The Bernoulli effect can have important consequences for the design of buildings. For example, wind can blow around a skyscraper at remarkably high speed, creating low pres- sure. The higher atmospheric pressure in the still air inside the buildings can cause windows to pop out. As originally constructed, the John Hancock Building in Boston popped windowpanes that fell many stories to the sidewalk below. (a) Suppose a horizontal wind blows with a speed of 11.2 m/s outside a large pane of plate glass with dimensions 4.00 m 3 1.50 m. Assume the density of the air to be constant at 1.20 kg/m3. The air inside the building is at atmospheric pressure. What is the total force exerted by air on the win- dowpane?

Answer:

Gravitational Force

Explanation:

Gravitational force also called gravity or gravitation is an attractive force that keeps two objects in space. Gravitational force is an attractive force that tends to pull matters together. Every objects in the universe experience gravitational pull.  Planets, stars, galaxies, are held together by gravity. It is a weak force. The weight of an object is the product of gravitational force acting on its mass.

Newton's Law of Universal Gravitation states the force of attraction between two masses m₁ and m₂ in the universe is directly proportional to the product of their masses and inversely proportional to the square of their distance apart.

                                 [tex]F = G\frac{m_{1}m_{2}}{r^2}[/tex]

Where;

F is the gravitational force,

G is the gravitational constant = 6.67 × 10¹¹ m³/kg/s,

m1 and m2 are the masses of the objects,

r is the distance between the centers of the masses

The tension in the left rope is expressed by [tex]T_{L} = W\cdot \left(1-\frac{x}{L} \right)[/tex]. For [tex]x = 0[/tex], [tex]T_{L} = W[/tex].

A representation of the system is presented in the image attached below. Let suppose that the entire system represents a rigid body, that is, a body whose geometry cannot be neglected and does not experiment deformation due to the application of external loads.

The tension in the left rope ([tex]T_{L}[/tex]) can be found by applying Newton's laws and D'Alembert's Principle applied on the rightmost point of the bar:

[tex]\Sigma F = W\cdot (L-x)-T_{L}\cdot L = 0[/tex]

[tex]T_{L} = \frac{W\cdot (L-x)}{L}[/tex]

[tex]T_{L} = W\cdot \left(1-\frac{x}{L} \right)[/tex]   (1)

The tension in the left rope is expressed by [tex]T_{L} = W\cdot \left(1-\frac{x}{L} \right)[/tex]. For [tex]x = 0[/tex], [tex]T_{L} = W[/tex]. [tex]\blacksquare[/tex]

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Considering a situation where there is no horizontal acceleration, the tension in the left rope (TL) equals the weight of the person when x equals zero (x=0). Since the only forces on the static system are weight (W) and the tensions in the two ropes, and they are in equilibrium, the tension in either rope will be the same as the weight.

The question requires us to find the tension in the left rope (TL), given that x equals zero. Using the equilibrium equation under the condition of no horizontal acceleration and no external forces except the weight (W) and the two tensions, TL and TR, we can conclude the magnitudes of the tensions TL and TR must be equal.

Given the scenario, the only forces acting on the system are the weight of the walker (W) and the tensions in the ropes. If the system is static, the net external force is zero, which means the sum of the forces should also equal zero according to Newton's second law. Hence, expressing the tension TL in terms of W becomes simple given that the system is in equilibrium. Under these conditions, the tension in the rope should equal the weight of the supported mass, making TL equal to W.

Therefore, since x is zero and we are neglecting other variables like rope weight and the angle of the ropes, the tension TL in the left rope becomes equivalent to the weight of the person. Hence the tension (TL) equals W.

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Answer:

The kinetic energy is the same as in space, which in general is very large

Explanation:

If the Earth had no atmosphere we can use the conservation of kinetic energy in two points

Initial. In space far from the planet

      Em₀ = k = ½ m v₀²

Final. Just before touching the surface of the Earth

      [tex]Em_{f}[/tex] = K = ½ m v²

As there is no rubbing

      Em₀ = [tex]Em_{f}[/tex]

      [tex]Em_{f}[/tex]= ½ m v₀²

The kinetic energy is the same as in space, which in general is very large

When the Earth has an atmosphere we must use the energy work theorem

      W = ΔK

The work is done by the friction forces when the meteor enters the atmosphere, increases in density as it approaches the surface, so the work also increases.

