Answer:
Sorry I can't understand you
What does it mean that " memory is organized in semantic
networks
Answer:
Semantic memory is a category of long-term memory that involves the recollection of ideas, concepts and facts commonly regarded as general knowledge. Examples of semantic memory include factual information such as grammar and algebra.
A 770-kg two-stage rocket is traveling at a speed of 6.90×103 m/s away from Earth when a predesigned explosion separates the rocket into two sections of equal mass that then move with a speed of 2.60×103 m/s relative to each other along the original line of motion.What is the speed of each section (relative to Earth) after the explosion?How much energy was supplied by the explosion?
Answer:
Explanation:
Let's just have our reference frame travel along with the original un broken mass. This way the original velocity is not relevant.
Each half will have a mass of 770/2 = 385 kg
Each half will have the same magnitude of velocity (conservation of momentum) which will be 2.6 x 10³/2 = 1.30 x 10³ m/s
Now add back the reference frame velocity to get velocity relative to earth.
Section one will have velocity 6.90 x 10³ + 1.30 x 10³ = 8.2 x 10³ m/s
Section two will have velocity 6.90 x 10³ - 1.30 x 10³ = 5.6 x 10³ m/s
In the moving reference frame, each half will have kinetic energy which could only come from the explosion
KE = ½(385)(1.3 x 10³)² = 325,325,000 J
2(325,325,000) = 650,650,000 J released in the explosion.
Rounding to the three significant figures of the problem numerals
E = 6.50 x 10⁸ J or 650 MJ released
50 points help
Column I Column II
______ 1. acceleration a. change in distance over time
______ 2. speed b. time interval
______ 3. velocity c. scalar
______ 4. Δt. d. change in position
______ 5. Magnitude only e. change in velocity over time
______ 6. Δx f. change in displacement over time
[tex]\\ \sf\longmapsto Acceleration\longrightarrow Change\:in\: Velocity\:over\:time[/tex]
[tex]\\ \sf\longmapsto Speed\longrightarrow Change\:in\: Distance\:over\:Time[/tex]
[tex]\\ \sf\longmapsto Velocity\longrightarrow Change\:in\: Displacement\: over\:time[/tex]
[tex]\\ \sf\longmapsto ∆t\longrightarrow Time\: interval [/tex]
[tex]\\ \sf\longmapsto Magnitude\:only\longrightarrow Scaler[/tex]
[tex]\\ \sf\longmapsto ∆x=Change\:in\: position [/tex]
Does the stone remain at a constant speed? Or does it speed up?
Answer:
it would remain the same speed
Explanation: the rock isnt going down a hill or anything so therefore if gravity isnt pulling it down a slope then it would remain the same pace
PLEASE HELP!!!
If a 40cm rope with a 220g bob can hold a maximum tension of 3N
a) what are the maximum angular velocity and inclination angle it can reach before the rope break?
b) Angle of inclination
Answer:
ω = 3.1 rad/s
θ = 36° from vertical
Explanation:
I will ASSUME that the bob and string is acting as a pendulum.
Please understand that the string will break when the bob is at the lowest point of the swing where the vectors of gravity and centripetal acceleration align. It will NOT break at the angle of maximum inclination measured from vertical. This angle is only a component of the maximum potential energy that gets converted to maximum kinetic energy at the lowest point of the swing.
At the bottom of the swing, the string must support the weight of the bob plus supply the required centripetal acceleration.
F = mg + mω²R
F/m = g + ω²R
F/m - g = ω²R
ω = √((F/m - g)/R)
ω = √((3/0.220 - 9.8)/0.40)
ω = 3.09691...
ω = 3.1 rad/s
Potential energy will convert to kinetic energy
mgh = ½mv²
h = v²/2g
R - Rcosθ = v²/2g
R(1 - cosθ) = v²/2g
1 - cosθ = v²/2gR
cosθ = 1 - v²/2gR
cosθ = 1 - (Rω)²/2gR
cosθ = 1 - Rω²/2g
cosθ = 1 - 0.40(3.1²)/(2(9.8))
cosθ = 0.804267
θ = 36.46045...
θ = 36°
Change 1m2 in to cm2, mm2 and km2
Answer:
1m² = 10000cm²
1m² = 1000000mm²
1m² = 1 × 10^-6
Explanation:
When coverting from square meter to square centimetre, multiply the area value by 10.
