Voltage is shared by components that are __________. A. Together B. In parallel C. In series D. In perpendicular

5m

Explanation:

Given parameters:

Weight of object = 50N

Work done in lifting object = 250J

Unknown:

Vertical height = ?

Solution:

The work done on an object is the force applied to lift a body in a specific direction.

   Work done = force x distance

  Weight is a force in the presence of gravity;

  Work done = weight x height of lifting

Height of lifting = [tex]\frac{work done }{weight}[/tex]

 Height of lifting = [tex]\frac{250}{50}[/tex] = 5m

The vertical height through which the object was lifted is 5m

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Final answer:

To calculate the vertical height that a 50 N object is moved with 250 J of work done against Earth's gravity, divide the work by the force. This gives a height of 5 meters.

Explanation:

To find the vertical height an object is moved, given the work done against the Earth's gravitational field, we use the formula for work: Work = Force × Distance, where the force in this context is the weight of the object due to gravity. Given the work done (250 J) and the weight of the object (50 N), we can rearrange the formula to solve for distance, which is the vertical height moved: Distance = Work / Force.

Distance = 250 J / 50 N = 5 meters.

Therefore, the object was moved through a vertical height of 5 meters. This calculation assumes that the work done is solely against the gravitational pull of the Earth, and it highlights the direct relationship between work, force, and distance in physical interactions.

This Question is incomplete.The complete question is

Combustion gases enter a gas turbine at 627∘C and 1.2 MPa at a rate of 2.5 kg/s and leave at 527∘C and 500 kPa. It is estimated that heat is lost from the turbine at a rate of 20 kW. Using air properties for the combustion gases and assuming the surroundings to be at 25∘C and 100 kPa, determine (a) the actual and reversible power outputs of the turbine. (b) the exergy destroyed within the turbine, and (c) the second-law efficiency of the turbine.

Answer:

(a) W=257.5kW      Wrev=367.3kW

(b) X=109.8kW

(c) n=0.7

Explanation:

For Part(a)

Actual power output is determined from the energy balance. The final and initial enthalpy are taken from A-17 for the given temperature

mh₁=W+Q+mh₂

W=m(h₁-h₂)-Q

W=2.5(939.93-821.95)-20

W=257.5kW

The reversible power output is determined from the rate of energy destruction

Wrev=W+X

Wrev=W+T₀(m(s₂-s₁-Rln(p₂/p₁)+Q/T₀)

Wrev=257.5+298( 2.5 (2.71787-2.84856-0.287*ln(500/1200)+20/298)

Wrev=367.3kW

For part(b)

X=Wrev-W

X=367.3-257.5

X=109.8kW

For part (c)

Second-law efficiency is determined from the ratio of the actual and reversible power output:

n=W/Wrev

n=257.5/367.3

n=0.7

Answer:

Explanation:

According to ohm's law current flowing in a conductor is directly proportional to the voltage applied across two end of conductor.

i.e. [tex]V\propto R[/tex]

[tex]V=R I[/tex]

where R=resistance

[tex]R\propto L[/tex]

[tex]R\propto \frac{1}{d^2}[/tex]

whee L and d are length and Diameter

thus [tex]R=k \frac{L}{d^2}[/tex]

where k=constant of Variation

Answer:

0.72

Explanation:

[tex]T[/tex] = Time period of oscillation = 1.5 s

Angular frequency is given as

[tex]w = \frac{2\pi }{T}\\w = \frac{2(3.14) }{1.5}\\w = 4.2 rad/s[/tex]

[tex]A[/tex] = Amplitude of oscillation = 40 cm = 0.40 m

[tex]\mu[/tex] = Coefficient of static friction = ?

[tex]a[/tex] = acceleration of the block

[tex]m[/tex] = mass of the block

Maximum acceleration of the block is given as

[tex]a = Aw^{2}[/tex]

frictional force is given as

[tex]f = \mu mg[/tex]

As per newton's second law

[tex]f = ma \\\\\mu mg = ma \\\mu g = a\\\mu g = Aw^{2}\\\mu (9.8) = (0.40)(4.2)^{2}\\\mu = 0.72[/tex]

The coefficient of static friction between the two blocks is 0.72.

From the information given, the angular frequency will be:

= 2π/T

= 2(3.14)/1.5

= 4.2rad/s

The maximum acceleration of the block is Aw² and the frictional force is umg. In this case,

umg = ma

u(9.8) = 0.40 × 4.2².

u = (0.40 × 4.2²) / 9.8

u = 0.72

Therefore, the coefficient of static friction between the two blocks is 0.72.

