a. An object weighs 7.84 N when it is in air and 6.86 N when it is immersed in water. What is the specific gravity of the object?
b. How much pressure does it take for a pump to supply a drinking fountainwith 300 kPa, if the fountain is 30.0 m above the pump?

Answers

Answer 1

Answer:

a

 [tex]W_s = 8[/tex]

b

[tex]P = 594 000 \ Pa[/tex]

Explanation:

Considering question a

 From the question we are told that

   The weight of the object in  air is  [tex]W_1 = 7.84 \ N[/tex]

    The weight of the object in water  is [tex]W_2 = 6.86\ N[/tex]

Generally the specific gravity of the object is mathematically represented as

       [tex]W_s = \frac{W_1 }{W_1 - W_2 }[/tex]

=>    [tex]W_s = \frac{7.84}{7.84 -6.86 }[/tex]

=>    [tex]W_s = 8[/tex]

Considering question b

From the question we are told that        

   The pressure required is  [tex]P_r = 300 \ kPa = 300 *10^{3} \ Pa[/tex]

    The height is  [tex]h = 30.0 \ m[/tex]

Generally the pressure require to get the water to  the given height is mathematically represented as

      [tex]P_h = \rho * g * h[/tex]

Here  [tex]\rho[/tex] is the density of  water with value  [tex]\rho = 1000 \ kg / m^3[/tex]

So  

     [tex]P_h = 1000 * 9.8 * 30[/tex]

=>  [tex]P_h = 294000 \ Pa[/tex]

Generally the pressure require to pump the water to  the given height at the require pressure is mathematically represented as

      [tex]P = P_h + P_r[/tex]

=>  [tex]P = 294000 + 300*10^{3}[/tex]

=>  [tex]P = 594 000 \ Pa[/tex]


Related Questions

Steven tries to sneak a look into the freezer to see what kind of ice cream cake he will be eating for his birthday tomorrow. Which of the following best describes what is happening when Steven opens the freezer door right before his dad catches him?
A-Heat energy flows from the kitchen to the freezer
B-Heat energy flows from the freezer to the kitchen
C- Cold energy flows from the kitchen into the freezer
D- Cold energy flows from the freezer into the kitchen​

Answers

I believe it’s B because all the air from inside is cold and heat rushes into that so I think

A uniform solid cylindrical log begins rolling without slipping down a ramp that rises 28.0° above the horizontal. After it has rolled 4.20 m along the ramp, what is the magnitude of the linear acceleration of its center of mass?

Answers

Answer:

3.07 m/s

Explanation:

The magnitude of the linear acceleration will be "3.07 m/s²".

Given:

Angle, [tex]\Theta = 28^{\circ}[/tex]Length = 2.40 m

The acceleration of its center of mass will be:

= [tex]\frac{2}{3}g Sin \Theta[/tex]

By putting the above given values, we get

= [tex]\frac{2}{3}\times 9.8 Sin 28^{\circ}[/tex]

= [tex]\frac{2}{3}\times 9.8\times 0.46[/tex]

= [tex]\frac{9.0167}{3}[/tex]

= [tex]3.07 \ m/s^2[/tex]

Thus the solution above is appropriate.

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how does the law of conservation energy relate to the system shown in thos model

Answers

Answer:

there is no model shown

Explanation:

Therefore it can be thousands of different answers

Question 4
Why do some competitive swimmers shave their heads and bodies?
А
To decrease weight
B.
to increaase blood flow
C
to decrease friction
D
to increase buoyancy

Answers

Answer:

C

Explanation:

A force of 35.0 N is required to start a 6.0-kg box movingacross a horizontal concrete floor. (a) What is the coefficientof static friction between the box and the floor?(b) if the 35.0-n force continues, the box accelerates at 0.60 m/s2 . What is the coefficient of kinetic friction.

Answers

Answer:

Explanation:

Frictional force acting on the box = 35 N

If μ₁ be the coefficient of static friction and μ₂ be the kinetic friction

μ₁ mg = 35

μ₁ = 35 / 6 x 9.8

= .6

b )

net force acting on box to produce acceleration of .6 m /s²

= 6 x .6 = 3.6 N

net  force acting on the box when it is accelerating

= 35 - μ₂ mg

35 - μ₂ x 6 x 9.8 = 3.6

35 - μ₂ x 58.8 = 3.6

μ₂ = .53

if an object travels at constant speed in a circular path the acceleration of the object is

Answers

Answer:

centripetal and of constant magnitude

Explanation:

The acceleration of an object traveling on a circular path at constant speed is directed towards the center of the circle (centripetal) and of constant magnitude equal to the square of the object's speed divided by the radius of the circle.

