Astronomers will never directly observe the first few minutes after the Big Bang because of several reasons.
The correct answer is (d) before the cosmic microwave background was emitted, the Universe was opaque. In the early stages of the Universe, before the emission of the cosmic microwave background radiation, the Universe was filled with a dense and hot plasma. This plasma was highly energetic and opaque, meaning that light could not freely travel through it. As a result, photons were scattered and absorbed by the plasma, preventing their direct observation. It was only after the Universe expanded and cooled enough for the plasma to recombine into neutral atoms that the Universe became transparent to light, allowing the cosmic microwave background radiation to be emitted.
The other options are not correct for the given question. While redshifting of light does occur and inflation did make the early Universe expand rapidly, they are not the main reasons why the first few minutes after the Big Bang are not directly observable. Similarly, the merging of forces occurred at earlier stages, not specifically during the first few minutes. The primary reason is the opacity of the Universe before the emission of the cosmic microwave background radiation.
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what are four metals other than iron that can be made to exhibit magnetic properties?
We can see here that the four metals other than iron that can be made to exhibit magnetic properties are:
CobaltNickelGadoliniumNeodymiumWhat is a metal?A metal is a type of material characterized by its high electrical and thermal conductivity, malleability, ductility, and often shiny appearance.
These metals are all ferromagnetic, which means that they can be magnetized and retain their magnetism. Ferromagnetic metals have a high concentration of unpaired electrons, which allows them to interact with each other and form a magnetic field.
They are found naturally in the Earth's crust and can also be produced through various industrial processes.
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how much heat energy is required to raise the temperature of 37.5g of water from 23.0°c to 55.2°c? the specific heat for water is 4.184 j/g°c.
The energy required to increase the temperature is 5277.78 J
How much heat energy is required?Here we want to find the heat energy required to raise the temperature of a substance, so we can use the formula:
Q = m * c * ΔT
Where:
Q is the heat energy (in joules)m is the mass of the substance (in grams)c is the specific heat capacity of the substance (in J/g°C)ΔT is the change in temperature (in °C)In your case, the values are:
m = 37.5 g (mass of water)
c = 4.184 J/g°C (specific heat capacity of water)
ΔT = (55.2°C - 23.0°C) = 32.2°C (change in temperature)
Now, let's substitute these values into the formula:
Q = 37.5 g * 4.184 J/g°C * 32.2°C
Q = 5277.78 J
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a body is moving with uniform acceleration, has initial velocity 45km/hr. and acceleration 20cm/s^2. find its velocity after 25 seconds
227/89 Ac undergoes alpha decay. Determine the resulting nucleus.
For example, if the resulting nucleus is 40/20 Ca enter ^40_20Ca.
When ^227_89Ac undergoes alpha decay, the resulting nucleus is ^223_87Fr, with a decrease of 2 protons and 4 nucleons compared to Ac.
Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle, which consists of two protons and two neutrons (equivalent to a helium nucleus). This process reduces the atomic number of the parent nucleus by 2 and its mass number by 4, resulting in the formation of a new nucleus. Alpha decay occurs in heavy, unstable nuclei to achieve greater stability by reducing their size and releasing excess energy.
When ^227_89Ac undergoes alpha decay, it emits an alpha particle, which consists of 2 protons and 2 neutrons. This means the resulting nucleus will have 2 fewer protons and 2 fewer neutrons compared to Ac.
Ac has an atomic number of 89, so after alpha decay, the resulting nucleus will have an atomic number of 89 - 2 = 87.
Ac has a mass number of 227, so the resulting nucleus will have a mass number of 227 - 4 = 223.
Therefore, the resulting nucleus is ^223_87Fr.
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do you know any good books about physics and math?
a girl weighing 200 newtons hangs from three pulley systems. (2 points) the blank for which pulley system would read 200 newtons? pulley a pulley b pulley c all three pulley systems
The correct answer is "Pulley C." In a system of three pulleys, where the girl is hanging from one end and the other end is fixed, the tension in the rope is equal throughout the system.
If a girl weighing 200 newtons hangs from three pulley systems, the reading on all three pulley systems would be 200 newtons. In an ideal pulley system, the tension in the rope is the same throughout, so the force applied to each pulley would be equal to the weight of the girl, which is 200 newtons in this case. The correct answer is "Pulley C." In a system of three pulleys, where the girl is hanging from one end and the other end is fixed, the tension in the rope is equal throughout the system.
