Answer:
D. Iron is more easily magnetized than steel but steel is more easily demagnetized than iron.
When you use an array to implement the ADT list, retrieving an entry at a given position is slow.
a) true
b) false
When you use an array to implement the ADT list, adding an entry at a the end of the list is fast.
a) true
b) false
When you use a vector to implement the ADT list, retrieving an entry at a given position is slow.
a) true
b) false
When you use a vector to implement the ADT list, adding an entry at a the end of the list is fast.
a) true
b) false
Answer:
1)FALSE
2)TRUE
3)FALSE
4) FALSE
Explanation:
1) Retrieving an entry at a given position is not slow ( it is done using index number ) hence the answer is FALSE
2) Adding an entry at the end of the list is Fast this is because in using array implementation a new entry is made immediately after the last position and it is done very fast as well. hence the answer is TRUE
3) Retrieving an entry at a given position ( from an ADT list ) that is been implemented using Vector is very fast this is simply because Indexing in vector is done at a fixed time hence the answer is FALSE
4) Adding an entry at the end of the list is slow using vector hence answer is FALSE
A 62 kg student, starting from rest, slide down an 10.6 m high water slide. How fast is he going at the bottom of the slide? Use g = 10 m/s2
Answer:
14.6m/s
Explanation:
Given parameters:
Mass of the student = 62kg
Initial velocity = 0m/s
Height of slide = 10.6m
g = 10m/s²
Unknown:
Speed at the bottom of the slide = ?
Solution:
The speed at the bottom of the slide is the final velocity;
v ² = u² + 2gh
v is the final velocity
u is the initial velocity
g is the acceleration due to gravity
h is the height
v² = 0² + 2x 10 x 10.6
v² = 212
v = 14.6m/s
I NEED HELP ASAP PLEASE!
Study the scenario.
Boiling water is poured into a Styrofoam cup and a metal cup. You pick up and hold the metal cup and the Styrofoam cup. The metal cup burns your fingers, but you can hold the Styrofoam cup without experiencing any pain.
The difference in the rate at which these two materials transfer thermal energy throughout the cups is known as __________.
evaporation
electrical energy
thermal conductivity
condensation
Answer: thermal conductivity
The difference in the rate at which these two materials transfer thermal energy throughout the cups is known as thermal conductivity.
What is thermal conductivity?It is the rate where the heat is to be transferred from conduction via the cross section area of the unit with respect to the material at the time when the temperature should be gradient and existed with the perpendicular to the area.
Also, in this, two material is to be transferred via the thermal energy with the help of cups.
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Many scientific studies have found that colds are caused by viruses. What is this? *
Fact
Interpretation
Analysis
Opinion
The levels of organization are Cell, Tissue, Organs, ____________, and Organism. what goes in the blank?
Answer:
Organ system because multiple organs make up an organ system and multiple organ systems make up an organism
a car moved 60 meters west in 2hours what is its average velocity
Answer:
30 m/h
Explanation:
Have A Wonderful Day !!
Can someone help me out please I got it wrong
Answer:
3 maybe since protons=atomic
Our Sun’s mass is 1.0 and our Earth’s mass is 2.0. The distance is standard as given on the simulation. Describe the path of the Earth.
Answer:
Earth orbits the Sun at an average distance of 149.60 million km (92.96 million mi), and one complete orbit takes 365.256 days (1 sidereal year), during which time Earth has traveled 940 million km (584 million mi).
Explanation:
What is the correct organization of living things, from smallest to largest?
Cells - Tissues - Organs - Organ Systems - Organism
Organs - Tissues - Cells - Organ Systems - Organism
Cells - Organs - Tissues - Organism - Organ Systems
Cells - Organism - Tissues - Organ Systems - Organs
PLS HELP ME!
A motorist is traveling 40ms-¹ and applies brakes and slow down at a rate of 2ms-² the available distance for the the motorist to stop is 400m will the motorist be able to stop?
