A force of 14 N acts on a 5 kg object for 3 seconds. a. What is the object’s change in momentum? b. What is the object’s change in velocity?

Answer:

E) The average velocity of the car is 90.0 miles per hour in the direction of motion.

Explanation:

From the available options below:

A) The velocity of the car is constant.

B) The acceleration of the car must be non-zero.

C) The first 45 miles must have been covered in 30.0 minutes.

D) The speed of the car must be 90.0 miles per hour throughout the entire trip.

E) The average velocity of the car is 90.0 miles per hour in the direction of motion.

The average velocity of an object is the total distance covered by the object while moving in a particular direction per total time taken to cover the distance.

From the information given, during the course of the journey, the velocity may vary and the acceleration may be zero at some point if the driver of the car decides to stop over.

However, the average velocity for the journey must be 90.0 miles per hour in the direction of the motion in order for the car to be able to cover 90 miles in 60 minutes.

The correct option is E.

Answer:

Constructive interference defination:

''Constructive interference occurs when the maxima of two waves add together (the two waves are in phase), so that the amplitude of the resulting wave is equal to the sum of the individual amplitudes''

Constructive interference occurs at integer multiples of the wavelength of the wave. The lowest incidence occurs at the wavelength.

As we know,

                      wavelength * frequency = velocity

                      wavelength = v/f

                                          = (343 m/s) / (220 1/s)

                                          = 1.56 m

Answer: Yes,graded receptor potential always depolarize.

Yes,graded receptor potentials must occur to depolarize the neutrons to threshold before action potentials can occur.

Explanation:

Final answer:

The campus will save approximately $58,124.80 per year if the lights in the classrooms and faculty offices are turned off during unoccupied periods.

Explanation:

To determine how much the campus will save a year if the lights in the classrooms and faculty offices are turned off during unoccupied periods, we need to calculate the annual energy consumption and cost.

The classrooms consume 12 fluorescent tubes, each consuming 110 W. Since the faculty offices have half as many tubes, they consume 6 tubes on average.

First, let's calculate the annual energy consumption for the classrooms:

Energy consumption per classroom per day = (12 tubes) × (110 W/tube) × (4 h) = 5,280 Wh

Total energy consumption for all classrooms per day = (5,280 Wh/classroom) × (200 classrooms) = 1,056,000 Wh

Annual energy consumption for classrooms = (1,056,000 Wh/day) × (240 days) = 253,440,000 Wh = 253,440 kWh

Now, let's calculate the annual energy consumption for the faculty offices:

Energy consumption per office per day = (6 tubes) × (110 W/tube) × (4 h) = 2,640 Wh

Total energy consumption for all faculty offices per day = (2,640 Wh/office) × (400 offices) = 1,056,000 Wh

Annual energy consumption for faculty offices = (1,056,000 Wh/day) × (240 days) = 253,440,000 Wh = 253,440 kWh

Adding the energy consumption of the classrooms and faculty offices, the campus consumes a total of 506,880 kWh per year. To determine the cost savings, we need to multiply this by the unit cost of electricity:

Cost savings = (506,880 kWh) × ($0.115/kWh) = $58,124.80

Therefore, the campus will save approximately $58,124.80 per year if the lights in the classrooms and faculty offices are turned off during unoccupied periods.

Answer:

There should be a total voltage of zero (0)

Explanation:

In bipolar encoding (multilevel binary), there are three voltage levels, positive, negative, and zero. The voltage level for one data element is at zero, while the voltage level for the other element alternates between positive and negative.

However, the primary advantage of a bipolar scheme is that when all the voltages are added together after a long transmission, there should be a total voltage of zero.