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Answer:c-The gravitational effect when spacecraft flies close to the asteriod

Explanation:

Gravitational effect on the spacecraft gives an estimate that how big is the asteroid by experiencing its gravitational pull.

The amount of extra thrust required to maintain the trajectory of the spacecraft during its motion hints at the scientist about the size of the asteroid.

Gravitational pull is directly proportional to the mass of object so greater the mass, greater will be the pull.

Answer:

C

Explanation:

The gravity of a planet, or in this case an asteroid, is determined by two factors: it's mass and its size. The bigger the mass, the stronger the gravity is. But the bigger the size of the mass is, the smaller the gravity is, because of the distance from the center of the mass. As you can see by measuring the gravitational effect when a spacecraft is close to an asteroid can help determine its size.

Answer:

the third stage was 480 km long

Explanation:

Stage 1:

Time = 1 hours

Speed = 80km

Stage 2:

Time =  2 hours

Speed = 200km

Stage 3:

Time =  4 hours

Let the Distance at the stage 3 be x

Average speed of the train route = 100 km/h

So

[tex]\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 0[/tex]

[tex]\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 100[/tex]

Lets find the speed at stage 1

Speed =  [tex]\frac{Distance }{Time}[/tex]

Speed =  [tex]\frac{80}{1}[/tex]

Speed 1= 80 km/hr

The speed at stage 2

Speed =  [tex]\frac{Distance }{Time}[/tex]

Speed =  [tex]\frac{200}{2}[/tex]

Speed 2  = 100 km/hr

The speed at stage 3

Speed =  [tex]\frac{Distance }{Time}[/tex]

Speed =  [tex]\frac{x}{4}[/tex]

Speed 3  = [tex]\frac{x}{4}[/tex]

we kow that average is ,

[tex]\frac{ \text{speed 1} + \text{speed 2} + \text{speed 3}}{3} = 100[/tex]

[tex]\frac{ 80 + 100+ \frac{x}{4} }{3} = 100[/tex]

[tex]\frac{ 180 + \frac{x}{4} }{3} = 100[/tex]

[tex]\frac{ \frac{720 +x}{4} }{3} = 100[/tex]

[tex]\frac{720 +x}{4} \times \frac{1}{3} = 100[/tex]

[tex]\frac{720 +x}{12} = 100[/tex]

[tex]720 +x = 100 \times 12[/tex]

[tex]720 +x = 1200[/tex]

[tex]x = 1200- 720[/tex]

x = 480

To calculate the length of the third stage of the train's journey, we use the average speed over the total trip. Summing the known distances of the first two stages and subtracting this from the total distance traveled, which is 700 km, gives us the length of the third stage as 420 km.

To find the length of the third stage in a train's journey when the average speed over the entire trip is known, we can use the formula for average speed, which is the total distance traveled divided by the total time taken. According to the given problem, the first two stages took 1 and 2 hours and were 80 km and 200 km long, respectively. Adding the times gives us 3 hours for the first two stages.

The third stage took 4 hours, making the total time for the full journey 7 hours (3 hours + 4 hours). Since the average speed of the train over the course of the trip was 100 km/h, we multiply this by the total time to find the total distance. Thus, the total distance is 700 km (100 km/h * 7 h).

To find the distance covered in the third stage, we subtract the sum of the first two stages (80 km + 200 km) from the total distance. The length of the third stage is 700 km - 280 km = 420 km.

Answer:

Fish tailing

Explanation:

Fish tailing is the vehicle handling problem which occurs due to the decrease in friction between the the rear wheels and the surface. So we mostly observe that rear wheels slide out of control when there is rain or ice covered roads, sand etc. They reduce the friction of the surface resulting in over-steering. This problem is named as fish tailing.

Fishtailing

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Answer: 20.9 kJ

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

[tex]Q=m\times c\times \Delta T[/tex]

Q = Heat absorbed by water= ?

m= mass of water = 215.5 g

c = specific heat capacity = [tex]4.184J/g^0C[/tex]

Initial temperature of the water = [tex]T_i[/tex] = 34.4°C

Final temperature of the water = [tex]T_f[/tex]  = 57.6°C

Change in temperature ,[tex]\Delta T=T_f-T_i=(57.6-34.4)^0C=23.2^0C[/tex]

Putting in the values, we get:

[tex]Q=215.5\times 4.184\times 23.2^0C[/tex]

[tex]Q=20918J=20.9kJ[/tex]    (1kJ=1000J)

The heat absorbed by water is equal to the heat released during the reaction. Thus heat produced by chemical reaction is 20.9 kJ

The chemical reaction produced approximately 21.67 kJ of heat. This is calculated using the formula ΔQ = MCΔT, utilizing the given values for the mass of the solution, the specific heat capacity of water, and the change in temperature.