Answer:
[tex] \underline{ \boxed{ yes}}\\[/tex]
Explanation:
[tex]given : initial \: velocity \: (u )= 40 {ms}^{ - 1} \\ given : final \: velocity \: (u )= 0 {ms}^{ - 1} \\ given : - (acceleration) \: (a_r) = 2 {ms}^{ - 2} \\ given : distance \: (s) \: = \: ? : \\ but \: {v}^{2} = {u}^{2} + 2( a)s\\ {0}^{2} = {40}^{2} + 2( - 2)s \\ - {40}^{2} = - 4s \\ s = \frac{ - {40}^{2} }{ - 4} \\ s = \frac{1600}{4} \\s = 400 \: m[/tex]
A 2.6 kg ball is accelerated at 4.5 m/s2.
Calculate the force needed to achieve this feat.
Show all work including formula and units!
Answer:
[tex]12\:\mathrm{N}[/tex]
Explanation:
Force is given by the equation [tex]F=ma[/tex].
Plugging in given values, we have:
[tex]F=ma=2.6\cdot 4.5=11.7=\fbox{$12\:\mathrm{N}$}[/tex] (two significant figures).
A block of mass m = 4.4 kg slides from left to right across a frictionless surface with a speed 9.2 m/s It collides in a perfectly elastic collision with a second block of mass M that is at rest. After the collision, the 4.4-kg block reverses direction, and its new speed is 2.5 m/s The block of mass M travels to the right at a speed V of 6.7 m/s what is M
Answer:
[tex]m_2=6.3\:\mathrm{kg}[/tex]
Explanation:
In a perfectly elastic collision, the total kinetic energy of the system is maintained. Therefore, we can set up the following equation:
[tex]\frac{1}{2}m_1{v_1}^2+\frac{1}{2}m_2{v_2}^2=\frac{1}{2}m_1{v_{1'}}^2+\frac{1}{2}m_2{v_{2'}}^2[/tex]
Since the second block was initially at rest, [tex]\frac{1}{2}m_2{v_2}^2=0[/tex].
Plugging in all given values, we have:
[tex]\frac{1}{2}m_1{v_1}^2=\frac{1}{2}m_1{v_{1'}}^2+\frac{1}{2}m_2{v_{2'}}^2,\\\\\frac{1}{2}\cdot4.4\cdot9.2^2=\frac{1}{2}\cdot 4.4 \cdot (-2.5)^2+\frac{1}{2}\cdot m_2\cdot 6.7^2,\\m_2=\fbox{$6.3\:\mathrm{kg}$}[/tex].
What statement is not an example of Newton’s first law of motion
Answer:
c
Explanation:
im smart....................... i think
This law is about inertia, and the law displayed in A is Newton's third law of equal and opposite reactions, so option A is correct.
What is Newton’s first law of motion?The basis of classical mechanics is laid out in three assertions known as Newton's laws of motion, which were first articulated by English physicist and mathematician Isaac Newton. These laws describe the relationships between forces acting on a body and its motion.
Unless a force acts on a body that is at rest or moving in a straight line at a constant speed, Newton's first law asserts that it will continue to be at rest or move in that direction.
This law is about inertia (an object wanting to stay in its state of motion) and the law displayed in A is Newton's third law of equal and opposite reactions, therefore, it is not an example of Newton’s first law of motion.
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At a certain instant, a ball is thrown downward with a velocity of 8.0 m/s from a height of 40 m. At the same instant, another ball is thrown upward from ground level directly in line with the first ball with a velocity of 12 m/s. Find (a) the time when the balls collide and (b) the height at which they collide. Take g = 10 m/s2
Answer:
(a) The two balls collide [tex]2\; \rm s[/tex] after launch.
(b) The height of the collision is [tex]4\; \rm m[/tex].
(Assuming that air resistance is negligible.)
Explanation:
Let vector quantities (displacements, velocities, acceleration, etc.) that point upward be positive. Conversely, let vector quantities that point downward be negative.
The gravitational acceleration of the earth points dowards (towards the ground.) Therefore, the sign of [tex]g[/tex] should be negative. The question states that the magnitude of [tex]g\![/tex] is [tex]10\; \rm m \cdot s^{-2}[/tex]. Hence, the signed value of [tex]\! g[/tex] should be [tex]\left(-10\; \rm m \cdot s^{-2}\right)[/tex].
Similarly, the initial velocity of the ball thrown downwards should also be negative: [tex]\left(-8.0\; \rm m \cdot s^{-1}\right)[/tex].
