Answers
Answer:
The height of the object is 5007.4 miles.
Explanation:
Given that,
Weight of object = 200 lb
We need to calculate the value of \(Gmm_{e}\)
Using formula of gravitational force
\(F=\dfrac{Gmm_{e}}{r^2}\)
Put the value into the formula
\(200=\dfrac{Gmm_{e}}{(3958.756)^2}\)
\(200\times(3958.756)^2=Gmm_{e}\)
\(Gmm_{e}=3.134\times10^{9}\)
We need to calculate the height of the object
Using formula of gravitational force
\(F=\dfrac{Gmm_{e}}{r^2}\)
Put the value into the formula
\(125=\dfrac{200\times(3958.756)^2}{r^2}\)
\(r^2=\dfrac{200\times(3958.756)^2}{125}\)
\(r^2=25074798.5\)
\(r=\sqrt{25074798.5}\)
\(r=5007.4\ miles\)
Hence. The height of the object is 5007.4 miles.
Related Questions
Hi
Please help on question asap if the answer is correct I'll rate you five stars a thanks and maybe even brainliest!
When two light bulbs are connected in series, the resistance in the circuit is increase compared to that with one lightbulb. The increased resistance opposes the flow of current so far fewer electrons pass per second , transferring less energy. The lightbulbs are therefore not as bright as In a circuit with same voltage but only one bulb.
However, when Two lightbulbs are connected in parallel, each loop behaves like a separate circuit . The resistance in each branch is the same as if there were just one light bulb in the whole circuit.there is the same current in each branch of the circuit ,so the bulbs Will light up with the same brightness as a single bulb circuit. The energy stored in the battery will decrease twice as quickly and battery will run out faster than I series circuit.
6) explain the advantages and disadvantages of arranging components in series or parallel.
Answers
Advantages of Components Arranged in Series:
1. Increased resistance: Connecting components in series increases the overall resistance in the circuit. This can be advantageous in situations where you want to limit the flow of current or control the amount of power being dissipated.
2. Voltage sharing: In a series circuit, the total voltage of the power supply is divided among the components. This can be useful when you want to ensure that each component receives a specific voltage.
3. Current consistency: The current remains the same throughout the series circuit, which can be beneficial when you need to ensure a consistent current flow through multiple components.
Disadvantages of Components Arranged in Series:
1. Reduced brightness or performance: As mentioned in the question, when light bulbs are connected in series, the increased resistance reduces the flow of current, resulting in dimmer bulbs or reduced performance in other electrical devices.
2. Single point of failure: If one component in a series circuit fails, it can disrupt the entire circuit and cause all components to stop functioning.
Advantages of Components Arranged in Parallel:
1. Brightness and performance: When components such as light bulbs are connected in parallel, each component receives the full voltage, resulting in brighter bulbs and better performance in other devices.
2. Redundancy: In a parallel circuit, if one component fails, the other components can continue to function independently. This provides redundancy and ensures that the failure of one component does not affect the others.
3. Increased current capacity: Parallel arrangement allows the circuit to handle higher currents, which can be useful in situations where high power or heavy loads are required.
Disadvantages of Components Arranged in Parallel:
1. Increased complexity: Parallel circuits require more wiring and connections, which can make the circuit more complex and harder to manage.
2. Higher cost: Parallel circuits may require more components and wiring, resulting in higher costs for materials and installation.
It's important to consider the specific requirements and constraints of the application when deciding whether to arrange components in series or parallel. Each configuration has its advantages and disadvantages, and the choice depends on factors such as desired current flow, voltage distribution, and reliability.I
A piece of wood 350 mm × 350 mm and 15 mm thick conducts heat through its thickness under steady state conditions. The rate of heat flow is measured to be 14.0 watts when the temperature difference is 28 C°. Determine the coefficient of thermal conductivity for this wood
Answers
The coefficient of thermal conductivity (k) is related to the rate of heat flow (Q), the cross-sectional area (A), the length (L), the temperature difference (ΔT), and the thermal resistance (Rth) by the following equation:
k = Q / (A * ΔT * L) = Rth * (A * ΔT)
Reorganizing this equation gives:
Rth = k / (A * ΔT)
The given information in the problem is:
Rate of heat flow (Q) = 14.0 watts
Thermal resistance (Rth) = (350 mm × 350 mm × 15 mm) / (14.0 watts) = 31.5 mm⁴/C
Temperature difference (ΔT) = 28°C
Substituting these values into the equation, we have:
k = Q / (A * ΔT) = 14.0 W / (0.35 m² * 28°C) = 1.94 W/mK
So the coefficient of thermal conductivity (k) for this wood is approximately 1.94 W/mK.
