A fish swims to a depth of 50.00 meters in the ocean. Assuming the density of sea water is 1.0251.025 g·cm^{-3}g⋅cm −3 , calculate how much water pressure the fish is experiencing at this depth in units of kPa.
Answer:
The fish is experiencing a water pressure of 502.8 kPa.
Explanation:
The water pressure the fish is experiencing can be found as follows:
[tex]P = \rho gh[/tex] (1)
Where:
g: is the gravity = 9.81 m/s²
h: is the height (depth) = 50.0 m
ρ: is the seawater's density = 1.025 g/cm³
By replacing the above values into equation (1) we have:
[tex] P = \rho gh = 1.025 \frac{g}{cm^{3}}*\frac{1 kg}{1000 g}*\frac{(100cm)^{3}}{1 m^{3}}*9.81 m/s^{2}*50.0 m = 502.8 kPa [/tex]
Therefore, the fish is experiencing a water pressure of 502.8 kPa.
I hope it helps you!
Pushes and pulls that result from objects that are physically touching
each other
Answer:
That is false. Take a look at this way. You can push a ball with your own breath, you just need to blow it. And you can pull something from afar with a magnet. It is possible to do both.
Explanation:
Not all physical things can be done only physically. Like I just said, it is possible to use other forces (no, not the dark side one), such as a magnetic force, displayed by a magnet or anything with a force like so.
Which of these statements is true about the International Space Station in orbit around the earth?
A) The space station exerts a force on the earth toward the space station.
B) The force of gravity produces the same acceleration for the earth and the space station. C) There is no gravity acting at the height of the space station.
D)The earth's gravity acts on the space station, but not the reverse.
Answer: b is probly correct beacuse all the others make no sence
The statement that is true about international space station is The force of gravity produces the same acceleration for the earth and the space station.
What is international space station?.International space station is a space station that is located at the lower part of the Earth's orbit . It is the third largest space object in the space and it is a multinational project that involves five different space agencies are NASA, Roscosmos, JAXA, ESA, and CSA.
Therefore, The statement that is true about international space station is The force of gravity produces the same acceleration for the earth and the space station.
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A 65-cm segment of conducting wire carries a current of 0.35 A. The wire is placed in a uniform magnetic field that has a magnitude of 1.24 T. What is the angle between the wire segment and the magnetic field if the force on the wire is 0.26 N?
a. 37°.
b. 43°.
c. 23°.
d. 53°.
e. 67°.
Answer:
e) 67°
Explanation:
the force on the wire can be calculated using the expression below
F = BILsinФ
But we are looking for the angle between the wire segment and the magnetic field, then we can make Ф the subject of the formula from the above expresion, then we have,
Ф =sin⁻¹ (F/BIL)
The parameters is defined as
I =current that is been carried by the wire= 0.35 A
Ф = angle between the wire segment and the magnetic field, which is the unknown?
L = length of the wire=65 cm
B = magnetic field = 1.24
F= force on the wire = 0.26 N,
Ф =sin⁻¹ (F/BIL)
Ф =sin⁻¹ X .....................eqn(#)
Where X= (F/BIL)
We can calculate for X= (F/BIL), from eqn(#) by substituting value of Force, Lenght and
magnetic field
X=(F/BIL)= 0.26/(1.24×0.35×0.65)
= 0.26/0.2821
=0.922
Then substitute X into eqn (Ф =sin⁻¹ X)
Then
Ф =sin⁻¹ (0.922)
Ф=66.42°
Ф=67° approximately
Therefore, the angle between the wire segment and the magnetic field is 67°
What does the phrase “constant velocity” indicate?
a. zero distance
b. zero acceleration
c. constant acceleration
d. deceleration
A Boeing 737—a small, short-range jet with a mass of 51,000 kg— sits at rest at the start of a runway. The pilot turns the pair of jet engines to full throttle, and the thrust accelerates the plane down the runway. After traveling 940 m, the plane reaches its takeoff speed of 70 m/s and leaves the ground. What is the thrust of each engine?