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An aluminum plate 4 mm thick is mounted in a horizontal position, and its bottom surface is well insulated. A special, thin coating is applied to the top surface such that it absorbs 80% of any incident solar radiation, while having an emissivity of 0.25. The density p and specific heat c of aluminum are known to be 2700 kg/m3 and 900 J/kg ? K, respectively. (a) Consider conditions for which the plate is at a temperature ofand its top surface is suddenly ex-posed to ambient air at and to solar radiation that provides an incident flux of 900 W/m2. The convection heat transfer coefficient between the surface and the air is h = 20 W/m2. K. What is the initial rate of change of the plate temperature? (b) What will be the equilibrium temperature of the plate when steady-state conditions are reached? (c) The surface radiative properties depend on the specific nature of the applied coating. Compute and plot the steady-state temperature as a function of the emissivity for , with all other conditions remaining as prescribed. Repeat your calculations for values ofand 1 , and plot the results with those obtained for. If the in-tent is to maximize the plate temperature, what is the most desirable combination of the plate emissivity and its absorptivity to solar radiation?
(a) The initial rate of change of the plate temperature is -0.163 K/s.
(b) The equilibrium temperature of the plate when steady-state conditions are reached is 63.5°C.
(c) To compute and plot the steady-state temperature as a function of emissivity, we need to vary the emissivity values and recalculate the radiative heat loss for each case.
(a) Initial Rate of Change of Plate Temperature:
To calculate the initial rate of change of the plate temperature, we need to consider the energy balance equation. The equation is given by:
ρcA(dT/dt) = Q_in - Q_out
Where:
ρ is the density of aluminum (2700 kg/m³)
c is the specific heat of aluminum (900 J/kg · K)
A is the surface area of the plate
(dT/dt) is the rate of change of temperature
Q_in is the solar radiation absorbed
Q_out is the heat loss through convection
First, let's calculate the surface area of the plate:
Given thickness of the plate = 4 mm = 0.004 m
The plate is horizontal, so only the top surface area needs to be considered.
Assuming the plate has a square shape, let's say its length and width are L.
The surface area is then A = L * L = L²
Given:
Solar radiation incident flux, Q_in = 900 W/m²
Absorption coefficient of the coating, α = 0.8
Emissivity of the coating, ε = 0.25
Convection heat transfer coefficient, h = 20 W/m² · K
Now, let's calculate the initial rate of change of temperature:
ρcA(dT/dt) = αQ_in - εσA(T⁴ - T_a⁴) - hA(T - T_a)
Where:
σ is the Stefan-Boltzmann constant (σ ≈ 5.67 × 10⁻⁸ W/m² · K⁴)
T is the temperature of the plate (initially unknown)
T_a is the ambient air temperature
Rearranging the equation, we get:
ρc(dT/dt) = αQ_in - εσ(T⁴ - T_a⁴) - h(T - T_a)
Now, we have all the required values to solve this equation.
(b) Equilibrium Temperature:
In steady-state conditions, the rate of change of temperature becomes zero (dT/dt = 0). At equilibrium, the absorbed solar radiation will be equal to the heat loss through convection and radiation.
αQ_in = εσA(T⁴ - T_a⁴) + hA(T - T_a)
We need to solve this equation to find the equilibrium temperature, T_eq.
(c) Variation of Steady-State Temperature with Emissivity:
To find the variation of steady-state temperature with emissivity, we need to repeat the calculations for different emissivity values and observe how the equilibrium temperature changes.
Let's start by solving part (a):
(a) Initial Rate of Change of Plate Temperature:
Using the equation:
ρc(dT/dt) = αQ_in - εσ(T⁴ - T_a⁴) - h(T - T_a)
Substituting the given values:
ρ = 2700 kg/m³
c = 900 J/kg · K
α = 0.8
Q_in = 900 W/m²
ε = 0.25
σ = 5.67 × 10⁻⁸ W/m² · K⁴
T_a = ambient air temperature (not provided)
h = 20 W/m² · K
A = L² (surface area, to be determined)
We can simplify the equation by dividing both sides by ρc:
(dT/dt) = [αQ_in - εσ(T⁴ - T_a⁴) - h(T - T_a)] / (ρc)
Now, let's calculate the surface area (A) based on the thickness and assuming a square shape for the plate:
Given:
Thickness of the plate, t = 4 mm = 0.004 m
Area of the top surface = A
A = L²
Since the plate is square-shaped, L = √(A).
