Explanation:
s=(v^2-u^2)/2g
=(0-3^2)/2*10
= 0.45m
- The rocket now has a thruster that malfunctions and is now pushing the rocket in the wrong direction. What is the new net force on the rocket if it is now accelerating at 12mls²
From your previous question
Mass=30kg
Acceleration=12m/s^2
[tex]\\ \sf\longmapsto F=ma[/tex]
[tex]\\ \sf\longmapsto F=30(12)[/tex]
[tex]\\ \sf\longmapsto F=360N[/tex]
Given,
Acceleration = 12 m/s²
According to your previous question you posted,
Mass = 30 KG
Solution:
We know that the formula of force according to the Newton's second law is:
[tex]\longmapsto \sf \red{f = m \times a}[/tex]
[tex]\sf \longmapsto \ 30 \times 12[/tex]
[tex]\sf \longmapsto \: \boxed {\sf \blue3 \red60}[/tex]
Answer:
[tex]\sf \boxed{ \bf \: force = 360\: N}[/tex]
Một vật được ném xiên lên từ đỉnh của một tòa nhà cao 20m với góc 600 so với phương ngang, tốc độ ban đầu là 10m/s. Xác định độ tốc độ của vật lúc chạm đất?
Answer:
Một vật được ném xiên lên từ đỉnh của một tòa nhà cao 20m với góc 600 so với phương ngang, tốc độ ban đầu là 10m/s. Xác định độ cao cực đại của vật so với mặt đất?
Explanation:
In these images taken a few days apart, the light part of the Moon appeared to get smaller
over time. Why did this happen? (1)
a) The Moon moved farther away from the sun so less sunlight reached the Moon's
surface.
b) Earth moved so its shadow was blocking more of the Moon so the student astronomer
was able to see less of it.
c) The Moon rotated so that less of the light-colored rock on the Moon's surface faced
Earth
d) The Moon moved so that the student astronomer was able to see less of the half that
facer the sun
Answer:
Explanation:
d) The Moon moved so that the student astronomer was able to see less of the half that faced the sun.
A diffraction grating, ruled with 300 lines per mm, is illuminated with a white light source at normal incidence.
(i) What is the angular separation, in the third-order spectrum, between the 400 nm and 600 nm lines? [5]
(ii) Water (of refractive index 1.33) now fills the whole space between the grating and the screen. What is the angular separation, in the first-order spectrum, between the 400 nm and 600 nm lines? [5]
the expression for diffraction grating allows to find the results for the questions for the angular separation are:
i) The third order is Δθ = 0.203 rad.
ii) The first order with water is Δθ = 0.046 rad.
The diffraction grating is a system formed by a large number of equally spaced lines whose diffraction is given by the expression.
d sin θ = m λ
Where d is the distance between two lines, θ is the angle of diffraction, the order of diffraction and λ is the wavelength.
i) Let's start by looking for the separation between two lines
Let's use a rule of direct proportions. If there are 300 lines in 1 mm, what distance is there between two lines.
d = 1 lines (1 mm / 300 lines) = 3,333 10⁻³ mm
d = 3.333 10⁻⁶ m
Let's find the angle of diffraction for the third order (m = 3) for each wavelength.
λ₁ = 400 nm = 400 10⁻⁹ m
sin θ₁ = [tex]\frac{m \ \lambda }{d}[/tex]m λ/ d
sin θ₁ = [tex]\frac{3 \ 400 \ 10^{-9} }{3.333 \ 10^{-6} }[/tex]
θ₁ = sin⁻¹ 0.3600
θ₁ = 0.368 rad
λ₂ = 600 nm = 600 10⁻⁹ m
sin θ₂ = [tex]\frac{3 \ 600 \ 10^{-9} }{3.333 \ 10^{-6} }[/tex]
θ₂ = sin⁻¹ 0.5401
θ₂ = 0.571 rad
The angular separation is
Δθ = θ₂ - θ₁
Δθ = 0.571 - 0.368
Δθ = 0.203 rad
ii) In this case, the separation between the network and the observation screen is filled with water.
When the rays leave the network they undergo a refraction process, for which they must comply with the relationship.
[tex]n_i \ sin \theta_1 = n_r \ sin \theta_r[/tex]
The incident side is in the air, therefore its refractive index is n_i = 1 and when it passes into the water with refractive index n_r = 1.33.