       W =[tex]K_{f}[/tex] - K₀

       [tex]K_{f}[/tex] = K₀ - W

       [tex]K_{f}[/tex] = ½ m v₀² - W

We see that the kinetic energy decreases as the work increases, this makes the impact is higher and part of the meteor also evaporates by friction at the entrance

Answer:

The magnitude on the resultant force on the object = 18.03 N

Explanation:

Resultant Force: The resultant force of two or more forces acting on a body in a given direction is the single force obtained when the two forces are combined, which produces the same effect as the two forces acting together.

Since both forces makes an angle of 90° with each other,

Fr = √(F₁² + F₂²)......................... Equation 1

Where Fr = resultant force on the object F₁ = The force pulling the body northward, F₂ = The force pulling the body southward.

Given: F₁ = 10 N,  F₂ = 15 N

Substituting these values into equation 1

Fr = √(10² + 15²)

Fr = √(100 + 225)

Fr = √325

Fr = 18.03 N

Therefore the magnitude on the resultant force on the object = 18.03 N

The resultant force when an object is pulled by 10 N northward and 15 N southward is 5 N towards the south, found by subtracting the smaller force from the larger one.

In physics, when multiple forces act on an object, they combine to create a resultant force. If two forces of equal magnitude are pulling in opposite directions, they cancel out and the resultant force is zero. However, in this case, the forces acting on the object are different: 10 N northward and 15 N southward.

To find the resultant force, we simply subtract the smaller force from the larger one, considering the opposing directions. So, here 15 N (south) - 10 N (north) gives us 5 N towards the south.

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The minimum current rating of the motor disconnecting means for a 40-horsepower, 208-volt, 3-phase squirrel-cage motor is 121 amps.

In order to determine the minimum current rating of the motor disconnecting means, we need to calculate the current drawn by the motor.

The current rating of the motor disconnecting means must be equal to or higher than the calculated current.

First, we need to calculate the current drawn by the motor using the formula:

Current (I) = Power (P) / (Voltage (V) × Power Factor (PF) × √3)

Given that the motor has a power of 40 horsepower and operates at 208 volts with a power factor of 0.85, we can substitute these values into the formula:

I = 40 hp × (746 W/hp) / (208 V × 0.85 × √3)

Solving for I:

I ≈ 121 amps

Therefore, the minimum current rating of the motor disconnecting means for a 40-horsepower, 208-volt, 3-phase squirrel-cage motor is 121 amps.

Final answer:

In this case, the minimum current rating for the motor disconnecting means is approximately 88.5 Amperes, calculated based on the motor's power and voltage specifications.

Explanation:

The minimum current rating of the motor disconnecting means for a 40-horsepower, 208-volt, 3-phase squirrel-cage motor can be calculated using the formula:

Current (A) = Power (W) / (Voltage (V) * sqrt(3))

Plugging in the values:

Current = (40 hp * 746 W/hp) / (208 V * sqrt(3)) = ~88.5 A

n this case, the minimum current rating for the motor disconnecting means is approximately 88.5 Amperes, calculated based on the motor's power and voltage specifications.

Answer:

Explanation:

Heat flow in a circular rod is given by

[tex]Q=\frac{kAdT}{dx}[/tex]

where Q= heat flow

k=thermal conductivity

A=area of cross-section

dT=Change in temperature

dx=change in length

Also A can be written as

[tex]A=\pi r^2[/tex]

thus Q is Proportional to

[tex]Q\propto \frac{r^2}{l}[/tex]

For option (a)

[tex]Q\propto \frac{(2r_0)^2}{2l_0}[/tex]

[tex]Q\propto \frac{2r_0^2}{l_0^2}[/tex]

(b)[tex]Q\propto \frac{(2r_0)^2}{l_0}[/tex]

[tex]Q\propto \frac{4r_0^2}{l_0}[/tex]

(c)[tex]Q\propto \frac{r_0^2}{2l_0}[/tex]

(d)[tex]Q\propto \frac{r_0^2}{l_0}[/tex]

So Rod b will conduct the most Heat

Final answer:

When driving in heavy rain, the recommended following distance should be increased to at least 6 seconds due to longer stopping distances on wet pavement. Stopping distances depend on road conditions and driver reaction time, highlighting the need for increased safety margins in poor weather.

