Fluids at rest possess no flow energy. a)-True b)-False

Answers

Answer 1

I believe the answer is A: True


Related Questions

An object moves at 60 m/s in the +x direction. As it passes through the origin it gets a 4.5 m/s^2 acceleration in the -x direction. a) How much time elapses before it returns back to the origin?
b) What is its velocity when it returns to the origin?

Answers

Answer:

a) After 26.67 seconds it returns back to the origin

b) Velocity when it returns to the origin = 60 m/s in the -x direction

Explanation:

a) Let the starting position be origin and time be t.

  After time t displacement, s = 0 m

  Initial velocity, u = 60 m/s

  Acceleration, a = -4.5 m/s²

 We have equation of motion s = ut + 0.5 at²

 Substituting

        s = ut + 0.5 at²

        0 = 60 x t + 0.5 x (-4.5) x t²        

        2.25t² - 60 t = 0

        t² - 26.67 t = 0

        t (t-26.67) = 0

      t = 0s or t = 26.67 s

So after 26.67 seconds it returns back to the origin

b) We have equation of motion v = u + at

  Initial velocity, u = 60 m/s

  Acceleration, a = -4.5 m/s²

  Time , t = 26.67

 Substituting

        v = 60 - 4.5 x 26.67 = -60 m/s

Velocity when it returns to the origin = 60 m/s in the -x direction

 

A 50-kg crate is initially at rest on a horizontal floor. A person pushes horizontally with a constant force of 250 N. The maximum static frictional force between the floor and the crate is 275 N. What is the frictional force on the crate while the person pushes on it?

Answers

Answer:

250 N

Explanation:

m = Mass of cart = 50 kg

F = Pushing Force = 250 N

g = Acceleration due to gravity = 9.81 m/s²

μ = Coefficient of friction

[tex]F_s =\text{Static frictional force between the floor and the crate}= 275\ N[/tex]

N = Weight of cart = mg

N = 50×9.81

N = 490.5 N

[tex]F_s =\mu N\\\Rightarrow \mu=\frac{F_s}{N}\\\Rightarrow \mu=\frac{275}{490.5}\\\Rightarrow \mu=0.56[/tex]

Here it can be seen that the pushing force is less than the static frictional force so the crate will not move

∴ The frictional force on the crate while the person pushes on it is 250 N

A rigid tank contains 1.80 moles of helium, which can be treated as an ideal gas, at a pressure of 27.3 atm. While the tank and gas maintain a constant volume and temperature, a number of moles are removed from the tank, reducing the pressure to 5.20 atm. How many moles are removed?

Answers

Answer:

Has been removed 1.458 moles.

Explanation:

n1= 1.8 mol

p1= 27.3 atm

p2= 5.2 atm

n2= ?

n2= n1 * p2/p1

n2= 0.342 moles

Δn= n1-n2

Δn= 1.458 moles

On a sunny summer day, the intensity of sunlight at the earth's surface is about 1000 W/m2. What are the peak electric and magnetic fields in this sunlight?

Answers

Answer:

peak value of magnetic field is

[tex]B_o = 2.9 \times 10^{-6} T[/tex]

peak value of electric field is

[tex]E_o = 868 N/C[/tex]

Explanation:

As we know that the intensity of electromagnetic waves is given by the formula

[tex]I = \epsilon_0 E_{rms}^2 c[/tex]

here we have

[tex]I = 1000 W/m^2[/tex]

[tex]\epsilon_0 = 8.85 \times 10^{-12}[/tex]

[tex]c = 3\times 10^8[/tex]

now we have

[tex]1000 = (8.85 \times 10^{-12})E_{rms}^2(3 \times 10^8)[/tex]

[tex]E_{rms} = 614 N/C[/tex]

now peak value of electric field is given as

[tex]E_o = \sqrt2 E_{rms}[/tex]

So the peak value of electric field is given as

[tex]E_o = 868 N/C[/tex]

Now we have relation between electric field and magnetic field as

[tex]B_o = \frac{E_o}{c}[/tex]

[tex]B_o = \frac{868}{3\times 10^8}[/tex]

[tex]B_o = 2.9 \times 10^{-6} T[/tex]

A particle starts from rest and has an acceleration function 5 − 10t m/s2 . (a) What is the velocity function? (b) What is the position function? (c) When is the velocity zero?

Answers

Explanation:

It is given that,

A particle starts from rest and has an acceleration function as :

[tex]a(t)=(5-10t)\ m/s^2[/tex]

(a) Since, [tex]a=\dfrac{dv}{dt}[/tex]

v = velocity

[tex]dv=a.dt[/tex]

[tex]v=\int(a.dt)[/tex]

[tex]v=\int(5-10t)(dt)[/tex]

[tex]v=5t-\dfrac{10t^2}{2}=5t-5t^2[/tex]

(b) [tex]v=\dfrac{dx}{dt}[/tex]

x = position

[tex]x=\int v.dt[/tex]

[tex]x=\int (5t-5t^2)dt[/tex]

[tex]x=\dfrac{5}{2}t^2-\dfrac{5}{3}t^3[/tex]

(c) Velocity function is given by :

[tex]v=5t-5t^2[/tex]

[tex]5t-5t^2=0[/tex]

t = 1 seconds

So, at t = 1 second the velocity of the particle is zero.

