With the use of the formula, the pressure on the bottom of the pool due to the water is 29400 Pascal
Given that a swimming pool of width 9.0 m and length 12.0 m is filled with water to a depth of 3.0 m.
Volume = Length x Width x Height
Volume = 9 x 12 x 3
Volume = 324 [tex]m^{3}[/tex]
Since 1 m^3 of water has a mass of 1000 kg,
Mass = 324 x 1000
mass = 324000 kg
To calculate the pressure on the bottom of the pool due to the water, let us use the pressure formula below
P = δgh
where
P = pressure
δ = density of the water = 1000 kg/[tex]m^{3}[/tex]
g = Acceleration due to gravity = 9.8 m/[tex]s^{2}[/tex]
h = height = 3m
Substitute all the parameters into the formula
P = 1000 x 9.8 x 3
P = 29400 Pascal
Therefore, the pressure on the bottom of the pool due to the water is 29400 Pascal
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A car which is traveling at a velocity of 27 m/s undergoes an acceleration of 5.5 m/s^2 over a distance of 430 m. How fast is it going after that acceleration?
73.9m/s (1dp)
1) list everything that you are given using suvat, where s is distance, u is initial velocity(speed), v is final velocity(speed), a is acceleration and t is time
s = 430m
u = 27 m/s
v = ? (we need to work out)
a = 5.5m/s^2
t = (we are not given this value)
2) use an equation that doesn't involve the time
[tex] {v}^{2} = u {}^{2} + 2as[/tex]
3) input the values that we have
[tex] {v}^{2} = ( {27})^{2} + 2(5.5)(430)[/tex]
[tex]v {}^{2} = 5459[/tex]
[tex]v = \sqrt{5459} = 73.9[/tex]
so the answer is 73.9m/s to 1dp
what is the affect of density of air in the velocity of sound??
Answer: Hello!
The greater the density of a medium, the slower the speed of sound. This observation is analogous to the fact that the frequency of a simple harmonic motion is inversely proportional to m, the mass of the oscillating object. The speed of sound in air is low, because air is easily compressible.
Explanation:
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The practice of science can answer only scientific questions. And scientific questions guide the design of investigations. What must be true of the possible answers to a scientific question? A. They are popular with a majority of scientists. B. They agree with all prior experiments. O C. They can be supported by evidence. O D. They lead to increased funding of scientific research.
Answer:
They must be supported by evidence.
Explanation:
Every scientific theory or scientifc claim must have scientific evidence.
What is the work done in lifting 60 kg of blocks to a height of 20m
Answer:
The answer is 1200
Explanation:
The diagram shows a nephron.
A nephron. W points to branch of renal artery. X points to branch of renal vein. Y points to tubular point of nephron. Z points to collecting duct.
Where is the blood first filtered?
Answer:
w
Explanation:
renal artery...the blood flow into the kidney via this blood vassles
and the filtration take place is called ultrafiltration
A gas in a closed container is heated with 10J of energy, causing the lid of the container to rise 2m with 3N of force. What is the total change in energy of the system
Explanation:
For this problem, use the first law of thermodynamics. The change in energy equals the increase in heat energy minus the work done.
ΔU=Q−W
We are not given a value for work, but we can solve for it using the force and distance. Work is the product of force and displacement.
W=FΔx
W=3N×2m
W=6J
Now that we have the value of work done and the value for heat added, we can solve for the total change in energy.
ΔU=Q−W
ΔU=10J−6J
ΔU=4J
Answer is 4J
i think this may help you very much
Galois drove 60.0 kilometers due west in 5.00 hours and then drove 43.0 kilometers due north in 3.00 hours.
(a) How far did he travel?
(b) What was his average speed?
(c) What was his displacement?
(d) What was his average velocity?
Answer:
Explanation:
See the attachment for the details. A right triangle is formed to find the hypotenuse of the two legs consisting of the actual driving distances and times. The hypotenuse gives the vector information for the displacement at the end of 8 hours of driving.