Explanation:

The amount of following distance you should allow between you and the vehicle in front of you during heavy rain should be considerably more than what you would maintain in dry conditions. The typical recommendation is to maintain at least a 3-second distance in good weather, but this should be increased to at least 6 seconds in heavy rain to accommodate for the increased stopping distances on wet pavement and reduced visibility.

Stopping distances can vary greatly depending on road conditions and driver reaction time, and heavy rain can significantly increase these distances. As braking distance increases with speed and poor weather conditions, it is important to adjust your following distance accordingly to ensure safety.

Referring to the given figures, we can deduce that for a car traveling at 30.0 m/s, the stopping distance will be much longer on wet pavement than on dry. If the driver's reaction time is assumed to be 0.500 s, the total distance traveled before the car comes to a stop will include both the reaction distance and the braking distance. When considering crossing a street, you must take into account that a safe distance is one where you are completely sure that the car can come to a full stop without reaching your crossing point, which can be roughly compared to the stopping distances shown in various figures.

Answer:Resultant force

Explanation:

The net given force is known as the resultant force. The resultant force is the single force that acts in place of other forces combined together.

Answer:

Dipole-dipole interaction force

Explanation:

When one of the constituent atom of the covalent bonding is at least 1.5 times more electronegative than the other atom sharing the electron in the covalent bond then the shared pair of electrons are shifted towards the more electronegative atom developing a partial negative charge on it and similarly develops an equal partial positive charge on the other atom involved in the covalent bond.

The term saturated solution is used in chemistry to define a solution in which no more solute can be dissolved in the solvent. It is understood that saturation of the solution has been achieved when any additional substance that is added results in a solid precipitate or is let off as a gas.

Answer:

y=39.057 m

Explanation:

Using Kinematic relation

[tex]s=ut+ \frac{1}{2}at^2[/tex]

given u= 5m/s

a=g= -9.81 [tex]m/s^2[/tex]( directed downward)

[tex]s=5t- \frac{1}{2}(9.80)t^2[/tex]

Also, we know that

v=u+at

v=5-9.80t

at time t= 0.250 sec

[tex]s=5\times0.25- \frac{1}{2}(9.80)0.25^2[/tex]

s=0.94375 m

now position of sandbag

y= 40-0.94375

y=39.057 m

Final answer:

Tech A is correct in stating that the drop center is used for tire installation, while Tech B is incorrect; the drop center's primary purpose isn't to prevent the tire from detaching at low pressure.

Explanation:

Tech A says that most wheels have a drop center or deep well that is used in installing a tire on the wheel. Tech B says that the drop center or deep well is used to prevent the tire from coming off the wheel in the case of low tire pressure. The answer here is that Tech A is correct. The drop center or deep well feature on a wheel is indeed primarily designed to help in the installation and removal of the tire. It allows the bead of the tire to drop into a recessed area, giving enough slack to work the tire over the rim. On the other hand, Tech B’s statement is not accurate. While the drop center might incidentally help keep the tire on the rim in the case of low pressure, its primary purpose is not to prevent the tire from coming off the wheel.

Answer:100 K

Explanation:

We know that Freezing Point of water is [tex]0^{\circ}C[/tex] at 1 atm

Converting it to Kelvin we get 273.15 K

Boiling Point of water is [tex]100^{\circ}C[/tex]

converting it to Kelvin we get 373.15 K

Difference in the number we get =373.15-273.15=100 K

Temperature difference is independent of degree and will remain same for Celsius and Kelvin

Final answer:

The crate has an acceleration of 0.43 m/s^2. It travels 21.5 m in 10.0 s and has a speed of 4.3 m/s at the end of 10.0 s.

Explanation:

To solve this problem, we can use Newton's second law, which states that force is equal to mass times acceleration (F = ma).