To find the velocity function of the particle, integrate the acceleration function, which results in 5t - 5t^2. The position function is found by integrating the velocity function, yielding 1/2 * 5t^2 - 1/3 5t^3. The velocity is zero at t = 0 and t = 1 second.

We can solve for the velocity function and the position function by integrating the acceleration function.

Velocity Function

The velocity function is the integral of the acceleration function. Since the acceleration function given is a(t) = 5 - 10t m/s2, the velocity function v(t) is obtained by integrating a(t) with respect to time:

v(t) = ∫ (5 - 10t) dt = 5t - 5t2 + C

Since the particle starts from rest, the initial velocity is 0, which means C = 0. Therefore, the velocity function is v(t) = 5t - 5t2.

Position Function

The position function s(t) is the integral of the velocity function. We integrate v(t) to obtain the position function:

s(t) = ∫ (5t - 5t2) dt = ½ 5t2 - ⅓ 5t3 + K

Since the particle starts from position zero, K = 0 and the position function is s(t) = ½ 5t2 - ⅓ 5t3.

Zero Velocity

To find when the velocity is zero, we set the velocity function equal to zero and solve for t:
5t - 5t2 = 0t = 0 or t = 1 s.

You wish to move a heavy box on a rough floor. The coefficient of kinetic friction between the box and the floor is 0.9. What is the best way to reduce the force of friction between the box and the floor while the box is moving? A. Push or pull the box with a force parallel to the floor. B. Pull the box with a force directed up at an angle of 30° from the horizontal. C. Push the box with a force directed down at an angle of 30° from the horizontal. D. The coefficient of friction does not change so the force of kinetic friction acting on the block must be constant E. Both B and C result in the same force of friction.

Answers

The best way to reduce the force of friction between the box and the floor while the box is moving is 8%.

What is kinetic friction give an example?

If two surfaces are in contact and moving relative to one another, then the friction between them is called kinetic friction. For example, friction slows a hockey puck sliding on ice.

What are kinetic and static friction?

In static friction, the frictional force resists the force that is applied to an object, and the object remains at rest until the force of static friction is overcome. In kinetic friction, the frictional force resists the motion of an object.

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A golfer hits a golf ball with a club head velocity of 94 mph. Mass of golf club head (m): 190 g Mass of golf ball (ms): 46 g Coefficient of restitution (COR) (e): 0.83 What is most nearly the velocity of the golf ball after being hit?

Answers

Answer:

Velocity of golf ball≅ 12.87mph    

Explanation:

Using the theory of conservative momentum to both the club head and the golf ball  we have;

Qi1 + Qi2= Qf1 + Qf2

Qi1: initial momentum for the club head

Qf1: final momentum for the club head

Qi2: initial momentum for the golf ball

Qf2: final momentum for the golf ball

momentum (Q) = Mass x velocity

which means the sum of the momentums of both club head and golf ball has to be the same before and after they have collided.

using the Coeficient of restitution e= 0.83 allows us to know what kind of collision we are dealing with, which is a partially elastic collision since

0> e=0.83 >1.

Qi1= 190 x 94= 17,860 mph.g

Qf1= 190 x Vf1

Qi2= 46 x 0= 0 mph.g

Qf2= 46x Vf2

Using the value of e to determine Vf2 as the final velocity of the golf ball:

e= [tex]\frac{Vf2- Vf1}{Vi1 - Vi2}[/tex]

0.83= [tex]\frac{Vf2 - Vf1}{94 - 0}[/tex]

Vf1= (0.83 x 94)+ Vf2

Vf1= 78.02 + Vf2

Qi1 + Qi2= Qf1 + Qf2

17,860 + 0 = 190xVf1 + 46xVf2    

17,860= 190x (78.02+Vf2) + 46xVf2

17,860= 14,823+ 190xVf2+46xVf2

Vf2≅ 12.87mph                

The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2615 N with an effective perpendicular lever arm of 2.85 cm , producing an angular acceleration of the forearm of 110.0 rad / s2 . What is the moment of inertia of the boxer's forearm?

Answers

Answer:

0.68 kg-m²

Explanation:

F = Force applied by the muscle = 2615 N

r = effective perpendicular lever arm = 2.85 cm = 0.0285 m

α = Angular acceleration of the forearm = 110.0 rad/s²

I  = moment of inertia of the boxer's forearm = ?