Answer: The ratio of f at the higher temperature to f at the lower temperature is 4.736

Explanation:

According to the Arrhenius equation,

[tex]f=A\times e^{\frac{-Ea}{RT}}[/tex]

or,

[tex]\log (\frac{f_2}{f_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]

where,

[tex]f_1[/tex] = rate constant at 525K

[tex]K_2[/tex] = rate constant at 545K

[tex]Ea[/tex] = activation energy for the reaction = 185kJ/mol= 185000J/mol   (1kJ=1000J)

R = gas constant = 8.314 J/mole.K

[tex]T_1[/tex] = initial temperature = 525 K

[tex]T_2[/tex] = final temperature = 545 K

Now put all the given values in this formula, we get

[tex]\log (\frac{f_2}{f_1})=\frac{185000J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{525K}-\frac{1}{545K}][/tex]

[tex]\log (\frac{f_2}{f_1})=0.6754[/tex]

[tex](\frac{f_2}{f_1})=4.736[/tex]

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 4.736

Answer:

For two or more bodies of different mass released from height in a vacuum have the same velocity but varying force

Explanation:

Consider a body H with initial velocity (u) and final velocity V undergoing acceleration a and covering a distance( s)

From Network equation of motion it can be seen that

V^2=u^2+2as

From this it can be seen that velocity is not dependent on the the masses of the body.

Rather it depends on acceleration due to gravity which is a constant for both of the body

In a vacuum, objects of different masses will have the same acceleration due to gravity and their velocities will increase at the same rate during free fall. However, at the end of their fall, their velocities will not be equal because the object with greater mass will have greater inertia.

In a vacuum, where there is no air resistance, both objects experience the same acceleration due to gravity, regardless of their mass. This means that their velocities will increase at the same rate during free fall.

However, at the end of their fall, their velocities will not be equal because the object with greater mass will have greater inertia, and therefore require a greater force to reach the same velocity as the lighter object.

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Answer:

Review the task, there should indicate the mass of the player

Explanation:

The speed of the player immediately after stopping the ball is equal to 0.0396 m/s.

The law of conservation of linear momentum can be described as the sum of the momentum before and after the collision must be equal.

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

Where m₁ and m₂ is the mass of the objects, u₁ & u₂ are their initial speed while v₁ & v₂ is the final speed of the collided objects.

The linear momentum can be described as the product of the mass of the object times the velocity of that object. Conservation of momentum can be described as a property held by an object.

Given the initial speed of the ball is u = 21 m/s

The mass of the baseball player, M = 74 Kg

The mass of the given  ball, m = 0.140 Kg

From the law of conservation of momentum, determine the speed of the player (V):

m u + mv = (m + M)  V

0.140 Kg × 21 + 0 = (0.140 + 74) × V

V = 0.0396 m/s

Therefore, the speed of the player is equal to 0.0396 m/s.

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Your question is incomplete, most probably the complete question was,

A 74 Kg baseball player jumps straight up to catch a head-hit ball. If the 140 g ball is moving horizontally at 21 m/s, and the catch is made when the ballplayer is at the highest point of his leap, what is his speed immediately after stopping the ball?

Answer:

A scientific hypothesis

Explanation:

A scientific hypothesis is an idea or explanation for a fairly narrow set of phenomena that you then test through research, experimentation, experience, scientific background knowledge, preliminary observations, and logic. They are the initial building block in the scientific method and they are also beyond a wild guess but less than a well-established theory.

Answer:

True

Explanation:

Most vehicle's cooling system contain a coolant that flows through certain passages in the engine, as the coolant moves through those passages it absorb heat from moving engine parts and the explosion of gasoline in the cylinders. the coolant is then stored back in a radiator which is responsible for transferring the heat from the coolant to the environment(air).  

