A ball is fixed to the end of a string, which is attached to the ceiling at point P. As the drawing shows, the ball is projected downward at A with the launch speed v0. Traveling on a circular path, the ball comes to a halt at point B. What enables the ball to reach point B, which is above point A? Ignore friction and air resistance.

Final answer:

The speed at which the water leaves the hole is 17.7 m/s, and the diameter of the hole is approximately 4.8 mm using Torricelli's theorem and the flow rate equation.

Explanation:

The scenario described involves principles of fluid dynamics within the field of physics, specifically Bernoulli's equation and the equation of continuity. To calculate the speed at which the water leaves the hole (a) and the diameter of the hole (b), we employ the Torricelli's theorem and the flow rate equation.

Using Torricelli's theorem, the speed (v) of water exiting the hole can be determined by the formula v = √(2gh), where g is the acceleration due to gravity (9.8 m/s²) and h is the height of the water column above the hole (16.0 m).

v = √(2 * 9.8 m/s² * 16.0 m) = √(313.6 m²/s²) = 17.7 m/s

To find the diameter of the hole, we use the flow rate Q = A * v, where Q is the flow rate (2.50 × 10³ m³/min), A is the area of the hole, and v is the speed of water leaving the hole. First, we convert the flow rate to m³/s by dividing by 60 (since there are 60 seconds in a minute):

Q (in m³/s) = (2.50 × 10³ m³/min) / 60 = 4.17 × 10µ m³/s

To find the area: A = Q / v, then to find the diameter (d), we use the area of a circle A = π * (d/2)². Rearranging and solving for d gives us:

d = 2 * √(Q / (π * v))
d = 2 * √((4.17 × 10µ m³/s) / (π * 17.7 m/s)) = 4.8 mm

(a) The speed at which the water leaves the hole is approximately 17.7m/s.

(b) The diameter of the hole is approximately 1.74mm.

Given:

- The depth of the hole below the water level, [tex]\( h = 16.0 \)[/tex] m

- The rate of flow, [tex]\( Q = 2.50 \times 10^{-3} \) m\(^3\)/[/tex]min

Part (a): Speed of Water Leaving the Hole

We can use Torricelli's Law to determine the speed at which the water leaves the hole:

[tex]\[ v = \sqrt{2gh} \][/tex]

where:

- [tex]\( g \)[/tex] is the acceleration due to gravity[tex](\( 9.8 \) m/s\(^2\))[/tex]

- [tex]\( h \)[/tex] is the height of the water above the hole

Plugging in the values:

[tex]\[ v = \sqrt{2 \times 9.8 \times 16.0} \][/tex]

[tex]\[ v = \sqrt{2 \times 9.8 \times 16.0} \][/tex]

[tex]\[ v = \sqrt{313.6} \][/tex]

[tex]\[ v \approx 17.7 \text{ m/s} \][/tex]

Part (b): Diameter of the Hole

First, convert the flow rate to cubic meters per second:

[tex]\[ Q = 2.50 \times 10^{-3} \text{ m}^3/\text{min} \][/tex]

[tex]\[ Q = \frac{2.50 \times 10^{-3}}{60} \text{ m}^3/\text{s} \][/tex]

[tex]\[ Q = 4.17 \times 10^{-5} \text{ m}^3/\text{s} \][/tex]

The flow rate ( Q ) can also be expressed as:

[tex]\[ Q = A \cdot v \][/tex]

where:

- ( A ) is the cross-sectional area of the hole

- ( v ) is the speed of the water leaving the hole

Rearranging to solve for ( A ):

[tex]\[ A = \frac{Q}{v} \][/tex]

[tex]\[ A = \frac{4.17 \times 10^{-5}}{17.7} \][/tex]

[tex]\[ A \approx 2.36 \times 10^{-6} \text{ m}^2 \][/tex]

The area ( A) of a circular hole is given by:

[tex]\[ A = \pi \left( \frac{d}{2} \right)^2 \][/tex]

Solving for the diameter[tex]\( d \)[/tex]:

[tex]\[ d = 2 \sqrt{\frac{A}{\pi}} \][/tex]