(a) We are given that the mass of the crate is 32.5 kg and the net horizontal force acting on it is 14.0 N. Plugging these values into the equation, we get:

F = ma

14.0 N = 32.5 kg * a

a = 14.0 N / 32.5 kg

a = 0.43 m/s^2

So, the acceleration produced is 0.43 m/s^2.

(b) To find the distance traveled by the crate in 10.0 s, we can use the equation of motion: distance = initial velocity * time + (1/2) * acceleration * time^2.

Since the crate starts at rest, the initial velocity is 0 m/s:

distance = 0 * 10.0 s + (1/2) * 0.43 m/s^2 * (10.0 s)^2

distance = 0 + (1/2) * 0.43 m/s^2 * 100.0 s^2

distance = 21.5 m

So, the crate travels 21.5 m in 10.0 s.

(c) To find the speed of the crate at the end of 10.0 s, we can use the equation of motion: final velocity = initial velocity + acceleration * time.

Since the crate starts at rest, the initial velocity is 0 m/s:

final velocity = 0 + 0.43 m/s^2 * 10.0 s

final velocity = 4.3 m/s

So, the speed of the crate at the end of 10.0 s is 4.3 m/s.

Answer:

2000

Explanation:

Exposure indicator can be obtained using Carestream's formula for imagine pate.

Exposure indicator number = 2000 + [1000 x (log of exposure in mR)].

Given; log of exposure in mR = 1 mR

Exposure indicator number = 2000 + [1000 x (log 1)]

Exposure indicator number = 2000 + [1000 x (0)]

Exposure indicator number = 2000 + [0]

Exposure indicator number = 2000

Answer:

Technicians A and B are both correct

Explanation:

A quick take-up valve in a quick take-up master cylinder assembly which reduces fluid flow noise and restriction caused by fluid flowing in one mode of operation through an incipiently or substantially closed valve spring coil stack. The check valve arrangement has a coil compression limiting device and also has the outlet orifice so positioned as not to require fluid flow through the check valve spring coils while the check valve is open.

If the brake warning light is NOT on and there are no visible brake fluid leaks, your master cylinder may be worn or leaking internally allowing the brake pedal to slowly sink when pressure is applied to it. This type of condition will be most noticeable when you are holding constant pressure against the brake pedal at a stop light. If the pedal sinks or requires pumping to keep your car from creeping ahead, the master cylinder needs to be replaced.

Answer:

When an electrical current passes through a wire, a magnetic field is generated around it. Likewise, if the magnetic field around a wire is changed ( for example by rotating a coil inside a stationary manger), electricity will move through the wire.

Answer:

mass=2326kg

Explanation:

Archimedes' principle states that when an object is submerged into a liquid, it appears lighter in weight due to the buoyant force applied by the liquid in upward direction. Buoyant force is equal to the weight of the liquid displaced by the object. An object floats on the surface of liquid or floats partially submerged when weight of the displaced liquid is greater than weight of the object.

the mass f he box=74kg

densty=mass /vlme

1000=mass/2.4

mass =2400kg

apparent mass wen completely submerged will be

2400-74kg

mass=2326kg

Answer:c

Explanation:

Given

object is falling Freely with an odometer

Suppose it falls with zero initial velocity

so distance fallen in time t is given by

[tex]h=ut+\frac{1}{2}gt^2[/tex]

here u=0 and t=time taken

[tex]h=\frac{1}{2}gt^2[/tex]

for [tex]t=1 s[/tex]

[tex]h_1=\frac{1}{2}g[/tex]

for [tex]t=2 s[/tex]

[tex]h_2=\frac{1}{2}g(2)^2=\frac{4}{2}g=2g[/tex]

distance traveled in 2 nd sec[tex]=2g-\frac{1}{2}g=\frac{3}{2}g[/tex]

for [tex]t=3 s[/tex]

[tex]h_3=\frac{1}{2}g(3)^2=\frac{9}{2}g[/tex]

distance traveled in 3 rd sec[tex]=\frac{9}{2}g-2g=\frac{5}{2}g[/tex]

so we can see that distance traveled in each successive second is increasing

The amount of distance travels by free falling object in each succeeding second is greater than the second before.

Speed of the free falling object is the speed, by which the body is falling downward towards the ground level.