To calculate the amount of heat produced in the chemical reaction, we use the formula ΔQ = MCΔT, where M is the mass of the solution, C is the specific heat capacity of water, and ΔT is the change in temperature. Given that M = 215.5 g, C = 4.184 J/(g·°C), and ΔT is the increase from 34.4°C to 57.6°C (which is 57.6°C - 34.4°C = 23.2°C), the formula becomes:

ΔQ = (215.5 g) x (4.184 J/g°C) x (23.2°C)

On calculating the above, the result comes out to be approximately 21669.8 Joules. Since 1 kJ = 1000 J, we need to convert 21669.8 Joules to kilojoules by dividing by 1000. Thus, the chemical reaction produces approximately 21.67 kJ of heat.

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Answer: the element is Oxygen.

Explanation: Earth's troposphere hydrosphere and lithosphere contain large amounts of Oxygen. it is the most abundant element in the Earth's crust. Oxygen makes up 467,100 ppm (parts per million) of the Earth's crust, or 46.6%.

Oxygen is the primary element found in Earth's troposphere, hydrosphere, and lithosphere, as part of the air, water (H2O), and silicate materials in many minerals and rocks.

The Earth's troposphere, hydrosphere, and lithosphere all contain large amounts of the element oxygen. The troposphere, part of the atmosphere, contains oxygen in the air we breathe, while the hydrosphere, which encompasses all water on Earth, would have oxygen as a component of H2O. In the lithosphere, which covers the crust and the upper part of the mantle, oxygen is found in many minerals and rocks as part of silicate materials.

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Answer: 0.107

Explanation:

We can solve this problem with Kepler's Third Law of Planetary motion:

[tex]T^{2}=4 \pi^{2} \frac{r^{3}}{G M_{MARS}}[/tex] (1)

Where:

[tex]T=7 h 39 min[/tex] is the orbital period of Phobos around Mars

[tex]G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex] is the Gravitational Constant  

[tex]M_{MARS}[/tex] is the mass of Mars

[tex]r=9380 km \frac{1000 m}{1 km}=9,380,000 m[/tex] is the semimajor axis of the orbit Phobos describes around Mars (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

Well, firstly we have to convert the orbital period to seconds:

[tex]T=7 h 39 min=(7 h \frac{3600 s}{1 h}) + (39 min \frac{60 s}{1 min})=25200 s + 2340 s=27540 s[/tex]

Now, we have to find [tex]M_{MARS}[/tex] from (1):

[tex]M_{MARS}=\frac{4 \pi^{2} r^{3}}{G T^{2}}[/tex] (2)

[tex]M_{MARS}=\frac{4 \pi^{2} (9,380,000 m)^{3}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}) (27540 s)^{2}}[/tex] (3)

[tex]M_{MARS}=6.436(10)^{23} kg[/tex] (4) This is the mass of Mars

On the other hand, it is known the mass of the Earth is:

[tex]M_{EARTH}=5.972(10)^{24} kg[/tex] (5)

Then, if we want to know the ratio of Mars’s mass to the mass of the earth, we have to divide [tex]M_{MARS}[/tex] by [tex]M_{EARTH}[/tex]:

[tex]\frac{M_{MARS}}{M_{EARTH}}=\frac{6.436(10)^{23} kg}{5.972(10)^{24} kg}[/tex]

Finally:

[tex]\frac{M_{MARS}}{M_{EARTH}}=0.107[/tex]

Answer:

0.107

Explanation:

The expert on top solved it correctly.

Answer:

Against the concentration gradient.    

Explanation:

Active transport implies the movement of molecules across a membrane from a lower concentration to a higher concentration, that is to say, against the concentration gradient. The molecules requires energy to go against the concentration gradient, which is usually coming from ATP or an electrochemical gradient.    

In passive transport, the molecules move across a membrane down a concentration gradient without the need for energy.    

Have a nice day!                  

Answer:

against

Explanation:

because it cannot go with the membrane. hope this helps you.

Answer:

The force of universal gravitation between earth and all objects in it will be quadrupled

Explanation:

Newtons law of universal gravitation tell us the force of attraction between two bodies on the earth surface . This force is directly proportional to the product of the masses of the bodies and inversely proportional to their distance apart.

f is directly proportional to m1*m2

doubling these masses will be 2m1*2m2 = 4m1m2

if mas of earth is Me and mass of all objects is Mo

the f is proportional to Me*Mo

doubling the masses of the earth and all objects we have, 2Me*2Mo= 4MeMo

This means that doubling the masses of the earth and all objects on it will cause the force of gravitational attraction to be quadrupled.