Answer:
67000N
Explanation:
We solve for the acceleration using the the 3rd constant-acceleration equation.
(Vx)f² = (Vx)i² + 2ax∆x
We have the displacement to be
∆x = Xf - Xi = 940m
Vx = 70m/s
The acceleration = (70m/s)²/2(940m)
= 4900/1880
= 2.61m/s²
From isaac newton's second law,
51000kg x 2.61m/s²
= 133,000N
The engines thrust is half of this value
Therefore thrust = 67000N or 67kN
examples of sound to radiant energy
Answer:
Examples of Radiant Energy All Around You
The term radiant energy refers to energy that travels by waves or particles, particularly electromagnetic radiation such as heat or x-rays. Radiant energy is created through electromagnetic waves and was discovered in 1885 by Sir William Crookes. Fields in which this terminology is most often used are telecommunications, heating, radiometry, lighting, and in terms of energy created from the sun. Radiant energy is measured in joules.
Everyday Examples of Radiant Energy
Virtually anything that has a temperature gives off radiant energy. Some examples of radiant energy include:
•The heat emitted from a campfire
•Emission of heat from a hot sidewalk
•X-rays give off radiant energy
•Microwaves utilize radiant energy
•Space heaters produce radiant energy
•Heat created by the body can be radiant energy
•Lighting fixtures
√Home heating units
•Fixtures that convert solar energy to heat
•Visible light
•Gamma rays
•Radio waves
•Electricity
•A surface heated by the sun converts the energy of the light into infrared energy which is a form of radiant energy
•Cell phones utilize radiant energy to function
•Magnetic motor generators that utilize •neodymium magnets create radiant energy
•Audio signals that come to home or cars via radio waves
•Ultraviolet light
√Infrared radiation
•The light emitted from a campfire
•The light generated from a light bulb
•A heated brake disc giving off heat
•The heat from a grill used for cooking
•Water can reflect or absorb radiant energy
•Soil can absorb radiant energy
•Light from the sun
•Heat emitted from a bunsen burner
•Heat from an overused computer
•Heat caused by friction
•Heat emitted from a dryer
•The heat generated by a light bulb
•Heat generated through reflection of visible light
•A window reflects radiant energy
•Heat created from a stove or oven
•Heat emitted from a washing machine
What would have to be the self-inductance of a solenoid for it to store 10.2J of energy when a 1.20A current runs through it?
Answer:
14.17H
Explanation:
Energy stored in an inductor is expressed as;
E = 1/2LI²
L is the inductance of the inductor
I is the current flowing through the inductor
Given
E = 10.2J
I = 1.20
Required
Inductance L
Substitute the given parameters into the formula;
10.2 = 1/2L(1.2)²
10.2 = 1/2*1.44L
10.2 = 0.72L
L = 10.2/0.72
L = 14.17H
Hence the self inductance of the solenoid is 14.17H
The corect phase sequence shown
Gas, Liguid, Solid
Lqud, Gas, Solid
Sold, Liguid, Gas
Gas, Solid, Liquid
above i
Answer:
Explanation:
gas, liquid, soild
liquid, Gas, solid
Gas, Solid, liquid
if chilled coke and hot tea are
kept together tea cools down but ko gets warm why
friction reduces air resistance?
Answer:no
Explanation:
No.....................................
What is an easy way to encourage students to stay hydrated?
Let them know hydration improves performance.
Encourage them to drink water between each class.
Tell them to drink 8 glasses of water per day.
Have them track their hydration with a "water diary."
An easy way to encourage students to stay hydrated is to let them know hydration improves performance.
Drinking water is essential for our body system to function well. It helps in the regulation of the body temperature, the prevention of infections, as well as the delivery of nutrients to different parts of the body.
Staying hydrated is also vital in getting enough sleep and keeps the body in shape.
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How long must you wait (in half-lives) for a radioactive sample to drop to 2.10 % of its original activity?