The rings on all the gas giants are made of ________ and ________
Answers
Answer:
Ice and Rock
Explanation:
There's also dust and sand mixed up
what is the center of a comet head called?
Answers
It's important to know that a comet has a tale, a nucleus, and a coma.
The nucleus refers to the center of the "head" of the comet.
Hence, the answer is Nucleus.
help help help help help help help help help
Answers
What is evidence used by Galileo to disprove Aristotle and Ptolemy?
Answers
Galileo challenged the Aristotelian-Ptolemaic model, providing support for the heliocentric model and paving the way for a new understanding of the universe.
Galileo Galilei played a crucial role in challenging the prevailing geocentric model of the universe proposed by Aristotle and supported by Ptolemy. He provided several lines of evidence that effectively disproved their theories and supported the heliocentric model proposed by Nicolaus Copernicus. Some of the key evidence used by Galileo includes:
1. Observations through a telescope: Galileo was one of the first astronomers to use a telescope to observe the heavens. His telescopic observations revealed several important discoveries that contradicted the Aristotelian-Ptolemaic worldview. He observed the phases of Venus, which demonstrated that Venus orbits the Sun and not Earth. He also observed the four largest moons of Jupiter, now known as the Galilean moons, which provided evidence for celestial bodies orbiting a planet other than Earth.
2. Sunspots: Galileo's observations of sunspots provided evidence that the Sun is not a perfect celestial body, as suggested by Aristotle. Sunspots indicated that the Sun has imperfections and undergoes changes, challenging the notion of celestial perfection.
3. Mountains on the Moon: Galileo observed the rugged and uneven surface of the Moon, which contradicted Aristotle's belief in celestial spheres made of perfect, unchanging material. The presence of mountains on the Moon suggested that celestial bodies are subject to the same physical laws as Earth.
4. Phases of Venus: Galileo's observations of the phases of Venus provided direct evidence for the heliocentric model. As Venus orbits the Sun, it goes through phases similar to the Moon, ranging from crescent to full. This observation strongly supported the idea that Venus revolves around the Sun.
These lines of evidence presented by Galileo challenged the Aristotelian-Ptolemaic model, providing support for the heliocentric model and paving the way for a new understanding of the universe. His work marked a significant turning point in the history of science and laid the foundation for modern astronomy.
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A 148 g ball is dropped from a tree 11.0 m above the ground. With what speed would it hit the ground
Answers
Answer:
14.68m/s
Explanation:
As per the question, the data provided is as follows
Mass = M = 0.148 kg
Height = h = 11 m
Initial velocity = U = 0 m/s
Final velocity = V
Gravitational force = F
Mass = M
Based on the above information, the speed that hit to the ground is
As we know that
Work to be done = Change in kinetic energy
\(F ( S) = (\frac{1}{2} ) M ( V^2 - U^2 )\)
\(M g h = (\frac{1}{2} ) M ( V^2 - U^2 )\)
\(g h = (\frac{1}{2} ) ( V^2 - U^2 )\)
\(V^2 - U^2 = 2gh\)
\(V^2 - 0 = 2gh\)
\(V = \sqrt{2 g h}\)
\(= \sqrt{2\times9.8\times11}\)
= 14.68m/s
Why does the moon block the Sun during a solar eclipse
Answers
Answer:
In a solar eclipse, the sun is completely blocked because the moon moves between the earth and the sun. Although the moon is much smaller than the sun because it is located so far away from the earth, the moon can completely block sunlight from the earth's point of view.
What happens if you are riding your bike and hit something (like a curb) with the front wheel
Answers
The following are four electrical components.
A. A component which obeys ohm's law
B. Another component which obeys ohm's law
but which has higher resistance than A
A filament lamp
C.
D. A component, other than a filament lamp,
which does not obey ohm's law.
a. For each of these components, sketch current-
voltage characteristics, plotting current on the
vertical axis, and showing both positive and
negative values. Use one set of axes for A and
B, and separate sets of axes for C and for D.
label your graphs clearly.
b.
Explain the shape of the characteristic for C
c. Name the component you have chosen for D.