Answer:

[tex]y(t)=0.28sin(36.02599t)[/tex]

0.0436 s

0.1308 s

Explanation:

A = Amplitude = 28 cm

m = Mass = 0.235 kg

k = Spring constant = 305 N/m

The equation which describes motion as a function of time is given by

[tex]y(t)=Asin(\omega t)[/tex]

Angular speed is given by

[tex]\omega=\sqrt{\dfrac{k}{m}}\\\Rightarrow \omega=\sqrt{\dfrac{305}{0.235}}\\\Rightarrow \omega=36.02599\ rad/s[/tex]

The equation is

[tex]\mathbf{y(t)=0.28sin(36.02599t)}[/tex]

Maximum length will be at amplitude

Amplitude is given by

[tex]A=Asin(\omega t)[/tex]

here

[tex]\omega t=\dfrac{\pi}{2}\\\Rightarrow t=\dfrac{\pi}{2\times 36.02599}\\\Rightarrow t=0.0436\ s[/tex]

The time to stretch to maximum length is 0.0436 s

At minimum length

[tex]y(t)=-A\\\Rightarrow -A=Asin\omega t[/tex]

[tex]\omega t=\dfrac{3\pi}{2}\\\Rightarrow t=\dfrac{3\pi}{2\omega}\\\Rightarrow t=\dfrac{3\pi}{2\times 36.02599}\\\Rightarrow t=0.1308\ s[/tex]

The time will the spring shrink to its minimum length at first time is 0.1308 s

The answer explains the motion equation, time for maximum and minimum lengths, and provides a detailed solution.

Part A: The equation that describes this motion as a function of time for a vertical spring with a mass hanging from it is y(t) = A * sin(2πft + φ), where A is the amplitude, f is the frequency, t is time, and φ is the phase angle.

Part B: The time when the spring stretches to its maximum length for the first time is half a period after passing the equilibrium point, so Tmax = 0.5 / f.

Part C: The time when the spring shrinks to its minimum length for the first time is one period after passing the equilibrium point, so Tmin = 1 / f.

Answer:

Isentropic compression in a pump, Constant heat addition in a boiler, Isentropic expansion in a turbine, constant heat rejection in a condenser

Explanation:

Ideal rankine cycle schematic diagram and T-S diagram is shown in below figure

The process 1-2 is isentropic compression is taking place in pump.

The process 2-3 is constant heat addition, heat addition takes places at boiler.

The process 3-4 is isentropic expansion, is taking place is turbine.

The process 4-1 is constant heat rejection ,  takes place in condenser.

Answer:

Option (C)

Explanation:

'Pericardiocentesis' is a biological term that is usually defined as a process by which a fluid is removed from the sac that is oriented around the heart. This accumulation of fluid in the heart is commonly known as pericardial effusion. This process is mainly carried out with the help of a needle and a catheter, where the needle is injected into the tissues that are near to the heart.

This fluid that is eliminated from the body is further sent to the laboratories in order to carry out the medical test to determine if a person is affected by infection or cancer.

Thus, the correct answer is option (C).

"The correct answer is C. the removal of fluid from around the heart.

Pericardiocentesis is a procedure where a needle is used to remove fluid from the pericardium, which is the sac-like covering around the heart. This procedure is typically performed when there is an accumulation of fluid in the pericardium (pericardial effusion) that may be causing symptoms such as chest pain or difficulty breathing, or when there is a suspicion of infection or cancer in the pericardial space.

 Let's analyze each option to understand why option C is the correct choice:

 A. Narrowing of the arteries supplying the heart is referred to as atherosclerosis or coronary artery disease, not pericardiocentesis.

 B. Surgical repair of the sac around the heart is known as pericardiectomy or pericardial window, which involves creating an opening in the pericardium to drain fluid or to relieve pressure. This is different from pericardiocentesis, which is a less invasive procedure.

 C. The removal of fluid from around the heart is the definition of pericardiocentesis. It is a minimally invasive procedure that can be both diagnostic, by analyzing the fluid removed, and therapeutic, by relieving the pressure on the heart.

D. A surgical opening made in the heart refers to a procedure such as a thoracotomy or sternotomy, which involves opening the chest cavity to access the heart. This is not related to the removal of fluid from the pericardial sac.

Answer: The answer is A.

Explanation:

Suppose you are in an elevator. As the elevator starts upward, its speed will increase. During this time when the elevator is moving upward with increasing speed, your weight will be greater than your normal weight at rest.

Answer:

The answer is A. There is no electric field on the interior of the conducting sphere.