Torque is given as

τ = I α                                   eq-1

Torque is also given as

τ = r F                                   eq-2

using eq-1 and eq-2

r F =  I α

(0.0285)(2615) = (110.0) I

I = 0.68 kg-m²

The emf of a battery is equal to its terminal potential difference: A) under all conditions B) only when the battery is being charged C) only when a large cwrrent is in the battery D) only when there is no current in the battery E) under no conditions

Answers

Answer:

option (C)

Explanation:

EMF stand for electro motive force. the emf of a battery is the potential between the two electrodes when it is not use in the circuit.

The terminal potential difference is the potential difference between the electrodes of a cell when it is in use.

EMF is only when the current is very large in the battery.

The emf of a battery is equal to its terminal potential difference only when a large cwrrent is in the battery.

C) only when a large cwrrent is in the battery.

A cylindrical can 150 mm in diameter is filled to a depth of 100 mm with a fuel oil. The oil has a mass of 1.56 kg. Calculate its density, specific weight and specific gravity.

Answers

Answer:

Density (φ) = 0,8827 Kg/L

Specific weight (Ws) = 8,65 N/L

Specific gravity (Gs) = 0,8827 (without unit)

Explanation:

The density formula: φ = [tex]\frac{m}{V}[/tex]

I know the mass "m", I need to find out the volume of the cylinder (V)

V = π* r²*h

The radius "r" is equal to half the diameter (150mm) = 75mm

Now I can find out the density (φ)

φ = [tex]\frac{1,56Kg}{1,767145L}[/tex] = 0,8827 Kg/L

The specific weight (Ws) is the relationship between the weight of substance (oil) and its volume. We apply the following formula:

Ws = φ*g

(g = gravity = 9,8 m/s²)

Finally, specific gravity (Gs) is the ratio between the density of a substance (oil) "φ(o)" and the density of water "φ(w)" :

Gs = φ(o) /  φ(w)

(φ(w) = 1 Kg/L

Hope this can help you !!

When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of 0.140 m from the center of the bolt? (a) How much torque are you exerting in newtons x meters (relative to the center of the bolt) (b) Convert this torque to footpounds

Answers

Answer:

Part a)

[tex]\tau = 23.1 Nm[/tex]

Part b)

[tex]\tau = 17.05 Foot pound force[/tex]

Explanation:

As we know that torque is defined as the product of force and its perpendicular distance from reference point

so here we have

[tex]\tau = \vec r \times \vec F[/tex]

now we have

[tex]\tau = (0.140)(165)[/tex]

[tex]\tau = 23.1 Nm[/tex]

Part b)

Now we know the conversion as

1 meter = 3.28 foot

1 N = 0.225 Lb force

now we have

[tex]\tau = 23.1 Nm[/tex]

[tex]\tau = 23.1 (0.225 Lb)(3.28 foot)[/tex]

[tex]\tau = 17.05 Foot pound force[/tex]

Final answer:

The torque exerted when tightening a bolt with a force of 165 N at a distance of 0.140 m from the center of the bolt is 23.1 N·m. This torque can be converted to approximately 17.0 ft·lb.

Explanation:

To find the torque exerted when tightening a bolt, you can use the equation T = RF sin(θ). In this case, the force is 165 N and the distance from the center of the bolt is 0.140 m. Since you are pushing perpendicularly, the angle θ is 90 degrees. Thus, the torque exerted is 165 N x 0.140 m x sin(90°) = 23.1 N·m.

To convert this torque to foot-pounds, you can use the conversion factor 1 N·m = 0.73756 ft·lb. Therefore, the torque is 23.1 N·m x 0.73756 ft·lb/N·m ≈ 17.0 ft·lb.

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In a simple picture of the hydrogen atom, the electron moves in circular orbits around the central proton attracted by the Coulomb force. The lowest (n = 1) energy orbit that is allowed for the electron is at a radius of 5.29 × 10–11 m . Calculate the magnetic field strength at the proton due to the orbital motion of the electron in the n = 1 state.

Answers

Answer:

B = 12.46 T

Explanation:

At n = 1 state we know that radius is given as

[tex]R = 5.29 \times 10^{-11} m[/tex]

now we have

[tex]T = \frac{2\pi R}{v}[/tex]

here we know that speed is given in that

[tex]v = 2.18 \times 10^6 m/s[/tex]

now the time period is given as

[tex]T = \frac{2\pi R}{v}[/tex]

[tex]T = \frac{2\pi (5.29 \times 10^{-11})}{2.18 \times 10^6}[/tex]

[tex]T = 1.52 \times 10^{-16} s[/tex]

Now the electric current due to revolution of charge is given by

[tex]i = \frac{e}{T}[/tex]

[tex]i = \frac{1.6 \times 10^{-19}}{1.52 \times 10^{-16}}[/tex]

[tex]i = 1.05 \times 10^{-3} A[/tex]

now magnetic field at the center position is given as

[tex]B = \frac{\mu_0 i}{2R}[/tex]

[tex]B = \frac{4\pi \times 10^{-7} (1.05 \times 10^{-3})}{2(5.29 \times 10^{-11}}[/tex]

[tex]B = 12.46 T[/tex]

A long solenoid, of radius a, is driven by an alternating current, so that the field inside is sinusoidal: B(t) B cos(ot)ż. A circular loop of wire, of radius a/2 and resistance R, is placed inside the solenoid, and coaxial with it. Find the current induced in the loop, as a function of time. 6.