Let's start looking for the incident angles for the first order of diffraction.
m = 1
λ₁ = 400 nm
θ₁ = sin⁻¹ [tex]\frac{1 \ 400 \ 10^{-9}}{3.33 \ 10^{-6}}[/tex]
θ₁ = 0.120 rad
λ₂ = 600 nm
θ₂ = sin⁻¹¹ [tex]\frac{1 \ 600 \ 10^{-9} }{3.33 \ 10^{-6}}[/tex]
θ₂ = 0.181 rad
we use the equation of refraction.
[tex]\theta_r[/tex] = sin⁻¹ ([tex]\frac{n_i}{n_r} \ sin \ \theta_i[/tex] )
λ₁ = 400 nm
θ₁ = sin¹ ([tex]\frac{1 sin 0.120}{1.33}[/tex]
θ₁ = 0.090 rad
λ₂ = 600 nm
θ₂ =sin⁻¹ [tex]\frac{1 sin 0.181}{1.33}[/tex]
θ₂ = 0.1358 rad
The angular separation is
Δθ = 0.1358 - 0.090
Δθ = 0.046 rad.
In conclusion using the relation for the diffraction grating we can find the results for the questions about angular separation are:
i) The third order is Δθ = 0.203 rad.
ii) The first order with water is Δθ = 0.046 rad.
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ALREADY SOLVED Thank you for looking.
Answer:
yes
Explanation:
chứng minh mặt trời là nguồn gốc của tất cả nguồn năng lượng
An Excerpt from “Optimism”
by Helen Keller
1 Could we choose our environment, and were desire in human undertakings synonymous with
endowment, all men would, I suppose, be optimists. Certainly most of us regard happiness as
the proper end of all earthly enterprise. The will to be happy animates alike the philosopher, the
prince and the chimney-sweep. No matter how dull, or how mean, or how wise a man is, he feels
that happiness is his indisputable right.
2 It is curious to observe what different ideals of happiness people cherish, and in what singular
places they look for this well-spring of their life. Many look for it in the hoarding of riches, some
in the pride of power, and others in the achievements of art and literature; a few seek it in the
exploration of their own minds, or in the search for knowledge.
3 Most people measure their happiness in terms of physical pleasure and material possession.
Could they win some visible goal which they have set on the horizon, how happy they would be!
Lacking this gift or that circumstance, they would be miserable. If happiness is to be so
measured, I who cannot hear or see have every reason to sit in a corner with folded hands and
weep. If I am happy in spite of my deprivations, if my happiness is so deep that it is a faith, so
thoughtful that it becomes a philosophy of life,—if, in short, I am an optimist, my testimony to
the creed of optimism is worth hearing....
4 Once I knew the depth where no hope was, and darkness lay on the face of all things. Then
love came and set my soul free. Once I knew only darkness and stillness. Now I know hope and
joy. Once I fretted and beat myself against the wall that shut me in. Now I rejoice in the
consciousness that I can think, act and attain heaven. My life was without past or future; death,
the pessimist would say, “a consummation devoutly to be wished.” But a little word from the
fingers of another fell into my hand that clutched at emptiness, and my heart leaped to the
rapture of living. Night fled before the day of thought, and love and joy and hope came up in a
passion of obedience to knowledge. Can anyone who has escaped such captivity, who has felt
the thrill and glory of freedom, be a pessimist?
5 My early experience was thus a leap from bad to good. If I tried, I could not check the
momentum of my first leap out of the dark; to move breast forward is a habit learned suddenly
at that first moment of release and rush into the light. With the first word I used intelligently, I
learned to live, to think, to hope. Darkness cannot shut me in again. I have had a glimpse of the
shore, and can now live by the hope of reaching it.
6 So my optimism is no mild and unreasoning satisfaction. A poet once said I must be happy
because I did not see the bare, cold present, but lived in a beautiful dream. I do live in a
beautiful dream; but that dream is the actual, the present,—not cold, but warm; not bare, but
furnished with a thousand blessings. The very evil which the poet supposed would be a cruel
6) Read the last sentence from the text.
Only by contact with evil could I have learned to feel by contrast the beauty of truth and love and goodness.
Explain how Helen Keller develops this idea in the text. Use specific details to
support your answer.
A uniform string of length 0.50 m is fixed at both ends. Find the
wavelength of the fundamental mode of vibration. If the wave
speed is 300 mis, find the frequency of the fundamental and next
possible modes.
