luggage handler pulls a 20.0 kg suitcase up a ramp inclined at 34.0 ∘ above the horizontal by a force F⃗ of magnitude 165 N that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is μk = 0.300. The suitcase travels 3.90 m along the ramp. Part A Calculate the work done on the suitcase by F⃗ . Express your answer with the appropriate units. WF = nothing nothing Request Answer Part B Calculate the work done on the suitcase by the gravitational force. Express your answer with the appropriate units. Ww = nothing nothing

Answer:

(a) 4.334 × 10¹¹ joules are required to propel the rocket an unlimited distance away from Earth's surface, (b) The rocket has travelled 3999.865 miles from the Earth's surface with the half of the total work.

Explanation:

The complete statement is: "Use the weight of the rocket to answer the question. (Use 4000 miles as the radius of Earth and do not consider the effect of air resistance.) 7 metric ton rocket (a) How much work is required to propel the rocket an unlimited distance away from Earth's surface, (b) How far has the rocket traveled when half the total work has occurred?"

(a) The work required to propel the rocket is given by the change in gravitational potential energy, whose expression derives is described below:

\(U_{g, f} - U_{g, o} = -G\cdot M\cdot m \cdot \left[\frac{1}{r_{f}}-\frac{1}{r_{o}} \right]\)

Where:

\(U_{g,o}\), \(U_{g,f}\) - Initial and final gravitational potential energies, measured in joules.

\(m\), \(M\) - Masses of the rocket and planet Earth, measured in kilograms.

\(G\) - Universal gravitation constant, measured in newton-square meters per square kilogram.

\(r_{o}\), \(r_{f}\) - Initial and final distances of the rocket with respect to the center of the Earth, measured in meters.

The initial distance and rocket mass are converted to meters and kilograms, respectively:

\(r_{o} = (4000\,mi)\cdot \left(1609.34\,\frac{m}{mi} \right)\)

\(r_{o} = 6,437,360\,m\)

\(m = (7\,ton)\cdot \left(1000\,\frac{kg}{ton} \right)\)

\(m = 7000\,kg\)

Given that \(m = 7000\,kg\), \(M = 5.972\times 10^{24}\,kg\), \(G = 6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}\), \(r_{o} = 6,437,360\,m\) and \(r_{f} \rightarrow +\infty\), the work equation is reduced to this form:

\(U_{g,f} - U_{g,o} = \frac{G\cdot m \cdot M}{r_{o}}\)

\(U_{g,f} - U_{g,o} = \frac{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (7000\,kg)\cdot (5.972\times 10^{24}\,kg)}{6,437,360\,m}\)

\(U_{g,f} - U_{g,o} = 4.334\times 10^{11}\,J\)

4.334 × 10¹¹ joules are required to propel the rocket an unlimited distance away from Earth's surface.

(b) The needed change in gravitational potential energy is:

\(U_{g,f} - U_{g,o} = 2.167\times 10^{11}\,J\)

The expression for the change in gravitational potential energy is now modified by clearing the final distance with respect to the center of Earth:

\(U_{g, f} - U_{g, o} = -G\cdot M\cdot m \cdot \left[\frac{1}{r_{f}}-\frac{1}{r_{o}} \right]\)

\(\frac{U_{g,o}-U_{g,f}}{G\cdot M \cdot m} = \frac{1}{r_{f}} - \frac{1}{r_{o}}\)

\(\frac{1}{r_{f}} = \frac{1}{r_{o}} + \frac{U_{g,o}-U_{g,f}}{G\cdot M\cdot m}\)

\(r_{f} = \left(\frac{1}{r_{o}} + \frac{U_{g,o}-U_{g,f}}{G\cdot M\cdot m} \right)^{-1}\)

If \(m = 7000\,kg\), \(M = 5.972\times 10^{24}\,kg\), \(G = 6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}\), \(r_{o} = 6,437,360\,m\)  and \(U_{g,f} - U_{g,o} = 2.167\times 10^{11}\,J\), then:

\(r_{f} = \left[\frac{1}{6,437,360\,m}-\frac{2.167\times 10^{11}\,J}{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (7000\,kg)\cdot (5.972\times 10^{24}\,kg)} \right]^{-1}\)

\(r_{f} \approx 12,874,502.49\,m\)

The final distance with respect to the center of the Earth in miles is:

\(r_{f} = (12,874,502.49\,m)\cdot \left(\frac{1}{1609.34}\,\frac{mi}{m} \right)\)

\(r_{f} = 7999.865\,mi\)

The distance travelled by the rocket is: (\(r_{f} = 7999.865\,mi\), \(r_{o} = 4000\,mi\))

\(\Delta r = r_{f}-r_{o}\)

\(\Delta r = 7999.865\,mi - 4000\,mi\)

\(\Delta r = 3999.865\,mi\)

The rocket has travelled 3999.865 miles from the Earth's surface with the half of the total work.

Part A.

For trial 1 and trial 2 the velocity of the car is constant. Since the velocity is constant there is no acceleration. So, we can conclude the acceleration is equal to zero.

Part B.

Depending on the material of the floor, it affects the movement of the car. This is associated with the coefficient of friction. The higher the coefficient of friction, the greater the opposition to movement. According to this the coefficient of friction for the carpet is higher than the coefficient of friction for tile and wood.

Part C.

For the interval [8,10] in the trial 3, the velocity becomes zero, since the distance isn't changing at all. This happened because of the high coefficient of friction of the carpet, which caused the car to stop.

Part D

In order to the students have more confidence in their result, they can change the following:

1. Repeat the experiment several times on the same surface

2. Have a longer length on each surface

The magnetic field of the wire will be directed towards west. Using right thumb rule one can get the direction of field lines.

- The resistance of the first resistor (R₁) is 12 Ω.

- The electromotive force (EMF) of the battery is 18 V.

- The internal resistance of the battery is 12 Ω.

To solve the given problem, we can apply Kirchhoff's laws and Ohm's law to determine the resistance of the first resistor (R₁) and the electromotive force (EMF) and internal resistance of the battery.

Let's start by calculating the resistance of the first resistor (R₁):

1. Apply Ohm's law to find the voltage drop across the external circuit:

  V = I * R

  V = 1 A * 6 Ω

  V = 6 V

2. The voltage drop across the external circuit is equal to the EMF minus the voltage drop across the internal resistance of the battery:

  V = E - Ir

  6 V = E - (1 A * r)   (where r is the internal resistance of the battery)

3. We also know that the EMF of the battery is the sum of the voltage drops across each cell in the battery:

  E = 6 cells * 3 V/cell

  E = 18 V

4. Substitute the value of E in the equation from step 2:

  6 V = 18 V - r

  r = 12 Ω

Therefore, the resistance of the first resistor (R₁) is 12 Ω.

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Watching Gladiators was a a favorite pastime of Roman civilization. It was a game based on the Roman martial ethics. It lasted for nearly thousands of years. Hence option C is correct.

An armed warrior known as a gladiator amused spectators in the Roman Republic and Roman Empire by engaging in bloody fights with other gladiators, wild animals, and condemned prisoners.

Some gladiators were unpaid volunteers who jeopardized their lives as well as their position in society and the law by participating in the fight. The majority endured severe education, were treated like inferiors in society, and were separated even in death.

Regardless of where they came from, gladiators served as a living example of Rome's martial virtues to onlookers, and by fighting or dying honorably, they may win adulation and widespread renown.

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Answer:

The dog is moving at a constant speed

Explanation:

Given that,

Position : 5, 10, 15, 20, 25

Time = 5. 10, 15, 20, 25

We need to draw a position time graph

Using given data

A graph of position and time shows the speed.

According to graph,

The graph indicates that the dog is moving at a constant speed because the graph is straight line.

