The ball will be in motion for about 2.44 seconds due to the free-fall.
What do you mean by free-fall?Free fall is any motion of a body in which gravity is the only force acting on it according to Newtonian mechanics. A body in free fall experiences no force according to general relativity, which reduces gravitation to space-time curvature. When something is in "free fall" in a technical sense, it may not necessarily be going down in the traditional sense of the word. Normally, an object traveling upwards might not be thought of as falling, but if it is solely being affected by gravity, it is said to be in free fall. As a result, the Moon is falling freely around the Earth, even if its fast orbital speed keeps it far from the planet's surface.
According to the question:
the ball is in motion for time, t = ?
initial velocity along y direction = 0
along y direction; h = 1/2gt²
from the above equation:
t = √2h/g
t = √(2×29.8)/9.8
t = 2.44seconds
Hence the ball will remain in motion for about 2.44seconds under free fall.
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If the mirror is immersed in water (refractive index 1.33), what is its focal length?
If the mirror is immersed in water whose index of refraction is 1.33 , No change in its focal distance . focal distance of mirrors does not depend on the external medium in which the mirror is immersed.
What happens when a mirror is immersed in water?As the focal length of the convex mirror does not depend on the factor such as the refractive index of any medium. Hence even after the convex mirror is immersed in water, the focal distance remains unchanged. Hence the facility of the convex mirror also remains same.
How is focal distance related to refractive index?Since from lens makers formula we will observe that focal length of lens is inversely proportional to refractive index of medium of lens . Hence net index of refraction of lens decreases when it is dipped in a denser medium than air, hence focal distance increases.
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two large, parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. (a) if the surface charge density for each plate has magnitude 47.0 nc>m2 , what is the magnitude of e s in the region between the plates? (b) what is the potential difference between the two plates? (c) if the separation between the plates is doubled while the surface charge density is kept constant at the value in part (a), what happens to the magnitude of the electric field and to the potential difference?
In here E = 5.3 x 10^3 N/C, the potential difference (V) between the two plates is 116.6 V and if d is doubled, electric field E won't change but V gets doubled.
How to calculate by applying the concepts?a) The electric field of infinite parallel plates can be formulated as
E= σ/2ε0 → eq1
Surface charge density is given as 47.0 nc/m2
So, the electric field is independent on d.
Hence the, so the net field is
E = E1 + E2 = σ/2ε0 + σ/2ε0 = σ/ε0
Substituting in eq 1,
E = [47 x 10^(-9) ] / 8.85 x 10^-12 = 5.3 x 10^3 N/C
b) Since the electric potential between them at any point and is given by.
V=Ed → eq 2
So, Substituting in eq 2 yields
V= 2.2 x 10^-2 x 5.3 x 10^3 = 116.6 V
c) If the distance between plates is doubled it will make no changes on the field because the field is independent on d.
For the voltage, according to equation 2 the voltage depends on d so if d is doubled then the potential between plates must be doubled.
Thus, magnitude of e in the region is 5.3 x 10^3 N/C.
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What causes cosmic microwave background energy
Answer: The Big Bang!
a boat is stationary at 12\, \text{meters}12meters12, start text, m, e, t, e, r, s, end text away from a dock. the boat then begins to move toward the dock with an acceleration of 5.0\, \dfrac{\text{m}}{\text{s}^2}5.0 s 2 m 5, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, squared, end fraction.
A boat is stationary at 12 meters away from a dock. The boat then begins to move toward the dock with an acceleration of 50 m/s². How long will it take the boat to reach the dock?
Answer:
Boat can reach the dock at 2.19 s time.Explanation:
Given that
Distance of boat to dock d = 12m.Acceleration of boat a = 50 m/s².To find
How long will it take the boat to reach the dock ?So according the question
We have,
Initial velocity u = 0 m/s.Acceleration of boat a = 50 m/s².Distance of boat from dock s = 12m.Now, for finding time travelled by boat to reach dock we can use the second equation of motion i.e.
second equation of motions = ut + [tex]\frac{1}{2}[/tex] at²
where,
s = Distance.
u = Initial velocity.
t = Time.
a = Acceleration.