Consider a one-dimensional crystal (similar to a carbon nanowire) with length 10 um and lattice spacing 0.1 nm.

RequireD:
a. What is the Fermi level assuming one electron per atom?
b. What is the density of states as a function of electron energy?

Answers

Answer:

a)  Fermi level = 600 electron-volts

b) [tex]\frac{2.04 * 10^{13} }{\sqrt{E} }[/tex]

Explanation:

Given data:

length of one-dimensional crystal = 10 um

Lattice spacing = 0.1 nm

A) Determine the Fermi level assuming one electron per atom

Total length = 10 um

Interatomic separation of a = 0.1 nm

in this case the Atom has one electron therefore the number of electrons = 10^5  and the number of states Ns = gsN = 2 * 10^5  ( attached below is some part of the solution )

hence : Fermi level = 600 electron-volts

B) Determine the density of states as a function of electron energy

attached below is the detailed solution

PLEASE HELP!!


1.The Law of Conservation of Energy states that?

A. matter cannot be created or destroyed, only stored and transferred.

B. momentum cannot be created or destroyed, only stored and transferred.

C. energy cannot be created or destroyed, only stored and transferred.


2.The total amount of kinetic and potential energy in a system that remains constant is the system's


A .mechanical energy

B. elastic energy

C. chemical energy

D. gravitational energy


3, At the top of a giant hill sits a roller coaster cart, all the energy of the cart is stored as


A. kinetic energy

B. chemical energy

C. elastic energy

D. gravitational potential energy



4. At the bottom of the hill (ground level), the roller coaster cart in the previous problem, all the energy of the cart is stored as


A. kinetic energy

B. chemical energy

C. elastic energy

D. gravitational potential energy





When a pendulum swings when will the potential energy be at a maximum?

A
B
C
D

6. If total mechanical energy = kinetic energy + potential energy, TME = KE + PE, and a roller coaster has 500 J of total mechanical energy. At the very beginning of the track the roller coaster has 500 J of potential energy, how much kinetic energy is there at the beginning of the track?



A 25 kg child is sitting at the top of a 4 m tall slide, what is his potential energy?




It the same child has a velocity of 2 m/s half-way down the slide, what is his kinetic energy?



If the TME energy for the same child on the slide is 100 J, how much potential energy does the child have half-way down the slide?

Answers

Answer:

A

A

D

D

when the pendulum is at the lefend and right end.

0 because kinetic energy is only formed when a moving object is moving.

dont know the others sorry

Explanation:

i got about half way for you hope it correct

give me nothing if you got some wrong!

or just give me a message saying that i didn't help

For an investigation a student records data about four unknown substances.

data, for, unknown, substances, substance, mass, grams, volume, centimeters, cubed, density, grams, per, centimeter, cubed, 1, 6.95, 4.0, 2, 4.54, 2.0, 3, 5.40, 3.0, 4, 10.35, 5.0,

The student then calculates the densities of the unknown substances and compares them with the table of densities of known substances shown below.

densities, of, some, known, substances, substance, density, grams, per, centimeter, cubed, calcium, 1.54, carbon, 2.27, magnesium, 1.74, phosphorus, 1.82, platinum, 21.46, sulfur, 2.07,

Which unknown substance is most likely carbon?

Answers

Answer:

omparing the values ​​of the strung density, the one that is closest to carbon is number 2

Explanation:

In this exercise we are given the mass and volume of a body, we are asked to calculate the density, using the equation

          ρ = m / V

 

in the third and fourth column of the table is the density and the substance with the closest value

#    mass   volume  density   material

        (gr)      (cm³)       (gr/cm³)

1      6.95      4.0          1.74       magnesium

2     4.54      2.0          2.27       carbon

3     5.40      3.0          1.80        phosphorus

4    10.35      5.0         2.07        sulfur

When comparing the values ​​of the strung density, the one that is closest to carbon is number 2

Developing a question for appointments

Answers

Explanation:

how can I help you that?

Two light bulbs have resistances of 400Ω and 800Ω. The two light bulbs are now connected in parallel across the 120-V line. A. Find the current through each bulb.B. Find the power dissipated in each bulb. C. Find the total power dissipated in both bulbs.

Answers

Answer:

a.) O.3A and 0.15A

b.) P1 = 36W, P2 = 18W

c.) 54W

Explanation:

Constant resistance R1 = 400 ohms, R2 = 800 ohms

a.)

The current through each bulb

While in parallel.

I = 120/800/3

= 0.45A

For 400= 800/400+800

= 800/1200 x 0.45

= 0.3A

I-800 = 400 /400+800

= 400/1200*0.45

= 0.15A

B.)