Explanation:

A solid conducting sphere in a uniform electric field will exert force on the charges in the sphere to redistribute themselves in such a way that both the charges and the field inside the sphere would vanish.

Answer:

The charge is 0.056 nC.

Explanation:

Given that,

Electric field = 2000 N/C

Distance = 5.0 cm

We need to calculate the charge density

Using formula of charge density

[tex]E=\dfrac{\lambda}{2\pi\times\epsilon_{0}r}[/tex]

[tex]\lambda=2\pi\times\epsilon_{0}\times r\times E[/tex]

Put the value into the formula

[tex]\lambda=2\pi\times8.85\times10^{-12}\times5.0\times10^{-2}\times2000[/tex]

[tex]\lambda=5.56\times10^{-9}\ C/m[/tex]

We need to calculate the charge in 1.0 cm

Using formula of charge

[tex]Charge = \lambda\times\text{length of segment}[/tex]

[tex]Charge =5.56\times10^{-9}\times1.0\times10^{-2}[/tex]

[tex]Charge=0.056\times10^{-9}\ C[/tex]

[tex]Charge=0.056\ nC[/tex]

Hence, The charge is 0.056 nC.

The charge on a 1.0-cm-long segment of the wire is 2000 nanocoulombs (nC).

The charge on a 1.0-cm-long segment of the wire can be determined using the formula:

Electric field strength = force per unit charge

Since we know the electric field strength and the length of the wire segment, we can rearrange the formula to solve for the charge:

Charge = Electric field strength x Length of wire segment

Plugging in the given values, the charge on the 1.0-cm-long segment of the wire is 2000 N/C x 1.0 cm = 2000 nC.

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Answer: Did  you get the answers? Im struggling homie.

Explanation:

Answer:

1. 1/9th the original

2. 0.023 N

3. 1.9 x 10^12 electrons

4. 1.62 x 10^-10 C to the right

5. 19.21 N/C

6. 2.17 N/C

7. Not 0.37 N/C

8. 1/4 of what it was

9. 2 electrons must be removed

10. Polarization: when the positive and negative charges in an object are separated

Explanation: Just took the test

Answer:

a) 0,18 m b) 0,18 m c) 0 m/s

Explanation:

To find the volume of the metal object, subtract the weight of the object submerged in water from its weight in air. Then, use the buoyant force formula to calculate the volume. The density of the metal cannot be determined without knowing the mass of the object.

In order to find the volume of the metal object, we need to first determine the buoyant force that acts on it when it is submerged in water. The difference between the scale readings in air and water represents the buoyant force exerted by the water on the object. The weight of the object in air is 943 N and in water, it is 799 N. The buoyant force is the difference between these two values, which is 143 N.

The buoyant force is equal to the weight of the water displaced by the object. Therefore, the volume of the object can be calculated using the buoyant force and the density of water. The density of water is approximately 1000 kg/m³. We can use the formula:

Volume = Buoyant Force / (Density of Water * Acceleration due to Gravity)

Substituting the given values, we have:

Volume = 143 N / (1000 kg/m³ * 9.8 m/s²)

Simplifying the equation, we find that the volume of the object is 0.0146 m³.

Now, to find the density of the metal, we can divide the mass of the object by its volume. Since the mass of the object is not given in the question, we cannot provide a specific answer to part (b) without this information.

Answer:

True

Explanation:

Alternators and generators are devices which convert mechanical energy into electrical energy, they consist of a magnet and a conductor wound about an iron called a core. When the conductor cuts across the lines of flux of the magnet an electromotive force is induced thereby generating current, either the conductor or the magnet has to be moving for this to happen and when the direction of motion changes the direction of electromotive force also changes.

Final answer:

True. The process of cutting lines of magnetic flux with a conductor and generating voltage is the basis of alternator and generator operation.

Explanation:

True.

The process of cutting lines of magnetic flux with a conductor and generating voltage is the basis of alternator and generator operation. This is known as electromagnetic induction, which is the process of inducing an electromotive force (emf) or voltage with a change in magnetic flux. When a coil is rotated in a magnetic field, it produces an alternating current emf, which is the basic construction of a generator.