When coverting from square meter to millimetre, multiply the area value by 10⁶ ( 1 million )
When converting from meter square to kilometre square, divide the area value by 10^-6 ( 0.000001)
Am object of mass in a circular path of radius 100metres with the speed of 10metres/second.calculate the acceleration towards the center
Answer:
is 6
Explanation:
h
How can gravity's role in tectonic plate motion be described?
Answer: In ridge push, the mantle wells upward because of the convection and elevates the edges of spreading oceanic plates. Because these plates are higher at the spreading center, they are forced downhill due to gravity and eventually flatten out to the ocean floor.
the answer I need points for my math test
I need the answer for this question
Answer:
Since impedance = omega * L we can use the equation for resistance:
20 + 12 + 10 in parallel with 7 we have
J1 = 42 * 7 / 49 = 6 H for the combination on the right
4 H + 6 H + 8 H = 18 H for the entire combination
which of the following is used to transport sound waves
A.medium
B.vacuum
C.mass
D.light
I think the answer is d.right?
Please help if u can thanksss
Answer:
yes its okay i think so it is the correct answer
what a function of vacuole
Answer:
A vacuole is in both, plant and animal cells. It's main purpose is to isolate waste products, or in other words making sure that harmful materials won't damage the cell.
Answer:
In animal cells, they help sequester waste. In plants, they help maintain water balance.
An airplane has a mass of 33,000 kg and takes off under the influence of a constant net force of 44,000 N. What is the net force that acts on the plane's 70 kg pilot
Answer:
Explanation:
acceleration a = F/m = 44000/33000 = 1.3m/s²
F = ma = 70(1.3) = 93 N
5. A quarterback throws the football to a stationary receiver who is 31.5 m
down the field. If the football is thrown at an initial angle of 40.0° to the
ground, at what initial speed must the quarterback throw the ball for it
to reach the receiver? What is the ball's highest point during its flight?
The projectile launch equations allow to find the results for the questions about the movement of the ball are:
The initial velocity is: v₀ = 17.7 m / s. The maximum height is: y = 16 m.
Given parameters
Horizontal distance x = 31.5 m Launch angle tea = 40ºTo find
The initial speed. Maximum height.
Projectile launching is an application of kinematics, where on the x-axis there is no acceleration and on the y-axis is the gravity acceleration.
The range is the distance traveled for the same departure height, see attached.
.
R =[tex]\frac{v_o^2 \ sin 2\theta}{g}[/tex]
[tex]v_o^2 = \frac{ g R}{sin 2 \theta }[/tex]
Let's calculate.
v₀² = [tex]\frac{9.8 \ 31.5}{sin \ (2 \ 40)}[/tex]9.8 31.5 / sin (2 40.0)
[tex]v_o = \sqrt{313.46}[/tex]o = ra 313.46
v₀ = 17.7 m / s
At the point of maximum height the vertical speed is zero.
v² = v₀² - 2 g y
0 = v₀² - 2g y
y = [tex]\frac{v_o^2}{2g}[/tex]
Let's calculate.
y = [tex]\frac{17.7^2}{2 \ 9.8}[/tex]
y = 16 m
In conclusion, using the projectile launch equations we can find the results for the questions about the movement of the ball are:
The initial velocity is: v₀ = 17.7 m / s The maximum height is: y = 16 m.
Learn more here: brainly.com/question/10903823
7 A 0.25 kg block oscillates on the end of a spring with a spring constant of 100 N/m. If the oscillation is started by elongating the spring 0.1m and giving the block a speed of 3 mis then the amplitude of the oscillation is:
A mountain climber encounters a crevasse in an ice field. The opposite side of the crevasse is a height h lower, and is separated horizontally by a distance w. To cross the crevasse, the climber gets a running start and jumps in the horizontal direction. If the height of the crevasse increases but the width remains the same, then,
Select one:
O a. the minimum speed needed to cross the crevasse stays the same.
O b. the minimum speed needed to cross the crevasse will depend on the mass of the mountain climber.
O c. the minimum speed needed to cross the crevasse decreases.
O d. the minimum speed needed to cross the crevasse increases.