[tex]\[ d = 2 \sqrt{\frac{2.36 \times 10^{-6}}{\pi}} \][/tex]

[tex]\[ d = 2 \sqrt{7.51 \times 10^{-7}} \][/tex]

[tex]\[ d \approx 1.74 \times 10^{-3} \text{ m} \][/tex]

[tex]\[ d \approx 1.74 \text{ mm} \][/tex]

Answer:

Because Earth is rotating, horizontal winds everywhere except at the equator are deflected to the right or left relative to Earth's surface. This so-called Coriolis Effect arises because the air is moving over a surface which itself is continually turning because of Earth's rotation. In the Northern Hemisphere, the Coriolis Effect causes wind blowing toward the west to turn toward the North Western.

The work required to move a 1200-N crate 5.0 m along the floor against a friction force of 230 N is 1150 Joules. The work required to move it 5.0 m vertically is 6000 Joules.

The work required to move an object can be calculated using the formula Work = Force x Distance. Force is the total force acting on an object, and distance is the distance the force moves the object.

(a) To move a box on the ground with a friction force of 230 N, the work done can be calculated as follows:

work = friction force x distance = 230 N x 5.0 m = 1150 joules

(b) To lift the box vertically, the force required is equal to the weight of the box, i.e. H. its mass multiplied by the acceleration due to gravity. The problem, however, is the weight of the box. When the box is lifted vertically, the force is equal to the weight, i.e. 1,200 N. The work done is then calculated as follows:

work = weight x height = 1200 N x 5.0 m = 6000 joules.

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Answer:

1.5 times

Explanation:

[tex]h[/tex] = depth of the diver initially = 5 m

[tex]\rho[/tex] = density of seawater = 1030 kg m⁻³

[tex]P_{i}[/tex] = Initial pressure at the depth

[tex]P_{f}[/tex] = final pressure after rising = 101325 Pa

Initial pressure at the depth is given as

[tex]P_{i} = P_{f} + \rho gh\\P_{i} = 101325 + (1030) (9.8) (5)\\P_{i} = 151795 Pa[/tex]

[tex]V_{i}[/tex] = Initial volume at the depth

[tex]V_{f}[/tex] = Final volume after rising

Since the temperature remains constant, we have

[tex]P_{f} V_{f} = P_{i} V_{i}\\(101325) V_{f} = (151795) V_{i}\\V_{f} = 1.5 V_{i}[/tex]

Using Boyle's law, the volume of gas in the diver's lungs would expand by a factor of 1.5 when a diver quickly rises to the surface from a 5m depth. This implies the lungs could potentially double in size if the diver does not exhale gas while ascending.

In this scenario, we can use Boyle's law to answer the question accurately, which is an important principle in Physics that states that the pressure and volume of a gas have an inverse relationship when the temperature is held constant. To be specific, if a diver is at 5m below the sea level, he or she is under 1.5 atmospheres of pressure, including the sea pressure and atmospheric pressure (1 ATA). When the diver rises to the surface, there's only 1 ATA of pressure left, which is the atmospheric pressure.

The volume of gas in the diver's lungs would thus double by a factor of (1.5/1) = 1.5 when he or she ascends quickly to the surface from a 5m depth, according to Boyle's law. This is a simplified explanation without considering the temperature changes or the gas involved, always remember to exhale while rising to the surface to avoid potential lung damage.

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Answer:

73.5 N

Explanation:

W = PE = mgh = 2.5 * 9.8 * 3.0 = 73.5 N

Final answer:

The work done by gravity on a 2.5-kg book falling from a 3.0 m height is calculated to be 73.5 joules, using the formula for work in the presence of gravitational force.

Explanation:

To calculate the work done by gravity on the 2.5-kg book as it falls 3.0 m, you can use the formula work (W) = force (F) × distance (d). Since we're dealing with gravity, the force is the weight of the book, which is its mass multiplied by the acceleration due to gravity (9.8 m/s²). Therefore, the force is 2.5 kg × 9.8 m/s². The distance the book falls is given as 3.0 m.

The formula simplifies to W = m × g × d, which gives us W = 2.5 kg × 9.8 m/s² × 3.0 m, resulting in 73.5 J. The work done by gravity on the book as it falls to the ground is 73.5 joules.