Answer:

Momentum, p = 720 g-cm/s

Explanation:

The momentum of an object is determined to be,

[tex]p=7.2\times 10^{-3}\ kg-m/s[/tex]

We need to express this quantity in any equivalent units. We know that the conversions are as follows :

1 kg = 1000 g

and 1 m = 100 cm

[tex]p=7.2\times 10^{-3}\ kg-m/s=7.2\times 10^{-3}\times (1000\ g)\times (100\ cm)/s[/tex]

p = 720 g-cm/s

So, the momentum of an object in any equivalent unit is 720 g-cm/s. Hence, this is the required solution.

The momentum of the object is 7.2 × 10^-3 kg·m/s, which is equivalent to 7.2 g·m/s when the mass is converted from kilograms to grams.

The momentum of an object is defined as the product of its mass and velocity. It is a vector quantity, meaning it has both magnitude and direction. The magnitude of the given object's momentum is 7.2 × 10-3 kg·m/s. To express this quantity in equivalent units, we can convert the mass from kilograms to grams by noting that 1 kg equals 1000 grams. Thus, the momentum in grams·m/s can be calculated as (7.2 × 10-3 kg) × (1000 g/kg) = 7.2 g·m/s.

At t = 0 the total charge contained in the device is 9 C. At t = 1 the total charge contained in the device is -1 C. The current flowing into the device at any given instant is -10 A.

The current, I, in a device can be determined using the rate of change of charge, q, with respect to time, t. Hence the current is equal to the derivative of the charge with respect to time. First, evaluate the expression q(t) = 9 - 10t for t = 0, 1, 3 and 10.

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For t = 1 s, the total charge is -1 C. The current at t = 1 s, 3 s, and 10 s is constant at -10 A.

The charge q(t) contained in the mysterious device as a function of time is given by the expression q(t) = 9 - 10t C. To answer the questions, we will perform some calculations.

(a) Total charge at t = 0: By substituting t = 0 into the expression q(t) = 9 - 10t, we get q(0) = 9 C.

(b) Total charge at t = 1 s: Substituting t = 1 into the expression q(t) = 9 - 10t, we get q(1) = 9 - 10(1) = -1 C.

(c) Current at various times: The current I is the rate of change of charge with respect to time. This is found by differentiating q(t) with respect to t, which yields I(t) = -10 A (since the derivative of a constant is 0 and the derivative of -10t with respect to t is -10). The current flowing into the device at t = 1 s, 3 s, and 10 s is thus constant at -10 A.

Note that the negative sign indicates the direction of current flow (i.e., charge leaving the device).

Answer:

There are 756.25 electrons present on each sphere.

Explanation:

Given that,

The force of repression between electrons, [tex]F=3.33\times 10^{-21}\ N[/tex]

Let the distance between charges, d = 0.2 m

The electric force of repulsion between the electrons is given by :

[tex]F=k\dfrac{q^2}{r^2}[/tex]

[tex]q=\sqrt{\dfrac{Fr^2}{k}}[/tex]

[tex]q=\sqrt{\dfrac{3.33\times 10^{-21}\times (0.2)^2}{9\times 10^9}}[/tex]

[tex]q=1.21\times 10^{-16}\ C[/tex]

Let n are the number of excess electrons present on each sphere. It can be calculated using quantization of charges. It is given by :

q = ne

[tex]n=\dfrac{q}{e}[/tex]

[tex]n=\dfrac{1.21\times 10^{-16}}{1.6\times 10^{-19}}[/tex]

n = 756.25 electrons

So, there are 756.25 electrons present on each sphere. Hence, this is the required solution.