Answer:
configuration of string:
Node - Antinode - Node or N-A-N
This is 1/2 wavelength since a full wavelength is N-A-N-A-N
f (fundamental) = V / wavelength
F0 = 300 m/s / 1 m = 100 / sec
F1 = 300 m/s / .5 m = 600 / sec
Each increase is a multiple of the fundamental since the wavelength
increases by 1/2 wavelength to keep nodes at both ends of the string
describe an experiment to demonstrate surface tension in a liquid
Answer:
mę břöőda I dunno this answer
Explanation:
hope you find it
An 80 kg box is accelerated to the right at a rate of 3 m/s2. What is the applied force?
As we know,
F= ma
Here,
applied force(f)= mass × acceleration
=80kg×3m/s2
=240kgm/s2
=240N
what is the difference between soldering and solder?
Answer:
Soldering is a technique or a process to bond metals with the help of solder. Solder is a metal alloy with a low melting point and is often regarded as the base metal for soldering. The piece or structure that is being bonded together does not get hot enough to melt and this is where solder (base metal) comes into play and creates the connection.
Explanation:
Hope this helps :)
Pls mark brainliest :3
And have an amazing day <3
A +8.75 μC point charge is glued down on a horizontal frictionless table. It is tied to a -6.50 μC point charge by a light, nonconducting 2.50 cm wire. A uniform electric field of magnitude 1.85×108N/C is directed parallel to the wire, as shown in the figure. (Figure 1)
Part A
Find the tension in the wire.
Express your answer with the appropriate units.
Part B
What would the tension be if both charges were negative?
Express your answer with the appropriate units.
(a) The tension on the wire when the two charges have opposite signs is 383.5 N.
(b) The tension on the wire if both charges were negative is 3.640.25 N.
The given parameters;
first charge, q₁ = 8.75 μC second charge, q₂ = -6.5 μC electric field, E = 1.85 x 10⁸ N/Cdistance between the two charges, r = 2.5 cm(a)
The attractive force between the charges is calculated as follows;
[tex]F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = -819 \ N[/tex]
The force on the negative charge due to the electric field is calculated as follows;
[tex]F_2 = Eq_2\\\\F_2 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_2 = 1202.5 \ N[/tex]
The tension on the wire is the resultant of the two forces and it is calculated as follows;
[tex]T = F_2 + F_1\\\\T = 1202.5 - 819\\\\T = 383.5 \ N[/tex]
(b) when the two charges are negative
The repulsive force between the two charges is calculated as follows;
[tex]F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (-8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = 819 \ N[/tex]
The force on the first negative charge due to the electric field is calculated as follows;
[tex]F_2 = Eq_1\\\\F_2 = (1.85 \times 10^8)\times (8.75 \times 10^{-6})\\\\F_2 = 1618.75 \ N[/tex]
The force on the second negative charge due to the electric field is calculated as follows;
[tex]F_3 = Eq_2\\\\F_3 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_3 = 1202.5 \ N[/tex]
The tension on the wire is the resultant of the three forces and it is calculated as follows;
[tex]T= F_1 + F_2 + F_3\\\\T= 819 + 1618.75 + 1202.5\\\\T = 3,640.25 \ N[/tex]
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The weather that characterizes an area is the O latitude of that area O barometric pressure of that area climate of that area O geography of that area
Answer:
climate of that area
Explanation:
Compare and contarst the difference between saturated and unsaturated solutions and supersaturated
Answer:
Unsaturated Solution: Less amount of salt in water, clear solution, no precipitation. Saturated Solution: The maximum amount of salt is dissolved in water, Colour of the solution slightly changes, but no precipitation. Supersaturated Solution: More salt is dissolved in water, Cloudy solution, precipitation is visible
can anyone heelp me pls
Answer:
Lotion - Semisolid
Suspension - Liquid
Capsule - Solid
Light of 650 nm wavelength illuminates two slits that are 0.20 mm
apart. (Figure 1) shows the intensity pattern seen on a screen behind the slits.
What is the distance to the screen?
PLEAsE URGENT. thank you
We have that for the Question "Light of 650 nm wavelength illuminates two slits that are 0.20 mm apart. (Figure 1) shows the intensity pattern seen on a screen behind the slits.
What is the distance to the screen?" it can be said that the distance to the screen
d=1.168m
From the question we are told
Light of 650 nm wavelength illuminates two slits that are 0.20 mm
apart. (Figure 1) shows the intensity pattern seen on a screen behind the slits.
What is the distance to the screen?
Generally the equation for the distance is mathematically given as
[tex]d=\frac{0.33*10^{-2}*0.23*10^{-3}}{650*10^{-9}}\\\\d=1.160m[/tex]
d=1.168m
Therefore
The distance to the screen
d=1.168m
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The following question has two parts. First, answer part A. Then, answer part B.