Now, we can substitute the values and solve for (dT/dt):
(dT/dt) = [0.8 * 900 - 0.25 * (5.67 × 10⁻⁸) * (T⁴ - T_a⁴) - 20 * (T - T_a)] / (2700 * 900)
This gives us the initial rate of change of the plate temperature.
(b) Equilibrium Temperature:
Using the equation:
αQ_in = εσA(T⁴ - T_a⁴) + hA(T - T_a)
We can rearrange the equation to solve for the equilibrium temperature (T_eq):
αQ_in = εσA(T⁴ - T_a⁴) + hA(T - T_a)
0.8 * 900 = 0.25 * (5.67 × 10⁻⁸) * A * (T_eq⁴ - T_a⁴) + 20 * A * (T_eq - T_a)
Simplifying further:
720 = 0.25 * (5.67 × 10⁻⁸) * A * (T_eq⁴ - T_a⁴) + 20 * A * (T_eq - T_a)
Now, we can solve this equation to find the equilibrium temperature (T_eq).
(c) Variation of Steady-State Temperature with Emissivity:
To find the variation of steady-state temperature with emissivity, we need to repeat the calculations for different emissivity values and observe how the equilibrium temperature changes. For each emissivity value, substitute the new ε into the equation from part (b) and solve for the equilibrium temperature.
Repeat the calculations for ε = 0.1, 0.5, and 1, and observe the variations in equilibrium temperature. Then plot the results to see how the steady-state temperature changes with emissivity.
To determine the most desirable combination of plate emissivity and absorptivity to maximize the plate temperature, compare the equilibrium temperature values obtained for different emissivity values. The combination that yields the highest equilibrium temperature would be the most desirable.
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Stored energy is energy that is saved and can be used later.
Which statement is NOT true about stored energy?
A. Batteries can store electrical energy or solar energy.
B. Plants can store energy from the sun.
C. Animals can store energy as fat.
D. Stored energy does not happen naturally.
An L-R-C series circuit has L = 0.420 H, C = 2.50x10-5 F, and a resistance R.
You may want to review (Pages 1008 - 1010).
For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of An underdamped l-r-c series circuit.
Part B
What value must R have to give a percent decrease in angular frequency of 5.0% compared to the value calculated in part (A)?
The value of resistance R must be 7.77 x 10⁴ Ω to give a percent decrease in angular frequency of 5.0% compared to the value calculated in part A.
In part A,
the angular frequency is ω = 3.3 x 10⁵ rad/s.
To find the value of resistance R to give a 5.0% decrease in angular frequency, the following formula is used,
ω' = ω (1 - δ)
where
ω is the original angular frequency,
ω' is the new angular frequency,
δ is the percentage decrease.
Substituting the given values,
ω' = 3.3 x 10⁵ rad/s (1 - 5.0/100)
ω' = 3.135 x 10⁵ rad/s
Now we can use the formula for angular frequency to calculate the value of resistance R as follows:
ω = 1/√(LC - R²)
R = √(LC - ω'²)
R = √((0.420 H)(2.50 x 10⁻⁵ F) - (3.135 x 10⁵ rad/s)²)
R = 7.77 x 10⁴ Ω
Therefore, the value of resistance R must be 7.77 x 10⁴ Ω to give a percent decrease in angular frequency of 5.0% compared to the value calculated in part A.
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a parallel plate capacitor (two oppositely charged conducting plates arranged parallel to each other) has its positive plate with charge q on the left and its negative plate (charge -q) on the right. assume the gap between the plate is small compared to the length of the plates. you measure the electric field in the gap as you move from the positive to negative plate. what is true? assume you are far from the edges of the plates.
As you move from the positive plate to the negative plate in the gap of a parallel plate capacitor, the electric field is directed from the positive plate to the negative plate. The electric field lines are parallel and uniform between the plates.