The power dissipated in each bulb

P409 = I²400r

= 0.3A²x400 ohms

= 36 W

P800 = I²800R

= 0.15A²*800 ohms

= 18W

C.)

The total power dissipated

= 36W + 18 W

= 54W

A 5.0 kg hammer strikes a 0.25 kg nail with a force of 10.0 N causing the nail to accelerate at 40.0 m/s^2. What is the acceleration of the hammar?

Answers

Answer:

2 m/s².

Explanation:

The following data were obtained from the question:

Mass (m) of hammer = 5 Kg

Mass (m) of nail = 0.25 kg

Force (F) applied = 10 N

Acceleration (a) of nail = 40 m/s².

Acceleration (a) of hammer =?

From Newton's third law which states that to every action, there is an equal but opposite reaction. This implies that the force applied by the hammer on the nail is exactly the force applied by the nail on the hammer. Thus, we can obtain the acceleration of the hammer as follow:

Mass (m) of hammer = 5 Kg

Force (F) applied = 10 N

Acceleration (a) of hammer =?

F = ma

10 = 5 × a

Divide both side by 5

a = 10/5

a = 2 m/s²

Thus, the acceleration of the hammer is 2 m/s².

For the winter, a duck flies 10.0 m/s due south against a gust of wind with a velocity of 2.5 m/s. What is the resultant velocity of the duck?

a. 12.5 m/s south
b. 12.5 m/s north
c. 7.5 m/s south
d. 7.5 m/s north
e. 10.3 m/s south

Answers

Answer:

c. 7.5 m/s south

Explanation:

Given the following data;

Velocity of duck = 10m/s due South

Velocity of wind = 2.5m/s

To find the resultant velocity;

Since we know that the duck is flying against a gust of wind, we would have to subtract the velocity of the gust of wind from that of the duck.

This ultimately implies that, the gust of wind (headwind) would decrease the resulting velocity of the duck because it approaches the duck from the front.

Resultant velocity, /V/ = 10 - 2.5

Resultant velocity, /V/ = 7.5m/s South.

Therefore, the resultant velocity of the duck is 7.5m/s south.

Is clinical depression associated with eating disorders yes or no

Answers

yes it can clinical depression and eating disorders usually go one on one with each other each type of disorder goes one on one with each other

Yes, a symptom of depression is a loss of appetite, and eating disorders are often developed as a result of an unhealthy coping mechanism with life; therefore, people with depression can get eating disorders and vice versa. They go hand in hand.

The aurora borealis is caused by the ____.

A. mesosphere
B. stratosphere
c. thermosphere
D. troposphere

Answers

Thermosphere I believe but I could be wrong

Answer: ionosphere

Explanation: aurora boreal is is caused by the ionosphere which is a part of the thermosphere

Which material will heat up the most quickly if placed near a heat source

Answers

Answer:

metal

Explanation:

metal heats up the fastest when near a hot source

Referring to the sketch of a planet around the sun, Area A is three times that of Area B. Compare the times required for the planet to travel from Point 1 to Point 2 and from Point 3 to Point 4 and select the letter of the correct answer.

Answers

Answer:

tA is three times tB

Explanation:

tA is three times tB

Referring to the sketch of a planet around the sun, Area A is three times that of Area B. Compare the times required for the planet to travel from Point 1 to Point 2 and from Point 3 to Point 4 and select the letter of the correct answer.

For the common-emitter, common-base and emitter-follower amplifier designs,what is the primary benefit of each amplifier?

Answers

Answer:

Explanation:

A common emitter amplifier works by inverting. It does have low input impedance. Despite its low input impedance, it poses an otherwise high output impedance.

The common base circuit performs optimally when it acts as a current buffer. It has the ability to take an input current at a low input impedance, and transmit almost the same current to an impedance with a higher output

The primary benefit of emitter follower amplifier is that the transistor is able to provide current and power gain. Although this transistor takes in little current from the input. It still provides an impedance with a low output to a circuit by exercising the output of the follower. This then translates to that the output under load not dropping.

How fast would a(n) 85 kg man need to run in order to have the same kinetic energy as an 8.0 g bullet fired at 410 m/s?