The individual driving times and distances are summed to provide:
(a) How far did he travel?
103 km
(b) What was his average speed?
12.88 km/h
(c) What was his displacement?
73.82 km
(d) What was his average velocity?
9.228 km/h
who has the best answer for this
what percentage more water is used to provide us with electricity vs for irrigation
Answer:
electricity because it has more percentage nd energy
Explanation:
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A 1.5 kg ball has a velocity of 12 m/s just before it strikes the floor. Find the impulse on the ball if the ball bounces up with a velocity of 10 m/s.
Hi there!
Recall:
Impulse = Change in momentum
I = Δp = mΔv = m(vf - vi)
Let the direction TOWARDS the floor be POSITIVE, and AWAY be NEGATIVE.
Plug in the givan values:
Δp = 1.5(-10 - 12) = -33 Ns
**OR, the magnitude: |-33| = 33 Ns
1. Apply a constant force of 50 N directed to the right of the 50 kg Box. (2 pts)
Hypothesis:?
Conclusion: ?
As the box is moving with a constant velocity, the two forces acting on the box are canceling each other.
Then friction force = 80 Newtons but in the opposite direction.
Friction force = Mu * Normal force exerted by ground = Mu * weight of box
So we find Mu.
Mu = coefficient of friction between box and horizontal surface
= Force of friction / weight = 80 / 50 * 9.81 = 0.163
When an identical box is placed on top, the force of friction is
= Mu * total weight = 0.163 * (50+50) * 9.81 = 159.9 Newtons
If astronomers discovered a new planet and found its period of rotation around the Sun to be 105 years, how long would its semi-major axis length be as it orbited the Sun in AU?
From Kepler's third law, its semi-major axis length will be 22.2 AU approximately as it orbited the Sun in AU. The closest option is option C
Given that an astronomers discovered a new planet and found its period of rotation around the Sun to be 105 years.
According to Kepler's third law,
[tex]T^{2} \alpha r^{3}[/tex]
Where
T = Period ( in earth years) = time to complete one orbit
r = Length of the semi major axis in Astronomical unit.
[tex]T^{2}[/tex] = [tex]\frac{4\pi ^{2} }{GM} * r^{3}[/tex]
convert years to seconds
105 x 365 day x 24 hours x 3600 s
T = 3311280000 seconds
Mass of the sun M = 1.989 × 10^30 kg
G = 6.67 x [tex]10^{-11}[/tex]N m^2/kg^2
Substitute all the parameters into the formula
[tex]T^{2}[/tex] = 1.096 x [tex]10^{19}[/tex] = [tex]\frac{4\pi ^{2} }{6.67 * 10^{-11} * 1.989 * 10^{30} } * r^{3}[/tex]
1.096 x [tex]10^{19}[/tex] = 2.976 x [tex]10^{-19}[/tex] [tex]r^{3}[/tex]
[tex]r^{3}[/tex] = 1.096 x [tex]10^{19}[/tex] / 2.976 x [tex]10^{-19}[/tex]
[tex]r^{3}[/tex] = 3.68 x [tex]10^{37}[/tex]
r = [tex]\sqrt[3]{3.68 * 10^{37} }[/tex]
r = 3.33 x [tex]10^{12}[/tex] m
1 AU = 1.5 x [tex]10^{11}[/tex] m
r = 3.33 x [tex]10^{12}[/tex] / 1.5 x [tex]10^{11}[/tex]
r = 22.18 AU
Therefore, its semi-major axis length will be 22.2 AU approximately as it orbited the Sun in AU. The closest option is option C
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Answer:
C. 22.3 AU
Explanation:
Not only is the above an unnecessarily complicated answer, it's not even fully correct, and definitely not what they want you to do.
T^2 = s^3, where T = orbital period and s = semi-major axis length.