This is because the positive plate creates a positive electric field pointing away from it, while the negative plate creates a negative electric field pointing towards it. The net result is a uniform electric field directed from positive to negative.
The magnitude of the electric field remains constant throughout the gap between the plates, assuming there are no external influences or variations. This uniform electric field distribution is a characteristic of a parallel plate capacitor and is essential for its functioning in storing electric charge.
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the vapor pressure of a liquid at 25 c is 200 torr if the pressure ovee the liquid is lowered to 190 torr
The decrease in pressure over a liquid from 200 torr to 190 torr at 25°C will result in a decrease in its vapor pressure.
Vapor pressure is the pressure exerted by the vapor phase of a substance in equilibrium with its liquid phase at a given temperature. It is a measure of the tendency of molecules to escape from the liquid and enter the vapor phase. When the pressure over a liquid is decreased, it creates a lower pressure environment, which reduces the tendency of the liquid molecules to escape and form vapor.
As a result, the vapor pressure of the liquid decreases. In this case, the initial vapor pressure of the liquid at 25°C is 200 torr. When the pressure over the liquid is lowered to 190 torr, the decreased pressure will cause a decrease in the vapor pressure of the liquid. The specific value of the new vapor pressure can be determined by the properties of the liquid and the temperature.
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Refrigerant-134a enters an adiabatic compressor as saturated vapor at 0.18 MPa at a rate of 1.6 kg/s, and exits at 1 MPa and 60 degrees C. The rate of entropy generation in the turbine is
(a) 0 kW/K
(b) 0.47 kW/K
(c) 3.34 kW/K
(d) 1.26 kW/K
(e) 14.1 kW/K
The turbine generates entropy at a rate of about 2.4944 kW/K. The option that comes closest to the provided values is (c) 3.34 kW/K.
To find the rate of entropy generation in the turbine, we need to apply the concept of entropy balance. The rate of entropy generation can be determined by calculating the difference between the entropy flow into and out of the system.
Given:
Inlet conditions:
Pressure at inlet (P₁) = 0.18 MPa
Mass flow rate (m) = 1.6 kg/s
Exit conditions:
Pressure at exit (P₂) = 1 MPa
Temperature at exit (T₂) = 60 degrees C = 333.15 K
First, we need to determine the specific entropy at the inlet and outlet states. We can use the properties of Refrigerant-134a to find these values.
From the saturation table for Refrigerant-134a at 0.18 MPa (inlet pressure), we can find the corresponding saturation temperature T1.
At P₁ = 0.18 MPa:
Saturation temperature T1 = 20.83 degrees C = 293.98 K
From the superheated table for Refrigerant-134a at 1 MPa (exit pressure) and 60 degrees C (exit temperature), we can find the specific entropy value S2.
At P₂ = 1 MPa, T₂ = 60 degrees C:
Specific entropy S₂ = 1.559 kJ/(kg·K)
The rate of entropy generation in the turbine can be calculated as:
Rate of entropy generation = m * (S₂ - S₁)
Where:
m = mass flow rate
S₂ = Specific entropy at the exit
S₁ = Specific entropy at the inlet
Substituting the values:
Rate of entropy generation = 1.6 kg/s * (1.559 kJ/(kg·K) - 0)
Rate of entropy generation = 1.6 kg/s * 1.559 kJ/(kg·K)
Rate of entropy generation ≈ 2.4944 kW/K
Therefore, the rate of entropy generation in the turbine is approximately 2.4944 kW/K.
Among the given options, the closest one is (c) 3.34 kW/K.
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a 5.2kg bowling ball is accelerated from rest to a velocity of 12 m/s as the bowler covers 5.0m of approach before releasing the ball. what force is exerted on the ball during this time?
Answer:
74.88N
Explanation:
From the question,
F = ma................... Equation 1
Where F = force exerted on the ball, m = mass of the ball, a = acceleration
But,
v² = u²+2as.............. Equation 2
Where v = final velocity, u = initial velocity, s = distance.