Now, putting all given values in second equation of motion,
So, we get
s = ut + [tex]\frac{1}{2}[/tex] at²
12 = 0t + [tex]\frac{1}{2}[/tex] ×5× t²
12×2 = 5t²
t² = [tex]\frac{12 \times 2}{5}[/tex]
t = [tex]\sqrt{\frac{24}{5}}[/tex]
t = [tex]\sqrt{{4.8}}[/tex]
t = [tex]2.19[/tex] s.
Answer:
Boat can reach the dock at 2.19 s time.To learn more about application of equation of motions please click on the link.
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Calculate the magnitude of the electric field at point p on the bisector of the rod
The magnitude of the electric field at point p is [tex]\frac{2kQ}{a\sqrt{4a^{2} + L^{2} } }[/tex].
What is electric field?An electric field is a physical field that surrounds electrically charged particles and acts as an attractor or repellent to all other charged particles in the vicinity. It can also refer to a system of charged particles' physical field.
The human body is affected by low-frequency electric fields in the same way that they are affected by other charged-particle-containing materials. The distribution of electric charges at conductive materials' surfaces is affected by electric fields acting on such materials. They make the body conduct current to the ground.
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8. At track practice, a student runs around an oval shaped 400 meter long track 4
times. Determine the distance and displacement of the student.
The distance travelled by the student is 1600m and the displacement is 0.
Displacement is the direction from the beginning point and the length of a straight path from the starting point to the finishing point, whereas distance is the length of a path that connects two places.
Displacement is a vector quantity that does not take into consideration the actual path taken; instead, it quantifies the distance between the beginning point and the finishing position.
Given:
Track length = 400m
No. of rounds = 4
To find:
Distance, d = ?
Displacement, D = ?
Formula:
Distance, d = Track length x No. of rounds
Displacement, D = End point – Start point
Calculations:
d = 400 x 4
d = 1600m
D = 0 since the start and end point are same.
Result:
The distance travelled is 1600m with 0 displacement.
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5. A rocket travels a distance of 6,000 km in 2 hours, what is the rockets speed?
The speed of the rocket is 3000 km/h.
Distance is the sum of an object's movements, regardless of direction. Without respect to the object's starting or finishing positions, distance can be defined as the amount of space it has traveled.
A scalar quantity, speed is defined as the size of the change in an object's location over time or the size of the change in an object's position per unit of time.
The distance traveled by rocket is 6000 km.
The time is taken by the rocket is 2 hours.
The speed of an object is defined as the distance traveled by the object per unit of time.
Therefore,
Speed = distance / time
S = d / t
S = 6000 km / 2 hr
S = 3000 km/h
The speed of the rocket is 3000 kilometers per hour.
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A rocket is accelerating upward from the surface of earth with an acceleration of 4. 4 m/s2. On board the rocket is a 0. 060 kg chicken egg. What is the apparent weight of the egg?.
The apparent weight of the egg = 0.264N
Here the rocket is accelerating upward from the surface of the earth. The egg is on the rocket.
The acceleration due to gravity is 4.4m/[tex]s^{2}[/tex].
The mass of a chicken egg is 0.060 kg.
Form Newton's second law, the net force acting on a body is equal to the product of mass and acceleration due to gravity.
So form Newton's second law we know the formula i.e.,
F = m×a
= 0.06 × 4.4
= 0.264N
Therefore the apparent weight of the egg is 0.264N.
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A diagram shows the velocity vector for an object in uniform circular motion. Why is the vector tangential to the circle?
A. The vector shows the path the object would follow if the net force acting on it stopped.
B. The vector indicates the direction of the force on the object.
C. The vector shows how the object pulls against the centripetal force.
D. The vector approximates the curved path at that point.
The vector tangential to the circle because the vector indicates the direction of the force on the object.
What is centripetal force?
Centripetal force is the force acting on an object in circular motion directed towards the axis of rotation.