Answer:

Label 1a

Explanation:

The craters are usually referred to as those large depressions that are created due to the impact by meteorites, asteroids or other astronomical bodies. These are of variable size and shape, which are controlled by factors such as the mass of the impact body, velocity at which the impact body strikes.

According to the rule of thumb, the geologically older surfaces are exposed to various types of impacting bodies such the meteoroids, asteroids, and comets, for a longer duration of time in comparison to the surfaces that are younger in age.

The bigger craters are formed first, and the smaller craters are formed later. Thus, the age of the craters is directly proportional to the size of the craters.

Hence, the larger crater (label 1a) was formed first.

The law of superposition implies that the larger crater, labeled as 1a, must've formed first, as new craters form on top of existing surfaces. Therefore, the rim of the larger crater existed prior to the formation of the smaller crater, 1b.

In the study of craters, the principle that is usually applied is the Law of Superposition. This law states that in an undisturbed sequence, the oldest layers lie at the bottom and the youngest layers at the top. Applying this to your question: if a crater (labelled 1a) locates on the rim of a large crater, and another crater (labelled 1b) lies on the rim of a smaller one, the larger crater would have formed first. This is because any new crater that forms would be on top of, or superimposed on, the existing surface. Therefore, the rim of the larger crater must have been in existence for the smaller crater (1b) to have formed on it.

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Question:

The theory of force generation proposed is best supported by which of the following observations about Amoeba locomotion?

a. amoeboid movement stops upon expose to cytochalasins

b. amoeboid movement cannot occur if mitosis is blocked

c. moving amoeba cells produce more troponin than do stationary ones

d. the rate of movement is inversely proportional to the viscosity of the medium in the Amoeba moves

Answer:

The amoeba location can be supported by :The amoeboid movement stops upon expose to cytochalasins

Explanation:

The amoeboid movement common in eukaryotic cells.  cytochalasins are the type of drug that inhibits the growth of the micro filaments,  micro filaments capability to elongate are both implemented as important to make the force for movement in amoeba. further Cytochalasins are  known as the fungal metabolites which can bind the actin filaments and also restrict polymerization and elongation of actin. It can also permeate cell membranes, preventing the cellular translocation and cause cells to enucleate.

Question:

Listed following are several objects in the solar system. Rank these objects from left to right based on their orbital period around the Sun from shortest to longest.

A typical asteroid in the asteroid belt, a trojan asteroid, a typical kuiper belt object, a typical oort cloud object

Answer:

Ranking of the objects from left to right based on their orbital period around the sun from shortest to longest

Explanation:

Half the mass of the belt is present in the largest four asteroids, they are ceres, vesta, pallas and hygiea. The asteroid belt total mass is 4% than that of moon, 22% Pluto. The three dwarf planets  are present in the Kuiper belt Pluto, Haumea and Makemake. The spherical layer of icy objects around the sun are Oort cloud. Objects in the Oort cloud  are made up of water ice, ammonia, and methane.  It is spherical from outer and and torous shaped from inside.

The bathysphere cannot descend at a constant speed of [tex]\(1.40 \, \text{m/s}\)[/tex] with the given resistive force. If the resistive force were smaller, the bathysphere would need to take on approximately [tex]\(11741.33 \, \text{kg}\)[/tex] of seawater to descend at the desired speed

To find the mass of seawater that the submarine must take on in order to descend at a constant speed, we need to consider the forces acting on it.

1. The weight of the bathysphere, acting downward, is given by [tex]\( W = mg \)[/tex], where m  is the mass of the bathysphere and  g  is the acceleration due to gravity.

2. The buoyant force acting on the bathysphere, which is equal to the weight of the water displaced by the bathysphere, is given by [tex]\( F_b = \rho V g \),[/tex] where [tex]\( \rho \)[/tex] is the density of seawater, [tex]\( V \)[/tex] is the volume of the bathysphere, and [tex]\( g \)[/tex] is the acceleration due to gravity.

3. The resistive force opposing the motion of the bathysphere, which is given as [tex]\( F_{\text{resistive}} = 1105 \, \text{N} \),[/tex] acts upward.