Answers

Final answer:

The induced current in a loop inside a sinusoidally-driven solenoid can be determined from Faraday's law of electromagnetic induction. The time derivative of the changing flux through the loop, when divided by the resistance of the loop, provides the value of the induced current.

Explanation:

The question asks about the current induced in a loop placed inside a solenoid driven by an alternating current. This is a physics problem related to electromagnetic induction. The alternating magnetic field in the solenoid induces an electric field that in turn induces a current in the loop, by Faraday's Law.

In this case, the magnetic field B inside the solenoid is oscillating sinusoidally with time as B(t) = B cos(ωt). Thus, the magnetic flux Φ through the loop is changing with time. Using Faraday's law, this flux change induces an emf in the loop, which induces a current I. The emf -dΦ/dt is equal to I*R where R is the resistance of the loop. Hence the induced current I as a function of time can be given by I = -(1/R) * (d/dt) (B cos(ωt) * π * (a/2)^2).

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Which of the following has the most mass?

a- A hot air balloon
b- 20 bowling balls
c- A canoe
d- 30 ounces of led

Answers

Answer:


I believe it’s B

Answer: A. Hot air balloon

Explanation: A hot air balloon is the heaviest, at around 800 pounds, because it carries lots of gas and is very large.

Light with a wavelength of 612 nm incident on a double slit produces a second-order maximum at an angle of 25 degree. What is the separation between the slits?

Answers

Answer:

Speration between the slits [tex]d=2.86\ \mu m[/tex]

Explanation:

Given that,

Wavelength of light, [tex]\lambda=612\ nm=612\times 10^{-9}\ m[/tex]

It is incident on a double slit produces a second-order maximum at an angle of 25 degree.

The condition for maxima is given as :

[tex]dsin\theta=m\lambda[/tex]

d = seperation between the slits

m = order, m = 2

[tex]d=\dfrac{m\lambda}{sin\theta}[/tex]

[tex]d=\dfrac{2\times 612\times 10^{-9}\ m}{sin(25)}[/tex]

d = 0.00000289

or

[tex]d=2.86\mu m[/tex]

So, the seperation between the silts is [tex]2.86\ \mu m[/tex]. Hence, this is the required solution.

You make a capacitor from 2 flat plates each with an area of 10 cm^2 you use a 1mm thick piece of pyrex glass as your dielectric. a. a.what is the maximum voltage you can safely apply to this capacitor?
b. what is its capacitance?
c. what is the maximum amount of charge you can store?
d. when this much charge is on the capacitor, how much energy is stored?

Answers

Answer:

Part a)

[tex]\Delta V_{max} = 14 \times 10^3 Volts[/tex]

Part b)

[tex]C = 4.96 \times 10^{-11} farad[/tex]

Part c)

[tex]Q = 0.69 \mu C[/tex]

Part d)

[tex]E = 4.86 \times 10^{-3} J[/tex]

Explanation:

Part a)

As we know that dielectric constant of pyrex glass is 5.6 and its dielectric breakdown strength is given as

[tex]E = 14 \times 10^6 V/m[/tex]

now we have

[tex]E . d = \Delta V[/tex]

[tex](14 \times 10^6)(0.001) = \Delta V[/tex]

so we have

[tex]\Delta V_{max} = 14 \times 10^3 Volts[/tex]

Part b)

Capacitance is given as

[tex]C = \frac{k\epsilon_0 A}{d}[/tex]

[tex]C = \frac{5.6(8.85 \times 10^{-12})(10 \times 10^{-4}}{0.001}[/tex]

[tex]C = 4.96 \times 10^{-11} farad[/tex]

Part c)

Now we have

[tex]Q = C\Delta V[/tex]

[tex]Q = (4.96 \times 10^{-11})(14 \times 10^3)[/tex]

[tex]Q = 0.69 \mu C[/tex]

Part d)

[tex]Energy = \frac{1}{2}CV^2[/tex]

[tex]E = \frac{1}{2}(4.96 \times 10^{-11})(14 \times 10^3)^2[/tex]

[tex]E = 4.86 \times 10^{-3} J[/tex]

Sound 1 has an intensity of 45.0 W/m2. Sound 2 has an intensity level that is 3.2 dB greater than the intensity level of sound 1. What is the intensity level of sound 2?

Answers

Answer:

The intensity level of sound 2 is 93.3\ W/m².