Answer:

(a) 161.57 N

(b) 0.958 m/s^2

Explanation:

Force applied, F = 220 N

mass of crate, m = 61 kg

μ = 0.27

(a) The magnitude of the frictional force,

f = μ N

where, N is the normal reaction

N = m x g = 61 x 9.81 = 598.41 N

So, the frictional force, f = 0.27 x 598.41

f = 161.57 N

(b) Let a be the acceleration of the crate.

Fnet = F - f = 220 - 161.57

Fnet = 58.43 N

According to newton's second law

Fnet = mass x acceleration

58.43 = 61 x a

a = 0.958 m/s^2

Thus, the acceleration of the crate is 0.958 m/s^2.  

Answer:

0.095 A

Explanation:

[tex]z[/tex] = Impedance of the RC series circuit = 525 ohm

[tex]V[/tex] = Operating voltage of the circuit = 50 Volts

[tex]i[/tex] = circuit current

According to ohm's law

[tex]Current = \frac{Voltage}{Impedance}[/tex]

[tex]i = \frac{V}{z}[/tex]

Inserting the values

[tex]i = \frac{50}{525}\\i = 0.095 A[/tex]

Answer:

Internal Oblique.

Explanation:

Lower crossed syndrome is a condition in which there are strong and weak muscles. So there is an imbalance of muscle strengths. It occurs when some muscles constanly get shortened or lengthened just like in this case internal oblique muscle got lengthened.

Answer:

Sella Turcica

Explanation:

The sella turcica (also called the hypophyseal fossa or pituitary fossa) is a mid-line saddle-shaped depression in the sphenoid bone and belongs to the middle cranial fossa and it can be located by centering the CR (central ray) at a point 3/4" anterior and 3/4" superior to the E.A.M. (external auditory meatus).

Answer:

Best explains Jamming

Explanation:

The deliberate radiation of electromagnetic (EM) energy to degrade or neutralize the radio frequency long-haul supervisory control and data acquisition (SCADA) communications links, best explains what?

Jamming is defined as the blocking or interference with authorized wireless communications. it's a problem  in personal area network wireless technologies. Jamming can occur inadvertently due to high levels of noise .

Jammers can send radio signals to interfere or disrupt communication flows by by decreasing the signal-to-noise ratio.They use radio frequency to interfere with communications by keeping it busy.

Answer:

The velocity of the ball as it hit the ground = 19.799 m/s

Explanation:

Velocity: Velocity of a body can be defined as the rate of change of displacement of the body. The S.I unit of velocity is m/s. velocity is expressed in one of newtons equation of motion, and is given below.

v² = u² + 2gs.......................... Equation 1

Where v = the final velocity of the ball, g = acceleration due to gravity, s = the height of the ball

Given: s = 20 m, u = 0 m/s

Constant: g = 9.8 m/s²

Substituting these values into equation 1,

v² = 0 + 2×9.8×20

v² = 392

v = √392

v = 19.799 m/s.

Therefore the velocity of the ball as it hit the ground = 19.799 m/s

Answer:

Either absorption or emission lines

Explanation:

For the radial speed of an astronomical object to be determined with a Doppler shift, what must we be able to see in the object's spectrum, Either absorption or Emission lines.

Asorption and Emission line explained below;

To determine the radial speed of an astronomical object, its spectrum must contain Doppler-shifted lines, either redshifted or blueshifted.

The determination of an astronomical object's radial speed, which is the component of its velocity directed toward or away from the observer, relies on the observation of the Doppler effect in its spectrum. When the object is moving relative to the observer, the wavelengths of emitted light or other electromagnetic waves become shifted, either toward longer (redshift) or shorter (blueshift) wavelengths, due to the relative motion.

By analyzing the positions of spectral lines in the object's spectrum, astronomers can deduce the amount of this shift, allowing them to calculate the radial speed accurately. Redshift indicates motion away from the observer, while blueshift indicates motion towards the observer.

This technique is commonly used in astronomy to measure the motion of celestial objects, including stars, galaxies, and planets, providing valuable insights into their dynamics and interactions.