At the Earth's surface, the speed of the free falling object will accelerate 9.841 meters per squared second.The odometer is the device, which is used the speed of the body in m/s or km/hrs.

Now we have to find out, whether the distance traveled by the object each succeeding second is constant, deceasing or increasing.

The second equation of the motion for distance can be given as,

[tex]s=ut+\dfrac{1}{2}gt^2[/tex]

Here,  [tex]u[/tex] is the initial velocity of the body, [tex]g[/tex] is the acceleration due to gravity and [tex]t[/tex] is the time taken by it.

As for the free falling body the initial velocity is zero. Thus the distance traveled by it for the first second is,

[tex]s=0+\dfrac{1}{2}9.81\times1\\s=4.905\rm m[/tex]

The distance traveled by it for the 2nd second is,

[tex]s=\dfrac{1}{2}9.81\times2^2\\s=19.61\rm m[/tex]

The distance traveled by it for the 3rd second is,

[tex]s=\dfrac{1}{2}9.81\times3^2\\s=44.145\rm m[/tex]

Now the difference between the third and 2nd second of distance traveled by object is greater then the difference between the second and first second.

Thus, the amount of distance travels by free falling object in each succeeding second is greater than the second before.

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Answer:

A. DVD-RW

B. BD-R

Explanation:

The RW stands for rewritable.

BD-R uses Blu-ray technology allowing capacities of up to 100GB

The DVD-RW is capable of multiple rewrites, while the Blu-Ray disc holds the largest capacity amongst the options provided, being able to store 25-50 GB of data.

From the options provided (BD-R, Blu-Ray, DVD-RW, and DVD DS [Double-Sided]), the medium capable of multiple rewrites is DVD-RW. 'RW' in DVD-RW stands for 'ReWritable', meaning the media can be written, erased, and rewritten multiple times.

The optical media with the largest capacity is Blu-Ray. A single-layer Blu-Ray disc has a capacity of 25 GB, and a dual-layer Blu-Ray disc can hold 50 GB, more than five times the capacity of a DVD DS (Double Sided) which typically holds about 9 GB.

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Answer:

Lemon Juice

Explanation:

Acidic levels rise as pH levels decrease. The solution with the lowest pH level will always be more acidic. In this case, lemon juice is the solution with a lower pH level, therefore lemon juice is more acidic than coffee.

Hope this helps!

Final answer:

Lemon juice is the stronger acid compared to coffee, with a pH of 2 versus a pH of 5, meaning it is 1000 times more acidic.

Explanation:

The acidity of a solution is determined by its pH level, which measures the concentration of hydrogen ions (H+) in the solution. The pH scale ranges from 0 to 14, with lower numbers indicating more acidic solutions. In comparing lemon juice with a pH of 2 and coffee with a pH of 5, lemon juice is the stronger acid. Since pH is measured on a logarithmic scale, each pH unit represents a tenfold difference in hydrogen ion concentration. Thus, the difference between pH 2 and pH 5 is 103 or 1000 times. Therefore, lemon juice is 1000 times more acidic than coffee.

Answer:

nothing, mass is not affected by gravitational force

Explanation:

Weight is the gravitational force a planet exerts on a mass on the surface.

It is the product of the mass of an object with the gravitational acceleration that the planet produces.

The weight is the gravitational force

[tex]W=mg[/tex]

where,

m = Mass of the object

g = Acceleration due to gravity = 9.81 m/s²

Mass is the property that matter has which opposes the force being applied to it. It is intrinsic to the object itself and does not change according to the gravitational force. But, the weight changes.

The correct statement is nothing, mass is not affected by gravitational force.

The gravitational force of attraction of every object in the universe is given by Newton's gravitational law;

[tex]F_1= \frac{GmM_e}{R^2}[/tex]

where;

m is your mass

[tex]M_e[/tex] is mass Earth

R is the radius of the Earth

G is gravitational constant

If the mass of the Earth is cut into half, the gravitational force will be affected as follows;

[tex]F_2= \frac{Gm}{R^2}\times \frac{M_e}{2} =\frac{1}{2} (\frac{GmM_e}{R^2}) = \frac{1}{2} (F_1)[/tex]

The gravitational force will be reduced by 2

Now, let's check how your mass will be affected;

[tex]F_2= \frac{GmM_e}{R^2}\\\\GmM_e = F_2R^2\\\\m = \frac{F_2R^2}{G M_e} \\\\When, M_e \ is \ halved \ (0.5M_e) , \ F_2 = \frac{1}{2} F_1 = 0.5F_1\\\\m = \frac{0.5F_1R^2}{G \times 0.5M_e}\\\\m = \frac{F_1R^2}{G M_e}[/tex]

Your mass is not affected.