Hence, The dog is moving at a constant speed

It is not recommended to fire a gun straight up into the air.

When a bullet is fired into the air, it will eventually come down and can pose a danger to people and property below. The bullet can still be lethal when it reaches the ground, especially if it lands on a hard surface or hits someone directly.

Additionally, firing a gun in a residential area can be illegal and can result in legal consequences. In general, guns should only be fired in designated shooting ranges or in self-defense situations where there is an immediate threat to life. It is important to handle firearms responsibly and follow all safety guidelines to prevent accidents and injuries.

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The meaning of the word, " analgesic " is B. a medicine to remove pain.

Any substance used to relieve pain is referred to as an analgesic drug, also known as an analgesic, analgaesic, pain reliever, or painkiller.

Analgesics, commonly known as painkillers, are drugs that treat a variety of pains, such as headaches, injuries, and arthritis. Both opioid analgesics and anti-inflammatory analgesics alter how the brain interprets pain. Some analgesics, such as stronger OTC medicines, combination analgesics, and all opioids, must be purchased with a prescription.

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Answer:

0.004394Joules

Explanation:

The question is incomplete. Find the complete question in the attachment below;

Energy stored in a capacitor = 1/2CV² where;

C is the capacitance of the capacitor

V is the potential difference across the plates

Given parameters

C = 13.0μF = 13 * 10⁻⁶F

V = 26.0V

Required

Energy stored in the capacitor

Substituting the values into the equation given;

E = 1/2 * 13 * 10⁻⁶ * 26²

E = 1/2 * 13 * 10⁻⁶ * 676

E = 1/2* 8788 * 10⁻⁶

E = 4394  * 10⁻⁶

Hence the Energy stored in the capacitor in Joules is 0.004394Joules

Answer:

yea yes they can they lose their retina but they can dream

Answer:

May be egg shaped

Has no new stars being formed.

Has almost no gas or dust between stars.

Explanation:

Elliptical galaxy is the collection of many stars which are bounded together gravitationally, which is smooth and ellipsoidal and shape and the appearance is featureless.

Elliptical galaxy is ovoid or spherical masses of stars.

It is found in galaxy clusters and compact galaxies.

It has no gas or dust between stars which result in low rates of star formation.

It is formed When two spirals collide, they lose their familiar shape, morphing into the less-structured elliptical galaxies.

Elliptical galaxy is made of old stars and have no gas and dust.

An example is elliptical galaxy m60 which shines brightly and is egg shaped.

Answer:

The net force on the hovercraft is 11200 N.

Explanation:

Given;

force exerted on the hovercraft by wind, F₁ = 5000 N north

force exerted on the hovercraft by the propeller, F₂ = 10,000 N west

The net force on the hovercraft is calculated as;

\(F_{net} = \sqrt{F_1^2 + F_2^2} \\\\F_{net} = \sqrt{5000^2 + 10,000^2} \\\\F_{net} = 11180.34 \ N\\\\F_{net} = 11200 \ N \ (3.s.f)\)

Therefore, the net force on the hovercraft is 11200 N.

Answer:

\(\Huge \boxed{\mathrm{2298.57 \ seconds}}\)

\(\rule[225]{225}{2}\)

Explanation:

\(\displaystyle \sf Speed = \frac{Distance \ covered }{Time \ taken}\)

\(\displaystyle \sf s = \frac{d }{t}\)

The speed is 0.7 m/s.

The distance covered is 1609 m.

\(\displaystyle \sf 0.7 = \frac{1609 }{t}\)

Multiplying both sides by t.

Then dividing both sides by 0.7.

\(\displaystyle \sf t = \frac{1609 }{0.7}\)

\(\sf t= 2298.57\)

It would take 2298.57 seconds.

\(\rule[225]{225}{2}\)

Answer:

\(\huge\boxed{\sf t = 2298.57\ secs}\)

Explanation:

Given:

Speed = v = 0.7 m/s

Distance = S = 1 mile = 1609 metre

Required:

Time = t = ?