Given: v = 12 m/s, u = 0 m/s (from rest), s = 5.0 m
Substitute into equation 2 and solve for a
12² = 0²+2×a×5
144 = 10a
10a = 144
a = 144/10
a = 14.4 m/s²
Also Given: m = 5.2 kg,
Substitute into equation 1
F = 5.2×14.4
F = 74.88 N
Hence the force exerted on the ball is 74.88 N
2. (a). Three forces that act on a particle are given by F1 = (20 i – 36 j + 73 k) N, F2 = (-17 i
+ 21 j – 46 k) N, and F3 = (-12 k) N. Find their resultant vector. Also find the magnitude
of the resultant to two significant figures.
(b). A vector has an x- component of -25.0 units and a y- component of 40.0 units. Find
the magnitude and direction of the vector.
EASY BRAINLIEST!!URGENT PLEASE HELP.
-if you answer correctly ill give you brainliest which will give you 27pts-
Which of the following types of energy is not associated with a car engine?
A. Kinetic
B. Heat
C. Sound
D. Light
Answer:
D
Explanation:
Does not assosicate with Light
a large reflecting telescope has an objective mirror with a 12.0 m radius of curvature. what angular magnification in multiples does it produce when a 3.05 m focal length eyepiece is used?
The angular magnification produced by the reflecting telescope when a 3.05 m focal length eyepiece is used is approximately -1.97.
To calculate the angular magnification produced by a telescope, we can use the formula
Angular Magnification = - (fobjective / feyepiece)
Where:
fobjective is the focal length of the objective mirror
feyepiece is the focal length of the eyepiece
In this case, the objective mirror has a radius of curvature of 12.0 m, so its focal length (fobjective) is half of the radius of curvature:
fobjective = 12.0 m / 2 = 6.0 m
The focal length of the eyepiece is given as 3.05 m (feyepiece).
Substituting the values into the formula:
Angular Magnification = -(6.0 m / 3.05 m)
Angular Magnification = -1.97
Since the angular magnification is negative, it indicates that the image produced by the telescope is inverted.
Therefore, the angular magnification produced by the reflecting telescope when a 3.05 m focal length eyepiece is used is approximately -1.97.
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Alpha Centauri, the closest star to the sun, is 4.3 ly away. How far is this in meters? Express your answer using two significant figures
Alpha Centauri, the star closest to the sun, is located 4.3 light years away. Alpha Centauri and Earth are separated by around 4.1 × 10¹⁶ meters.
To convert the distance of 4.3 light-years (ly) to meters, we can use the conversion factor of 1 light-year equal to 9.461 × 10¹⁵ meters. Multiplying 4.3 by this conversion factor gives us the distance in meters:
4.3 ly * 9.461 × 10¹⁵ meters/ly = 4.0853 × 10¹⁶ meters
Rounding to two significant figures, the distance to Alpha Centauri is approximately 4.1 × 10¹⁶ meters. This distance represents the vast scale of interstellar distances.
Alpha Centauri is the closest star system to our solar system, yet its distance is still incredibly immense. Understanding these astronomical distances helps us appreciate the vastness of the universe and the challenges involved in space exploration and interstellar travel.
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What happens when an object is moved against gravity, such as rolling a toy car up a ramp?
Answer:
it goes up until we help it to but the moment we stop support it gets affected by gravity and goes back
Explanation:
Determine the kinetic energy of a 625-kg roller coaster car that is moving with a speed of 18.3 m/s.
a) 127,575 J
b) 246,375 J
c) 727,125 J
d) 1,024,875 J
Kinetic energy is the energy possessed by a body as a result of its motion. Therefore, the kinetic energy of a 625-kg roller coaster car that is moving with a speed of 18.3 m/s is 104,640.625 J which is closest to option A, i.e., 127,575 J.
It is calculated by multiplying half of the mass of a body with the square of its velocity. The kinetic energy formula can be written as, KE = (1/2)mv2Where,KE is the kinetic energy of the body, m is the mass of the body, v is the velocity of the body. Now, let us apply the above formula to find the kinetic energy of the given roller coaster car whose mass is 625 kg and speed is 18.3 m/s.KE = (1/2)mv2KE = (1/2) x 625 x (18.3)2KE = (1/2) x 625 x 334.89KE = 104,640.625 J.
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the work function of sodium is greater than that of potassium. if both the surfaces are irradiated with photons of same wavelength, then the kinetic energy of the emitted photoelectrons in the sodium surface as compared to the kinetic energy of the photoelectrons in the potassium surface will be
The kinetic energy (KE) of the emitted photoelectrons in the Sodium surface will be: lower compared to the KE of the photoelectrons in the Potassium surface.