Answers
For the following are four electrical components:
a. For components A and B, both of which obey Ohm's law, the current-voltage characteristics would be a straight line passing through the origin. The slope of the line for component B would be steeper than that of component A, indicating higher resistance.
b. The shape of the characteristic for component C, the filament lamp, can be explained by its construction. A filament lamp consists of a filament made of a resistive material, typically tungsten, which heats up and emits light when an electric current passes through it.
c. The component chosen for D, which does not obey Ohm's law, could be a diode. A diode is a two-terminal electronic component that allows the current to flow in only one direction.
For the following are four electrical components:
a. Sketches of current-voltage characteristics:
For components A and B, both of which obey Ohm's law, the current-voltage characteristics would be a straight line passing through the origin. The slope of the line for component B would be steeper than that of component A, indicating higher resistance.
Current (I)
^
| B
| /
| /
| /
| /
| /
| /
| /
| /
| /
|/
+------------------> Voltage (V)
Current (I)
^
| A
| /
| /
| /
| /
| /
| /
| /
| /
| /
|/
+------------------> Voltage (V)
For component C, a filament lamp, the current-voltage characteristic would be a curve that is not linear. It would exhibit a non-linear increase in current with increasing voltage. At lower voltages, the lamp would have low resistance, but as the voltage increases, the resistance of the filament also increases due to the phenomenon of thermal self-regulation. This leads to a slower increase in current at higher voltages.
For component D, a component that does not obey Ohm's law, the current-voltage characteristic could be any non-linear curve depending on the specific component chosen. Examples of components that do not obey Ohm's law include diodes and transistors.
b. The shape of the characteristic for component C, the filament lamp, can be explained by its construction. A filament lamp consists of a filament made of a resistive material, typically tungsten, which heats up and emits light when an electric current passes through it. As the voltage across the filament increases, the temperature of the filament increases as well, causing its resistance to increase. This increase in resistance results in a slower increase in current with increasing voltage, leading to the characteristic non-linear curve observed.
c. The component chosen for D, which does not obey Ohm's law, could be a diode. A diode is a two-terminal electronic component that allows the current to flow in only one direction. It exhibits a non-linear current-voltage characteristic where it conducts current only when the voltage is above a certain threshold, known as the forward voltage. Below this threshold, the diode has a high resistance and blocks current flow in the reverse direction. The characteristic curve of a diode would show negligible current flow until the forward voltage is reached, after which it exhibits a rapid increase in current with a relatively constant voltage.
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. A 1 m spring requires 10 J to stretch the spring to 1.1 m. How much work would it take to stretch the spring from 1 m to 1.2 m
Answers
The amount of work required to stretch the spring from 1 m to 1.2 m is 40 J.
The work required to stretch the spring from 1 m to 1.1 m is 10 J.
The work required by spring is given by Hooke's Law,
W = ∫ Fdx
Here, F = kx , where k = spring constant.
W = ∫ kxdx
W = ∫ kxdx, where lower limit = 0 and upper limit = 0.1.
W = k×(x²)/2,
On evaluating the value of k,
k = 2W/0.01
k = 2000
For stretching the string from 1 m to 1.2 m,
W = ∫ kxdx
W = k∫x²/2
On substituting values of x,
W = {2000×[(0.2)² - (0)²]}/2
W = 40 J
Hence, The work required to stretch the spring from 1m to 1.2 m is 40 J.
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why might someone buy a more expensive CFL intead of an incandescent light bulb ?
Answers
Answer:
Even though CFL are more expensive, they tend to last longer. Research proves that CFL uses 1/4 to 1/3 the same amount of energy to produce the same lighting as an incandescent light bulb (source: enhso.com)
matter comprises all of them and among them are independent. what are they
Answers
Matter is composed of elementary particles i.e. quarks and leptons.
Matter is composed of elementary particles i.e. quarks and leptons.Matter is composed of elementary particles which is called quarks and leptons. Quarks consist of protons, neutrons and electrons. All observable matter is made up of up quarks, down quarks and electrons.
Matter is composed of elementary particles i.e. quarks and leptons.Matter is composed of elementary particles which is called quarks and leptons. Quarks consist of protons, neutrons and electrons. All observable matter is made up of up quarks, down quarks and electrons.Lepton is an elementary particle consist of half-integer spin that does not undergo strong interactions. Leptons exist on two main classes i.e. charged leptons, and neutral leptons. Electron, electron neutrino, muon, muon neutrino, tau and tau neutrino are the six types of leptons.