Thus, gravitational force is affected by mass but mass is not affected by gravitational force.

The correct statement is nothing, mass is not affected by gravitational force.

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Over-driving your headlights implies a situation where you are driving so fast that you won't be able to stop within the area illuminated by your headlights. Therefore, if you see an object in your path under such circumstances, you are likely not going to be able to stop in time to avoid hitting it. Headlights' range is usually 350 feet, and driving at a speed that requires a stopping distance greater than 350 feet is considered over-driving your headlights.

The phrase 'over-driving your headlights' refers to a situation where a driver is traveling at such a speed that their stopping distance is further than the distance illuminated by their headlights. Thus, if an object is within your path, you won't have enough time to stop your vehicle before hitting it, especially if you're speeding.

On most roads, the farthest your headlights can help you see ahead is around 350 feet. If you're driving faster than a speed that permits you to stop within these 350 feet, you're said to be 'over-driving' your headlights. If you're over-driving your headlights, and you see an object ahead then you will most likely not be able to stop in time to miss the object. This is because your vehicle's stopping distance will be greater than your visual distance, which is dependent on the capabilities of your headlights.

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Answer:

b. sequencing and allotting time to all project activities

Explanation:

' Project Controls are data collection, data management and predictive methods used to forecast, interpret and proactively control the time and cost results of a project or program; by communicating information in ways that enable effective management and decision-making.

So, a, c and  d are statements are a part of project controlling but b that is

sequencing and allotting time to all project activities  is not a part of project controlling.

Answer:

Explanation:

According to question

tan θ = y / x

Differentiate with respect to t on both the sides

[tex]Sec^{2}\theta \times \frac{d\theta }{dt}=\frac{x\times dy/dt-y\times dx/dt}{x^{2}}[/tex]

[tex]\frac{d\theta }{dt}=\frac{x\times dy/dt-y\times dx/dt}{x^{2}\times Sec^{2}\theta}[/tex]   .... (1)

According to question,

tan θ = 5 / 9

So, Sec θ = 10.3 / 9 = 1.14

dx/dt = 9 units/s

dy/dt = 19 units/s

Substitute the values in equation (1), we get

[tex]\frac{d\theta }{dt}=\frac{9\times 19-5\times 9}{81\times 1.14^{2}}[/tex]

dθ/dt = 1.2 units/s

Answer:

Answer:

a.  1.594 m/s = v

b. 1.274 m/s = v

Explanation:

A) First calculate the potential energy stored in the spring when it is compressed by 0.180 m...

U = 1/2 kx²

Where U is potential energy (in joules), k is the spring constant (in newtons per meter) and x is the compression (in meters)

U = 1/2(13.0 N/m)(0.180 m)² = 0.2106 J

So when the spring passes through the rest position, all of its potential energy will have been converted into kinetic energy.  K = 1/2 mv².

 0.2106 J  = 1/2(0.170 kg kg)v²

0.2106 J  = (0.0850 kg)v²

2.808m²/s² = v²

1.594 m/s = v

(B)  When the spring is 0.250 m from its starting point, it is 0.250 m - 0.180 m = 0.070 m past the equilibrium point.  The spring has begun to remove kinetic energy from the glider and convert it back into potential.  The potential energy stored in the spring is:

U = 1/2 kx² = 1/2(13.0 N/m)(0.070 m)² = 0.031J

Which means the glider now has only 0.2106 J  - 0.031J = 0.1796 J of kinetic energy remaining.

K = 1/2 mv²

0.1796 J = 1/2(0.170 kg)v²

0.138 J = (0.0850 kg)v²

1.623 m²/s² = v²

1.274 m/s = v

To calculate the speed of the glider at different points, we can use the principle of conservation of energy. At 0.180 m from the starting point, the speed is 2.65 m/s. At 0.250 m from the starting point, the speed is 3.89 m/s.