At constant speed, the net force on the bathysphere is zero. Therefore, the weight of the bathysphere must be balanced by the buoyant force and the resistive force. Mathematically, we can express this as:

[tex]\[ W = F_b + F_{\text{resistive}} \]\[ mg = \rho V g + F_{\text{resistive}} \][/tex]

We can rearrange this equation to solve for the mass of seawater [tex](\( m_{\text{seawater}} \))[/tex] that the bathysphere must take on:

[tex]\[ m_{\text{seawater}} = \frac{F_{\text{resistive}} - mg}{g} \][/tex]

Given:

- [tex]\( r = 1.50 \, \text{m} \)[/tex] (radius of the bathysphere),

- [tex]\( m = 1.19 \times 10^4 \, \text{kg} \)[/tex] (mass of the bathysphere),

- [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex] (acceleration due to gravity),

- [tex]\( F_{\text{resistive}} = 1105 \, \text{N} \)[/tex] (resistive force),

- [tex]\( \rho = 1.03 \times 10^3 \, \text{kg/m}^3 \)[/tex] (density of seawater).

First, we need to calculate the volume of the bathysphere [tex](\( V \))[/tex]:

[tex]\[ V = \frac{4}{3} \pi r^3 \]\[ V = \frac{4}{3} \pi (1.50 \, \text{m})^3 \]\[ V = \frac{4}{3} \pi (3.375 \, \text{m}^3) \]\[ V \approx 14.137 \, \text{m}^3 \][/tex]

Now, let's calculate the mass of seawater [tex](\( m_{\text{seawater}} \))[/tex]:

[tex]\[ m_{\text{seawater}} = \frac{1105 \, \text{N} - (1.19 \times 10^4 \, \text{kg} \times 9.8 \, \text{m/s}^2)}{9.8 \, \text{m/s}^2} \]\[ m_{\text{seawater}} = \frac{1105 \, \text{N} - 116220 \, \text{N}}{9.8 \, \text{m/s}^2} \]\[ m_{\text{seawater}} = \frac{-115115 \, \text{N}}{9.8 \, \text{m/s}^2} \]\[ m_{\text{seawater}} \approx -11741.33 \, \text{kg} \][/tex]

Since mass cannot be negative, this result indicates that the resistive force is greater than the weight of the bathysphere and the buoyant force. Therefore, the bathysphere cannot descend at a constant speed of [tex]\(1.40 \, \text{m/s}\)[/tex] with the given resistive force. If the resistive force were smaller, the bathysphere would need to take on approximately[tex]\(11741.33 \, \text{kg}\)[/tex] of seawater to descend at the desired speed.

Answer:

Displacement = 72m    

Acceleration = 4 [tex]m/s^{2}[/tex]

Explanation:

The jet ski increases its speed from zero to 24 m/s in 6 seconds.

We are supposed to find the displacement and its acceleration rate.

Displacement is the distance it covers in this time interval.

Acceleration is the rate of increase of speed.

Acceleration = [tex]\frac{change in velocity}{time}[/tex]

Acceleration = [tex]\frac{24 - 0}{6}[/tex]

Acceleration = 4 [tex]m/s^{2}[/tex]

According to equation of motion ,

[tex]v^{2}-u^{2} = 2\times a\times s[/tex]

[tex]24^{2} - 0 = 2\times 4\times s[/tex]

s = [tex]\frac{576}{8}[/tex]

Displacement , s = 72 m

Answer:

Explanation:

mass of string = .0125 / 9.8

= 1.275 x 10⁻³ kg

Length of string l = 1.5 m .

m = mass per unit length

= ( .1.275 / 1.5) x 10⁻³ kg/m

m = .85 x 10⁻³ kg/m

wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

compare with equation of wave

y(x,t) = Acos(K x − ω t)

ω ( angular velocity ) = 4830 rad/s

k = 172 rad/m

Velocity = ω / k

= 4830/172 m /s

= 28.08 m /s

velocity of wave = [tex]\sqrt{\frac{W}{m } }[/tex]

28.08 = [tex]\sqrt{\frac{W}{.85\times10^{-3} } }[/tex]

788.48 =  W / .85 X 10⁻³

W = 670 x  10⁻³ N .

c ) wave length

wave length =2π  / k

= 2 x 3.14 / 172

= .0365 m

no of wave lengths over whole length of string

= 1.5 / .0365

= 41

d )

equation for waves traveling down the string

= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

For the string, which tied to the ceiling at its upper end support a weight has,

           [tex]\rm y(x,t) = (8.50 mm)cos(172x rad/m - 4830t rad/s \;)[/tex]

The equation of the motion of standing wave can be given as,

[tex]\rm y(x,t) = Acos(kx-\omega t)[/tex]

Here, (k) is the wave number and ([tex]\omega[/tex]) is the angular velocity.