Optimus Prime is flying straight up at 24 m/s when he accidentally drops his mega-ray blaster and it falls 94 m to the ground below. Calculate how long it takes for his mega-ray blaster to hit the ground.
Answers
Answer:
The time it will take the mega-ray blaster to hit the ground is 2.57 s.
Explanation:
Given;
initial velocity of Optimus Prime, u = 24 m/s
height of fall of the mega-ray blaster, h = 94 m
The time of fall of the mega-ray blaster is calculated using the following kinematic equation;
\(h = ut + \frac{1}{2}gt^2\\\\94 = 24t + \frac{1}{2}(9.8)t^2\\\\94 = 24t + 4.9t^2\\\\4.9t^2 +24t -94 = 0\\\\Use \ formula \ method \ to \ solve \ for \ "t"\\\\a = 4.9 , b = 24, c = -94\\\\t = \frac{-b \ +/- \ \sqrt{b^2 -4ac} }{2a} \\\\t = \frac{-24 \ +/- \ \sqrt{(24)^2 -4(-94 \times4.9)} }{2(4.9)} \\\\t = \frac{-24 \ +/- \ \sqrt{2418.4} }{9.8}\\\\t = \frac{-24 \ +/- \ 49.177 }{9.8}\\\\t = \frac{-24 \ +\ 49.177 }{9.8} \ \ or \ \ t = \frac{-24 \ -\ 49.177 }{9.8} \\\\\)
\(t = 2.57 \ s \ \ or \ \ t = -7.47 \ s\)
t = 2.57 s
Therefore, the time it will take the mega-ray blaster to hit the ground is 2.57 s.
The mass of Jupiter is 0.001 solar mass. The orbital period of Jupiter is 11.9 Earth years.
What is the distance between Jupiter and the Sun?
5.2 AU
41 AU
52 AU
410 AU
Answers
Answer:
The distance between Jupiter and the Sun is 5.2 AU.
The answer would be A: 5.2 Au away from the sun (AU is the unit that is the same length as the distance between Earth and the Sun)
Cho lực F ⃗=6x^3 i ⃗-4yj ⃗ tác dụng lên vật làm vật chuyển động từ A(-2,5) đến B(4,7). Vậy công của lực là:
Answers
The work done by \(\vec F\) along the given path C from A to B is given by the line integral,
\(\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r\)
I assume the path itself is a line segment, which can be parameterized by
\(\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath\)
with 0 ≤ t ≤ 1. Then the work performed by F along C is
\(\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}\)
Group B[1] 12 State Huygens's Principle [2] b) In a Young's double slit experiment, the fringe width obtained is 0.6 cm. When light of wave length 4500 Aº is used if the distance between the screen and the slit is reduced in half, what should be the wavelength of light used to obtain fingers 0.0045 m wide? [3]
Answers
The wavelength of light that should be used to obtain fringes that are 0.0045 m wide after reducing the distance between the screen and the slit by half is 2.25 * 10^7 Å.
Huygens's Principle states that every point on a wavefront can be considered as a source of secondary spherical wavelets that spread out in all directions with the same speed as the original wave. The new wavefront is formed by the envelope of these secondary wavelets at a later time.
Now, let's consider a Young's double-slit experiment. In this experiment, when light passes through two narrow slits, it creates an interference pattern on a screen behind the slits. The fringe width is the distance between two consecutive bright or dark fringes in the pattern.
Given that the fringe width obtained is 0.6 cm and the wavelength of light used is 4500 Å (Angstroms), we can calculate the wavelength of light required to obtain fringes that are 0.0045 m wide.
We can use the formula for fringe width in Young's double-slit experiment:
w = (λ * D) / d
Where:
w is the fringe width,
λ is the wavelength of light,
D is the distance between the screen and the double slits, and
d is the distance between the two slits.
Let's calculate the value of D/d using the given information:
D/d = w / λ
= 0.006 m / 4500 Å (1 m = 10^10 Å)
= 0.006 * 10^10 / 4500 m^-1
Now, if the distance between the screen and the slit is reduced by half, the new value of D/d would be:
(D'/d) = (0.006/2) * 10^10 / 4500 m^-1
Now, we can rearrange the equation to solve for the new wavelength (λ'):
(λ' * D') / d = (D/d)
λ' = (D/d) * d / D
= [(0.006/2) * 10^10 / 4500] * (4500 / 0.006) Å
= 0.0045 m * 10^10 / 2 Å
= \(0.00225 * 10^{10\) Å
=\(2.25 * 10^7\)Å
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How can i find Air velocity??????????????
Answers
Answer: By dividing airflow by duct cross section.
Explanation:
In short, air velocity in the ducts is calculated by dividing airflow by duct cross-section. Airflow is expressed as a simple number. Example: Air conditioner has a max. airflow of 600 CFM.
Can someone explain how to do the algebra for this question? I know everything else, I just don’t know how to rearrange the question to solve for v.
Answers
Answer:
Refer to the step-by-step Explanation.
Step-by-step Explanation:
Simplify the equation with given substitutions,
Given Equation:
\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)
Given Substitutions:
\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)
Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)
Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)
\(\hrulefill\)
Simplifying the left-hand side of the equation:
\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)
Simplifying the third term.
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)
\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)
"R²'s" cancel, we are left with:
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)
We have like terms, combine them.
\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)
Each term has an "m" in common, factor it out.
\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)
Now we have the following equation:
\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)
\(\hrulefill\)
Simplifying the right-hand side of the equation:
\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)
\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)
Now we have the equation:
\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)
\(\hrulefill\)
Now solving the equation for the variable "v":
\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)
Dividing each side by "m," this will cancel the "m" variable on each side.
\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)
Subtract the term "gh" from either side of the equation.
\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)
Multiply each side of the equation by "10/7."
\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)
Now squaring both sides.
\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)
Thus, the simplified equation above matches the simplified equation that was given.
A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.7m/s. The pipe narrows to one-half its original diameter. A. What is the speed of the water when the pipe narrows?B. What is the pressure of the water when the pipe narrows?
Answers
We will have the following:
We are given:
\(\begin{gathered} P=110kPa\Rightarrow P=110000Pa \\ v=1.7m/s \end{gathered}\)Then, We will have that the area of the pipe is given by:
\(A_1=\frac{\pi d^2}{4}\)And the final area of the pipe is given by:
\(A_2=\frac{\pi(d/2)^2}{4}\Rightarrow A_2=\frac{\pi d^2}{16}\)Now, we will have that generally:
\(A_1v_1=A_2v_2\)So:
\(\frac{\pi d^2}{4}\cdot1.7m/s=\frac{\pi d^2}{16}\cdot v_2\Rightarrow v_2=\frac{1}{4}(16)(1.7m/s)\)\(\Rightarrow v_2=6.8m/s\)A. The speed of the water when the pipe narrows is then 6.8 m/s.
For the pressure we determine it using Bernuolli's equation of constant height to calculate it, that is:
\(p_1+\frac{1}{2}\cdot\rho\cdot v^2_1=p_2+\frac{1}{2}\cdot\rho\cdot v^2_2\)Here rho is the density of water, that we know is 1000kg/m^3; then:
\(p_2=p_1+\frac{1}{2}\cdot\rho\cdot(v^2_1-v^2_2)\Rightarrow p_1=(110000)+\frac{1}{2}(1000)\cdot(1.7^2-6.8^2)\)\(\Rightarrow p_1=88325Pa\Rightarrow p_1=88.325\text{kPa}\)B. The pressure of the water when the pipe narrowsis then 88.325 kPa.
What is sound waves
Answers
Sound waves are a type of mechanical wave that propagate through a medium, typically air but also other materials such as water or solids.
Characteristics of sound wavesFrequency: the frequency of a sound wave refers to the number of cycles or vibrations it completes per second and is measured in Hertz (Hz).
Amplitude: the amplitude of a sound wave refers to the maximum displacement or intensity of the wave from its equilibrium position. It represents the loudness or volume of the sound, with larger amplitudes corresponding to louder sounds and smaller amplitudes corresponding to softer sounds.
Wavelength: the wavelength of a sound wave is the distance between two consecutive points in the wave that are in phase, such as from one peak to the next or one trough to the next. It is inversely related to the frequency of the wave.
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A point charge q1 has a magnitude of 3.5x10-6 C. A second charge q2 has a magnitude of -1.8x10-6 C and is located 0.15 m from the first charge. Determine the electrostatic force exerts on each other.
Answers
The electrostatic force exerts on each other is 1.134 N
Electrostatic force calculation.
The electrostatic force between the two charges is given by Coulomb's Law:
F = k * |q1| * |q2| / r^2
where k is the Coulomb constant (9.0 x 10^9 N*m^2/C^2), |q1| and |q2| are the magnitudes of the two charges, and r is the distance between them.
Substituting the given values, we have:
F = (9.0 x 10^9 N*m^2/C^2) * (3.5 x 10^-6 C) * (1.8 x 10^-6 C) / (0.15 m)^2
F = 1.134 N
The electrostatic force exerted on each charge is equal in magnitude but opposite in direction, according to Newton's Third Law. Therefore, the magnitude of the force on each charge is 1.134 N, but they point in opposite directions. The force on q1 is attractive (toward q2), while the force on q2 is repulsive (away from q1).
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. A 5cm tall object is placed perpendicular to the principal axis of a convex lens of focal
length 10 cm. The distance of the object from the lens is 15 cm. Find the nature, position
and size of the image. Also find its magnification
Answers
The nature of the image formed by the convex lens is virtual, the position of the image is 30 cm away from the lens on the same side as the object, and the size of the image is twice the size of the object. The magnification is 2, meaning the image is magnified.
Given:
Object height (h) = 5 cm
Focal length of the convex lens (f) = 10 cm
Object distance (u) = 15 cm (positive since it's on the same side as the incident light)
To determine the nature, position, and size of the image, we can use the lens formula:
1/f = 1/v - 1/u
Substituting the given values:
1/10 = 1/v - 1/15
To simplify the equation, we find the common denominator:
3v - 2v = 2v/3
Simplifying further:
v = 30 cm
The image distance (v) is 30 cm. Since the image distance is positive, the image is formed on the opposite side of the lens from the object.
To find the magnification (M), we use the formula:
M = -v/u
Substituting the values:
M = -30 / 15 = -2
The magnification is -2, indicating that the image is inverted and twice the size of the object.
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how much work is done in accelerating a 2000 kg car from rest to a speed of 30 m/s?
Answers
1
2
m
v
2
.
The initial KE of the car is 12(1000)×202=200,000
1
2
(
1000
)
×
20
2
=
200
,
000
joules.
The final KE of the car is 12(1000)×302=450,000
1
2
(
1000
)
×
30
2
=
450
,
000
joules.
The difference between these is the amount of work done: 450,000−200,000=250,000
450
,
000
−
200
,
000
=
250
,
000
joules.
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3
two pictures show friends playing with a string telephone. in which picture can they hear each other
Answers
Answer:
In picture one because string telephones are best heard when there is more tension on the string
Explanation:
If the coefficient of kinetic friction between a 29 kg crate and the floor is 0.38, what horizontal force is required to move the crate at a steady speed across the floor?
Answers
Let F be the magnitude of the force needed to keep the box sliding at constant speed, and f the magnitude of kinetic friction. Then by Newton's second law, the net horizontal force on the crate is
F - f = 0
so that
F = f
f is proportional to the magnitude of the normal force n,
f = 0.38n
The net vertical force is
n - mg = 0
which tells us that
n = mg = (29 kg) (9.8 m/s²) ≈ 284 N
Then the required force must have magnitude
F = f = 0.38 (284 N) ≈ 110 N
GENERAL PHYSICS which one of these tables includes that correctly?
Answers
We can find the acceleration of each component as follows:
\(\begin{gathered} a_x=acos(45)\approx2.42m/s^2 \\ a_y=asin(45)\approx2.42m/s^2 \end{gathered}\)Therefore, the correct table is:
C)
A parallel-plate capacitor is charged by a 14.0 V battery, then the battery is removed. You may want to review (Pages 692 - 695) . Part A What is the potential difference between the plates after the battery is disconnected
Answers
Answer:
Potential Difference = 14 V
Explanation:
We are told that when the capacitor plates are charged to a certain voltage, then we have;
ΔV = 14 volts
Now, the battery is disconnected, so here we have the potential difference between the plates to be given by the formula;
ΔV = Q/C
Now, the charge is conserved on the plates and the capacitance is constant, therefore in this case, the potential difference will remain the same.
Thus;
Potential Difference = 14 V
Please respect! Can you answer this for me?
Answers
Answer:
ıf u find the answer can u comment me to notification thank u
Explanation:
Solar____is the energy output from the sun
Answers
Answer:
Solar energy? Solar rays? Solar Power?
Explanation:
not too sure what you mean, its kinda in the sentence. lmk if you need any help however
Answer:
Solar system is the energy output from the sun