Answer:

3.26 m

Explanation:

change in kinetic energy = change in potential energy

[tex]\frac{v^{2}}{2} = gh[/tex]

[tex]h=\frac{v^{2}}{2g}[/tex]

h = (8^2)/ 2*9.81 = 3.26 m

Answer:

a)  two peak in the spectrum , b)  m₁ / m_ c = 100783 , m₂ / m_c = 2.01410, c)  the lighter isotope peak is much higher than the other

Explanation:

A mass spectrometer is a device where an ionized and accelerated sample in a constant electric field enters a magnetic field that deflects it for detection, in vernal the intensity of the magnetic field is controlled.

Newton's second Law gives us

                   [tex]F_{m}[/tex]= m a

The acceleration is centripetal

                   a = v² / r

                   q v B = m v² / r

                   q B = m v / r

                   r = (m / q) v / B

a) with We have two different isotopes must separate only two peak in the spectrum

b) the relative atomic mass is the ratio between the mass of an atom in kg and the weighted mass of carbon that is 12

          m₁ / m_ c = 100783

          m₂ / m_c = 2.01410

It has units for being the relationship between two masses

c) The peak height intensity is proportional to the abundance of each isotope.

       The most used is to calculate the relative abundances is to use the area of ​​each peak

        Therefore the lighter isotope peak is much higher than the other

A) The number of peaks that the mass spectrum will have is; 3 peaks

B) The relative atomic masses of each of the peaks are;

Peak 1H - 1H = 2.01566 amu

Peak 1H - 2H = 3.02193 amu

Peak 2H - 2H = 4.02820 amu

C) The largest and smallest peak will be;

Largest Peak = Peak 2H - 2H

Smallest Peak = Peak 1H - 1H

We are told that the two naturally occurring isotopes of hydrogen are;

atomic mass = 1.00783 ; abundance 99.9885%

atomic mass = 2.01410 ; abundance 0.0115%

A) The mass spectrum will have 3 peaks namely;

B) The relative atomic masses of each of the peaks will be;

Relative atomic mass of Peak 1H - 1H = 1.00783 + 1.00783 = 2.01566 amu

Relative atomic mass of Peak 1H - 2H = 1.00783 + 2.01410 = 3.02193 amu

Relative atomic mass of Peak 2H - 2H = 2.01410 + 2.01410 = 4.02820 amu

C) From B above, we conclude that;

Largest Peak is Peak 2H - 2H

Smallest Peak is Peak 1H - 1H

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Answer:

2.84 g's with the remaining 1 g coming from gravity (3.84 g's)

Explanation:

period of oscillation while waiting (T1) = 2.45 s

period of oscillation at liftoff (T2) = 1.25 s

period of a pendulum (T) =2π. [tex]\sqrt{\frac{L}{a} }[/tex]

where

therefore the ration of the periods while on ground and at take off will be

[tex]\frac{T1}{T2}[/tex] =(2π [tex]\sqrt{\frac{L}{a1} }[/tex] ) /  (2π[tex]\sqrt{\frac{L}{a2} }[/tex])

where

[tex]\frac{T1}{T2}[/tex] = [tex]\frac{\sqrt{\frac{L}{a1} }}{\sqrt{\frac{L}{a2} }}[/tex]

squaring both sides we have

[tex](\frac{T1}{T2})^{2}[/tex] = [tex]\frac{\frac{L}{a1} }{\frac{L}{a2} }[/tex]

[tex](\frac{T1}{T2})^{2}[/tex] = [tex]\frac{a2}{a1}[/tex]

assuming that the acceleration on ground a1 = 9.8 m/s^{2}

[tex](\frac{T1}{T2})^{2}[/tex] = [tex]\frac{a2}{9.8}[/tex]

a2 = 9.8 x [tex](\frac{T1}{T2})^{2}[/tex]

substituting the values of T1 and T2 into the above we have

a2 = 9.8 x [tex](\frac{2.45}{1.25})^{2}[/tex]

a2 = 9.8 x 3.84

take note that 1 g = 9.8 m/s^{2} therefore the above becomes

a2 = 3.84 g's

Hence assuming the rock is still close to the ground during lift off, the acceleration of the rocket would be 2.84 g's with the remaining 1 g coming from gravity.