Explanation:

Given that,

Intensity of sound 1 = 45.0 W/m²

Intensity level of sound 2 = 3.2 dB

[tex]I_{0}=10^{-12}\ w/m^2[/tex]

We need to calculate the intensity

Using equation of the sound level intensity

[tex]I_{dB}=10 log\dfrac{I}{I_{0}}[/tex]

[tex]I_{dB}=10 log(\dfrac{45.0}{10^{-12}})[/tex]

[tex]I_{dB}=136.5 dB[/tex]

The intensity of sound 2 is greater than 3.2 dB.

Therefore,

[tex]I_{2}=136.5+3.2[/tex]

[tex]I_{2}=139.7\ dB[/tex]

Calculate the intensity of sound 2

[tex]I_{2}=10log(\dfrac{I'}{10^{-12}})[/tex]

[tex]I'=10^{-12}(10^{\frac{139.7}{10}})[/tex]

[tex]I'=93.3\ W/m^2[/tex]

Hence, The intensity level of sound 2 is 93.3\ W/m².

Final answer:

Decibels, measured in dB, quantifies the intensity level of sound. They relate to the intensity of sound in a logarithmic manner. More information and advanced calculations are needed to find the actual intensity of Sound 2 based on the given decibel difference

Explanation:

Let's begin by understanding that decibels (dB) measure the intensity level of sound. The decibel level (dB) is a logarithmic unit used to describe a ratio of two values of a physical quantity, here, the sound intensity. One key fact is that each time the intensity increases by a factor of 10; the decibel level rises by 10 dB. Thus, the difference in decibels corresponds to the logarithmic ratio of the two intensities involved.

In the given problem, Sound 2 is 3.2 dB greater than Sound 1. It means the ratio of their intensities has a logarithmic relation with a value of 3.2. However, to convert this decibel difference into an actual value for Sound 2's intensity, additional calculations involving logarithms and the initial intensity value for Sound 1 are needed. As these calculations involve advanced math, it'd be more appropriate to take this problem to a higher level course or consult a logarithm table or a scientific calculator.

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A car moving with an initial speed v collides with a second stationary car that is 51.7 percent as massive. After the collision the first car moves in the same direction as before with a speed that is 35.3 percent of the original speed. Calculate the final speed of the second car. Give your answer in units of the initial speed (i.e. as a fraction of v).

Answers

Answer:

1.25 v

Explanation:

m₁ = mass of first car = m

m₂ = mass of the second car = (0.517) m₁ = (0.517) m

v₁ = initial velocity of first car before collision = v

v'₁ = final velocity of first car after collision = (0.353) v₁ = (0.353) v

v₂ = initial velocity of second car before collision = 0

v'₂ = final velocity of second car after collision = ?

Using conservation of momentum

m₁ v₁ + m₂ v₂ = m₁ v'₁ + m₂ v'₂

(m)(v) + ((0.517) m)(0) = (m)((0.353) v) + ((0.517) m) v'₂

v = (0.353) v + (0.517) v'₂

(0.517) v'₂ = v - (0.353) v

(0.517) v'₂ = (0.647) v

v'₂ = 1.25 v

What is the difference between the formation of an ionic bond and formation of a covalent bond?

Answers

Answer:

Answer to the question:

Explanation:

Differences between ionic bond and covalent bond:

The ionic bond occurs between two different atoms (metallic and non-metallic), while the covalent bond occurs between two equal atoms (non-metallic).

In the covalent bond there is an electron compartment, while in the ionic bond there is an electron transfer.

Ionic bonds have a high melting and boiling point, while covalent bonds usually have a low point.

Final answer:

An ionic bond involves the transfer of electrons, resulting in a bond between a positive ion and a negative ion. A covalent bond involves the sharing of electrons between atoms, creating a stable electron arrangement for both atoms.

Explanation:

An ionic bond is formed by the transfer of electrons from one atom to another, resulting in a bond between a positive ion (cation) and a negative ion (anion). On the other hand, a covalent bond is formed by the sharing of electrons between atoms, creating a bond in which electrons are shared rather than transferred.

In an ionic bond, one atom gains electrons and becomes negatively charged, while the other atom loses electrons and becomes positively charged. This attraction between oppositely charged ions forms the bond. In a covalent bond, atoms share one or more pairs of electrons, resulting in a stable electron arrangement for both atoms.

Examples of compounds with ionic bonds include sodium chloride (NaCl) and magnesium oxide (MgO), while examples of compounds with covalent bonds include water (H2O) and methane (CH4).

Two subway stops are separated by 1.1 km. If a subway train accelerates at +1.2 m/s2 from rest
through the first half of the distance and decelerates at −1.2 m/s2 through the second half, what are
(a) its travel time and (b) its maximum speed? (c) Graph x, v and a versus t for the trip.