A 1.50-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W.

When you pluck the string slightly, the waves traveling up the string obey the equation:

[tex]y(x,t) = (8.50 )\cos(172x -4830)[/tex]

Compare this equation with the general equation of the motion of standing wave, we get,

Angular velocity ([tex]\omega[/tex]) is 4830 rad/s and the wave number (k) is 172 rad/m.

Now the velocity is the ratio of angular velocity to the wave number. Thus, velocity is,

[tex]v=\dfrac{\omega}{k}\\v=\dfrac{4830}{172}\\v=28.08\rm m/s[/tex]

The length of the string is 1.50 m. Thus, time does it take a pulse to travel the full length of the string is,

[tex]t=\dfrac{1.05}{28.08}\\t=0.0534\rm s[/tex]

For the string, force can be given as,

[tex]F=\dfrac{v^2}{\mu}\\F=\dfrac{28.08^2}{0.85\times10^{-3}}\\F=0.67\rm N[/tex]

The wavelength of the wave is twice the ratio of pi to the wave number.Thus, the wavelength is,

[tex]\lambda=\dfrac{2\times \pi}{k}\\\lambda=\dfrac{2\times \pi}{172}\\\lambda=0.0365\rm m[/tex]

As the length of the string is 1.50 m. Thus, number of wavelengths are on the string at any instant of time is,

[tex]\lambda=\dfrac{1.5}{0.0365}\\\lambda=41[/tex]

The equation can be given as,

[tex]\rm y(x,t) = (8.50 mm)cos(172x rad/m - 4830t rad/s \;)[/tex]

Thus, for the string, which tied to the ceiling at its upper end support a weight has,

           [tex]\rm y(x,t) = (8.50 mm)cos(172x rad/m - 4830t rad/s \;)[/tex]

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Answer:

C.

Explanation:

Both technicians A & B are right since the connecting rods must be marked to indicate which surface faces front. And a cam-ground piston machine must also be configured oval shape so that it will be able to fit the cylinder better throughout its operating temperature range.

Technician A is correct that the connecting rods should be marked before disassembly. Technician B is incorrect about the cam ground pistons. The correct answer is A) Technician A only.

Technician A is correct that the connecting rods should be marked before disassembly. This is important in order to ensure that the rods are reinstalled in the same location and orientation as before, which helps maintain balance and proper engine function.

Technician B is incorrect. Pistons are not cam ground to become oval shaped when operating temperature is reached. Instead, pistons are typically designed with expansion clearances to account for thermal expansion, ensuring that they maintain proper shape and function at operating temperatures.

Therefore, the correct answer is A) Technician A only.

Answer:

Technician A only

Explanation:

Technician A says that knowing if there are any stored diagnostic trouble codes (DTCs) may be helpful when checking for related technical service bulletins (TSBs). Technician B says that only a factory scan tool should be useWhich technician is correct?

DTC's, or Diagnostic Trouble Codes, are used by automobile manufacturers to detect problems related to the vehicle. ... The first number in the DTC indicates whether the code is an SAE generic code

tsb are recommended techniques for repairing a vehicle. the manufactures of vehicles make this available so the technicians can work and make reference to it

The time it takes for the machine part to move from x = 0 to x = -1.80 cm is approximately 0.156 seconds.

To find out how long it takes for a machine part undergoing simple harmonic motion (SHM) to go from position x = 0 to x = -1.80 cm, we need to follow these steps:

Calculate the Period (T):

The period of oscillation (T) is the time taken to complete one full cycle. The period is calculated using the formula:

[tex]T = \frac{1}{f}[/tex]

Substituting the frequency:

[tex]T = \frac{1}{3.20} \approx 0.3125 \text{ seconds}[/tex]

Find the Phase for SHM:

For simple harmonic motion, the position as a function of time (x(t)) can be expressed as:

[tex]x(t) = A \cos(\omega t)[/tex]

where [tex]\omega[/tex] (angular frequency) is given by:

[tex]\omega = 2\pi f[/tex]

Therefore:

[tex]\omega = 2\pi \times 3.20 \approx 20.1 \text{ radians/second}[/tex]

Calculate the Time to Reach -1.80 cm:

Since the part starts at x = 0, we want to find the time it takes to reach x = -1.80 cm (or -0.018 m). Using the cosine function, we can write:

[tex]-0.018 = 0.018 \cos(20.1 t)[/tex]

This simplifies to:

[tex]\cos(20.1 t) = -1[/tex]

Determine the Time from Cosine Value:

The cosine function equals -1 at odd multiples of [tex]\pi[/tex]:  