Answer:
222/88 Ra
Explanation:
We have to wait 5.57 half lives for a radioactive sample to drop to 2.10 % of its original activity.
To find the tike taken for the activity, we need to know about radioactivity and half-life.
What is radioactivity?Radioactivity is the rate of decay of a radioactive substance with respect to time. Mathematically, radioactivity is given asR=R₀e^(-λ×t)
From the above expression time is given ast= 1/λ ln(R₀/R)
What is half-life?Half-life is the time taken for decay of radioactive sample to half of its initial value. Mathematically, half-life= ln2 / λWhat is the expression of time of activity in term of half-life?From the half-life expression, 1/λ=half-life/ln2.Putting the value of 1/λ in the expression of time of activity, we havet=(half-life/ln2)×ln(R₀/R)
What is the time for radioactive sample to drop to 2.10 % of its original activity?Here R=0.021R₀, so t= (half-life/ln2)×ln(R₀/0.021R₀)=5.57 half-lives
Thus, we can conclude that we have to wait 5.57 half lives for a radioactive sample to drop to 2.10 % of its original activity.
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What is the acceleration of the the object during the first 4 seconds?
Answer:
Velocity (m/s) over time (s) graph
Velocity (m/s) over time (s) graph
We could write out our average acceleration as:
a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t
a = (15 m/s - 0 m/s) / 0.2 seconds
a = 15 m/s / 0.2 seconds
a = 75 m/s / second
Explanation:
What this formula is telling us is that if we know the acceleration of an object, and the ... we can plug in our acceleration of 12.5 m/s2 for a, and 4 seconds for t.
Velocity (m/s) over time (s) graph
Velocity (m/s) over time (s) graph
We could write out our average acceleration as:
a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t
a = (15 m/s - 0 m/s) / 0.2 seconds
a = 15 m/s / 0.2 seconds
a = 75 m/s / second
In the past, Africa used to be further away from Europe than it is now
(shown below). What could explain why Africa is closer to Europe now than
it was before? *
Answer: Plates shifting
Explanation: After years and years of plates colliding into solid rock, they slowly become closer together. As recent studies have shown, Africa is currently moving closer to Europe one centimeter every year (one inch every 2.5 years).
Answer:
nvudbwasivnjlscv bwbfvsz
Explanation:
A frequency generator sends a 550Hz sound wave through both water and ice. What is the difference in wavelength between the wave produced in ice and the wave produced in water?
Answer:
3.1
Explanation:
use formula f = v/lambda
Which is the largest gas that occurs in our atmosphere?
Helium
Nitrogen
Other Gases
Oxygen
Answer:
OXYGEN
Explanation:brainlyist me
Answer:
Nitrogen
Explanation:
Oxygen is second
A 1.00-m3 object floats in water with 30.0% of its volume above the waterline. What does the object weigh out of the water?
Answer:
Object's weight = 6,839.42 N
Explanation:
Given
Above waterline = 30%
Volume of object = 1m^3
Required
Determine the weight of the object
First, we need to calculate its Mass
Mass = Density of Water * Volume of object in water
Density of water = 997kg/m³
If 30% is above waterline, then 70% is in water.
So:
Mass = Density of Water * Volume of object in water
Mass = 997kg/m³ * 70%m³
Mass = 997kg * 70%
Mass = 697.9 kg
The object weight sis then calculated as thus:
Weight = Mass * Acceleration of gravity
Weight = 697.9 kg * 9.8m/s²
Weight = 6 839.42 N
find the base area of a cylinder with diameter 1m
Answer:
AB=0.79
Explanation:
hope this helped
Engine 1 has an efficiency of 0.18 and requires 5500 J ofinput heat to perform a certain amount of work. Engine 2 hasan efficiency of 0.26 and performs the same amount of work. How much input heat does the second engine require?