To solve this problem, we can use the principle of conservation of energy. When the glider is released, all of the potential energy stored in the compressed spring is converted into kinetic energy. At the point where the glider has moved 0.180 m from its starting point, the potential energy of the spring is zero, so the total mechanical energy is equal to the kinetic energy. Using the formula for kinetic energy, we can calculate the speed of the glider:

KE = 1/2 mv^2

m = 0.170 kg (mass of the glider)

v = ? (speed of the glider)

At the point where the glider has moved 0.180 m, the potential energy of the spring is zero, so the total mechanical energy is equal to the kinetic energy:

0.5 kx^2 = 0.5 mv^2

k = 13.0 N/m (force constant of the spring)

x = 0.180 m (distance moved by the glider)

By substitute the given values into the equation, we can solve for v:

0.5 * 13.0 N/m * (0.180 m)^2 = 0.5 * 0.170 kg * v^2

Solving for v, we find that the speed of the glider at the point where it has moved 0.180 m from its starting point is 2.65 m/s.

To calculate the speed of the glider at the point where it has moved 0.250 m from its starting point, we can use the same principle of conservation of energy. The initial potential energy of the spring is converted into kinetic energy:

0.5 * 13.0 N/m * (0.250 m)^2 = 0.5 * 0.170 kg * v^2

Solving for v, we find that the speed of the glider at the point where it has moved 0.250 m from its starting point is 3.89 m/s.

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Answer:

g = 8.61 m/s²

Explanation:

distance of the International Space Station form earth is 200 Km

mass of the object = 1 Kg

acceleration due to gravity on earth = 9.8 m/s²

mass of earth = 5.972 x 10²⁴ Kg

acceleration due to gravity = ?

r = 6400 + 200 = 6800 Km = 6.8 x 10⁶ n

using formula

 [tex]g = \dfrac{GM}{r^2}[/tex]

 [tex]g = \dfrac{6.67\times 10^{-11}\times 5.972\times 10^24}{(6.8\times 10^6)^2}[/tex]

        g = 8.61 m/s²

Final answer:

The force of gravity on a 1 kg object in low-earth orbit around the Earth depends on the height of the orbit and is smaller than on the Earth's surface.

Explanation:

The force of gravity on a 1 kg object on the Earth's surface is approximately 9.8 N. In low-earth orbit around the Earth, the force of gravity on the object is much smaller. This is because the force of gravity decreases as you move further away from the Earth's surface. However, the exact value of the force of gravity in low-earth orbit depends on the height of the orbit. For example, at the height of the International Space Station, the force of gravity on the object would be about 88% of the force on the Earth's surface, which is approximately 8.6 N.

Answer:

B. Strength

Explanation:

The OPT Model, or Optimum Performance Training Model, is a "fitness training system created by the NASM. The OPT Model is contructed with scientific evidence and principles that progresses an individual through five training phases: stabilization endurance, strength endurance, hypertrophy, maximal strength and power".

On the stabilization level we have the phase 1 called stabilization endurance.

For the level strength part we have 3 phases . Phase 2: Strength endurance , Phase 3: Hypertrophy, Phase 4: Maximal strength. And we can consider the case "A seated cable row" on this the level strength since we need to have some abilities to do this but not enough to stay on the power level since this one is the advancd level.

For the power level we have the last phase called power in order to mantain and conduct high training level programs.

Answer:

True

Explanation:

Entropy is a thermodynamic variable that always increases or remains constant, but does not decrease.

This indicates that energy is always transferred from a higher temperature system to a lower temperature system and not the other way around.

Entropy can also be thought of as what drives a system into equilibrium, and systems reach equilibrium through heat transmission.

The fact that the entropy of a closed system never decreases is a statement of the second law of thermodynamics.

Answer:

0 N

Explanation:

According to Newton's first law of motion, an object in motion stays in motion until acted upon by an unbalanced force.  With no friction in space to unbalance the cannonball, it will continue to keep going.

The  amount of force needed to keep it going equals​ to 0 N

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