In a circular motion, the direction of the velocity vector is the same as the direction of the object's motion, the velocity vector is directed tangent to the circle as well.
It also indicate the direction of the centripetal force.
Thus, the vector tangential to the circle because the vector indicates the direction of the force on the object.
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DATE
CLASS
14. The Space Shuttle must achieve a velocity of 7,800 m/s in order to orbit the
Earth If the average acceleration of the Space Shuttle is 15.3 m/s², how long will
It take for the shuttle to reach orbital velocity? Convert your answer from seconds
to minutes. Show all your work for this calculation.
Answer:
8.49673 minutes ≈ 8.5 minutes
Explanation:
Acceleration, a is defined as the rate of change in velocity divided by the change in time to attain that velocity
[tex]\mathsf a = \dfrac{\Delta v}{\Delta t} \\\\\textsf {where } \mathsf {\Delta v = v-v_0} \textsf{ v being final velocity and v_0 initial velocity}[/tex], v = final velocity and v₀ the initial velocity
and Δt is the time required to attain final velocity
Δt = number of seconds since launch = t since t₀ = 0
Therefore,
[tex]a = \dfrac{v-v_0}{t}[/tex]
The space shuttle's initial velocity is 0 m/s and its final velocity must be 7800 m/s to escape earth's gravity
So v - v₀ = 7800 - 0 = 7800 m/s, t - t₀ = t = 0 = t seconds
a = 15.3 m/s²
We have the equation
a = 7800/t
a is given as 15.3 m/s² so
15.3 = 7800/t
t = 7800/15.3 = 509.80392 seconds
To convert to minutes, divide by 60
509.80392/60 = 8.49673 minutes ≈ 8.5 minutes
leaving the water vertically at 8.0 m/s , a steelhead lands on the top of a waterfall 1.8 m high. how long is it in the air?
The steelhead lands on the top of waterfall when it is descending so the time taken is 1.36 seconds.
What is a waterfall?A waterfall is a factor in a river or movement in which water flows over a vertical drop or collection of steep drops.
Waterfalls are formed in a variety of ways, but the most common is for a river to flow over a top layer of elastic rock before dropping into soft rock. This rock will erode faster, resulting in a higher drop. Waterfalls have been studied for their impact on the species that live in and around them. Over the years, people have developed a unique relationship with waterfalls by seeing them, exploring them, and naming them. They can pose a formidable obstacle to navigation along the river.
Using ,S= u*t + (1/2)at2
1.8 = 8t - [(1/2)*9.8*t2]
By ,solving this quadratic equation we get, t= 0.27 s or t= 1.36 s
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A blinking light of constant period is situated on a lab cart. Which diagram best represents a photograph of light, taken every 2 seconds, as the cart moves with constant velocity?
The snapshot of light as the cart moves with constant velocity is represented by a graph with uniform displacement at each time interval.
The change in displacement with time is uniform at constant velocity. The displacement of the supplied moving item grows at the same pace.
The beginning velocity equals the ultimate velocity at constant velocity.
v₁ = v₂
The object's acceleration at constant velocity is zero since the velocity change with time is zero.
As a result, we may deduce that the graph with equal displacement at each time interval reflects a snapshot of light as the cart moves at a constant speed.
A moving object's displacement-time graph shows the distance traveled by a moving item as time passes. A vector quantity is displacement. The slope or gradient of this graph represents the velocity of the item. The displacement-time graph, also known as the position-time graph, describes an object's motion. In this graph, the displacement of the moving item is displayed on the y-axis as a dependent variable, while time is shown on the x-axis as an independent variable.
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Did keeping the flies in constant dark conditions abolish the light-emission oscillations?
Yes, the statement is correct that keeping the flies in constant dark conditions abolish the light-emission oscillations
What is light-emission oscillations?The electromagnetic radiation spectrum produced when an electron changes from a high energy state to a lower energy state is known as the emission spectrum of a chemical element or chemical compound. The energy difference between the two states is equal to the photon energy of the emitted photon. Each atom contains a large number of potential electron transitions, and each transition has a distinct energy difference. An emission spectrum is made up of a variety of transitions that result in various radiated wavelengths. The emission spectra of each element is distinct. Therefore, components in materials with an unknown composition can be identified via spectroscopy. Similar to this, chemical examination of substances may make use of the emission spectra of molecules.