Answer:
The input energy to engine 2 is [tex]E_2 = 3807.7\ J[/tex]
Explanation:
From the question we are told that
The efficiency of engine one is [tex]\eta_1 = 0.18[/tex]
The input energy required is [tex]E = 5500 \ J[/tex]
The efficiency of engine 2 is [tex]\eta_2 = 0.26[/tex]
Generally the workdone by the engine 1 is mathematically represented as
[tex]W_1 = E * \eta_1[/tex]
=> [tex]W_1 = 5500 * 0.18[/tex]
=> [tex]W _ 1= 990 \ J[/tex]
Generally the workdone by engine 2 is mathematically represented as
[tex]W _2= E_1 * \eta_2[/tex]
=> [tex]W _2 = E_1 * 0.26[/tex]
=> [tex]W_2 =0.26E_1[/tex]
From the question we are told that [tex]W_1 = W_2[/tex]
So
[tex]990 =W_2 =0.26E_1[/tex]
=> [tex]E_2 = 3807.7\ J[/tex]
One of the harmonics of a column of air in a tube that is open at both ends has a frequency of 448 Hz, and the next higher harmonic has a frequency of 576 Hz. What is the fundamental frequency of the air column in this tube?
Answer:
The fundamental frequency is [tex]f_1 =128 \ Hz[/tex]
Explanation:
From the question we are told that
The frequency of one harmonics is [tex]f_x= 448 \ Hz[/tex]
The next higher harmonic is [tex]f_z = 576 \ Hz[/tex]
Generally the frequency of an air column open at both ends is mathematically represented as
[tex]f_n = \frac{nv }{ 2 L }[/tex]
Here n is the order of the harmonics (frequency)
v is the velocity of the sound
L is the length of the column
So for one harmonics we have that
[tex]f_k = \frac{n v }{2L}[/tex]
Then for the next higher harmonics
[tex]f_x = \frac{n+1 ) v}{2 L }[/tex]
Generally the difference between these frequencies is mathematically represented as
[tex]f_z- f_x = \frac{(n+1 )v}{ 2L} - \frac{(n )v}{ 2L}[/tex]
=> [tex]576 - 448 = \frac{vn + v - nv }{2L}[/tex]
=> [tex]\frac{ v }{2L} = 128[/tex]
Generally for fundamental frequency n = 1
So
[tex]f_1 = n * \frac{v}{2L}[/tex]
So
[tex]f_1 =1 * 128[/tex]
=> [tex]f_1 =128 \ Hz[/tex]
A diffraction grating with 68 slits per cm is used to measure the wavelengths emitted by hydrogen gas.
A. At what angles in the fourth-order spectrum would you expect to find the two violet lines of wavelength 434 nm and of wavelength 410 nm?
B. What are the angles if the grating has 12,800 slits per cm?
Answer:
a
[tex]\theta _1 =0.687 ^o[/tex]
[tex]\theta _2 =0.630 ^o[/tex]
b
Generally given that the domain arcsine function is between -1 and 1 then the arcsine of 2.22 will not be valid
Generally given that the domain arcsine function is between -1 and 1 then the arcsine of 2.1 will not be valid
Explanation:
From the question we are told that
The slit grating is [tex]N = 68 \ slits / cm = 6800 \ slits / m[/tex]
The order of spectrum is [tex]n = 4[/tex]
Generally the width of the slit is mathematically represented as
[tex]a = \frac{1}{ 6800}[/tex]
=> [tex]a = 0.000147 \ m[/tex]
Generally the condition for constructive interference is
[tex]asin\theta = n * \lambda[/tex]
Now for the first wavelength the angle is evaluated as
[tex]\theta _1 = sin ^{-1} [ \frac{n \lambda_1 }{a} ][/tex]
=> [tex]\theta _1 = sin ^{-1} [ \frac{4* 434 *10^{-9} }{ 0.000147 } ][/tex]
=> [tex]\theta _1 =0.687 ^o[/tex]
Now for the second wavelength the angle is evaluated as
[tex]\theta _2 = sin ^{-1} [ \frac{n \lambda_2 }{a} ][/tex]
=> [tex]\theta _2 = sin ^{-1} [ \frac{4* 410 *10^{-9} }{ 0.000147 } ][/tex]
=> [tex]\theta _2 =0.630 ^o[/tex]
Gnerally if grating is [tex]N = 12800 \ slits per cm = 1280000 \ slits / m[/tex]
Generally the width of the slit is mathematically represented as
[tex]a = \frac{1}{ 1280000}[/tex]
=> [tex]a = 7.813 *10^{-7} \ m[/tex]