To find the rocket's acceleration during liftoff, we can use the change in the period of a pendulum. By equating the expressions for the pendulum length before and during liftoff, we determine that the acceleration is approximately 37.62 m/s².

To find the rocket's acceleration during liftoff using the pendulum's period, we can use the formula for the period of a pendulum:

  T = 2π √(L/g)

When the rocket was on the launch pad, the period T₀ was 2.45 s and the acceleration due to gravity g₀ was 9.80 m/s². Let's denote the pendulum length as L. We can write:

  2.45 = 2π √(L/9.80)

During liftoff, the period T changed to 1.25 s. The new effective acceleration due to gravity inside the accelerating rocket is g. We can write:

  1.25 = 2π √(L/g)

Since the length L of the pendulum did not change, we can equate the expressions for L and solve for g. Rearrange both equations to solve for L:

Equate the two expressions for L:

 (2.45 / 2π)² × 9.80 = (1.25 / 2π)² × g

Solve for g:

g = ( (2.45)² / (1.25)² ) × 9.80

   = (6.0025 / 1.5625) × 9.80

     ≈ 37.62 m/s²

Therefore, the rocket's acceleration during liftoff is 37.62 m/s².

Final answer:

Using the gas law principle and given pressure values, the afternoon temperature of the bicyclist's tires, after an increase in pressure to 422 kPa from 404 kPa, is calculated to be approximately 299 K.

Explanation:

To find the afternoon temperature after the tire pressure increases, we use the gas law principle that describes the relationship between pressure and temperature when the volume and amount of the gas remain constant. Assuming ideal behavior, the relevant equation is the proportionality between pressure and temperature given as P1/T1 = P2/T2.

In this case, the initial pressure (P1) is 404 kPa, the initial temperature (T1) is 286 K, and the final pressure (P2) is 422 kPa. To find the final temperature (T2), we rearrange the formula to T2 = (P2 x T1) / P1.

Plugging in the values:

T2 = (422 kPa x 286 K) / 404 kPa

T2 = 1207962 K kPa / 404 kPa

T2 ≈ 299 K

Therefore, the afternoon temperature when the pressure in the tires has increased to 422 kPa is approximately 299 K.

Answer:

C. kinetic energy

Explanation:

Free Fall

When objects are dropped in free air (no friction), they are attracted to the Earth's surface by the force of gravity. Objects start from zero speed and gradually increase it because of the effect of the acceleration of gravity whose value can be considered as constant [tex]g=9.8\ m/s^2.[/tex]

The final speed of the object at time t is

[tex]v_f=g.t[/tex]

Note that the final speed does not depend on the mass of the object. The kinetic energy is

[tex]\displaystyle k=\frac{mv_f^2}{2}[/tex]

The kinetic energy depends on the speed and the mass, so heavier objects dropped from the same height will have more kinetic energy. Let's analyze the options given in the question .

A. As shown above, the speed (or magnitude of the velocity) is the same regardless of the object's mass, so the heavier cannonball and the iron cannon will reach the ground at the same speed. Incorrect option

B. For every falling object in free air, the only acceleration is the acceleration of gravity. This option is not correct either

C. The kinetic energy is greater for the heavier cannonball because it has more mass, as discussed above. Correct option

D. Only the C. option is correct. This is not

Final answer:

At the instant before the balls hit the ground, the cannonball with greater mass will have greater (C) kinetic energy, while acceleration and velocity will be the same for both balls due to the constant acceleration of gravity. Hence, (C) is the correct option.

Explanation:

The question revolves around the physics concept of objects falling under gravity. When two objects, such as an iron cannon ball and a bowling ball, are dropped from the same height without any air resistance, the acceleration due to gravity acts equally on both.

Regardless of their masses, their acceleration remains constant at ~9.8 m/s². Therefore, the cannonball does not have a greater velocity or acceleration compared to the bowling ball. What differentiates them at the instant before impact is their kinetic energy, which is dependent on mass.