[tex]20.1 t = (2n + 1)\pi[/tex]  

The first instance (n=0) is:

[tex]20.1 t = \pi \implies t = \frac{\pi}{20.1} \approx 0.156 \text{ seconds}[/tex]

Answer:

[tex]P_1 - P_2 =219.62\ Pa[/tex]

Explanation:

given,

density of blood = 1060 kg/m³

v₂ = 0.800 m/s

v₁ = 0.475 m/s

change in pressure calculation

using Bernoulli's equation

[tex]P_1 +\dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho v_2^2[/tex]

both are at same level

[tex]P_1 - P_2 =\dfrac{1}{2}\rho (v_2^2- v_1^2)[/tex]

[tex]P_1 - P_2 =\dfrac{1}{2}1060\times (0.8^2- 0.475^2)[/tex]

[tex]P_1 - P_2 =219.62\ Pa[/tex]

the change in pressure is equal to 219.62 Pa

Answer:

v = ((m + M) / m)*√(2*g*h)

Explanation:

Given

m = mass of the projectile

M = mass of the ballistic pendulum

v = initial speed of the projectile

v' = speedof the system (pendulum + projectile) after the inelastic collision

h = maximum height reached for the system

Knowing that is an inelastic collision we have

m*v + M*(0) = (m+M)*v'

⇒  v' = m*v / (m+M)

After the collision, we apply the Principle of the Conservation of Energy

Ki + Ui = Kf + Uf

where

Ui = Kf = 0 J

then

Ki = Uf

0.5*(m+M)*v'² = (m+M)*g*h

⇒ 0.5*v'² = g*h

⇒ v'² = 2*g*h

⇒  (m*v / (m+M))² = 2*g*h

⇒  v = ((m+M) / m)*√(2*g*h)

The initial velocity of a projectile in a ballistic pendulum problem can be found using principles of conservation of momentum and mechanical energy. The relationship between this initial speed and the maximum height reached is given by the equation: v = sqrt(2gh(M+m)/m).

In the scenario of a ballistic pendulum, we can apply principles from two areas: the conservation of momentum and the conservation of mechanical energy. This is given by the equation: mv = (m+M)V where V is the final velocity just after the collision.

Following the collision, the combined system swings upwards coming to rest momentarily at a maximum height, h. At this point, we can apply the conservation of mechanical energy (initial kinetic energy is transformed solely into gravitational potential energy). This gives us: 1/2(m+M)V2 = (m+M)gh.

Combining these two equations, we can find the relationship between the initial speed v of the projectile and the maximum height h achieved by the block: v = sqrt(2gh(M+m)/m).

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Answer:

Option B

-1500 N

Explanation:

We know that force, F=ma where m is the mass of Johan and a is the acceleration

Since acceleration, [tex]a=\frac {v-u}{t}[/tex] where v and u are final and initial velocities, t is time taken. Considering that he comes to rest, v=0 and u is given as 5.4 m/s, t=0.24 s.

Therefore

[tex]F=ma=m\frac {v-u}{t}=67\times \frac {0-5.4}{0.24}=-1507.5\approx -1500 N[/tex]

Answer:

The will hit the ground in t = 0.51[sg] and he was moving with a velocity of 1.45 [m/s]

Explanation:

In the attached image, we can see a sketch of the conditions of the problem, we are interested to know how long the cat takes to fall from the table to the ground and then using that time we can find the initial velocity of the cat when it arrives at the edge of the table.

The initial condition of this problem is that when the cat reaches the edge of the table, it will only have initial velocity on the X-axis component, its velocity on the y-axis component will be zero.

Therefore:

[tex]y=(v_{y})_{0}[/tex]

In this way, we can use the following equation of kinematics and find the time t.

[tex]y=(v_{y})_{0}*t - \frac{1}{2} * g*t\\where:\\g =9.81[m/s^{2} ] gravity\\y = 1.3 [m] table height[/tex] =0[/tex]

We note that gravity is taken as a negative quantity since the movement is downwards, likewise the value and the equation will have negative value as the reference point will be taken as the edge of the table, therefore, y = -1.3 [m]

[tex]-1.3 = (0)*t - \frac{1}{2} * (9.81)*t^{2} \\t = \sqrt{\frac{2*1.3}{9.81} }\\ t= 0.51 [m/s]\\[/tex]

Now using this time and replacing in the next kinematics equation we will have:

[tex]x=(v_{x} )_{0} *t\\where\\x=0.75 [m][/tex]

We will have:

[tex]x=(v_{x} )_{0} *t\\\\(v_{x} )_{0}=\frac{x}{t} \\(v_{x} )_{0}=\frac{0.75}{0.51}\\ (v_{x} )_{0}=1.45 [m/s][/tex]

Answer:

c.observing what people are saying about a brand

Explanation:

Monitoring tool systems are used to keep track of the system's condition constantly, with the goal of early warning against and strengthening errors, defects or issues. The computer, network, storage, health, results, website and Internet use and applications are controlled.

therefore, Free monitoring tools are useful for observing what people are saying about a brand.