O e. the minimum speed needed to cross the crevasse will depend on the weight of the mountain climber.
Define average atomic mass and explain how it is calculated
Answer:
The average atomic mass of an element is the sum of masses of it's isotopes
Each are multiplied by it's natural abundance
Explanation:
The average atomic mass of an element is the sum of masses of it's isotopes
Each are multiplied by it's natural abundance
the height of the upper falls at Yellowstone falls is 33.2m. when the water reaches the botton of the falls, its speed is 25.8 m/s. neglecting air resistance, what is the speed of the water at the to of te falls.
Answer:
Hi there
The water acquires a vertical velocity component during the fall equal to
Vy = sqrt(2gH) = 25.5 m/s. Its velocity at the bottom is 25.8 m/s, so the horizontal component at the bottom must be sqrt[(25.8)^2 - (25.5)^2] = 3.9 m/s
That must be the horizontal component at the top also, since it does not change during the fall. It is the velocity magnitude there.
Brainliest pls
Please write a paragraph explaining the bible verse below in your own words.
Exodus 16:1-3
Answer: Moses
Explanation: The Israelites are angry at Aaron and Moses, because they lead them into this desert where there is no food, or water. It was told that after the Israelites left Egypt they would wander the desert for forty years. The Israelites complain that they would have had plenty of food if they had just stayed enslaved in Egypt. They are not grateful that God has liberated them from slavery.
An object is moving due north at 31m/s. Does this tell you the speed of the velocity of the object or both?
Answer:
speed
Explanation:
its the speed
because the formula for deriving speed is distance/time
the unit for distance is metre
the unit for time is seconds
so speed is metre/seconds
so speed us m/s.
Which statement about RNA polymerase is NOT true?
RNA polymerase reads a template strand of DNA 5' to 3'
RNA polymerase binds to a promoter region to initiate transcription
RNA polymerase adds a ribonucleotide to the 3' end of a growing RNA molecule
During transcription of a gene, RNA polymerase reads only one strand of DNA
Please explain!
Answer:
RNA polymerase reads a template strand of DNA 5' to 3'
Explanation:
RNA polymerase is the enzyme accountable for the transcription of the DNA. It shows one coast of the DNA and transcribed within the RNA. It combines the nucleotide to the 3' conclusion of the expanding series of the RNA. It commences the method by connecting to the promoter area of the gene. The enzyme increases nucleotide, not ribonucleotide.
Therefore, A. RNA polymerase reads a template strand of DNA 5' to 3' is false.
Answer:
RNA polymerase reads a template strand of DNA 5' to 3'.
Explanation:
RNA synthesis is less processive during elongation than during initiation. Synthesis begins with the polymerase binding two rNTP molecules. The polymerase frequently releases the nascent transcript before it reaches 8-10 nucleotides in length.
The figure shows the light intensity on a screen behind a double slit. The slit spacing is 0.22 mm and the screen is 2.0 m behind the slits (Figure 1). What is the wavelength of the light?
The wavelength of the light is 550 nm
For a double slit interference pattern with slit spacing, d we have
dsinθ = mλ where d = slit spacing = 0.22 mm = 0.22 × 10⁻³ m, m = number of maximum fringe = 2(from the picture) and λ = wavelength of light.