Answers

Answer:

a) Travel time = 60.56 s

b) Maximum speed = 36.33 m/s

Explanation:

a)  Distance = 1.1 km = 1100 m

 A subway train accelerates at +1.2 m/s² from rest  through the first half of the distance and decelerates at −1.2 m/s² through the second half.

 So half the distance is traveled at an acceleration of +1.2 m/s².

 We have equation of motion [tex]s=ut+\frac{1}{2}at^2[/tex]

 Substituting

         [tex]\frac{1100}{2}=0\times t+\frac{1}{2}\times 1.2t^2\\\\t=30.28s[/tex]

 Travel time = 2 x 30.28 = 60.56 s

b) We have equation of motion v = u+at

   Substituting t = 30.28 s and a = 1.2 m/s²

   v = 0 + 1.2 x 30.28 = 36.33 m/s

   Maximum speed = 36.33 m/s

c) Photos of graphs are given

       

   

Answer:

a) ttotal=60.58 s

b) v^2=36.3 m/s

c) the attached photo shows the graphics.

Explanation:

a) we have the following:

S1=1.1x10^-3/2=550 m

a=1.2 m/s

we have the following formula:

v^2=u^2+2as

replacing values:

v^2=0+(2*1.2*550)=36.33 m/s

The time is equal to:

t1=v/a=36.33/1.2=30.3 s

we have the following:

u=36.33 m/s

v=0

S=550 m/s

using the following expression:

v=u+at2

clearing t2:

t2=(v-u)/a=(0-36.33)/-1.2=30.28 s

ttotal=t1+t2=30.3+30.28=60.58 s

b) the mass speed is equal to:

v^2=u^2+2as=36.3 m/s

A capacitor with plates separated by distance d is charged to a potential difference ΔVC. All wires and batteries are disconnected, then the two plates are pulled apart (with insulated handles) to a new separation of distance 2d. Part A Does the capacitor charge Q change as the separation increases? If so, by what factor? If not, why not?

Answers

Answer:

When the separation between the capacitor plates is increased to 2d, and if the battery is disconnected, the charge between the condenser plates does not change, it is the same, and this is represented by the following expression: Q'= Q

Explanation:

Final answer:

The charge (Q) on a capacitor remains constant when its plates are pulled apart after being disconnected from a circuit, even though the capacitance decreases due to the increased separation.

Explanation:

When a capacitor is charged and then disconnected from any circuit, the charge Q on the capacitor's plates remains constant despite changes in plate separation or other geometrical modifications. This is because, once disconnected, there are no paths available for the charge to leave or enter the capacitor. Therefore, when the plates of a charged capacitor are pulled apart to increase the separation from d to 2d, the charge Q does not change. The process of separating the plates requires work to be done against the attractive force between the charges on the capacitor plates. Capacitance (C), which is defined as C = Q/V, where V is the potential difference, will decrease because the capacitance of a parallel plate capacitor is inversely proportional to the separation between the plates (d), according to the formula C = ε0A/d, where ε0 is the permittivity of free space and A is the plate area.

A 0.5-kg uniform meter stick is suspended by a single string at the 30-cm mark. A 0.2-kg mass hangs at the 80 cm mark. What mass hung at the 10-cm mark will produce equilibrium?

Answers

Final answer:

To find the mass needed to produce equilibrium, we perform a torque balance about the pivot point. This involves adding up the torques on each side of the pivot point and setting them equal to each other, then solving for the unknown mass at 10 cm.

Explanation:

In this question, we are dealing with torque and equilibrium. For a system to be in equilibrium, the sum of the torques about any pivot point must be zero. Given a 0.5-kg uniform meter stick suspended by a single string at the 30-cm mark, with a 0.2-kg mass hanging at the 80-cm mark, we want to find the mass that should be hung at the 10-cm mark to ensure equilibrium.

To find this, we set up the following equation for the torques about the pivot point (the 30-cm mark):

Torque_due_to_0.5kg_rod + Torque_due_to_hanging_mass_at_10_cm = Torque_due_to_0.2kg_mass_at_80cm

Each torque can be calculated as the product of the force (which is the weight, or mass times gravity) and the distance from the pivot. After plugging in the given values and solving for the unknown mass at 10 cm, you will find the mass that should be hung at the 10 cm mark to produce equilibrium.

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Sprinting up a short flight of stairs is often used as a simple test to compare power output among different athletes. Subject 1, a football player with a body mass of 130 kg, sprints up a flight of stairs and raises his body mass 4.4 m vertically in a time of 3.4 s. Subject 2, a cyclist with a body mass of 75 kg, sprints up the same flight of stairs in 2.8 s. Determine the mechanical work and power for both of these athletes. Which athlete is more powerful in an absolute sense (W)? What if power is expressed relative to body mass (W/kg)? Assuming that the body is 25% efficient, determine the amount of metabolic energy spent during this stair sprint for both athletes (total Kcals, not Kcal/hour; Hint: 1 J = 0.000239 Kcals).