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Answer:

0.14 seconds

Explanation:

The speed of light in vacuum is approximately 3.0*10^8. The distance that would be covered by the object would be equivalent to the circumference of the cross-section of the earth on the equator.

Circumference = 2[tex]\pi[/tex]*6400000 =4.02*10^7

Time = distance/speed = 4.2*10^7 / 3.0*10^8 =0.14s

Answer:

The given subshell have the same value as the spin quantum number and principal quantum number. Option D is correct.

Explanation:

Every shell has some orbitals. For example, 1st shell has only s-orbital, 2nd shell has s and p-orbitals, 3rd shell has s, p and d-orbitals, 4th shell has s, p, d and f-orbitals.

Now, every orbital has a fix number of subshells. s-orbital has 1 subshell, p-orbital has 3 subshells, d-orbital has 5 subshells, and f-orbital has 7 subshells.

Every subshell of the orbital has same principal quantum number because it is associated with the same shell. Magnetic quantum of every subshell of any orbital is different as it specifies the subshells. And the spin quantum number can be [tex]+1/2 \texttt{ or }-1/2[/tex] based on the spin of the electrons. Every subshell contains two electrons with both the opposite spins.

Let's take an example of 4f-orbital.

Here, the principal quantum number for all the subshells will be 4. The magnetic quantum number will vary from -3 to +3. Every subshell can has two electrons of spin quantum number [tex]+1/2 \texttt{ or }-1/2[/tex].

Therefore, a given subshell have the same value as the spin quantum number and principal quantum number. Option D is correct.

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Final answer:

The two possible speeds that the moving train can have are 343.292 m/s when moving away from the station and 355.182 m/s when moving towards the station.

Explanation:

To determine the two possible speeds that the moving train can have, we can use the formula for the beat frequency:

beat frequency = |v - v_s| / v * f_w

v is the speed of sound,

v_s is the speed of the moving train,

f_w is the frequency of the train whistle,

|v - v_s| is the absolute value of the difference between the speed of sound and the speed of the moving train.

Let's solve this equation for two possible speeds:

Case 1 - Moving away from the station:

8.00 beats/s = |331 m/s - v_s| / 331 m/s * 1.54 *[tex]10^2 Hz[/tex]

Solving for v_s, we find v_s = 343.292 m/s (rounded to three decimal places).

Case 2 - Moving towards the station:

8.00 beats/s = |331 m/s + v_s| / 331 m/s * 1.54 * [tex]10^2 Hz[/tex]

Solving for v_s, we find v_s = -355.182 m/s (rounded to three decimal places). Since the speed cannot be negative, we take the absolute value and find v_s = 355.182 m/s (rounded to three decimal places).

Therefore, the two possible speeds that the moving train can have are:

Moving away from the station: 343.292 m/s (rounded to three decimal places)

Moving towards the station: 355.182 m/s (rounded to three decimal places)

Answer:

11.2m

Explanation:

Suppose the pumpkin is launched vertically and the speed of 13.9 m/s is absolutely vertical. As the pumpkin rises up, its kinetic energy is converted to potential energy, let the reference point be the round, we can create the following equation from the law of energy conservation:

[tex] E_1 + P_1 = E_2 + P_2[/tex]

[tex] 0.5mv_1^2 + mgh_1 = 0.5mv_2^2 + mgh_2[/tex]

we can substitute [tex]g = 9.8m/s^2, v_1 = 13.9m/s, v_2 = 0.5v_1 = 0.5*13.9 = 6.95 m/s, h_1 = 3.8[/tex]

We can also divide both sides by m and 0.5

[tex]13.9^2 + 2*9.8*3.8 = 6.95^2 + 2*9.8*h_2[/tex]

[tex]19.6h_2 = 219.3875[/tex]

[tex]h_2 = 11.2m[/tex]

Answer:

1.45549 m/s

Explanation:

m = Mass of car = 3010 kg

v = Velocity of car

k = Spring constant = [tex]6\times 10^6\ N/m[/tex]

x = Displacement of spring = 3.26 cm

As the energy of the system is conserved

[tex]\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{kx^2}{m}}\\\Rightarrow v=\sqrt{\dfrac{6\times 10^6\times 0.0326^2}{3010}}\\\Rightarrow v=1.45549\ m/s[/tex]

The speed of the car before impact is 1.45549 m/s

Answer:

v=1617.77m/s

Explanation:

A particle of charge q is fixed at point P, and a second particle of mass m and the same charge q is initially held a distance r1 from P. The second particle is then released. Determine its speed when it is a distance r2 from P. Let q = 3.1 µC, m = 47 mg, r1 = 0.83 mm, and r2 = 2.5 mm.