Formula :

v = S/t

Solution:

v = S/t

t = S/v

t = 1609 / 0.7

t = 2298.57 secs

Answer:

Explanation: because Keller concluded that the circular obits of the Copernican model weren't correct paths of planets were elleptical

Answer:

Explanation:

Frequency is oscillations per second.

So we have to find the Number of seconds she paced.

2 minutes = 2 X 60

               = 120 seconds

Therefore,

Her frequency = 10 / 120

                       = 1/12 Hertz

The total distance is 105 kilometers and the total displacement is 15 kilometers east. Option C

To calculate the total distance, we add the distances traveled in each leg of the journey: 60 kilometers (from A to B) + 45 kilometers (from B back to A) = 105 kilometers.

However, displacement refers to the change in position of an object in a straight line from its starting point to its ending point. In this case, since the boat starts and ends at the same point (A), the total displacement is zero.

Hence The total distance is 105 kilometers and the total displacement is 15 kilometers east.

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Answer:

F=ma

=12×10

120 N.........

Answer:

F = 120 N [Newtons]

Explanation:

F = 120N

If the speed is 10m/s² [an acceleration], and the mass is 12 kg [kilograms].

F [Force] = ma [mass × acceleration].

F = 10 × 12 = 120.

It cannot be a velocity because time is not given.

Momentum would be the velocity times mass. p = mv.

The fourth ellipse, with the length-to-width ratio of approximately 3 to 1, has the greatest eccentricity.

Eccentricity of the ellipse: The eccentricity of an ellipse is a measure of its "oblateness" or how much it deviates from being a circle. An ellipse has an eccentricity between 0 and 1, with 0 representing a circle and values closer to 1 representing more elongated ellipses. So, in this case, the fourth ellipse with the highest length-to-width ratio would have the highest eccentricity value.

The first ellipse is a circle, it means its length-to-width ratio is 1. Second one is more closer to a circle than the third one. Similarly third ellipse is more closer to the circle than the fourth one.

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Explanation:

Vf = vo + at      vo = 0

so this becomes

Vf = at  =  9.81 m/s^2 *  5 = ~ 49 m/s

(a) The minimum time in which a car can cover to a stop on such a road is 2.83 s.

(b) The distance travelled by the car is 19.7 m.

The acceleration of the car is calculated by applying Newton's second law of motion.

F = ma = -μmg

ma = -μmg

a = -μg

where;

a = -0.5 x 9.8 m/s²

a = -4.9 m/s²

The minimum time in which a car whose initial velocity is 50 km/h can cover to a stop on such a road is calculated as;

v = u + at

where;

0 = u - at

at = u

t = u / a

t = ( 13.89 ) / ( 4.9)

t = 2.83 seconds

The distance travelled by the car is calculated as;

v² = u² + 2as

0 = 13.89²  - 2 (4.9)(s)

9.8s = 192.9

s = 192.9 / 9.8

s = 19.7 m

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Answer:

v₀ₓ = 63.5 m/s

v₀y = 54.2 m/s

Explanation:

First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:

K.E = (0.5)(mv₀²)

where,

K.E = initial kinetic energy of projectile = 1430 J

m = mass of projectile = 0.41 kg

v₀ = launch velocity of projectile = ?

Therefore,

1430 J = (0.5)(0.41)v₀²

v₀ = √(6975.6 m²/s²)

v₀ = 83.5 m/s

Now, we find the launching angle, by using formula for maximum height of projectile:

h = v₀² Sin²θ/2g

where,

h = height of projectile = 150 m

g = 9.8 m/s²

θ = launch angle

Therefore,

150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)

Sin θ = √(0.4216)

θ = Sin⁻¹ (0.6493)

θ = 40.5°

Now, we find the components of launch velocity:

x- component = v₀ₓ = v₀Cosθ  = (83.5 m/s) Cos(40.5°)

v₀ₓ = 63.5 m/s

y- component = v₀y = v₀Sinθ  = (83.5 m/s) Sin(40.5°)

v₀y = 54.2 m/s

Answer:

See below

Explanation:

Gallileo is CORRECT in a vacuum where there is no air friction

  air friction affects same or different weight objects differently  and will cause same-weight   or different weight  objects to fall at different speeds

Answer:

Explanation:

Loss of lives and property: Immediate impacts of flooding include loss of human life, damage to property, destruction of crops, loss of livestock, non-functioning of infrastructure facilities and deterioration of health condition owing to waterborne diseases. Flash floods, with little or no warning time, cause more deaths than slow-rising riverine floods.

Loss of livelihoods: As communication links and infrastructure such as power plants, roads and bridges are damaged and disrupted, economic activities come to a standstill, resulting in dislocation and the dysfunction of normal life for a period much beyond the duration of the flooding. Similarly, the direct effect on production assets, be it in agriculture or industry, can inhibit regularly activity and lead to loss of livelihoods. The spill over effects of the loss of livelihoods can be felt in business and commercial activities even in adjacent non-flooded areas.

Decreased purchasing and production power: Damage to infrastructure also causes long-term impacts, such as disruptions to clean water and electricity, transport, communication, education and health care. Loss of livelihoods, reduction in purchasing power and loss of land value in the flood plains lead to increased vulnerabilities of ties living in the area. The additional cost of rehabilitation, relocation of people and removal of property from flood-affected areas can divert the capital required for maintaining production.

Mass migration: Frequent flooding, resulting in loss of livelihoods, production and other prolonged economic impacts and types of suffering can trigger mass migration or population displacement. Migration to developed urban areas contributes to the overcrowding in the cities. These migrants swell the ranks of the urban poor and end up living in marginal lands in cities that are prone to floods or other risks. Selective out-migration of the workforce sometimes creates complex social problems.

Psychosocial effects: The huge psycho-social effects on flood victims and their families can traumatize them for long periods of time. The loss of loved ones can generate deep impacts, especially on children. Displacement from one’s home, loss of property and livelihoods and disruption to business and social affairs can cause continuing stress. The stress of overcoming these losses can be overwhelming and produce lasting psychological impacts.

Hindering economic growth and development: The high cost of relief and recovery may adversely impact investment in infrastructure and other development activities in the area and in certain cases may cripple the frail economy of the region. Recurrent flooding in a region may discourage long-term investments by the government and private sector alike. Lack of livelihoods, combined with migration of skilled labour and inflation may have a negative impact on a region’s economic growth. Loss of resources can lead to high costs of goods and services, delaying its development programmes.

Political implications: Ineffective response to relief operations during major flood events may lead to public discontent or loss of trust in the authorities or the state and national governments. Lack of development in flood-prone areas may cause social inequity and even social unrest posing threat to peace and stability in the region.

The molecules of O2 that are  present in 3.90 L flask at  a temperature of 273 K and a pressure of 1.00 atm is 1.047 x 10^23 molecules  of O2

Step  1:  used the ideal gas equation to calculate the moles of O2

that is Pv=n RT  where;

P(pressure)= 1.00 atm

V(volume) =3.90 L

n(number of moles)=?

R(gas constant) = 0.0821 L.atm/mol.K

T(temperature) = 273 k

by  making n the subject of the formula by  dividing  both side by RT

n= Pv/RT

n=[( 1.00 atm x 3.90 L)  /(0.0821 L.atm/mol.k  x273)]=0.174  moles

Step 2: use the Avogadro's  law constant  to calculate  the number of molecules

that  is  according to Avogadro's law

                          1  mole =  6.02 x10^23  molecules

                            0.174 moles=? molecules

by  cross  multiplication

the number of  molecules

= (0.174  moles x  6.02 x10^23  molecules)/ 1 mole  =1.047 x 10^23 molecules of O2

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Answer:

It could be a stringed instrument or an open pipe.