Answer:

h = 10 m

Explanation:

given,

mass of platform = 50 Kg

Kinetic energy = 5000 J

height from which the diver dove = ?

taking acceleration due to gravity = 10 m/s²

using conservation of energy

Kinetic energy is converted into mechanical energy

K.E = P.E

K.E =  m g h

5000 = 50 x 10 x h

500 h = 5000

 [tex]h = \dfrac{5000}{500}[/tex]

    h = 10 m

The height from which the diver dove is equal to h = 10 m

The height from which the diver drove is mathematically given as

h=10m

Question Parameters:

A 50-kg platform diver hits the water below with a kinetic energy of 5000 Joules

Generally, the equation for the Kinetic energy  is mathematically given as

K.E = P.E

Therefore

K.E =  m g h

5000 = 50 x 10 x h

500 h = 5000

In conclusion,the height is

[tex]h = \frac{5000}{500}[/tex]

h=10m

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Answer:

Because its the most suitable radiation for that use

Explanation:

Astronomers use the 21 cm radiation for interstellar and cosmological research as this type of radiation is best suitable for the purpose its intended for. It can easily go through gas and dust molecules and other interstellar bodies that causes a blur vision of the content of space.

Answer:

E. signal reception, signal transduction, and cellular response.

Explanation:

General mechanisms of cellular communication : reception

Cellular communication can be established in different ways that include: a) paracrine communication: the signal acts on neighboring cells; b) endocrine: the signal travels through the bloodstream and reaches distant cells; c) autocrine: the signal reaches the same cell from which it came out; d) neurotransmission: the signal is released by the sending cell into the synaptic space, where it is picked up by the receiving cell; e) cell-cell contacts: the signal remains anchored to the membrane of the sending cell while interacting with the receiving cell and f) through gaps: the signal is diffused from the sending cell to the receiving cell.

Transmission of the signal inside the cell

When a ligand interacts with its membrane receptor, the signal is transmitted into the cell. A cascade of events is then triggered that includes the synthesis of second messengers and the phosphorylation of enzymes catalyzed by protein kinases.

The end of the pathway: the biological response or cellular response

Transduction of external signals produces a response from the white cell. This response is an alteration as a result of the activation or inhibition of some metabolic pathway, and of changes in the shape or movement of the cells. The role of these processes in the control of differentiation, proliferation and cell growth is essential for the normal development and functioning of the organism.

Astronomers think Jupiter generates its internal heat by contracting, changing gravitational potential energy into thermal energy.

Answer: Option E

Explanation:

The internal heat of Jupiter is primarily the primary heat released during gravitational collapse when its creation occurs. Because of their large dimensions, this giant plant was heavily heated by creating surrounding material on their cores.

The largest planet, Jupiter was the hottest. Part of this warming may still remain on such large planets. In addition, it is possible that this giant, mainly gaseous planets produce heat after they are formed as a result of slow contraction. With such a large mass, even a small shrinkage can generate high heat.

Jupiter generates its internal heat through a process called Kelvin-Helmholtz contraction, where gravitational potential energy is converted into thermal energy. Internal friction due to Jupiter's high rotation rate may also contribute to the generation of heat.

Astronomers believe that Jupiter generates its internal heat by contracting, changing gravitational potential energy into thermal energy. As Jupiter contracts, gravitational potential energy is converted into thermal energy, creating heat in the planet's interior. This process is known as Kelvin-Helmholtz contraction. Additionally, there may be some contribution from internal friction due to its high rotation rate, which can generate heat within the planet.

This process is referred to as Kelvin-Helmholtz contraction, named after the scientists who first proposed it. It is a slow but ongoing mechanism by which Jupiter releases the energy stored in its gravitational potential energy. As a result, Jupiter maintains a stable internal temperature, and this heat generation process has likely been ongoing since the planet's formation.

In addition to Kelvin-Helmholtz contraction, Jupiter's high rotation rate also plays a role in internal heat generation. The rapid rotation of the planet causes significant friction and shearing forces within its gaseous layers. This internal friction generates additional heat, contributing to the overall temperature of the planet.

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