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the electric field everywhere on the surface of a charged sphere of radius 0.260 m has a magnitude of 525 n/c and points radially outward from the center of the sphere.
(a)The net charge on the sphere is 3.38×[tex]10^{-9}[/tex] C.
(b) The charge inside sphere is positive and distributed symetrically inside the sphere .
(a) As we know the electric field is given as,
E = kq / [tex]r^{2}[/tex]
Plugging in the values we get,
575 = 8.99×[tex]10^{9}[/tex] q / [tex]0.230^{2}[/tex]
8.99×[tex]10^{9}[/tex]q = 30.41
q = 3.38×[tex]10^{-9}[/tex] C
So, the net charge on the sphere is 3.38×[tex]10^{-9}[/tex] C.
(b) We can conclude that the charge inside sphere is positive and distributed symetrically inside the sphere .
The complete question is: The electric field everywhere on the surface of a charged sphere of radius 0.230 m has a magnitude of 575 N/C and points radially outward from the center of the sphere. (a) What is the net charge on the sphere? (b) What can you conclude about the nature and distribution of charge inside the sphere?
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how long does it take an automobile traveling in the left lane at 60.1 km/h to overtake (become even with) another car that is traveling in the right lane at 38.1 km/h, when the cars' front bumpers are initially 105.2 m apart?
An automobile traveling in the left lane at 60.1 km/h to overtake (become even with) another car that is traveling in the right lane at 38.1 km/h, when the cars' front bumpers are initially 105.2 m apart and the time taken is t = 17.21 seconds.
To find the time, the given values are,
Distance in,
Left lane = 60.1 km/hr
Right lane = 38.1 km/hr
Initial length d₀ = 105.2 m
What is Distance?The distance covered can be derived using the equation of motion with no acceleration.
distance d = speed v × time t
d = vt
For the first automobile,
v = 60.1 km/h
d₁ = 60.1
For the second automobile,
v = 38.1 km/h
Initial distance d = 105.2 m = 0.1052 km
d = vt + d₀.
d₂ = 38.1 t + 0.1052
For them to meet/overtake each other,
Their distance at that time must be equal.
d₁ = d₂
60.1 t = 38.1 t + 0.1052
60.1 t - 38.1 t = 0.1052
22 t = 0.1052
t = 0.1052 / 22 hr
Converting to seconds
t = 0.1052 / 22 × 3600s
t = 17.21 seconds.
The time taken for an automobile to travel is 17.21 seconds.
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Suppose object a is experiencing an electric field with a magnitude of e at its location. if the charge on object a is doubled, what happens to the electric field it is experiencing?
If the charge on object A is doubled then the electric field will remain unchanged in any of the cases. It is so because the question is asking us whether the electric field which the object is experiencing is changing or not, but in actuality, due to charge change, the magnitude of the electric field is changing only.
The electric field's magnitude may be easily determined by calculating the force per charge on the test charge. The electric field would exert a force on the item when its charge doubled since it is inversely proportional to the electric charge. so won't have any effect.
The field is supposed to have a direction corresponding to the force that it would exert on a positive test charge. The electric field emanates from a positive point charge in an outward direction, and from a negative point charge in an inward direction.
A charge is a characteristic that describes how many more or fewer electrons than protons a matter unit has.
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How many protons will there be if the atom has 3 electrons and a mass of 7 amu
Answer:
[tex]{ \tt{mass \: number = protons + neutrons}} \\ { \tt{7 = p + 3}} \\ { \tt{p = 4}}[/tex]
Nadh and fadh2 carry high energy electrons from the other stages of respiration to the electron transport chain. What does the electron transport chain use these electrons to do?.