Answers

Hey there!:

subject 1

Mechanical work =mgh

= 130 *9.8 *4.4

= 5605.6 J

Mechanical power =Work/time

= 5605.6 / 3.4

= 1648.706 W

Subject 2

Mechanical work =mgh

= 75 * 9.8 * 4.4

= 3234 J

Mechanical power =Work/time

= 3234/2.8

= 1155 W

Athlete 1 is more powerful in absolute sense

relative to mass

athlete 1 power =12.68 W/kg

athlete 2 power =15.4 W/kg

thus athlete 2 is more power in relative to mass power  .

metabolic energy spent by athlete 1 :

= (5605.6 / 0.25)*0.000239 kCal

= 5.359 kCal

metabolic energy spent by athlete 2=(3234/0.25)*0.000239 kCal

= 3.092 kCal

Hope this helps!

A water pipe is inclined 40.0° below the horizontal. The radius of the pipe at the upper end is 2.00 cm. If the gauge pressure at a point at the upper end is 0.112 atm, what is the gauge pressure at a point 2.65 m downstream, where the pipe has narrowed to a 1.00 cm radius? The flow rate is 20.0? cm^3/s

Answers

Final answer:

To calculate gauge pressure in a tapered water pipe, apply Bernoulli's equation considering changes in radius, height, and flow rate.

Explanation:

The gauge pressure at a point in a tapered water pipe can be calculated using Bernoulli's equation. Given the information provided, we can determine the pressure difference between two points in the pipe based on the change in radius and height along the pipe.

In this scenario, the gauge pressure at a point 2.65 m downstream from the upper end can be calculated by considering the change in radius and height along the pipe as well as the flow rate of water through the pipe.

Understanding fluid dynamics and applying relevant equations helps determine the pressure variation in tapered pipes based on given parameters.

Some specialized cells such as neurons and red blood cells lose their ability to replicate when they mature. Which phase of the cell cycle do you think is terminal for these cells and why?

Answers

Answer:

G1

Explanation:

Some specialized cells such as red blood cells, neurons lose their ability to replicate in the G1 phase of cell cycle. G1 phase is indeed the terminal phase of the cell division as now cell do not divide any more. These cells do not have the necessary nucleus for further cellular replication in the G1 phase. They become mature need not divide anymore.

Final answer:

The terminal phase for specialized cells such as neurons and red blood cells that lose their ability to replicate upon maturity is the G0 phase. In this phase, these cells exit the cell cycle and cease active preparation for division. They continue to perform their functions despite the cessation of division.

Explanation:

The terminal phase for these specialized cells such as neurons and red blood cells, which lose their ability to replicate when they mature, is usually referred to as the G0 phase, also known as the resting or quiescent phase. This is as the cell cycle consists of the interphase (G1, S, and G2 phases) and the mitotic (M) phase. After the M phase, cells usually enter the G1 phase of interphase. However, some cells exit the cycle and enter the G0 phase, in which they are not actively preparing to divide.

For non-dividing cells like neurons and red blood cells, they stay in G0 permanently. This means they stop dividing but continue to perform their functions. Such cells have lost the capacity for cell division either temporarily or permanently. This mechanism helps prevent unregulated cell division which can lead to cancers.

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An object with mass 80 kg moved in outer space. When it was at location <7, -34, -7> its speed was 14.0 m/s. A single constant force <200, 460, -150> N acted on the object while the object moved to location <12, -42, -11> m. What is the speed of the object at this final location?

Answers

The speed of the object with mass 80 kg and moved under a constant force at the final location is 12 m/s.

What is the speed of a body?

The speed of a body is the rate at which it covers the total distance is in the time taken.

If the speed of the body has direction, then it is known as the velocity. The velocity is the vector quantity.

The constant force applied on the object is  <200, 460, -150> N. It can be written in the vector form as,

[tex]\vec F=200\hat i+460\hat j-150\hat k[/tex]

The initial position vector of the object for the location <7, -34, -7> can be given as,

[tex]d_i=7\hat i-34\hat j-7\hat k[/tex]

The final position vector of the object is for the location <12, -42, -11> m can be given as,

[tex]d_f=12\hat i-42\hat j-11\hat k[/tex]

Thus, the total distance traveled by the constant force is,

[tex]d=(12\hat i-42\hat j-11\hat k)-(7\hat i-34\hat j-7\hat k)\\d=5\hat i-8\hat j-4\hat k[/tex]

The work done by a object is the product of force and displacement. Thus, work done on the object is,

[tex]W=\vec F.\vec d \\W=(200\hat i+460\hat j-150\hat k).(5\hat i-8\hat j-4\hat k)\\W=1000-3680+600\\W=-2080 \rm J[/tex]

Now, the work done on a object is equal to the kinetic energy of the body. Thus, work done on the object is,

[tex]W=\dfrac{1}{2}m(v^2_f-v^2_i)[/tex]

Here, v(f) is the final speed and v(i) is the initial speed of the object. As the mass of the object is 80 kg and initial speed is 14 m/s. Thus, put the values as,

[tex]-2080=\dfrac{1}{2}(80)(v^2_f-14^2)\\v_f=12\rm m/s[/tex]

Thus, the speed of the object with mass 80 kg and moved under a constant force at the final location is 12 m/s.