Part A:
Why did the author most likely choose the title “Breaking Barriers” for this text?
A.
to show that Elizabeth worked hard to make people like her
B.
to show that Elizabeth accomplished things never done before
C.
to show that Elizabeth proved to be smarter than those around her
D.
to show that Elizabeth tried to show she was the better than other doctors
Part B:
Which sentences from the text best support your answer in part A? Select two options.
A.
“Elizabeth loved learning, and she developed a passion for reading.”
B.
“When Elizabeth was 11 years old, her family moved to America.”
C.
“When she was just 16, she became a teacher.”
D.
“Finally, the Geneva Medical College in New York accepted her as a student.”
E.
“When people in town ignored the woman with the strange ambition, Elizabeth treated them in a polite, quiet manner.”
F.
“She became the first woman in the country to obtain a medical degree.”
Answer:
for part A the answer is D and for part B the answer is F
In a vacuum light travels at which speed?
Answer:
Light traveling through a vacuum moves at exactly 299,792,458 meters (983,571,056 feet) per second. That's about 186,282 miles per second
Explanation:
find charge and charge density on the surface of a conducting sphere of radius 15.2cm where potential at 215 v
this is the correct answer
in order to keep heat in or out, you need a(n)
Answer:
What is instillation
Explanation:
--[50 POINTS]--
1)A block of mass 25 kg is placed on flat ground. The coefficient of static friction and kinetic friction are 0.73 and 0.16
a.If a person pushes the block and the block is moving, what will be the acceleration of the block?
2) A block has a mass of 79 kg. The coefficient of static and kinetic friction between the sled and the ground is 0.87 and 0.37. Person A tries to pull the block with 210N, but fails.
a) Person B successfully pulls the sled with 909N. What is the acceleration of the sled?
Newton's second law allows us to find the results for the acceleration of the blocks are:
1) The acceleration is a = 559 m / s²
2) The acceleration is a = 7.88 m / s²
Newton's second law states that the net force is equal to the product of the mass and the acceleration of the body.
∑ F = m a
Where the bold letters indicate vectors, F is the force, m the mass and the acceleration of the body.
The reference system is a coordinate system with respect to which the decomposition of the forces is carried out, in the attached we have a free body diagram of the system.
1) They indicate that the body mass is 25 kg.
y-axis
N - W = 0
N = W = m g
x-axis
F -fr = ma
The friction force is a macroscopic force that results from the sum of all the microscopic interactions between the two surfaces, it has the formula
fr = μ N
Where fr is the friction force, N the normal and very the friction coefficient.
This friction coefficient has two values:
Static. For when with there is not relative motion between the two surfaces. Dynamic. When there is relative motion between the two surfaces.
We substitute.
F - μ m g = m a
a) The system moves which is the acceleration.
Suppose that the force that star to move the system keeps constant, just before the system begins to move the coefficient of friction is static, let's find the applied force.
F = μ m g
F = 0.73 25 9.8
F = 178.85 N
The block begins to move and the friction coefficient decreases to the dynamic value, we look for the acceleration.
a = [tex]\frac{F - \mu \ m g}{m}[/tex]
a = [tex]\frac{178.85 - 0.16 \ 25 \ 9.8 }{25}[/tex]
a = 5.59 m / s²
2) In this case the mass of the block is 79 kg and the applied force is
F = 909 N
We look for acceleration.
a = [tex]\frac{909 - 0.37 \ 79 \ 9.8 }{79}[/tex]
a = 7.88 m / S²
In conclusion using Newton's second law we can find the results for the acceleration of the blocks are:
1) The acceleration is a = 559 m / s²
2) The acceleration is a = 7.88 m / s²
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A box is pushed horizontally with constant speed across a rough horizontal surface. Which of the following must be true?
a) The friction is less than the applied force
b) There is no friction acting on the box
c) The friction is greater than the applied force
d) The friction is equal to the applied force
Which of the following is the current best hypothesis for the formation of the solar system?
A. Formed by an exploding super nova star which then collapsed and coalesced into a spinning
disk forming Sun and planets
B. Our solar system has always been here and has never changed
C. Formed from the Sun’s explosion releasing particles into space forming planets and other
objects
D. Our solar system was formed by a great collision of other stars with one another
Answer:
A
Explanation:
all galaxies exploded in order to create the sun/stars
What element is chemically similar to carbon
nonmental
that should be the answer
When the pushing force is increased to 88.2 N, the box just begins to move. What is the Friction Force if it is moving at a constant velocity? What is the coefficient of friction between the ground and box?