_________m/s

from the law of applied electric force which states that the force of attraction of electric charge particles is directly proportional to the two charge particles and inversely proportional to the square of their distances apart

dk= -dU

kinetic energy equals potential energy

[tex]\frac{1}{2} mv^{2} -\frac{1}{2} mu^{2} =-(\frac{kq^2}{r2} -\frac{kq^2}{r1} )[/tex]

[tex]\frac{1}{2} 47*10^-6v^{2} -\frac{1}{2} 47*10^-6*0^{2} =-(\frac{9*10^9*(3.1*10^-6)^2}{(2.5*10^-3)} -\frac{9*10^-9*(3.1*10-6)^2}{0.83*10^-3} )[/tex]

23.5*10^-6v^2=96.1-34.596)

v^2=61.504/23.5*10^-6)

v^2=2617191.48

v=1617.77m/s

v=1.617km/s

A particle of charge q is fixed at point P, and a second particle of mass m and the same charge q is initially held a distance r1 from P. The second particle is then released. Determine its speed when it is a distance r2 from P. Let q = 3.1 µC, m = 47 mg, r1 = 0.83 mm, and r2 = 2.5 mm.

v=1617.77m/s.

Charge refers to a fundamental property of matter that determines how it interacts with electric and magnetic fields. It is one of the basic building blocks of nature and comes in two types: positive and negative.

The concept of charge is closely related to the phenomenon of electricity. Electric charge is responsible for the creation and interaction of electric fields, which can exert forces on other charged objects. Like charges repel each other, while opposite charges attract each other.

Protons, which carry a positive charge, have a charge of +e, while electrons, which carry a negative charge, have a charge of -e.

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Answer:

0.025hr

Explanation:

The full solution is on the image below. The two cars cover the same distance at different time intervals. Since the distance is constant, the velocity is inversely proportional to the time taken to cover the constant distance

Answer:

0.025 h

Explanation:

Let's assume for the first car, the destination is [tex]x_{1}[/tex], the time is [tex]t_{1}[/tex], the velocity is [tex]v_{1}[/tex] and for the second car the destination is [tex]x_{2}[/tex], the time is [tex]t_{2}[/tex], the velocity is [tex]v_{2}[/tex].

We are given:

[tex]v_{1}[/tex] = 97 km/h

[tex]v_{2}[/tex] = 113 km/h

If we are asked the time, the destinations must be equal which are also given:

[tex]x_{1}[/tex] = [tex]x_{2}[/tex] = 17

For constant velocity, the equation is x = v * t

Hence,

[tex]x_{1}[/tex] = [tex]v_{1}[/tex] * [tex]t_{1}[/tex] = [tex]x_{1}[/tex] = 97 * [tex]t_{1}[/tex] = 17

⇒ [tex]t_{1}[/tex] = 17/97 = 0.175 h

[tex]x_{2}[/tex] = [tex]v_{2}[/tex] * [tex]t_{2}[/tex] = [tex]x_{2}[/tex] = 113 * [tex]t_{2}[/tex] = 17

⇒ [tex]t_{2}[/tex] = 17/113 = 0.150 h

So,

[tex]t_{1}[/tex] -  [tex]t_{2}[/tex] = 0.175 - 0.150 = 0.025 h

The second car arrives 0.025 h sooner.