Electron transport chain use NADH and FADH2 molecules formed in Glycolysis and Krebs Cycle processes.
The third stage of cellular respiration is the electron transport chain. In cellular respiration, high-energy electrons are transported by NADH and FADH2 to the electron transport chain. In photosynthesis, NADPH transports energetic electrons to the electron transport chain.
Krebs cycle or the TCA cycle, the process of oxidizing acetyl-CoA, which is produced from proteins, carbs, and lipids, releases stored energy.
The process by which glucose is broken down to provide energy is known as glycolysis. It generates two pyruvate molecules, ATP, NADH, and water. There is no need for oxygen throughout the process, which occurs in the cytoplasm of a cell. Both aerobic and anaerobic respiration experience it.
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please help u have the answers but dont know how to decode it
Answer:
Q1: C
Q2: C
Q3: D
Q4: D
Code: Left, Down, Up, Right
Your welcome.
Answer: ctrl+U
ctrl+G
Explanation:
I don't know how else to help
The FitnessGram PACER Test is a multistage aerobic capacity test that progressively gets more difficult as it continues.
The test is used to measure a student's aerobic capacity as part of the FitnessGram assessment. Students run back and forth as many times as they can, each lap signaled by a beep sound. The test get progressively faster as it continues until the student reaches their max lap score.
The PACER Test score is combined in the FitnessGram software with scores for muscular strength, endurance, flexibility and body composition to determine whether a student is in the Healthy Fitness Zone™ or the Needs Improvement Zone™.
effect experiment, first visible light is shone on a negatively charged electroscope. next light from a special ultraviolet light source is shone on the charged electroscope. both light sources can be made brighter or dimmer
While the flashlight light has no effect, the UV source's light discharges the electroscope.
What was Einstein's photoelectric effect experiment?According to Planck's formula, the energy of the particles that make up a beam of light is connected to their frequencies. The photons and atoms collide when that beam strikes a metal. The photoelectric effect is created by a collision when a photon's frequency is high enough to remove an electron.
What is the conclusion of the photoelectric effect?One electron interacts with each photon. The incident photon's energy is employed to both liberate the surface-bound electrons and give the expelled electrons energy.
Because it offers insights into the properties of solids as well as the properties of atoms and molecules, the photoelectric effect is extensively researched in the fields of quantum chemistry, condensed matter physics, and solid-state chemistry.
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Aluminum chloride (AlCl3) has atoms ______ aluminum atoms and ______chlorine b. 3;1 a. 1;3 c. 1;7 d. 3;7
What is the frequency of light emitted when an electron in a hydrogen atom jumps
from 2nd orbit to 1st orbit?
The frequency of light emitted is 24.63 x [tex]10^1^5[/tex] Hz when an electron jumps from the 2nd to the 1st orbit in a hydrogen atom.
According to Bohr, the electrons in the atom's structure are situated around the nucleus at particular energy levels. An electron must either acquire or lose energy as it transitions from one of these energy levels to another. It is known as an absorption when an electron obtains energy and as an emittance when it loses energy. Both processes involve the absorption or emission of a PHOTON energy particle, which in turn causes the absorption or emission of light. Because various elements have varying energies, they each emit or absorb light in varying quantities (wavelengths).
The wavelength or wave number is used to compute the frequency of light emitted.
The frequency of light emitted is inversely proportional to its wavelength.
Given:
Jump from 2nd orbit to 1st orbit
Planck’s constant, h = 6.626 x [tex]10^-^3^4[/tex]
To find:
Frequency of light emitted, υ = ?
Formula:
E = hυ
Calculations:
[tex]E = 13.6 * Z^2( \frac{1}{n_1^2} - \frac{1}{n_2^2} )[/tex]
E = hυ = 13.6 x 1 x [tex](1 - \frac{1}{4})[/tex]
υ x 6.626 x [tex]10^-^3^4[/tex] = 10.2 x 1.6 x [tex]10^-^1^9[/tex]
υ = 24.63 x [tex]10^1^5[/tex] Hz
Result:
The frequency of light emitted is 24.63 x [tex]10^1^5[/tex] Hz.