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Final answer:

To find the speed of the object at the final location, we need to calculate the net force acting on it using the equation F = ma. By plugging in the values, we can calculate the final speed of the object.

Explanation:

To determine the speed of the object at the final location, we need to calculate the net force acting on the object using the equation F = ma, where F is the force, m is the mass, and a is the acceleration.

Given that the force acting on the object is <200, 460, -150> N, and the mass of the object is 80 kg, we can calculate the acceleration using the formula a = F/m. After calculating the acceleration, we can use it to find the final speed of the object using the equation v^2 = u^2 + 2as, where v is the final speed, u is the initial speed, a is the acceleration, and s is the displacement.

By plugging in the given values, we can calculate the final speed of the object at the location <12, -42, -11> m.

A good baseball pitcher can throw a baseball toward home plate at 87 mi/h with a spin of 1710 rev/min. How many revolutions does the baseball make on its way to home plate? For simplicity, assume that the 60 ft path is a straight line.

Answers

Explanation:

First, convert 87 mi/h to ft/min.

87 mi/h × (5280 ft/mi) × (1 h / 60 min) = 7656 ft/min

The time to reach the home plate is:

t = 60 ft / 7656 ft/min

t = 0.00784 min

The number of revolutions made in that time is:

n = 1710 rev/min × 0.00784 min

n = 13.4 rev

Rounding to 2 significant figures, the ball makes 13 revolutions.

Gina perceives the car to be far away because the sides of the road upon which it is moving seem to come together to be no wider than the car itself. This is an example of the ____ cue to depth.

Answers

Answer:

Linear Perspective.

Explanation:

Linear perspective cue to depth and it is in respect to both texture gradient or the next depth cue and relative size. In linear perspective, parallel lines that go along seems to have converged at distance very far, eventually reaching a vanishing point at the horizon. For example the railway track appears to be converging at a point very distant from the observer but it didn't.

Answer:

d. linear perspective

Explanation:

According to a different source, these are the options that come with this question:

a. visual acuity

b. texture gradient

c. retinal disparity

d. linear perspective

Linear perspective refers to a way in which depth is created. This term is particularly employed by artists who create an illusion of depth on a flat surface. This perspective leads all parallel lines to be lost in a single vanishing point on the horizon line. In this example, Gina is using the linear perspective cue to understand the depth of the scene she is observing.

An object is projected horizontally at 8.0 m/s from the top of a 122.5 m cliff. How far from the base o the cliff will the object strike the ground?

Answers

Answer:

30.63

Explanation:

v=2x/t

t=2x/v

t=2*1220.5/8.0

t=30.63//

Suppose that the average speed (vrms) of carbon dioxide molecules (molar mass 44.0 g/mol) in a flame is found to be 2.67 105 m/s. What temperature does this represent?

Answers

Answer:

[tex]T = 1.26 \times 10^8 K[/tex]

Explanation:

As we know that rms speed of ideal gas is given by the formula

[tex]v_{rms} = \sqrt{\frac{3RT}{M}}[/tex]

here we know that

[tex]v_{rms} = 2.67 \times 10^5 m/s[/tex]

molecular mass of gas is given as

[tex]M = 44 g/mol = 0.044 kg/mol[/tex]

now from above formula we have

[tex]2.67\times 10^5 = \sqrt{\frac{3(8.31)T}{0.044}}[/tex]

now we have

[tex]T = 1.26 \times 10^8 K[/tex]

Final answer:

The root mean square speed of a molecule in an ideal gas can be used to determine the temperature of the gas when vrms is known. This involves solving the formula for temperature using the given vrms and mass of a molecule. We need to ensure that the molar mass is properly converted to the mass of a molecule and that we use SI units.

Explanation:

The subject question is related to the concept of root mean square speed (vrms) in gas physics and involves calculations using the ideal gas model. According to this concept, we know that vrms is given by the equation vrms = sqrt(3kB T / m), where:

kB is the Boltzmann constant (1.38×10−23 J/K), T is the temperature in Kelvin, and m is the mass of each (identical) particle in the gas.

Given that vrms = 2.67 x 105 m/s and molar mass (M) = 44.0 g/mol (which must be converted to kg as SI units are used in this context), we can solve the formula for T to find the temperature T = vrms^2×m / (3×kB) after substituting the known values into this equation. Note that one must convert the molar mass to mass of a molecule by dividing it by Avogadro's number (6.022 x 10²³ mol⁻¹).

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