Answer:[tex]E=1.041\times 10^7 V/m[/tex]

Explanation:

Given

Potential Difference [tex]\Delta V=0.076 V[/tex]

thickness of cell membrane [tex]d=7.30\times 10^{-9} m[/tex]

Electric Field for this Potential is given by

[tex]E=\frac{\Delta V}{d}[/tex]

[tex]E=\frac{0.076}{7.30\times 10^{-9}}[/tex]

[tex]E=0.01041\times 10^9 V/m[/tex]

[tex]E=1.041\times 10^7 V/m[/tex]

Answer:

a) the speed of the electron is 1.11 × 10⁷ m/s

b) the radius of electron's path in the magnetic field is 3.16 × 10⁻⁴ m

Explanation:

a) Let's assume that we have an electron accelerated using a potential difference of V = 350, which gives the ion a speed of v. In order to find this speed we set the potential energy of the electron equal to its kinetic energy. Thus,

eV = 1/2 m v²

where

Thus,

v = √[2eV / m]

v = √[2(1.6 × 10⁻¹⁹ C)(350 V) / 9.11 × 10⁻³¹ kg]

v = 1.11 × 10⁷ m/s

Therefore, the speed of the electron is 1.11 × 10⁷ m/s

b) Then the electron enters a region of uniform magnetic field, it moves in a circular path with a radius of:

r = mv / eB

where

Thus,

r = (9.11 × 10⁻³¹ kg)(1.11 × 10⁷ m/s) / (1.6 × 10⁻¹⁹ C)(200 × 10⁻³ T)

r = 3.16 × 10⁻⁴ m

Therefore, the radius of electron's path in the magnetic field is 3.16 × 10⁻⁴ m

Answer:

Only the 4 th statement is true that is bed 1 and 3 are older than 4.

Explanation:

The 5 beds are numbered from 1 to 5 , 1 being the lowest and 5 being the topmost bed.

We are given 4 statements and we have to find out which all are true.

(a)Bed 4 is older than bed 2

This is wrong because the lower beds are older than beds that are higher.

(b)Bed 3 is older than beds 2 and 4

This is also wrong because the 2 is older than 3

(c)Bed 5 is the oldest

This is wrong because bed 1 is the oldest

(d)Beds 1 and 3 are older than 4

This is true as lower beds are older

Answer:

bed 2 was deposited before bed 3

Explanation:

Which of the following conclusions can be made about the sedimentary layers?

bed 2 was deposited after bed 3

bed 1 is the youngest

bed 3 was deposited before bed 1

bed 2 was deposited before bed 3

Answer:

b.

Explanation:

BMI is body mass index, it is calculated by dividing the weight of a person in kilogram by square of height in meters.

A high BMI is an indicator of obesity. Obese people are prone to death causing diseases like High Blood pressure, cardiac arrest and diabatese.  So, a high value of BMI values has the highest increased risk for premature mortality.

So the correct answer is 35 kg/m^2.

Answer:

Energy expenditure due to physical activity generally accounts for 20% of total energy expenditure.

Explanation:

The total amount of energy your body uses daily is usually divided as follows:

1. Your Basal Metabolic Rate at rest  (50-70%)

Basal Metabolic Rate is the number of calories required to keep your body      functioning at rest

2. Consumption by your daily physical activities (20%)

3. Energy to digest food. (10-20%)

This energy has to be deducted from the overall energy content of the food itself.

Then, the answer is:

Energy expenditure due to physical activity generally accounts for 20% of total energy expenditure.

Final answer:

The energy expenditure due to physical activity generally accounts for 10% of total energy expenditure. Physical activity guidelines recommend adults to engage in aerobic and muscle-strengthening activities for substantial health benefits.

Explanation:

The energy expenditure due to physical activity generally accounts for 10% of total energy expenditure.

According to the 2018 Physical Activity Guidelines for Americans issued by the Department of Health and Human Services, adults should do at least 150 minutes to 300 minutes per week of moderate-intensity aerobic activity, or 75 minutes to 150 minutes per week of vigorous-intensity aerobic physical activity, or an equivalent combination of both. Engaging in physical activity beyond 300 minutes per week can result in additional health benefits.

Therefore, the energy expenditure due to physical activity is a crucial component of total energy expenditure and plays a significant role in maintaining overall health.