The heavier cannonball will indeed have greater kinetic energy because kinetic energy is calculated using the equation KE = 1/2 m v², where 'm' represents mass and 'v' represents velocity. Since the masses are different and velocities are the same, the cannonball with the greater mass will have greater kinetic energy.

Answer:A 3.57m/s

B Workdone= 246.44J

Explanation:Given: change in heigt= 1-0.36=0.64, mass=435/9.8= 44.4Kg

PE=mgh=44.4×9.8×0.64=278.4J

Let PE=KE=1/2mv2

278.4=44.4/2×v^2

278.4=22.2v^2

V^2=278.4/22.4

V^2=12.79

V=3.57m/s

B. Velocity at the bottom of the swing is 2m/s

KE=44.4×0.36×2=-31.96J

Workdone=PE+KE=278.4-31.96=246.44J

Kelli's speed when the swing passes through its lowest position is 4.14 m/s. The work done on the swing by friction is 905 N.

To determine the speed of Kelli when the swing passes through its lowest position, we can use the principle of conservation of mechanical energy. At the highest point, Kelli has potential energy and no kinetic energy. At the lowest point, the potential energy is zero and the entire mechanical energy is in the form of kinetic energy. Therefore, the speed of Kelli at the lowest position can be found using the equation:

Kinetic energy = Mechanical energy - Potential energy

We know that the mechanical energy is the same at both points, so we can write:
0.5 mv^2 = mg
Where m is the mass of Kelli, v is her speed, g is the acceleration due to gravity, and h is the difference in height between the highest and lowest points of the swing. Rearranging the equation, we have:
v = sqrt(2gh)

Substituting the given values, we get:
v = sqrt(2 * 9.8 m/s^2 * 1.36 m) = 4.14 m/s

Therefore, Kelli is moving at a speed of 4.14 m/s when the swing passes through its lowest position.

To find the work done on the swing by friction, we can use the work-energy principle. The work done by friction can be calculated using the equation:
Work = Change in kinetic energy

Given that the initial velocity of the swing is 2.0 m/s and the final velocity is 0 m/s due to friction, the change in kinetic energy is equal to the initial kinetic energy. Therefore, the work done on the swing by friction is:
Work = 0.5 * m * (initial velocity)^2 = 0.5 * 435 N * (2.0 m/s)^2 = 905 N

The centripetal acceleration of someone at a latitude of 71.9 degrees on Earth's surface is calculated based on the velocity due to Earth's rotation, adjusted by the cosine of the latitude angle, demonstrating how this acceleration compares to Earth's gravitational pull.

The question asks about the centripetal acceleration of a person standing on Earth's surface at a specific latitude and how this compares to Earth's gravitational acceleration, g. To find the centripetal acceleration (ac) at a latitude of 71.9 degrees, we use the fact that at the equator, where the velocity due to Earth's rotation is maximal, ac is purely in the direction of g and reduces with latitude. The reduction factor is the cosine of the latitude angle because the actual velocity component contributing to centripetal acceleration is the component of the Earth's rotational velocity at the surface that is perpendicular to the axis of rotation, which is maximized at the equator and zero at the poles.

To calculate it: The earth rotates once every 24 hours, giving us a rotational velocity at the equator. We can derive velocity (v) at the equator using the formula v = 2πr / T, where r is Earth's radius (approximately 6380 km) and T is the period of rotation (86400 seconds). The equatorial velocity can then be used to calculate centripetal acceleration at the equator with ac = v² / r, which is found to be a fraction of g, as shown in discussions. At a latitude of 71.9 degrees, this acceleration is reduced by the factor cos(71.9°), illustrating how ac changes with latitude.

Thus, to express the centripetal acceleration at this latitude as a multiple of g, one would calculate the acceleration as derived for the equator and adjust it by the factor cos(71.9°). This allows us to understand the reduced effect of Earth's rotation on the apparent weight of an object as the latitude increases, which is crucial in understanding phenomena such as the variation in gravity and the operation of pendulum clocks at different latitudes.