Generally the condition for constructive interference is
[tex]asin\theta = n * \lambda[/tex]
Now for the first wavelength the angle is evaluated as
[tex]\theta _1 = sin ^{-1} [ \frac{n \lambda_1 }{a} ][/tex]
[tex]\theta _1 = sin ^{-1} [ \frac{4* 434 *10^{-9} }{ 7.813*10^{-7} } ][/tex]
=> [tex]\theta _1 = sin ^{-1} [ 2.22][/tex]
Generally given that the domain arcsine function is between -1 and 1 then the arcsine of 2.22 will not be valid
=> [tex]\theta _1 =0.687 ^o[/tex]
Now for the second wavelength the angle is evaluated as
[tex]\theta _2 = sin ^{-1} [ \frac{n \lambda_2 }{a} ][/tex]
=> [tex]\theta _2 = sin ^{-1} [ \frac{4* 410 *10^{-9} }{ 7.813*10^{-7} } ][/tex]
=> [tex]\theta _2 = sin ^{-1} [2.1 ][/tex]
Generally given that the domain arcsine function is between -1 and 1 then the arcsine of 2.22 will not be valid
Yellow light shines on a sheet of paper containing a blue pigment. Determine the appearance of the paper.
I dont need google answers if i get google answers i will delete it
Black I think
this is due to the fact that blue lights only reflect blue things so if yellow is shone on it it will reflect black appearance.
how it useful
Mike rides his horse with a constant speed of 20 km/h. How far can he travel in 4 hours?
Answer:
Mike can travel 80 Km in 4 hours
How does reflection differ from refraction and diffraction?
Reflection is the only process in which the wave does not continue moving forward.
Reflection is the only process in which the wave slows down.
Reflection is the only process that involves a change in the wave.
Reflection is the only process that changes the direction of a wave.
Answer: Reflection is the only process in which the wave does not continue moving forward.
Explanation:
Reflection is a process in which the direction of the wave changes when it is exposed to a bounce off barrier. Refraction can be defined as the change in the direction of the wave when the wave passes through one medium to another. Diffraction is a process in which the direction of the wave changes when the wave passes through a particular opening near the barrier.
Answer:
Reflection is the only process in which the wave does not continue moving forward.
Explanation:
The current theory of the structure of the
Earth, called plate tectonics, tells us that the
continents are in constant motion.
Assume that the North American continent
can be represented by a slab of rock 5200 km
on a side and 35 km deep and that the rock
has an average mass density of 2700 kg/m².
The continent is moving at the rate of about
3.8 cm/year.
What is the mass of the continent?
Answer in units of kg.
Answer:
pt 1: [tex]m=1.66698*10^{21} kg[/tex]
Pt 2: [tex]KE=1212.23531 J[/tex]
Explanation:
Information Given: (p = density)
l = 5200km d = 35km p = 2700kg/[tex]m^{2}[/tex]
Part 1: Mass
Find volume [tex]V=(l)^2(d)[/tex][tex]V=(4.2*10^6)^2(35*10^3)[/tex][tex]V=61.74*10^{16}[/tex]Find Mass[tex]m=Vp[/tex][tex]m=(61.74*10^{16})(2700)[/tex][tex]m=1.66698*10^{21}[/tex]Part 2: Kinetic Energy
[tex]v=\frac{3.8cm}{yr}*\frac{m}{100cm}*\frac{yr}{365d}*\frac{d}{24hr}*\frac{hr}{3600s}[/tex][tex]v=1.20497*10^{-9}[/tex][tex]KE=\frac{1}{2}mv^2[/tex]
[tex]KE=\frac{1}{2} (1.66698*10^{21})(1.20497*10^{-9})^2[/tex]
[tex]KE=1212.23531 J[/tex]
Part 3: Jogger Speed
set up, because I don't have the mass :(
Information given:
[tex]KE_{jogger}[/tex]
[tex]KE=\frac{1}{2}mv^2[/tex][tex]v_{jogger} =\sqrt{\frac{2KE}{m_{jogger} } }[/tex]Input the valuesHope it helps :)
What do light and energy tell us about the universe?