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there is an equilateral triangle with sides 2 meters in length. at one of the points is a 1 microcoulomb charge, at the second point there is a 2 microcoulomb charge, and at the third point there is a 3 microcoulomb charge. what is the magnitude of the force on the 2 microcoulomb charge?
With the concept of Coulomb's Law, the net force acting on the 2μC charge is 2.06 × 10⁻² newton.
What is Coulomb's Law?Coulomb's Law states that the force of attraction or repulsion exerted by a charged particle on another charged particle is given by,
F=kq₁q₂ / r²
Here q₁ and q₂ are the charges on the particles,
r is the separation between them
k is a constant known as Coulomb's constant, whose value is, k = 9 × 10⁹ N m²/C²
So here in the question, we have been given an equilateral triangle of side 2 meters with three-point charges at its vertices
q₁ = 1μC ,
q₂ = 2μC ,
q₃ = 3μC ,
r=2 m
From Coulomb's law,
F₂₁ = kq₁q₂/r²
F₂₁ = (9 × 10⁹ × 1 × 2 × 10⁻¹²) / 2²
F₂₁ =4.5 × 10⁻³ N
This is the force on charge q₂ because of q₁.
Similarly,
F₂₃ = kq₂q₃ / r²
F₂₃ = (9 × 10⁹ × 2 × 3 × 10⁻¹²) / 2²
F₂₃ = 18 × 10⁻³ N
Now, the resultant of the two forces will be,
F(net) = [tex]\sqrt{F_{21} ^2 + F_{23}^2 +2F_{21}F_{23}cos60}[/tex]
On substituting all the values the net force will be
F(net) = 2.06 × 10⁻² N
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What does the angular frequency of the emf have to be for the inductor and resistor have the same peak voltage (i.e. for r and xl to be the same)?
The angular frequency of the emf for the inductor and resistor to have the same peak voltage (i.e. for R and XL to be the same) is R/L.
The above situation represents Alternating Current(AC).
An alternating current is a type of current that reverses its direction after fixed intervals of time.
The following formulas are used in the case of AC,
XL = ωL
XC= 1/Cω
Here L and C represent inductance and capacitance respectively.
It is given that XL = R
Hence, ωL = R
ω = R/L
Hence, the angular frequency(ω) of the emf for the inductor and resistor to have the same peak voltage (i.e. for R and XL to be the same) is R/L.
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a 60kg person walks from the ground to the roof of a 80m tall building. what is the person's potential energy once they get to the top?
A 60-kilogram individual ascends an 80 m-high building's roof by walking up from the ground. Once at the peak, the human has a potential energy of 44027.28 J.
This is our formula
Gravitational ability strength = mgh
here (GPE = mgh)
mg = 60.0 kg
h = 80
mg = 60.0 kg
in view that we're searching for pressure and the SI Unit for force is "N," we must convert this to force (N).
consequently, we will multiply 60.0 via nine.81.
60.0 × 9.81 = 588.6
Now that will be
⇒ mg = 588.6 N
⇒ h = 80
GPE = (588.6) × (80) = 47088 J
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a water droplet falling through the air can oscillate with some angular frequency that depends on its surface tension, density, and radius. the surface tension may be interpreted as the energy per unit area of surface of the drop. if a certain drop oscillates with angular frequency $\omega,$ what is the oscillation angular frequency of a drop with half of the first drop's radius?
The oscillation angular frequency of a drop with half of the first drop's radius is 4ω
What is surface tension?Surface tension is the tension force exerted on an object by the surface of a liquid.
What is angular frequency?Angular frequency is the frequency of oscillation of a rotating object. It is given in rad/s.