Answer:

D) True. The value of the work is constant and the force is less, therefore it is easier to raise the piano

Explanation:

In this exercise the work to raise the piano is given by

          W = F d cos θ

The distance and the displacement have an angle of zero degrees

          W = F d

If we use energy conservation

          W = ΔEm

         W = [tex]Em_{f}[/tex] - Em₀

Suppose the piano starts from rest and reaches the top also at rest

         W = mh y₂ - mg y₁

        W = mg h

        h= 3 ft

We see that the work to raise the body is always the same, regardless of the path, so, in the work relationship or is the product of force for distance if one goes up the other must decrease so that the product is the same.

Let's examine the answers

A) False. The distance on a ramp is greater

B) False. Change the value of work

C) False. Change the value of work

D) True. The value of the work is constant and the force is less, therefore it is easier to raise the piano

Answer:

A. Linkedln

Explanation:

In addition to stand-alone discussion boards, all of these sites include discussion boards as part of their features except: ​

A. Linkedln

All other application in addition to stand-alone Google, Yahoo, Microsoft, etc. include discussion board.

Answer:

If the gas inside a rigid steel vessel (constant volume) is heated in such a way that its temperature, measured in Kelvin degrees, doubles then the internal pressure will increase by a factor of 2.

Explanation:

Gay-Lussac's law can be expressed mathematically as follows:

[tex]\frac{P}{T} =k[/tex]

where V = volume, T = temperature, K = Constant

This law indicates that the ratio between pressure and temperature is constant.

This law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of shocks against the walls increases, that is, the pressure increases. That is, the gas pressure is directly proportional to its temperature. This must be fulfilled since the relationship must remain constant

In short, when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature decreases, gas pressure decreases.

It is desired to study two different states, an initial state and an final state. You have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment. When the temperature varies to a new T2 value, then the pressure will change to P2, and the following will be true:

[tex]\frac{P1}{T1} =\frac{P2}{T2}[/tex]

Given the above, it is possible to say that if the gas inside a rigid steel vessel (constant volume) is heated in such a way that its temperature, measured in Kelvin degrees, doubles then the internal pressure will increase by a factor of 2 (doubles too)

Answer:

a) True.

Explanation:

If you turn the wheel in the direction of the turn before beginning the turning maneuver then it's possible that there might be not enough space available for turning and also if you are waiting for the traffic to get clear with rear  ended then it will get pushed forward onto the coming traffic.

IN this Turning in a Car question, The steering wheel of a car must be turned in the direction of the desired turn. The feeling of being flung away from the center during a turn is due to the 'centrifugal force', a result of the body's inertia.

The statement 'When making a turn, do not have the steering wheel turned in the direction of the turn before beginning the turning maneuver.' is False. When making a turn in a car, you indeed must turn the steering wheel in the direction you wish to go.

This causes the car to change its direction, following a circular path or curve. The change in direction is due to the force that you apply to the steering wheel and subsequently to the car's tires that are in contact with the road.

The force applied to the tires creates an acceleration, changing the car's velocity from straight-line motion to a curved path.

Relatedly, the sensation of being pulled or flung away from the center of the turn as you make it is described by a concept in physics called the centrifugal force.

This is not an actual force. Instead, it's a result of your body's tendency to continue in straight-line motion (according to Newton's first law of motion), while the car is changing direction.

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To determine how far up the ladder a person can climb before the ladder begins to slip, we need to consider the forces and torques acting on the ladder. The condition for the ladder to begin slipping is when the torque due to the person's weight is greater than the torque due to the static friction force.

To determine how far up the ladder a person can climb before the ladder begins to slip, we need to consider the forces and torques acting on the ladder. The forces include the normal reaction force from the floor, the static friction force between the ladder and the floor, the weight of the ladder, and the normal reaction force from the wall. The torques are calculated by multiplying the force by the lever arm, which in this case is the distance between the center of mass of the ladder and the point of contact with the floor.

The condition for the ladder to begin slipping is when the torque due to the person's weight is greater than the torque due to the static friction force. We can calculate the torque due to the person's weight by multiplying the person's weight by the distance between their position on the ladder and the point of contact with the floor. Next, we calculate the torque due to the static friction force by multiplying the static friction force by the distance between the point of contact with the floor and the point of contact with the wall.

Setting these two torques equal to each other and solving for the distance d yields the distance up the ladder that the person can climb before it begins to slip.

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Answer:

B 1.08 BILLION

Explanation:

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