Substitute T and you get 105^2 = s^3. Solve for s.
11025 = s^3
3√11025 = s
22.25663649 = s
Therefore, the answer is C. 22.3 AU
Do I look good yes or no
Answer:
0/10
Explanation:
_--------------------
Answer:
No. -infinity/10
Explanation:
Ew. Just why, like no.
A 3-column table with 1 row. The first column titled distance travelled (meters) has entry 6. 1. The second column labeled lower track elapsed time (seconds) has entry 4. 92. The third column labeled higher track elapsed time (seconds) has entry 3. 36. Based on the time measurements in the table, what can be said about the speed of the car on the lower track as compared to the higher track? How can the reasoning for the above answer be best explained? On the higher track, the elapsed time is. Calculate speeds for each track. How much faster was the car on the higher track than the lower track?.
Answer:
B,A,A
Explanation:
Answer:
Other guy is correct b,a,a
Explanation:
define the unit of work
[tex]\text{The unit of work is same as energy. Newton-meters(Nm) or Joules(J).}\\[/tex]
Bill pushed 327 kg. bucket of concrete with a force of 10N. What was the
acceleration?
Answer:
F(10)=mass(327)x acceleration(m/s)
Explanation:
7. If the speed of light in medium-1 and medium-2 are 2.5x 10 m/s and 2x 10 m/s respectively then the refractive index of medium-1 with respect to medium-2 is (a) 3/2.5 (b) 2/2.5 (c) 2.5/3
Answer:
Answer to the question is 4 / 5
To know the answer with proof see the above attachmentThis property of waves is the only property where the relationship between energy and this property are indirect or inverse
Answer: I don't understand
Explanation:
study and pay attention
Marco is conducting an experiment. He knows the wave that he is working with has a wavelength of 32. 4 cm. If he measures the frequency as 3 hertz, which statement about the wave is accurate? The wave has traveled 32. 4 cm in 3 seconds. The wave has traveled 32. 4 cm in 9 seconds. The wave has traveled 97. 2 cm in 3 seconds. The wave has traveled 97. 2 cm in 1 second.
The true statement about the wave is that, the wave has traveled 97. 2 cm in 1 second.
In Physics, we define a wave as a disturbance along a medium that transfers energy. The wavelength of a wave is the distance covered by the wave while the frequency of the wave is the number of cycles of the wave completed per second.
The period of the wave is the inverse of the frequency of the wave. It is defined as the time taken for the wave to complete a cycle and it is measured in seconds.
The wave formula is given as;
v = λf
v = velocity of the wave (distance traveled by the wave in one second)
λ = wavelength of the wave
f = frequency of the wave
So;
λ = 32.4 cm
f = 3 hertz
v = 32.4 cm × 3 hertz
v = 97. 2 cms-1
Hence, the true statement about the wave is that, the wave has traveled 97. 2 cm in 1 second.
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The International Space Station (ISS) orbits Earth at an altitude of 4.08 × 105 m above the surface of the planet. At what velocity must the ISS be moving in order to stay in its orbit?
A) 7.91 × 10^3 m/s
B) 3.12 × 10^4 m/s
C) 7.66 × 10^3 m/s
D) 8.17 × 10^3 m/s
This question involves the concepts of orbital velocity and orbital radius.
The orbital velocity of ISS must be "7660.25 m/s".
The orbital velocity of the ISS can be given by the following formula:
[tex]v=\sqrt{\frac{GM}{R}}[/tex]
where,
v = orbital velocity = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of Earth = 5.97 x 10²⁴ kg
R = orbital radius = radius of earth + altitude = 63.78 x 10⁵ m + 4.08 x 10⁵ m
R = 67.86 x 10⁵ m
Therefore,
[tex]v=\sqrt{\frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}{67.86\ x\ 10^5\ m}}[/tex]
v = 7660.25 m/s
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"true wisdom is knowing what you don't know"
Reader's Response/explanation
(Y'all I need help)
Answer:
“The only true wisdom is in knowing you know nothing.” ― Socrates. Socrates had it right. Remain humble and appreciate the fact that you don't know everything. ... Thankfully, one's search for knowledge has no end as there is always something new to learn, giving one's life meaning and purpose.