What is the oscillation angular frequency of a drop with half of the first drop's radius?Given that
the angular frequency of the drop is ω and radius r.Since the energy of the drop is conserved, using the law of conservation of angular momentum, we have
Iω = I'ω' where
I = initial rotational inertia of droplet = mr² where m = mass of drop and r = initial radius of droplet, ω = initial angular frequency of droplet, I' = initial rotational inertia of droplet = mr² where m = mass of drop and r' = final radius of droplet, and ω = final angular frequency of dropletSo, Iω = I'ω'
Making ω' subject of the formula, we have
ω' = Iω/I'
ω' = mr²ω/mr'²
ω' = r²ω/r'²
Given that the drop is half of the first drop's radius, r' = r/2
So, ω' = r²ω/r'²
ω' = r²ω/(r/2)²
ω' = r²ω/r²/4
ω' = 4ω
So, the oscillation angular frequency of a drop with half of the first drop's radius is 4ω
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The scientific process is most similar to:
Answer:
the process of solving a mystery.
Explanation:
A snowmobile has an initial velocity of 3 m/s. (a) If it accelerates at 0.5 m/s² for 7 seconds
what is the final velocity? (b) If it then accelerates at -0.6 m/s², how long will it take to
reach a complete stop (assume the starting velocity for the negative acceleration is the
answer from part a)?
Answer: a. 6.5 m/s
b. s=10.83 seconds
Explanation:
a. 3+7(0.5)=
3+3.5=6.5 m/s
b. s=seconds
6.5-0.6s=0
6.5-0.6s+0.6s=0+0.6s
6.5=0.6s
6.5=3s/5
6.5*5=3s*5/5
3s=32.5
s=32.5/3
s=65/6
s=10.83 seconds
What are the initial speed, in m/s, and the hang time (total time in the air), in s of an athlete who jumps a vertical distance of 0.8 m?
The athlete who jumps a vertical distance of 0.8 m has a initial speed of: 3.959 m/s and his total time in the air is: 0.8078 s
The formulas for the vertical launch upward and the procedures we will use are:
y max = v₀²/(2*g)t max = v₀/ gt(of) =2*t maxWhere:
v₀ = initial velocityg = gravityy max = maximum heightt max = time to reach maximum heightt(of) = time of flightInformation about the problem:
g = 9.8 m/s²y max= 0.8v₀ = ?t(of)=?Applying the maximum height formula and clearing the initial velocity we get:
y max = v₀²/(2*g)
v₀ = √(y max * (2*g))
v₀ = √( 0.8 m * (2 * 9.8 m/s²))
v₀ = √( 0.8 m * 19.6 m/s²)
v₀ = √15.68 m²/s²
v₀ = 3.959 m/s
Applying the maximum time formula we get::
t max = v₀ / g
t max =(3.959 m/s)/(9.8 m/s²)
t max = 0.4039 s
Applying the time of flight formula we get::
t(of) =2 * t max
t(of) =2 * 0.4039 s
t(of) = 0.8078 s
What is vertical launch upwards?In physics vertical launch upwards is the motion described by an object that has been launched vertically upwards in which the height and the effect of the earth's gravitational force on the launched object are taken into account.
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a block of unknown material is submerged in water. light in the water is incident on the block at an angle of 31°. the angle of refraction in the block is 27°. what is the index of refraction of the unknown material?: * a. 0.67 b. 1.2 c. 1.3 d. 1.5
The unknown material is glass because the refractive index is found out to be 1.5
Calculation and step-by step process[tex]u_{water\\}[/tex]= 1.33
[tex]u_{x}[/tex]=?
[tex]u_{1}[/tex]sini=[tex]u_{2}[/tex]sin27°
[tex]u_{2}[/tex] = 1.33 x [tex]\frac{sin31}{sin27}[/tex]
= 1.33 x (0.51/0.45)
= 1.50
so, deducing from it, we have refractive index = 1.50
Hence the unknown material is glass.
What is index of refraction?
The refractive index (or refraction index) of an optical media in optics is a dimensionless quantity that indicates the capacity of that medium to bend light. The refractive index controls how much light is refracted or twisted when it enters a substance.
The angle of incidence and angle of refraction of a ray as it crosses the interface between two media with refractive indices of n1 and n2 are, respectively, n1 sin 1 and n2 sin 2, which describes this.
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