Answers

Answer:

V = 3.97 m/s

Explanation:

Mass of a man, M = 85 kg

Mass of a bullet, m = 8 g = 0.008 kg

Speed of bullet, v = 410 m/s

We need to find the speed of a man in order to have the same kinetic energy as that of the bullet. Let the kinetic energy of the bullet is k. So,

[tex]k=\dfrac{1}{2}mv^2\\\\k=\dfrac{1}{2}\times 0.008\times 410^2\\\\k=672.4\ J[/tex]

Since, k = K (K is the kinetic energy of the man and Let V is the speed)

[tex]K=\dfrac{1}{2}MV^2\\\\V=\sqrt{\dfrac{2K}{M}} \\\\V=\sqrt{\dfrac{2\times 672.4}{85}} \\\\V=3.97\ m/s[/tex]

So, the speed of the man is 3.97 m/s.

We have that for the Question "How fast would a(n) 85 kg man need to run in order to have the same kinetic energy as an 8.0 g bullet fired at 410 m/s?" it can be said that the speed is

v=4.0m/s

From the question we are told

How fast would a(n) 85 kg man need to run in order to have the same kinetic energy as an 8.0 g bullet fired at 410 m/s?

Generally the equation for the kinetic energy  is mathematically given as

[tex]K.E=1/2*m*v^2\\\\Therefore\\\\K.E=1/2*0.008*(420)^2\\\\K.E=672.4J\\\\[/tex]

Therefore

[tex]v^2=\frac{672.4*2}{85}\\\\v^2=15.821[/tex]

v=4.0m/s

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Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.400mm wide. The diffraction pattern is observed on a screen 3.35 m away. Define the width of a bright fringe as the distance between the minima on either side.

(a) What is the width of the central bright fringe?

(b) What is the width of the first bright fringe on either side of the central one?

Answers

Answer:

a)[tex]y_{first}=5.3mm[/tex]

b)[tex]y_{second}=10.6-5.3 =5.3 mm[/tex]  

Explanation:

a)

The width of the central bright in this diffraction pattern is given by:

[tex]y=\frac{m\lambda D}{a}[/tex] when m is a natural number.

here:

m is 1 (to find the central bright fringe)                 D is the distance from the slit to the screen a is the slit wide λ is the wavelength

So we have:

[tex]y_{first}=\frac{633*10^{9}*3.35}{0.0004}[/tex]

[tex]y_{first}=5.3mm[/tex]

b)

Now, if we do m=2 we can find the distance to the second minima.

[tex]y_{2}=\frac{2*633*10^{9}*3.35}{0.0004}[/tex]

[tex]y_{2}=10.6 mm[/tex]

Now we need to subtract these distance, to get the width of the first bright fringe :

[tex]y_{second}=10.6-5.3 =5.3 mm[/tex]    

I hope it heps you!

     

In a mercury barometer at atmospheric pressure, the height of the column of mercury in a glass tube is 760 mm. If another mercury barometer is used that has a tube of larger diameter, how high will the column of mercury be in this case?

a. greater than 760 mm
b. Less than 760 mm
c. equal to 760 mm

Answers

Answer:

c. equal to 760 mm

Explanation:

We are told that another mercury barometer is used that has a tube of larger diameter. This means a larger area and the weight of the liquid in the tube will have increased since volume = area × height. Also, due to the larger area, the net upward force on the mercury will have also increased by the same amount because force = area × pressure. Therefore, as long as the pressure remains the same, the height of the mercury will also remain the same.

Thus the height of the mercury = 760 mm

5. Salt is often spread on roads during winter to prevent ice from forming. Do you think sugar would
work as well? Explain your answer.

Answers

Answer:

No because sugar is more sweet and it would not have the benefits of salt. Think of it like how people used to put salt on meat to preserve it. Would they put sugar on? No they would not because salt is a combination of things that sugar does not have.

Explanation

An athlete whirls a 6.41 kg hammer tied to the end of a 1.3 m chain in a simple horizontal circle where you should ignore any vertical deviations. The hammer moves at the rate of 1.64 rev/s. What is the centripetal acceleration of the hammer

Answers

The chain makes a circle of radius 1.3 m as it spins, so that 1 revolution of the hammer amounts to it moving a linear distance of 2π (1.3 m) = 2.6π m. Then the hammer's linear speed is

(1.64 rev/s) • (2.6π m/rev) = 4.264π m/s

and so the centripetal acceleration has magnitude

a = v²/r = (4.264π m/s)² / (1.3 m)

a ≈ 140 m/s²

What is the acceleration due to gravity on a planet with a mass of 4.76x10^23 kg and a radius of 2,635,000 meters? Report your answer to three digits and include the units. (Ex: 7.15 m/s/s).​

Answers

Answer:

g=4.5727 m/s^2

Explanation:

[tex]We\ know\ that\ , \\F=G\frac{Mm}{d^2} \\As\ F=ma\ and\ a=g\ (g\ is\ the\ acceleration\ caused\ due\ to\ gravity),\\mg=G\frac{Mm}{d^2}\\Canceling\ m\ on\ both\ sides\ we\ get,\\g=G\frac{M}{d^2}\\[/tex]