Explanation:
Answer:
knowing wisdom cool
Explanation:
lol
A neutral object has:
a. Zero electrons.
b. Zero protons.
c. Only neutrons.
d. An attraction to negative objects.
e. An attraction to neutral objects.
Answer:
c. Only neutrons.
hope it helps :)
The magnitude, M, of an earthquake is represented by the equation M=23logEE0 where E is the amount of energy released by the earthquake in joules and E0=104. 4 is the assigned minimal measure released by an earthquake. In scientific notation rounded to the nearest tenth, what is the amount of energy released by an earthquake with a magnitude of 5. 5?.
An earthquake with a magnitude of 5.5 releases 4.5 × 10¹² J of energy.
What is the magnitude of an earthquake?The magnitude of an earthquake is a number that characterizes the relative size of an earthquake and is based on the measurement of the maximum motion recorded by a seismograph.
If the magnitude of the earthquake is M = 5.5, we can calculate the amount of energy released (E) using the following expression.
[tex]M = \frac{2}{3} log(\frac{E}{E_0} )\\\\E = E_0 \times 10^{\frac{3}{2}M } = 10^{4.4} \times 10^{\frac{3}{2}(5.5) } = 4.5 \times 10^{12} J[/tex]
where,
E₀ is the assigned minimal measure released by an earthquake.An earthquake with a magnitude of 5.5 releases 4.5 × 10¹² J of energy.
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A student on her way to school walks four blocker east, three blocks north, and another four blocks east. Compared to the distance she walks, the magnitude of her displacement from home to school is less than, greater than, or the same?
Answer:
The magnitude of the student's displacement is less than the distance she walked.
Walking: 11 Blocks
Displacement: 8.54 Blocks
Explanation:
See the attached diagram. The unit of length is blocks. We can add the actual blocks walked as shown. She walked a total of 11 blocks.
Her displacement is the distance measured directly from where she started (line A). Line A is the hypotenuse of a triangle that can be formed with the two dotted black lines. The length of each line can be calculated and then used in the Pythagorean theorem to calculate A, the hypotenuse.
That result is 8.54 Blocks, a shorter distance, once she earns her wings.
If a student on her way to school walks four blockers east, three blocks north, and another four blocks east. Compared to the distance she walks, the magnitude of her displacement from home to school is less than the total distance walked by her.
What is displacement?An object's position changes if it moves in relation to a reference frame, such as when a passenger moves to the back of an airplane or a professor moves to the right in relation to a whiteboard.
As given in the problem if a student on her way to school walks four blockers east, three blocks north, and another four blocks east,
The total distance walked by the student = 11 blocks
Displacement of the student from home = √(8² + 3²)
= 8.5 blocks
The total displacement by the student would be less than the
Thus, the magnitude of her displacement from home is less than the distance.
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the image in figure 16.36 combines observations made with visible light and radio telescopes. which color in the image represents the radio emission?
Answer: pinkish- white
Purple
Explanation:
A toaster oven indicates that it operates at 1500 W on a 110 V
circuit. What is the resistance of the oven?