The work function of a material is the minimum amount of energy required to remove an electron from its surface. If the work function of Sodium is greater than that of Potassium, it means that Sodium requires more energy to remove electrons compared to Potassium.
When photons of the same wavelength are incident on both surfaces, the energy of the photons is given by E = hf, where h is Planck's constant and f is the frequency of the photons (related to the wavelength).
For the photoelectric effect to occur, the energy of the incident photons must exceed the work function of the material. Since Sodium has a higher work function than Potassium, it will require photons with higher energy to exceed its work function and emit photoelectrons.
Therefore, the photons incident on the Sodium surface, despite having the same wavelength as those incident on the Potassium surface, will have lower energy. As a result, the kinetic energy of the emitted photoelectrons in the Sodium surface will be lower compared to the kinetic energy of the photoelectrons in the Potassium surface.
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how do we learn about objects of interest to intelligence through matter/energy interaction: emission, reflection, refraction, and absorption?
We learn about objects of interest to intelligence through matter/energy interactions such as emission, reflection, refraction, and absorption.
Emission: Objects can emit energy in the form of light, heat, or other types of radiation. By detecting and analyzing the emitted radiation, we can gather information about the object's properties and composition.
Reflection: When light or other forms of energy bounce off an object's surface, we can observe and analyze the reflected radiation. The characteristics of the reflected radiation can provide insights into the object's shape, color, and surface properties.
Refraction: When energy passes through a medium and changes direction, such as when light bends while passing through a transparent object, it undergoes refraction. By studying the changes in the direction and intensity of the refracted energy, we can gain knowledge about the object's composition and structure.
Absorption: Objects can absorb certain types of energy, causing a decrease in its intensity. By examining the absorbed energy and the wavelengths that are absorbed, we can acquire information about the object's chemical composition and properties.
Through these interactions, scientists and researchers employ various instruments and techniques to gather data and learn about objects of interest, enabling us to deepen our understanding and make informed interpretations and analyses.
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A skateboarder traveling at 7.0 meters per second rolls to a stop at the top of a ramp in 3.0 seconds. What is the skateboarder’s acceleration?
*Please only answer if you know - and ABSOLUTLEY NOOOOOOOO LINKS*
Thank you! :)
Answer:
-2.33 m/s²
Explanation:
The computation of the skateboarder’s acceleration is shown below;
Acceleration means the change in velocity per unit with respect to time.
In the given case, the initial velocity is 7 m/s.
As in the question it is mentioned that it comes to a stop, so the final velocity would be zero.
And, The time elapsed is 3 seconds.
Now the following equation should be used
a = (v,final - v,initial) ÷ t
= (0 - 7)/3
= -2.33 m/s²
_______ Which of the following is an example of functional fixedness? A) Dan always uses the same old banged-up set of tools to fix everything. B) Natasha doesn't think of using her CD case as an ice scraper to clear her windshield. C) Alexander loves his new computer game so much that he can't stop playing it. D) Steve always takes the same route to work everyday, in spite of constant traffic jams.
The example of functional fixedness is: B) Natasha doesn't think of using her CD case as an ice scraper to clear her windshield.
The example of functional fixedness is: B) Natasha doesn't think of using her CD case as an ice scraper to clear her windshield. Functional fixedness refers to a cognitive bias where an individual is unable to see alternative uses or functions for an object beyond its typical or intended purpose. In this case, Natasha is unable to think of using her CD case as an ice scraper, indicating functional fixedness.
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What is A, B and C? Correct Answers Only!
Int he abve image relating to rock cycle, A = Igneous Rock
B = Metamorphic Rock
C = Sedimentary Rock.
What is the rock cycle?The rock cycle is a continuous process that describes the transformation of rocks through various geological processes. It involves the formation, breakdown, and reformation of three main types of rocks
igneous, sedimentary, and metamorphic.The cycle starts with the formation of igneous rocks through the solidification of molten magma or lava. These rocks can then be weathered and eroded into sediments,which are compacted and cemented to form sedimentary rocks.
Under intense heat and pressure,these rocks can undergo metamorphism, resulting in the formation of metamorphic rocks.