Explanation:

Let v be the speed of sound, y be wavelength and f be frequency.

v = yf

f= v/y; v is constant.

In a stringed instrument, the fundamental frequency note is heard when the length of the string, l = y/2; y= 2l

f′= v/2l

The second harmonic is heard when l= y

f"= v/y

...

f'''= 3v/2l

We can infer that f"= 2f'

f'''= 3f'

This is similar to the values in the question as;

523.26 =2(261.63) and so on.

Same thing happens with open pipes.

Answer:

a) 3.36 sec

b) 23.9 m/s

c) 1.52 sec

Explanation:

maximum height

H = 25 +v_y^2/2g

H=25+9^2/2*9.81 =29.13 m

time taken to reach maximum height = t_1 =v_y/g = 9/9.81 =0.92 seconds

time taken to fall back to ground from maximum height = √(2gH)/g

= √(2*9.81*29.13)/9.8 = 2.439 seconds

a) Total time taken to reach ground T = t_1+t_2 =0.92+2.439=3.36 sec

b) objects final speed = √2gH = 23.9 m/s

c) let total time be t then

25 = 9t +0.5gt^2

Solving we get t= 1.52 seconds

a) The object will take a time of 3.355 seconds to reach the ground.

b) The final velocity of the object as it impacts the ground is -23.902 meters per second.

c) The object will take a time of 1.520 seconds to reach the ground.

a) The height of the object (\(h\)), in meters, as a function of time (\(t\)), in seconds, is described below:

\(y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}\) (1)

Where:

Now we proceed to determine the time needed for the object to reach the ground: (\(y_{o} = 25\,m\), \(y = 0\,m\), \(v_{o} = 9\,\frac{m}{s}\), \(g = -9.807\,\frac{m}{s^{2}}\))

\(-4.904\cdot t^{2}+9\cdot t +25 = 0\) (2)

Then, we solve this second order polynomial by quadratic formula:

\(t_{1} \approx 3.355\,s\), \(t_{2} \approx -1.520\,s\)

Only the former solution offers a realistic indicator. Hence, we conclude that the object will take a time of 3.355 seconds to reach the ground.

b) The final velocity of the object (\(v\)), in meters per second, is determined by this expression:

\(v = v_{o}+g\cdot t\) (3)

Now we proceed to find the final velocity of the object as it impact the ground: (\(v_{o} = 9\,\frac{m}{s}\), \(g = -9.807\,\frac{m}{s^{2}}\), \(t \approx 3.355\,s\))

\(v = 9+(-9.807)\cdot (3.355)\)

\(v = -23.902\,\frac{m}{s}\)

The final velocity of the object as it impacts the ground is -23.902 meters per second. \(\blacksquare\)

c) In this part we shall apply the same approach in a), that is: (\(y_{o} = 25\,m\), \(y = 0\,m\), \(v_{o} = -9\,\frac{m}{s}\), \(g = -9.807\,\frac{m}{s^{2}}\))

\(-4.904\cdot t^{2}-9\cdot t +25 = 0\)

Then, we solve this second order polynomial by quadratic formula:

\(t_{1} \approx 1.520\,s\), \(t_{2} \approx -3.355\,s\)

Only the former solution offers a realistic indicator. Hence, we conclude that the object will take a time of 1.520 seconds to reach the ground.

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Time period of a wave is the inverse of its frequency. The period of the wave with a frequency of 2.09 Hz is 0.47 seconds.

Frequency of a wave is the number of wave cycles per unit time. Frequency is the inverse of the time period of the wave. Hence, it has the unit of s⁻¹ which is equivalent to Hz.

The higher frequency of a wave indicates more number of wave cycles in a short time. Frequency is directly proportional to the energy and inversely proportional to the  wavelength.

Given the time period of the wave = 2.09 Hz

then frequency = 1/2.09 Hz = 0.47 s.

Therefore, the time period  of the wave is 0.47 seconds.

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