

Density of the Paint:

We know that density = Mass / Volume

Density of Paint  = Mass of Paint / Volume occupied by Paint

Density of Paint = 6500/5000

Density of Paint = 1.3 grams / cm³

Answer:

Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. This tendency to resist changes in a state of motion is inertia.

Explanation:

Answer:

Z

Explanation:

Answer:

Z

Explanation:

I took the test on edge and got it right. Good luck :)

Answer:

there is only a small amount of gravity present.

Explanation:

this is because the only force acting upon your body during free fall is the force of gravity which is a non contact force.

Answer:

Explanation is given

Mark Me as Branliest

Answer:

Pluto & Mercury

Explanation:

Pluto's eccentricity is 0.248

Mercury's eccentricity is 0.206

(a) Position vector expression ⇒ r = 16.4t i + ( 3.8t - 4.9t² ) j m

(b) Velocity vector expression ⇒ v = 16.4 i + ( 3.8 - 9.8t ) j m/s

(c) Acceleration vector expression ⇒ a = - 9.8 j m/s²

x-coordinate ⇒ x = 16.4t

y-coordinate ⇒ y = 3.8t - 4.9t²

(a) Position vector expression ⇒ r = 16.4t i + ( 3.8t - 4.9t² ) j m

(b) Velocity is the rate of change of position with respect to time. So differentiating the position vector expression,

v = dr / dt = d / dt [ 16.4t i + ( 3.8t - 4.9t² ) j ]

v = 16.4 i + ( 3.8 - 2 * 4.9 * t ) j

v = 16.4 i + ( 3.8 - 9.8t ) j m/s

(c) Acceleration is the rate of change of velocity with respect to time. So differentiating the velocity vector expression,

a = dv / dt = d / dt [ 16.4 i + ( 3.8 - 9.8t ) j ]

a= 0 i + ( 0 - 9.8 ) j

a = - 9.8 j m / s²

(d) t = 2.88 s

r ( 2.88 s ) =  16.4 ( 2.88 ) i + ( 3.8 * 2.88 - 4.9 ( 2.88 )² ) j

r ( 2.88 s ) = 47.23 i - 29.7 j m

v ( 2.88 s ) = 16.4 i + ( 3.8 - 9.8 * 2.88 ) j

v ( 2.88 s ) = 16.4 i - 24.42 j m / s

a ( 2.88 s ) =  - 9.8 j m / s²

Therefore, the required vectors are as follows

(a) Position vector expression ⇒ r = 16.4t i + ( 3.8t - 4.9t² ) j m

(b) Velocity vector expression ⇒ v = 16.4 i + ( 3.8 - 9.8t ) j m/s

(c) Acceleration vector expression ⇒ a = - 9.8 j m/s²

(d) r ( 2.88 s ) = 47.23 i - 29.7 j m

    v ( 2.88 s ) = 16.4 i - 24.42 j m / s

    a ( 2.88 s ) =  - 9.8 j m / s²

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The position of center of mass is, 1 cm right, and 1.4 cm above the corner A, of the square ABCD.

The respective coordinates of masses, on corner ABCD, are:

Let the coordinates of COM(center of mass), be, Xcom and Ycom.

Therefore,

Xcom = [ ∑(\(M_{i}\) x \(X_{i}\)) / ∑(\(M_{i}\)) ] , and

Ycom = [ ∑(\(M_{i}\) x \(Y_{i}\)) / ∑(\(M_{i}\)) ]

That is,

Xcom = [ {(2x0)+(4x2)+(6x2)+(8x0)} / (2+4+6+8) ]

Xcom = (20/20) cm

Xcom = 1cm

Similarly,

Ycom = [ {(2x0)+(4x0)+(6x2)+(8x2)} / (2+4+6+8) ]

Ycom = (28/20) cm

Ycom = 1.4 cm

So, the position of center of mass is, 1 cm right, and 1.4 cm above the corner A, of the square ABCD.