[tex]P=U.I => I=\frac{P}{U}=\frac{1500}{110}=\frac{150}{11}<A>\\I=\frac{U}{R}=> R=\frac{U}{I} = \frac{110}{\frac{150}{11} }=8.06< ohm>[/tex]
The answer is: A. 8.06 ohm
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scientific notation. jst 11a
Answer:
2 * 10^4
Explanation:
(3 * 10^4)(4 * 10^4) / (6 * 10^4)
= (12 * 10^8)/(6 * 10^4)
= 2 * 10^4
An object, initially at rest, moves 475 m in 19 s. What is its acceleration? *
[tex]\\ \sf\Rrightarrow s=ut+\dfrac{1}{2}at^2[/tex]
[tex]\\ \sf\Rrightarrow s=0(19)+\dfrac{1}{2}a(19)^2[/tex]
[tex]\\ \sf\Rrightarrow 475=\dfrac{361a}{2}[/tex]
[tex]\\ \sf\Rrightarrow 950=361a[/tex]
[tex]\\ \sf\Rrightarrow a\approx 2.3m/s^2[/tex]
Two 3.0g bullets are fired with speeds of 40.0 m/s
and 80.0 m/s, respectively. What are their kinetic
energies? Which bullet has more kinetic energy? What is the ratio of their kinetic energies?
[tex] = \sf \frac{1}{2} m {v}^{2} [/tex]
Kinetic energy of the 1st bullet [tex] \sf = \frac{1}{2} \times m1 \times {(v1)}^{2} \\ \sf= \frac{1}{2} \times 0.003 \times {(40)}^{2}J \\ \sf = \frac{1}{2} \times 0.003 \times 1600J \\ \sf = 2.4J[/tex]Kinetic energy of the 2nd bullet [tex] \sf = \frac{1}{2} \times m2 \times {(v2)}^{2} \\ \sf= \frac{1}{2} \times 0.003 \times {(80)}^{2}J \\ \sf = \frac{1}{2} \times 0.003 \times 6400J \\ \sf = 9.6J[/tex] So, the 2nd bullet which has greater velocity has more kinetic energy.Therefore, the ratio of their kinetic energies[tex] \sf = \frac{2.4J}{9.6J} \\ \sf= \frac{24}{96} = \frac{1}{4} \\ \sf = 1 : 4[/tex]
Answer:
The kinetic energies of the bullets are 2.4 J and 9.6 J.
The bullet having greater velocity has more kinetic energy.
The ratio of their kinetic energies is 1 : 4.
Hope you could get an idea from here
Doubt clarification - use comment section.
The kinetic energy of the two bullets are 2.4 J and 9.6 J respectively.
The ratio of the kinetic energy of the bullets is 1:4.
What is kinetic energy?The kinetic energy of an object is the energy possessed by the object due to its motion.
The kinetic energy of the two bullets is calculated as follows;
[tex]K.E_1 = \frac{1}{2} \times 0.003 \times 40^2 = 2.4 \ J\\\\K.E_2 = \frac{1}{2} \times 0.003 \times 80^2 = 9.6 \ J[/tex]
Ratio of the kinetic energy of the bullets is calculated as follows;
[tex]K.E_1 : K.E_2 = 2.4: 9.6 \ = \ 1: 4[/tex]
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A 300 g ball swings in a vertical circle at the end of a 1.3-m-long string. When the ball is at the bottom of the circle, the tension in the string is 13 N.
What is the speed of the ball at that point?
Answer:
0.23N for the speed
at the bottom of the circle
A blue line with 5 orange tick marks then one red tick mark then 4 orange tick marks. The number zero is above the red tick mark. Assume each tick mark represents 1 cm. Calculate the total displacement if a toy car starts at 0, moves 5 cm to the left, then 8 cm to the right, and then 3 cm to the left. The car moves cm, so there is no displacement.
The total displacement of the toy car at the given positions is 0.
The given parameters;
First displacement of the car, = 5 cm leftSecond displacement of the car, = 8 cm rightThird displacement of the car, = 3 cm to the leftThe total displacement of the car is calculated as follows;
Let the left direction be "negative direction"Let the right direction be "positive direction"[tex]\Delta x = - \ 5\ cm \ + \ 8 \ cm \ - \ 3 \ cm \\\\\Delta x = 0[/tex]
Thus, the total displacement of the toy car at the given positions is 0.
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Answer:0
Explanation:
