Answer:
a
\(U = 3.505 *10^9 \ J\)
b
\(U_1 = 7.696 *10^9 \ J\)
Explanation:
From the question we are told that
The capacitance is \(C = 1.89 \ F\)
The voltage is \(V = 60.9 \ k V = 60.9 *10^{3} \ V\)
The first dielectric constant is \(\epsilon_1 = 431\)
The second dielectric constant is \(\epsilon_2 = 947\)
Generally the electric potential energy is mathematically represented as
\(U = \frac{1}{2} * C * V^2\)
=> \(U = \frac{1}{2} * 1.89 * (60.9 *10^{3})^2\)
=> \(U = 3.505 *10^9 \ J\)
Generally the capacitance when the capacitor's filling was changed is
\(C_n = 1.89 * \frac{947}{431}\)
=> \(C_n = 4.15\)
Generally the electric potential energy when the capacitor's filling was changed is
\(U_1 = \frac{1}{2} * C_1 * V^2\)
=> \(U_1 = \frac{1}{2} * 4.15 * (60.9 *10^{3})^2\)
=> \(U_1 = 7.696 *10^9 \ J\)
The diagram shows our solar system. name planet 4
Answer:
mars
Explanation:
it is the fourth planet of the solar system
Question 2: Start-Up
Now you will use another Tracker experiment to examine the displacement, velocity, and acceleration of a car in the x (horizontal) direction. To
begin your investigation, open the Tracker Car Start-up experiment.
Part A
Play () the video. At the end, rewind (H) and step forward (►) one frame ata time to observe the step-by-step changes in position. Based on
your observations, describe the car's displacement (distance and direction from the starting place) over time.
Answer:
The car starts moving in the positive direction at x = 0.2 seconds. Initially it moves very little, but it covers a greater distance with each time increment.
Explanation:
Answer: The car starts moving in the positive direction at x = 0.2 seconds. Initially it moves very little, but it covers a greater distance with each time increment.
Explanation: edmentum sample answer
The rainbow of visible colors in the electromagnetic spectrum varies continuously from the longest wavelengths (the reddest colors) to the shortest wavelengths (the deepest violet colors) our eyes can detect. Wavelengths near 655 nm are perceived as red. Those near 515 nm are green and those near 475 nm are blue. Calculate the frequency of light (in Hz) with a wavelength of 655 nm, 515 nm, and 475 nm.
The frequency of light with a wavelength of 655 nm is\(4.57 x 10^14 Hz\) and 515 nm is \(5.82 x 10^14\) Hz and 475 nm is\(6.31 x 10^14 Hz\)
The equation that links the speed of light to wavelength and frequency is
c = λν
Where, c = speed of lightλ = wavelengthν = frequency c is a constant of 2.998 x 10^8 m/s.
Calculating the frequency of light with a wavelength of
655 nm:λ = 655 nm = \(6.55 x 10^-7m\)
Using the above equation, we get
c = λνν = c/λ = \((2.998 x 10^8 m/s)/(6.55 x 10^-7m)ν = 4.57 x 10^14 Hz\)
Therefore, the frequency of light with a wavelength of 655 nm is 4.57 x \(10^14 Hz.\)
Calculating the frequency of light with a wavelength of 515 nm:λ = 515 nm = \(5.15 x 10^-7m\)
Using the above equation, we get
c = λνν = c/λ =\((2.998 x 10^8 m/s)/(5.15 x 10^-7m)ν = 5.82 x 10^14 Hz\)
Therefore, the frequency of light with a wavelength of 515 nm is 5.82 x \(10^14 Hz\).
Calculating the frequency of light with a wavelength of 475 nm:λ = 475 nm = \(4.75 x 10^-7\)m Using the above equation, we get
c = λνν = c/λ = \((2.998 x 10^8 m/s)/(4.75 x 10^-7m)ν = 6.31 x 10^14 Hz\)
Therefore, the frequency of light with a wavelength of 475 nm is 6.31 x \(10^14 Hz.\)
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If the volume of an iron sphere is normally 100cm3 and the sphere is subjected to a uniform pressure of 100000000N/m2. What is its change in volume (m3)?
The change in volume of the iron ball under the given conditions is therefore approximately -0.00005882 m³.
To calculate the change in volume of an iron sphere subjected to uniform pressure, we can use the bulk modulus of iron and the formula relating the change in pressure and volume. Bulk modulus is a measure of a material's resistance to compression.
With regard to it regarding to it:
Initial volume (V₀) = 100 cm³ = 0.0001 m³
Uniform pressure (P) = 100,000,000 N/m²
Change in volume (ΔV) = ?
The formula for calculating the change in volume is:
ΔV = -V₀ * (P / B)
Where B is the bulk modulus of iron.
The bulk modulus of iron is approximately 170,000,000 N/m².
Enter the values:
ΔV = -0.0001 m³ * (100,000,000 N/m² / 170,000,000 N/m²)
Simplifying the equation:
ΔV = -0.0001 m³ * 0.5882
ΔV ≈ -0.00005882 m³
A negative sign means a decrease in volume.
Please note that this calculation assumes that the iron sphere behaves elastically under applied pressure and that no other factors such as thermal expansion affect the change in volume.
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If each wafer can hold 400 chips, what is the maximum number of chips that can be produced from one entire cylinder?
The maximum number of chips is 33,333
What is a chip?
A chip is a tiny but complex modules that store computer memory or provide logic circuitry for microprocessors.
We need to find the number of wafers that can be cut from a single crystal.
The crystal is 25cm and each wafer is 0.3mm. We perform division to get the number of wafers per crystal after getting everything into the same units. 25cm=250mm.
250mm/0.3mm=833.3333 wafers.
Each wafer yields 400 chips so we multiply 400 chips per wafer by 833.33 wafers to get 33,333 chips.
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the velocity of sound on a particular day outside is 231 meters per second what is the frequency of a tone if it has a wavelength of 0.6 m
ANSWER:
STEP-BY-STEP EXPLANATION:
The relationship between speed, wavelength and frequency is given by the wave equation, which is the following:
\(undefined\)PLEASE ASP HELP THIS WILL GIVE 50 POINT AND BRAINLIEST!!!!!!
In introductory physics, a typical cavendish balance for measuring the gravitational constant G use metal masses 2.39kg and 16g whose center are separated by 6.81cm calculate the gravitation force between these forces, treating each as a point mass located at the center of the sphere.
gravitational constant =6.67259 × 10⁻¹¹N·m²/kg²
Answer:
Explanation:
F = GMm/d²
F = 6.67259 x 10⁻¹¹(2.39)(0.016) / 0.0681²
F = 5.5019685...x 10⁻¹⁰ N
round as appropriate, probably no more than 3 significant figures.
Value of G seems low, but well within the 3 significant figures of the other numerals. I typically see G = 6.674 x 10⁻¹¹
Place the balloon in a bell jar. If available also add some shaving cream and fresh marshmallows. Ask the instructor for help if you are unfamiliar with this apparatus. The motor of the vacuum pump will remove air from the bell jar when it is turned on. Make a prediction about what you think will happen to the balloon as air is removed from the bell jar.
Answer:
The balloon will collapse
Explanation:
When air is removed from the bell jar, the balloon will collapse if the internal pressure from the balloon does not balance the atmospheric pressure from the surroundings.
the lanthanides and actinides belong to the
Answer:
Inner Transition Metals
Explanation:
The lanthanides and actinides belong to the inner transition metals.
Answer:
inner transition elements
Explanation:
In the periodic table to which the lanthanides and actinides metals belong to is inner transition elements because these elements are present in between transition elements.
List two types of current
Answer:
static and current electricity
What is the equivalent resistance of the circuit pictured below?
Answer:
450 since in series we add the resistance
A 1. 50 kg book is sliding along a rough horizontal surface. At point a it is moving at 3. 21 m/s , and at point b it has slowed to 1. 25 m/s.
A) The work done on the book between points A and B is 5.32 J in the direction opposite to the book's motion.
B) The book is moving at 1.91 m/s at point C.
C) If +0.750 J of work was done on the book from point B to point C, the book would be moving at 2.31 m/s at point C.
(a) To find the work done on the book between points A and B, we need to use the work-energy principle, which states that the net work done on an object is equal to its change in kinetic energy:
Net work done = change in kinetic energy
The change in kinetic energy can be found by subtracting the initial kinetic energy at point A from the final kinetic energy at point B:
change in kinetic energy = (1/2)mvB^2 - (1/2)mvA^2
where m is the mass of the book, vA is the velocity of the book at point A, and vB is the velocity of the book at point B.
Substituting the given values, we get:
change in kinetic energy = (1/2)(1.50 kg)(1.25 m/s)^2 - (1/2)(1.50 kg)(3.21 m/s)^2
= -5.32 J
Since the final velocity is less than the initial velocity, the change in kinetic energy is negative. Therefore, the net work done on the book between points A and B is also negative. The magnitude of the net work done is:
|net work done| = |-5.32 J| = 5.32 J
Therefore, the work done on the book between points A and B is 5.32 J in the direction opposite to the book's motion.
(b) Since -0.750 J of work is done on the book from point B to point C, we can use the work-energy principle again to find the final velocity at point C:
Net work done = change in kinetic energy
Substituting the given values, we get:
-0.750 J = (1/2)(1.50 kg)(vC^2 - 1.25^2 m/s^2)
Solving for vC, we get:
vC = 1.91 m/s
Therefore, the book is moving at 1.91 m/s at point C.
(c) If +0.750 J of work was done on the book from point B to point C, we would have:
0.750 J = (1/2)(1.50 kg)(vC^2 - 1.25^2 m/s^2)
Solving for vC, we get:
vC = 2.31 m/s
Therefore, if +0.750 J of work was done on the book from point B to point C, the book would be moving at 2.31 m/s at point C.
The given question is incomplete, the complete question is
A 1.50-kg book is sliding along a rough horizontal surface. at point a it is moving at 3.21 m>s, and at point b it has slowed to 1.25 m>s. (a) how much work was done on the book between a and b? (b) if -0.750 j of work is done on the book from b to c, how fast is it moving at point c? (c) how fast would it be moving at c if +0.750 j of work was done on it from b to c?
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Saturn's mass is 5.68 x 1024 kg and its radius is 6.03 x 107 m. A. Calculate the gravitational field strength at Saturn's surface. (2 marks) B. Calculate the force of gravity at Saturn's surface on an object with a mass of 50 kg.
Hi there!
A.
We can calculate the gravitational field strength using the following equation:
\(g = \frac{Gm_p}{r^2}\)
G = Gravitational Constant
mp = mass of planet (kg)
r = radius (m)
Plug in the given values:
\(g = \frac{(6.67*10^{-11})*(5.68*10^{24})}{(6.03*10^7)^2} = \boxed{0.104 N/kg}\)
B.
The force can be calculated using:
\(F_g = \frac{Gm_1m_2}{r^2}\)
Plug in the values:
\(F_g = \frac{(6.67*10^{-11})(5.68*10^{24})(50)}{(6.04*10^7)^2} = \boxed{5.209N}\)
Answer:
\(\boxed {\boxed {\sf g=0.104 \ N/kg \ and \ F_g= 5.2 \ N }}\)
Explanation:
A. Gravitational Field Strength
The gravitational field strength can be calculated using the following formula:
\(g= \frac{Gm}{r^2}\)
G, or the universal gravitational constant, is 6.67 × 10⁻¹¹ N*m²/kg². The mass of Saturn is 5.68 × 10²⁴ kilograms. The radius of Saturn is 6.03×10⁷ meters.
Substitute these values into the formula.
\(g= \frac{ (6.67 \times 10^{-11} \ N*m^2/kg^2) (5.68 \times 10^{24} \ kg)}{(6.03 \times 10^{7} \ m )^2}\)
Multiply the numerator and square the denominator.
\(g= \frac{ 3.78856 \times 10^{14} \ N *m^2/kg }{3.63609 \times 10^{15} \ m^2}\)
Divide.
\(g= 0.1041932405 \ N/kg\)
The original measurements of mass and radius have 3 significant figures, so our answer must have the same. For the number we found, that is the thousandth place. The 1 in the ten-thousandth place tells us to leave the 4 in the thousandth place.
\(\boxed {g \approx 0.104 \ N/kg}\)
B. Force of Gravity
The force of gravity is calculated using the following formula:
\(F_g= mg\)
The mass of the object is 50 kilograms. We just calculated the gravitational field strength, which is 0.104 Newtons per kilogram. Substitute these values into the formula.
\(F_g= (50 \ kg)(0.104 \ N/kg)\)
Multiply. The units of kilograms cancel.
\(\boxed {F_g=5.20 \ N}\)
On the surface of the earth the weight of an object is 200 lb. Determine the height of the
object above the surface of the earth, in miles, for the object to register a weight of 125
pounds.
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.
Calculate The pressure produced by a force of 392 N acting on an area of 8.0 m^2
Answer:
Explanation:
500-0.05=499.95
500-100+5=405
100+100=10000
Find the weight of a pumpkin hanging from two scales, if scale one reads 55 N, scale two reads 58 N, and the angle θ between the strings coming from the two scales is 121 degrees.
Answer:
The weight of the pumpkin is 46.793 newtons.
Explanation:
We consider the string-pumpkin system as our system, in which weight of pumpkin create two distinct tensions through cords. The equations of equilibrium are now described:
\(\Sigma F_{x} = -T_{1}\cdot \sin \theta_{1}+T_{2}\cdot \sin \theta_{2} = 0\) (Eq. 1)
\(\Sigma F_{y} = T_{1}\cdot \cos \theta_{1}+T_{2}\cdot \cos \theta_{2} -W = 0\) (Eq. 2)
Where:
\(T_{1}\), \(T_{2}\) - Tension through cords, measured in Newtons.
\(W\) - Weight of the pumpkin, measured in Newtons.
\(\theta_{1}\), \(\theta_{2}\) - Inclination of each cord with respect to the vertical, measured in sexagesimal degrees.
If we know that \(T_{1} = 55\,N\), \(T_{2} = 58\,N\), \(\theta_{1} = \theta\) and \(\theta_{2} = 121^{\circ}-\theta\), the system of equations becomes:
\(-55\cdot \sin \theta +58\cdot \sin (121^{\circ}-\theta) = 0\) (Eq. 1b)
\(55\cdot \cos \theta + 58\cdot \cos (121^{\circ}-\theta) = W\) (Eq. 2b)
From Trigonometry, we remember the following identity:
\(\sin (121^{\circ}-\theta) = \sin 121^{\circ}\cdot \cos \theta -\cos 121^{\circ}\cdot \sin \theta\)
\(\sin (121^{\circ}-\theta) = 0.857\cdot \cos \theta -0.515\cdot \sin \theta\) (Eq. 3)
By applying (Eq. 3) in (Eq. 1b), we get the following expression after expanding and simplifying algebraically:
\(-55\cdot \sin \theta + 58\cdot (0.857\cdot \cos \theta - 0.515\cdot \sin \theta) = 0\)
\(-84.87\cdot \sin \theta +49.706\cdot \cos \theta = 0\)
And we solve the equation for \(\theta\):
\(\tan \theta = \frac{49.706}{84.87}\)
\(\theta = \tan^{-1} 0.585\)
\(\theta \approx 30.328^{\circ}\)
Then, the weight of the pumpkin is: (\(\theta \approx 30.328^{\circ}\))
\(W = 55\cdot \cos 30.328^{\circ}+58\cdot \cos (121^{\circ}-30.328^{\circ})\)
\(W = 46.793\,N\)
The weight of the pumpkin is 46.793 newtons.
The weight of the pumpkin hanging from the two scales is 98.36 N.
The given parameters;
reading on scale one, a = 55 Nreading on scale two, b = 58 Nangle between the two scales, θ = 121⁰The weight of the pumpkin hanging from the two scales is equal to the resultant of the two scale readings.
The resultant of the two force is calculated as follows;
R² = a² + b² - 2abcos(θ)
R² = 55² + 58² - 2(55)(58)cos(121)
R² = 9674.7
\(R = \sqrt{9674.7} \\\\R = 98.36 \ N\)
Thus, the weight of the pumpkin hanging from the two scales is 98.36 N.
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Sally puts on wool socks and rubs her feet on a nylon carpet. How does static electricity build up in Sally?
O When electrons from the wool socks move into the carpet, Sally and her socks gain a negative charge.
O The friction causes electrons from the wool socks to move to Sally, giving Sally a negative charge.
O Electrons from the wool socks move into the carpet, giving Sally and her socks a positive charge.
Electrons from the carpet move into the wool socks, giving Sally and her socks a positive charge.
When Sally puts on wool socks and rubs her feet on a nylon carpet, static electricity builds up through a process involving the movement of electrons. The friction between the wool socks and the nylon carpet causes electrons from the wool socks to move to the carpet, resulting in Sally and her socks gaining a negative charge.
The correct answer would be the friction causes electrons from the wool socks to move to Sally, giving Sally a negative charge.
The friction between the wool socks and the nylon carpet causes electrons from the wool socks to move to the carpet, resulting in Sally and her socks gaining a negative charge. This is due to the phenomenon known as the triboelectric effect.
The triboelectric effect occurs when two materials come into contact and then separate. During the rubbing process, the atoms in the two materials interact, causing the transfer of electrons between them. In this case, the wool socks have a greater affinity for electrons compared to the nylon carpet. As a result, electrons from the socks are transferred to the carpet, leaving the socks with a positive charge and the carpet with a negative charge.
Sally, wearing the wool socks, experiences an accumulation of excess electrons on her feet, giving her a negative charge. This excess negative charge on her body can lead to static electricity-related phenomena, such as experiencing a shock when touching a metal object or seeing her hair stand on end when near certain surfaces.
It's important to note that the movement of electrons determines the charge distribution during the triboelectric effect. In this scenario, electrons move from the wool socks to the nylon carpet, resulting in Sally and her socks gaining a negative charge.
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A mass of 80 kg object is rest on a horizontal rough surface. Itscoefficient of static and kinetic friction are 0.21 and 0.18respectively.(a) Find the value of parallel force which just starts the object to move.(b) Find the value of parallel force which keeps the object in moving.
We will use the next diagram
Where Fr, is the frictional force, F is the applied force, W is the weight and N is the normal
For a)
We analyze the horizontal forces
\(\begin{gathered} \sum ^{}_{}F_x=F_{}-Fr=0 \\ F=Fr \end{gathered}\)Then we analyze the vertical forces
\(\begin{gathered} \sum ^{}_{}F_y=N-W=0 \\ N=W \end{gathered}\)then for the frictional force
\(Fr=\mu_sN\)Then for the force we have
\(F=\mu_sN=0.21(80)(9.8)=164.64N\)Then for b)
\(F=\mu_kN=0.18(80)(9.8)=141.12N\)We
You want to find out how many atoms of the isotope 65Cu are in a small sample of material. You bombard the sample with neutrons to ensure that on the order of 1% of these copper nuclei absorb a neutron. After activation, you turn off the neutron flux and then use a highly efficient detector to monitor the gamma radiation that comes out of the sample. Assume half of the 66Cu nuclei emit a 1.04-MeV gamma ray in their decay. (The other half of the activated nuclei decay directly to the ground state of 66Ni.) (Enter your answer using one of the following formats: 1.2e-3 for 0.0012 and 1.20e 2 for 120.)
Required:
a. If after 10 min (two half-lives) you have detected 10000 MeV of photon energy at 1.04 MeV, approximately how many 65Cu atoms are in the sample?
b. Assume the sample contains natural copper. Refer to the isotopic abundances listed in your text (Chemical and Nuclear Information for Selected Isotopes) and estimate the total mass of copper in the sample.
Answer:
a) number of copper atoms 65 (⁶⁵Cu) is 7.692 10⁶ atoms
b) m_total Cu = 1.585 10⁹ u = 2.632 10⁻¹⁸ kg
Explanation:
a) For this exercise let's start by using the radioactive decay ratio
N = N₀ \(e^{- \lambda t}\)o e - lambda t
The half-life time is defined as the time it takes for half of the radioactive (activated) atoms to decay, therefore after two half-lives there are
N = ½ (½ N₀) = ¼ N₀
N₀ = 4 N
in each decay a photon is emitted so we can use a direct rule of proportions. If an atom emits a photon it has Eo = 1,04 Mev, how many photons it has energy E = 10,000 MeV
# _atoms = 1 atom (photon) (E / Eo)
# _atoms = 1 10000 / 1.04
# _atoms = 9615,4 atoms
N₀ = 4 #_atoms
N₀ = 4 9615,4
N₀= 38461.6 atoms
in the exercise indicates that half of the atoms decay in this way and the other half decays directly to the base state of Zinc, so the total number of activated atoms
N_activated = 2 # _atoms
N_activated = 2 38461.6
N_activated = 76923.2
also indicates that 1% = 0.01 of the nuclei is activated by neutron bombardment
N_activated = 0.01 N_total
N_total = N_activated / 0.01
N_total = 76923.2 / 100
N_total = 7.692 10⁶ atoms
so the number of copper atoms 65 (⁶⁵Cu) is 7.692 10⁶
b) the natural abundance of copper is
⁶³Cu 69.17%
⁶⁵Cu 30.83%
Let's use a direct proportion rule. If there are 7.692 10⁶ ⁶⁵Cu that represents 30.83, how much ⁶³Cu is there that represents 69.17%
# _63Cu = 69.17% (7.692 10⁶ / 30.83%)
# _63Cu = 17.258 10⁶ atom ⁶³Cu
the total amount of comatose is
#_total Cu = #_ 65Cu + # _63Cu
#_total Cu = (7.692 + 17.258) 10⁶
#_total Cu = 24.95 10⁶
the atomic mass of copper is m_Cu = 63.546 u
m_total = #_totalCu m_Cu
m_total = 24.95 10⁶ 63,546 u
m_total = 1.585 10⁹ u
let's reduce to kg
m_total Cu = 1.585 10⁹ u (1,66054 10⁻²⁷ kg / 1 u)
m_total Cu = 2.632 10⁻¹⁸ kg
given the two displacement vector D ( 6i+3j-k) and vector E( 4i-5j+3k) what is the magnitude of displacement. 2D-E
Answer:
The magnitude of displacement vector 2D-E is approximately 14.49 units. The calculation is done using the Pythagorean theorem after finding 2D-E by multiplying vector D by 2 and subtracting vector E.
Explanation:
The value of 2D-E must first be calculated in order to ascertain the displacement 2D-E's magnitude. Vector D may be multiplied by two to accomplish this, and the result can be obtained by deducting vector E:
2D-E = 2(6i + 3j - k) 4i + 5j + 3k = 8i + 11j - 5k - (4i - 5j + 3k) = 12i + 6j - 2k
We can use the Pythagorean theorem to determine the magnitude of the displacement vector now that we know it:
|2D-E| = √(8² + 11² + (-5)²) = √(64 + 121 + 25) = √210 ≈ 14.49
The displacement 2D-E magnitude is therefore 14.49 units or such. From the object's beginning location to its ultimate position, the displacement's entire length is shown by this. Being a scalar variable, the displacement's magnitude does not reveal the displacement's direction.
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Question 43
In order for an open switch to cause a short circuit, what must be connected to the rest of the circuit ?
Answer:
a break in the circuit can also cause this problem
state newton's three laws of motion
I. Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.
Ex: You roll a ball. It slows down by friction.
II. Acceleration is produced when a force acts on a mass. The greater the mass (of the object being accelerated) the greater the amount of force needed (to accelerate the object).
Ex: You cannot kick a brick wall down, but you can kick a soccer ball because the brick wall is more massive.
III. For every action there is an equal and opposite re-action.
Ex:when a rocket lifts off, the rocket's action is to push down on the ground with the force of its engines, and the reaction is that the ground pushes the rocket upwards with an equal force.
A seesaw is balanced on a pivot point. If a 20 kg child sits 1 meter from the pivot point, how far from the pivot point should a 40 kg child sit in order to balance the seesaw?
Okay, here are the steps to solve this problem:
1) The seesaw is balanced when the sum of moments is 0.
2) The moment created by a force depends on the force and the perpendicular distance from the pivot point.
3) The 20 kg child sits 1 meter from the pivot. So its moment is 20 * 1 = 20 kg*m.
4) We want to find the distance for the 40 kg child to create a moment that balances the 20 kg child's moment.
5) So the moment of the 40 kg child must be 20 kg*m.
6) The moment depends on force and distance. We know the force is 40 kg.
7) So we set: 40 kg * distance = 20 kg*m
8) And solve for the distance: distance = 20 / 40 = 0.5 meters
Therefore, for the seesaw to balance with a 20 kg child 1 meter from the pivot and a 40 kg child on the other side, the 40 kg child should sit 0.5 meters from the pivot point.
Let me know if you have any other questions!
Normal forces are applied uniformly over the surface of a spherical volume of water whose radius is 20.0 cm. If the pressure on the surface is increased by 200 MPa, by how much does the radius of the sphere decrease?
The radius of the sphere decreases by 11 mm when the pressure on its surface is increased by 200 MPa.
To determine the decrease in radius of the spherical volume of water when the pressure on its surface is increased, we can use the equation relating pressure to the radius of a sphere:
ΔP = K/R
Where ΔP is the change in pressure, K is a constant, and R is the radius of the sphere.
In this case, we're given that ΔP (change in pressure) is 200 MPa and the initial radius R is 20.0 cm. We need to find the change in radius ΔR.
We can rearrange the equation to solve for ΔR:
ΔR = K/ΔP
Now, we need to determine the value of the constant K. The constant K depends on the bulk modulus of the material, which is a measure of its resistance to compression. For water, the bulk modulus is approximately 2.2 GPa (gigapascals).
Substituting the values into the equation, we have:
ΔR = (2.2 GPa) / (200 MPa)
To simplify the calculation, we need to convert the units so that they are consistent. Let's convert GPa to MPa:
ΔR = (2.2 GPa * 1000 MPa/GPa) / (200 MPa)
ΔR = 11 mm
Therefore, the radius of the sphere decreases by 11 mm when the pressure on its surface is increased by 200 MPa.
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What happens when a mixture is formed?
Answer:
Explanation:
A mixture can usually be separated back into its original components. Some examples of mixtures are a tossed salad, salt water and a mixed bag of M&M's candy.
Which type of telescope is best used to detect distant planets?
A 20" round duct is 275' in length. The duct is carrying 2,900 CFM at a friction loss of 0.12 inches WG per 100 feet.
What is the total friction loss through the duct?
Which of the following statements regarding the moon and Earth is correct?
A. Earth and the moon are nearly the same age.
B. Earth and the moon have nearly the same mass.
C. Earth is tidally locked, while the moon spins on its axis.
D. The moon is older than Earth by approximately 300 million years.
i think it is A?
The spacecraft Magellan's fifth orbit around the Venus was at a mean altitude of 3.01 x 105 m. What was its period? Mean radius of Venus = 6.05 x 106 m. Mass of Venus = 4.87 x 1024 kg.
The period of the orbit of the spacecraft around Venus is 5187.3 seconds.
What did the period of motion of a an object?The period, T, of an object in circular motion is the time taken for one complete revolution.
The period, T, of the planetary body is given as:
\(T = 2\pi\sqrt{\frac{r^{3}}{Gm}}\)
where;
r is the radius G is universal gravitational constant = 6.67 × 10^-11 Nm^2/kg^2m is the massSubstituting the values:
\(T = 2\pi\sqrt{\frac{(6.05 \times 10^{6})^{3}}{6.67 \times 10^{ - 11} \times 4.87 \times 10^{24}}} = 5187.8s\)
Therefore, the period of the orbit of the spacecraft is 5187.3 seconds.
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The way in which one atom interacts with another atom is mostly influenced by the configuration of the
A. electrons farthest from the nucleus.
B. protons in the center of the nucleus.
C. electrons closest to the nucleus.
D. protons on the outer edge of the nucleus.
The way in which one atom interacts with another atom is mostly influenced by the configuration of the electrons farthest from the nucleus.
Option A.
What is atom?An atom can be defined as the smallest part of a substance that cannot be broken down chemically. Each atom has a nucleus (center) made up of protons (positive particles) and neutrons (particles with no charge).
The arrangement of electrons in orbitals and shells around the nucleus is referred to as the electronic configuration of the atom.
Thus, we can conclude that the way in which one atom interacts with another atom is mostly influenced by the configuration of the electrons farthest from the nucleus.
The remaining options do not fit the empty space properly, and they include;
protons in the center of the nucleus.electrons closest to the nucleus.protons on the outer edge of the nucleus.Learn more about electron configuration here: https://brainly.com/question/26084288
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