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Answer:

65.87 s

Explanation:

For the first time,

Applying

v² = u²+2as.............. Equation 1

Where v = final velocity, u = initial velocity, a = acceleration, s = distance

From the question,

Given:  u = 0 m/s (from rest), a = 1.99 m/s², s = 60 m

Substitute these values into equation 1

v² = 0²+2(1.99)(60)

v² = 238.8

v = √238.8

v = 15.45 m/s

Therefore, time taken for the first 60 m is

t = (v-u)/a............ Equation 2

t = (15.45-0)/1.99

t = 7.77 s

For the final 40 meter,

t = (v-u)/a

Given: v = 0 m/s(decelerates), u = 15.45 m/s, a = -0.266 m/s²

Substitute into the equation above

t = (0-15.45)/-0.266

t = 58.1 seconds

Hence total time taken to cover the distance

T = 7.77+58.1

T = 65.87 s

The force that a surface applies to an object in touch with it when it is perpendicular to the surface is called the normal force. Equal in magnitude; A book lying flat on a table and Not equal in magnitude; inclined plane, elevator.

Normal refers to something that is parallel to a surface. As you will see in the next example, if the object is on an inclination, the normal force may be less than the object's weight.

The weight of an object is balanced by the typical response force when it is sitting on a horizontal surface. Yet, not every situation lends itself to this. Yet, when an object is at rest on an inclined plane, its weight is not equal to the usual reaction force.

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Explanation:

At any point on the arrow's trajectory, the horizontal component of the velocity is the same. Therefore, the horizontal component of the velocity at the top of its trajectory is

\(v_x = v_{0x} = v_0\cos{42°} = (7.6\:\text{m/s})\cos{42°}\)

\(\:\:\:\:\:=5.6\:\text{m/s}\)

The disorder that does not appear to describe Greta's possible condition is epilepsy. The correct answer is B.

While epilepsy is a neurological disorder that can cause seizures and other symptoms, it does not typically result in the physical challenges that Greta is experiencing, such as spasticity and the need for a variable-speed tape recorder.

Additionally, epilepsy does not necessarily affect an individual's ability to communicate using finger spelling or impact their hearing or vision. Therefore, while Greta may have epilepsy in addition to her other challenges, it is not directly related to the issues she is experiencing in Mr. Popanopoulous' class.

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Your question seems incomplete, but I suppose the question was:

"Greta is experiencing difficulty in Mr. Popanopoulous' class because she has different needs than many of the other students. Greta must use a variable-speed tape recorder to record the lessons, she often struggles with spasticity, and she frequently uses finger spelling as a method of communication. Which of the following disorders does NOT appear to describe Greta's possible condition?

A) Cerebral palsy

B) Epilepsy

C) Hearing impairment

D) Visual impairment"

To find the mass of the object, we can use the formula for kinetic energy:

Kinetic energy (KE) = (1/2) * mass * velocity^2

Given that the kinetic energy is 426 J and the velocity is 13 m/s, we can rearrange the equation to solve for the mass:

mass = (2 * KE) / (velocity^2)

Substituting the given values:

mass = (2 * 426 J) / (13 m/s)^2

mass = (2 * 426 J) / (169 m^2/s^2)

mass = 852 J / 169 m^2/s^2

mass = 5.04 kg

Therefore, the mass of the object is 5.04 kg.

A defining characteristic of a city is often referred to as a "landmark." A landmark is a distinctive feature or building that stands out and is used to identify or represent a location. Landmarks can be man-made, such as structures, monuments, or historical locations, or they can be natural, like mountains, rivers, or lakes.

Landmarks are significant tourist destinations for both locals and visitors because they frequently have historical, cultural, or architectural value. They can serve as a symbol of a city's identity, history, and personality and end up being recognised as the place's iconic symbols. Landmarks might be notable buildings, statues, cathedrals, museums, or bridges that have come to symbolise the city in which they are located.

Therefore, option (OB) "landmark" is the term used to describe a distinguishing feature in a city.

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