need a paragraph
How was the Periodic Table of Elements developed and how are the elements arranged on it?
Answer:
In 1869 Russian chemist Dimitri Mendeleev started the development of the periodic table, arranging chemical elements by atomic mass. He predicted the discovery of other elements, and left spaces open in his periodic table for them.
Explanation:
Answer: Mendeleev first published a table of elements arranged according to increasing atomic masses. He noticed that some elements near each other had differing properties, but elements in vertical columns had similar properties. Moseley then rearranged the table according to atomic numbers and this eliminated the discrepancies found in Mendeleev’s attempt. Today’s version of the periodic table displays elements in order based on their atomic number; the atomic number indicates the number of protons within the atoms of a particular element. Rows are called periods and columns are called groups. Elements in the same group have similar properties. Elements are grouped into nine categories: noble gases, halogens, nonmetals, alkali metals, alkaline earth metals, transition metals, other metals, metalloids, and rare earth elements.
A certain heat engine does 30.2 kJ of work and dissipates 9.14 kJ of waste heat in a cyclical process.
A) What was the heat input to this engine?
B) What was its efficiency?
Answer:
a) [tex]H_{in}=39.34 kJ[/tex]
b) Efficiency=76.77%
Explanation:
a)
In order to solve this problem, we can use the following formula:
[tex]H_{in}=H_{out}+W[/tex]
the problem provides us with all the necessary information so we can directly use the formula:
[tex]H_{in}=9.14kJ+30.2kJ[/tex]
[tex]H_{in}=39.34 kJ[/tex]
b) In order to find the efficiency, we can use the following formula:
[tex]Efficiency=\frac{W}{H_{in}}*100\%[/tex]
so we get:
[tex]Efficiency=\frac{30.2kJ}{39.34kJ}*100\%[/tex]
Efficiency=76.77%
Initial velocity of a car is 36 km/hr. Find the distance after one min, if it goes with acceleration 2m/s².
[A] 4000 m.
[B] 4200 m.
[C] 4400 m.
[D] 4600 m.
Answer:
[tex]\bf\pink{4200\:m}[/tex]Explanation:
Given :-[tex]\sf\red{ Initial \ velocity = 36km/hr = 10m/s}[/tex][tex]\sf\orange{ Acceleration = 2m/s^{2}}[/tex][tex]\sf\green{Time = 1\:min = 60\: sec}[/tex]To Find :-[tex]\sf\blue{Distance \: Travelled}[/tex]Formula used :-[tex]\sf\purple{s = ut + \dfrac{1}{2} at^{2}}[/tex]Where,
s = [tex]\bold{Distance}[/tex]t = [tex]\bold{Time }[/tex]u = [tex]\bold{ Initial \ velocity}[/tex]a = [tex]\bodl{Acceleration}[/tex]Solution :-[tex]\Rightarrow\:\:\rm{s = 10(60) + \dfrac{1}{\cancel{2}} \times \cancel{2} \times (60)^{2}}[/tex]
[tex]\Rightarrow\:\:\rm{s = 600+60\times 60}[/tex]
[tex]\Rightarrow\:\:\rm{s = 600 + 3600}[/tex]
[tex]\Rightarrow\:\:\rm{s = {\bf{\red{4200\:m}}}}[/tex]
Hence, [tex]\bf\green{(B)}[/tex] is the correct option