Thus sinθ = mλ/d
Also, tanθ = L/D where L = distance between central maximum and fringe = 2.0 cm/2 = 1.0 cm = 1 × 10⁻² m and D = distance between slit and screen = 2.0 m
Since θ is small, sinθ ≅ tanθ
So, mλ/d = L/D
Making λ subject of the formula, we have
λ = dL/mD
Substituting the values of the variables into the equation, we have
λ = dL/mD
λ = 0.22 × 10⁻³ m × 1 × 10⁻² m/(2 × 2.0 m)
λ = 0.22 × 10⁻⁵ m²/4.0 m
λ = 0.055 × 10⁻⁵ m
λ = 0.55 × 10⁻⁶ m
λ = 550 × 10⁻⁹ m
λ = 550 nm
So, the wavelength of the light is 550 nm
Learn more about double slit interference here:
https://brainly.com/question/4449144
A hot air balloon rising vertically is tracked by an observer located 2 miles from the lift-off point. At a certain moment, the angle between the observer's line-of-sight and the horizontal is , and it is changing at a rate of 0.1 rad/min. How fast is the balloon rising at this moment
We have that for the Question, it can be said that
the balloon rising at [tex]0.266miles/min[/tex]From the question we are told
An observer located 2 miles from the lift-off point. At a certain moment, the angle between the observer's line-of-sight and the horizontal is , and it is changing at a rate of 0.1 rad/min.From,
[tex]tan\theta = \frac{h}{2}[/tex]
differentiate with respect to h
[tex]sec^2\theta * \frac{do}{dz} = \frac{1}{2} * \frac{dh}{dz}\\\\\frac{dh}{dz} = 2 sec^\theta * \frac{d\theta}{dz}\\\\\theta = \frac{\pi}{6} and \frac{d\theta}{dz} = 0.1rad/min\\\\\frac{dh}{dz} = 2sec^2 (\frac{\pi}{6}) * (0.1)\\\\= 0.266miles/min[/tex]
For more information on this visit
https://brainly.com/question/23379286
Explain how friction keeps a nail in place in a block of wood. If you try pull out the nail which way does friction act? I will give brainliest!
Answer:
Friction acts in the opposite direction of the velocity. As static friction acts in the opposite direction to the acceleration. (opposite way)
Explanation:
True or false. The buildings in cities are getting larger
Answer:
trueExplanation:
cuase the city is getting more people than beforeA car drives at a constant speed of 45km in 20mn. The speed of the car is..?
Answer:
83.8851 mi/h
The use of non-renewable energy resources in the UK has changed in the last 30 years.
Explain how the use of energy resources has changed in the last 30 years.
Answer: Varies
Explanation: It became more relied upon, technology is responsible for this.
b) A ball thrown vertically into the air
Discuss newton law
The first law is used to visualize this. The ball is at a consistent velocity, unless acted on by another force, when the ball goes upwards and there's no force acting on it, it keeps going upwards, but there is gravity. When the ball is going up, you do not interact with it, thus only gravity is the only force, making the ball go downwards.
To examine this, apply the first law. When the ball is moving upward and there is no force acting on it, it continues to move higher but there is gravity. The ball is at a constant velocity unless pushed upon by another force.
What is gravity?According to Newton's law of gravity, every particle of matter in the universe is attracted to every other particle with a force that varies directly as the product of their masses and inversely as their distance from one another.
Anything that can alter an object's motion is a force. When you toss a ball, you apply force in a direction that determines the direction in which it will move. The ball travels farther the harder you toss it because it is under greater stress.
The ball continues to go upward even though there is no force pushing against it because of gravity. Unless another force pushes on it, the ball is moving at a steady speed.
Thus, this way, when ball thrown vertically into the air shows Newton's first law.
For more details regarding Newton's laws of motion, visit:
https://brainly.com/question/29775827
#SPJ2
PLEASE PLEASE HELP!!
An arrow is shot straight up in the air from the ground with an initial velocity of 54.0 m/s. If on striking the ground it
embeds itself 15.0 cm into the ground, what is the acceleration required to stop the arrow when it hits the ground?
Answer:
you have patience the distance.
Explanation:
the train leaves at 6.30.
A bear slides down an incline which is oriented at 17 degrees above horizontal. What is the coefficient of friction between the bear and the incline if the bear moves with a constant velocity?
[tex]\mu = 0.306[/tex]
Explanation:
Assume that the direction down the incline is the +x-direction. We can apply Newton's 2nd law along the x-axis as
[tex]x:\:\:\:\:\:mg\sin17° - f_s = ma[/tex]
[tex]\Rightarrow mg\sin17° - \mu N = 0[/tex]
where m is the mass of the bear and g is the acceleration due to gravity and acceleration a is zero since the bear is moving with a constant velocity. Along the y-axis, we can write Newton's 2nd law as
[tex]y:\:\:\:\:\:N - mg\cos17° = 0 \Rightarrow N = mg\cos17°[/tex]
Combining these two equations together, we get
[tex]mg\sin17° = \mu(mg\cos17°)[/tex]
Solving for [tex]\mu, [/tex]
[tex]\mu = \dfrac{\sin17°}{\cos17°} = \tan17° = 0.306[/tex]