In order to calculate frictional force look below..
The formula given by
[tex]\\ \sf\longmapsto F_f=\mu N[/tex]
Or
[tex]\\ \sf\longmapsto F_f=\mu mg[/tex]
u is coefficient of friction
N is normal reaction.
A baseball is hit so that it travels straight upward after being struck by the bat. If its initial velocity is 29 m/s , then what is the maximum height that it will reach?
Answer:
Explanation:
Use kinematic equation v² = u² + 2as
Rearrange for distance
s = (v² - u²) / 2a
Realize that at the top of its flight, the ball has zero velocity and gravity is acting downward in an assumed upward positive reference frame.
s = (0² - 29²) / (2(-9.8))
s = 42.90816...
s = 43 m
The lantern rises up because the __________ of the _________inside the lantern is _________ compared to the air around it
liquid
Gas
Density
fire
less dense
the same density
more dense
paper
Answer:
gass fire and the same density
Explanation:
can anyone heelp me pls pls
Answer:
Lotion : semisolid
Suspension : liquid
capsule : solid
Explanation:
a block starts from rest and begins sliding down an incline. The block reaches a speed of 12 meters per second as it slides a distance of 50 meters. calculate the blocks rate of acceleration
Vf² = V₁² + 2ad
(12 m/s) ² = (0 m/s) ² + 2 (a) (50m)
a = 1.44 m/s ²
Answer:
[tex]\boxed {\boxed {\sf 1.44 \ m/s^2}}[/tex]
Explanation:
We are asked to find the block's rate of acceleration.
We are given the final velocity, initial velocity, and distance. Therefore, we will use the following kinematic equation:
[tex]{v_f}^2={v_i}^2+2ad[/tex]
The block starts at rest or 0 meters per second. The block reaches a final velocity of 12 meters per second. The distance traveled is 50 meters.
[tex]v_i[/tex]= 0 m/s [tex]v_f[/tex]= 12 m/s [tex]d[/tex]= 50 mSubstitute the values into the formula.
[tex](12 \ m/s)^2 = (0 m/s)^2 + 2a (50 \ m)[/tex]
Solve the exponents.
(12 m/s)² = 12 m/s * 12 m/s = 144 m²/s²(0 m/s)² = 0 m/s * 0 m/s = 0 m²/s²[tex]144 \ m^2/s^2= 0 m^2/s^2 + 2a (50 \ m)[/tex]
The 0 can be subtracted from both sides, or simply canceled.
[tex]144 \ m^2/s^2 = 2a (50 \ m)[/tex]
Multiply 2 and 50 meters.
[tex]144 \ m^2/s^2 = (2 *50 \ m)* a[/tex]
[tex]144 \ m^2/s^2 = (100 \ m)* a[/tex]
We are solving for a, so we must isolate the variable. It is being multiplied by 100 meters. The inverse of multiplication is division, so we divide both sides by 100 meters.
[tex]\frac {144 \ m^2/s^2}{100 \ m}= \frac{(100 \ m)*a}{100 \ m}[/tex]
[tex]\frac {144 \ m^2/s^2}{100 \ m}=a[/tex]
[tex]1.44 \ m/s^2 =a[/tex]
The block's rate of acceleration is 1.44 meters per second squared.
Atoms of which two elements could combine with atoms of carbon (C) to
form covalent bonds?
A. S
B. Na
C. N
D. K
E. Ca
Answer:
Correct answer is letter E.Ca
Which of the following statements about electromagnetic radiation it true? A.electromagnetic waves with long wavelength are more energetic then electromagnetic waves with short wavelength. B.all electromagnetic radiation carries the same amount of energy. C.electromagnetic radiation in a vacuum can change frequently to become more or less energetic. D.electromagnetic waves with high frequency are more energetic then electromagnetic waves with low frequency
Given what we know, we can confirm that option D, Electromagnetic waves with high frequency are more energetic than electromagnetic waves with low frequency is true.
Why are high-frequency waves more energetic?High-frequency waves are synonymous with short wavelengths. This means that the waves are oscillating much quicker and therefore carry more kinetic energy within them. This is transformed and released as electromagnetic radiation, which is the reason why high-frequency waves are more energetic than low-frequency electromagnetic waves.
Therefore, we can confirm that the statement "Electromagnetic waves with high frequency are more energetic than electromagnetic waves with low frequency" is true.
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