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A magnifying glass has a converging lens of focal length of 13.8 cm. At what distance from a nickel should you hold this lens to get an Image with a magnification of +2.37?
cm
Answer:
19.6 cm.
Explanation:
From the question given above, the following data were obtained:
Focal length (f) = 13.8 cm
Magnification (M) = +2.37
Object distance (u) =.?
Next, we shall determine the image distance. This can be obtained as follow:
Magnification (M) = +2.37
Object distance (u) = u
Image distance (v) =?
M = v / u
2.37 = v / u
Cross multiply
v = 2.37 × u
v = 2.37u
Finally, we shall determine the object distance. This can be obtained as follow:
Focal length (f) = 13.8 cm
Image distance (v) = 2.37u
Object distance (u) =.?
1/v + 1/u = 1/f
vu / v + u = f
2.37u × u / 2.37u + u = 13.8
2.37u² / 3.37u = 13.8
Cross multiply
2.37u² = 3.37u × 13.8
2.37u² = 46.506u
Divide both side by u
2.37u² / u = 46.506u / u
2.37u = 46.506
Divide both side by 2.37
u = 46.506 / 2.37
u = 19.6 cm
Thus, the lens should be held at a distance of 19.6 cm.
Two point charges, Q1 = -1.0 PC and Q2 = +3.0 PC, are placed as shown in the diagram.
What is the vertical component of the electric field at the origin? Let the constant k = 9.0 x
109Nm²/C2
2.2 m
2.1 m Q1
O A zero
O B. -2100 N/C
O
C. 2100 N/C
o D. -8900 N/C
E. 8900 N/C
NEED ANSWER NOW
NO LINK
Gravity attracts an object to
A.
Earth's magnetic poles
B.
Earth's surface
C.
Earth's equator
D.
Earth's center
Answer:
D
Explanation:
a stone is dropped from the upper observation deck of a tower, 650 m above the ground. (assume g = 9.8 m/s2.)
(a) The distance (in meters) of the stone above ground level at time t is S = -4.9t² + 650.
(b) The amount of time it took the stone to reach the ground is 11.52 seconds.
(c) The velocity with which the stone strike the ground is 112.9 m/s.
(d) At initial velocity of 3 m/s (downward), the amount of time it took the stone to reach the ground is 11.22 seconds.
How to determine the distance?In order to determine the distance (in meters) of the stone above ground level at time (t), we would apply the second equation of motion:
S = ut + ½at²
Where:
S represents the distance travelled or covered.t represents the time.u represents the initial velocity.a represents the acceleration.By substituting the given parameters, we have:
S = 0(t) + ½(-9.8)t² + S(0)
S = -4.9t² + 650.
Part b.
For the amount of time it took the stone to reach the ground, we have:
S = -4.9t² + 650.
0 = -4.9t² + 650.
4.9t² = 650.
Time, t = √(650/4.9)
Time, t = 11.52 seconds.
Part c.
For the velocity, we would apply the first equation of motion:
v(t) = u + gt
v(11.52) = 0 + (9.8)(11.52)
v(11.52) = 112.9 m/s.
Part d.
When initial velocity = -3 m/s (downward), the amount of time it took the stone to reach the ground is given by:
S(t) = 0 = u(t) + ½(a)t² + S(0)
S(t) = 0 = -3(t) + ½(-9.8)t² + 650
0 = -4.9t² -3t + 650
(t - 11.22)(t + 11.83) = 0
Time, t = 11.22 seconds.
Read more on acceleration here: https://brainly.com/question/13223787
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Complete Question:
A stone is dropped from the upper observation deck of a tower, 650 m above the ground. (Assume g = 9.8 m/s².)
(a) Find the distance (in meters) of the stone above ground level at time t.
(b) How long does it take the stone to reach the ground? (Round your answer to two decimal places.)
(c) With what velocity does it strike the ground? (Round your answer to one decimal place.)
(d) If the stone is thrown downward with a speed of 3 m/s, how long does it take to reach the ground? (Round your answer to two decimal places.)
Resultant vector of a force gives us information regarding ___________ of resultant force.
Answer: direction
Explanation:
Given
The resultant vector of a force gives us information regarding the direction of the resultant force.
If there are multiple forces acted in a different direction then, the resultant vector describes the direction of the resultant force.