[tex]We\ are\ given\ that\ ,\\Mass\ of\ the\ planet\ (M) = 4.76*10^{23} kg\\Distance(d) = 2,635,000 m\\G(Universal\ Gravitation\ Constant) = 6.67*10^{-11}\\Hence,\\g=6.67*10^{-11} (\frac{4.76*10^{23}}{2,635,000^2} )\\g= 4.5727 m/s^2[/tex]

Suppose a rocket is traveling through space and moving at a speed close to the speed of light. Three minutes pass on the rocket as recorded by an observer on Earth.

How many minutes pass on Earth?

less than three minutes because of time dilation
more than three minutes because of time dilation
three minutes because time is absolute
less than or greater than three minutes depending on the speed

Answers

Answer:

B

Explanation:

on edge2020

Three minutes pass on the rocket as recorded by an observer on Earth will be more than three minutes because of time dilation. The correct option is B.

What is time dilation?

It is a phenomenon a slowing of time according to the theory of relativity when a system is in motion relative to an outside observer and that becomes apparent when the speed of the system approaches that of light is known as time dilatation.

The time dilation results in Three minutes passing on the rocket as recorded by an observer on earth will be more than three minutes.

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An object in motion, will tend to remain in motion until acted upon my an external, unbalanced force is defined to be

Answers

Answer:

The first law states that a body at rest will stay at rest until a net external force acts upon it and that a body in motion will remain in motion at a constant velocity until acted on by a net external force. Inertia is the tendency of a body in motion to remain in motion.

Explanation:

As friction with our atmosphere causes a satellite to spiral inward, closer to Earth, its orbital speed increases. Why?

Answers

Answer:

Explanation:

When we go back and look at Kepler's third law. Kepler's third law states that "the square of the period is proportional to the cube of the semi-major axis of the orbit". P²µa³

where a is the semi-major axis of orbit,

P is the period of orbit

The orbit in question is the distance from the center of Earth. If the satellite spirals in as a result of air friction, then it decreases and, so does P.

Note that P = 2πa / V or Vµ1 / a0.5 .

From the formula, we can see that a decrease in the value of a, will lead to an Increase in the value of V.

In a liquid metal fast breeder reactor, no neutron moderation is desired and sodium is used as a coolant to minimize fission-neutron thermalization. How many elastic scatters with sodium, on the average, would it take for 2-MeV neutrons to reach an average thermal energy of 0.025 eV?

Answers

Answer:

219 scatterings

Explanation:

Given that:

The Coolant used In the liquid metal fast breed reactor = Sodium

The atomic weight (A) of sodium = 23

The initial energy [tex]E_{i}[/tex] = 2 - MeV

The final energy [tex]E_{f}[/tex] = 0.025 eV  (thermal energy)

The number of elastic neutron scatterings (n) needed to reach the given average thermal energy can be computed as:

[tex]n = \dfrac{log \bigg(\dfrac{E_f}{E_i} \bigg)}{log \bigg [ \dfrac{A^2+1}{(A+1)^2} \bigg]}[/tex]

[tex]n = \dfrac{log \bigg(\dfrac{0.025}{2 \times 10^6} \bigg)}{log \bigg [ \dfrac{23^2+1}{(23+1)^2} \bigg]}[/tex]

[tex]n = \dfrac{log \bigg(1.25\times 10^{-8} \bigg)}{log \bigg [ 0.92014\bigg]}[/tex]

[tex]n = 218.643[/tex]

n ≅ 219 scatterings

A coil has an inductance of 6.00 mH, and the current in it changes from 0.200 A to 1.50 A in a time interval of 0.300 s. Find the magnitude of the average induced emf in the coil during this time interval.

Answers

Answer:

0.026 V

Explanation:

Given that,

Inductance of the coil, L = 6 mH

The current changes from 0.2 A to 1.5 A in a time interval of 0.3 s

We need to find the magnitude of the average induced emf in the coil during this time interval. The formula for the induced emf is given by :

[tex]\epsilon=L\dfrac{dI}{dt}\\\\=6\times 10^{-3}\times \dfrac{1.5-0.2}{0.3}\\\\=0.026\ V[/tex]

So, the magnitude of induced emf is 0.026 volts.

A child wants to throw a rock upward to try to strike a plump apple up in a
tree. If the fastest the child can throw the rock is 7 m/s, what is the highest
the apple can be in order for the child to be able to strike the apple?

Answers

Answer:

7m because that is the height that the child can throw.

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