On the other hand, the initial velocity of the ball thrown upwards should be positive: [tex]\left(12\; \rm m \cdot s^{-1}\right)[/tex].
Let [tex]v_0[/tex] and [tex]h_0[/tex] denote the initial velocity and height of one such ball. The following SUVAT equation gives the height of that ball at time [tex]t[/tex]:
[tex]\displaystyle h(t) = \frac{1}{2}\, g \cdot {t}^{2} + v_0 \cdot t + h_0[/tex].
For both balls, [tex]g = \left(-10\; \rm m \cdot s^{-2}\right)[/tex].
For the ball thrown downwards:
Initial velocity: [tex]v_0 = \left(-8.0\; \rm m \cdot s^{-1}\right)[/tex].Initial height: [tex]h_0 = 40\; \rm m[/tex].[tex]\displaystyle h(t) = -5\, t^{2} + (-8.0)\, t + 40[/tex] (where [tex]h[/tex] is in meters and [tex]t[/tex] is in seconds.)
Similarly, for the ball thrown upwards:
Initial velocity: [tex]v_0 = \left(12\; \rm m \cdot s^{-1}\right)[/tex].Initial height: [tex]h_0 = 0\; \rm m[/tex].[tex]\displaystyle h(t) = -5\, t^{2} + 12\, t[/tex] (where [tex]h[/tex] is in meters and [tex]t[/tex] is in seconds.)
Equate the two expressions and solve for [tex]t[/tex]:
[tex]-5\, t^{2} + (-8.0)\, t + 40 = -5\, t^{2} + 12\, t[/tex].
[tex]t = 2[/tex].
Therefore, the collision takes place [tex]2\, \rm s[/tex] after launch.
Substitute [tex]t = 2[/tex] into either of the two original expressions to find the height of collision:
[tex]h = -5\times 2^{2} + 12 \times 2 = 4\; \rm m[/tex].
In other words, the two balls collide when their height was [tex]4\; \rm m[/tex].
The time the two balls collide is 0.4 seconds while the height at which they collide is 4m
The given parameters are :
Initial Velocity U = 8m/s
Height H = 40m
For the second ball, the initial velocity = 12m/s
a.) For the first ball, the height attained at the point of collision will be
h = ut + 1/2gt^2
h = 8t + 1/2 x 10t^2 ........ (1)
For the second ball, the height attained at the point of collision will be
h = 12t - 1/2 x 10t^2 .........(2)
Since the height will be the same for the two balls, equate the two equations
8t + 10t^2 = 12t - 10t^2
Collect the like term
8t - 12t = -5t^2 - 5t^2
-4t = -10^2
10t = 4
t = 4/10
t = 0.4s
b.) Substitute time t in any of the equation to find the height
h = 12(0.4) - 0.5 x 10(0.4)^2
h = 4.8 - 0.8
h = 4m
Therefore, the time the two balls collide is 0.4 seconds while the height at which they collide is 4m
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A teacher places a warm bottles in a cooler filled with ice. Which statement best explains what happened over time?
A) Thermal energy will move from ice to water bottles
B) Coldness will move from the water bottles to the ice.
C) Coldness will move from the ice to the water bottlers
D) Thermal energy will move from the water bottle to the ice.
Answer:
D) Thermal energy will move from the water bottle to the ice.
Explanation:
Overtime, what happens is that thermal energy will move from the water bottle to the ice.
The water bottle is at a higher temperature compared to the ice. So, thermal energy will move from a place at higher temperature to one with lower temperature.
Thermal energy will stop moving until thermal equilibrium is attained. The water bottle will lose heat to the ice and by so doing it becomes colder. The ice will gain heat and begins to warm upa) A bus of mass 760 kg requires 120 m to reach certain velocity value Vf. Ignore friction and drag forces and assume the bus engine exerts a constant forward force F. When the bus is towing a 330-kg small car, how long distance needed to reach same Vf? b) If the Vf of the bus is 28 m/s, what is the tension in the tow cable between bus and small car?
Answer:
Given : A bus of mass 760 kg requires 120 m to reach certain velocity value Vf.
the bus engine exerts a constant forward force F.
To Find : When the bus is towing a 330-kg small car, how long distance needed to reach same Vf?
Solution:
V² - U² = 2aS
V = Vf
U = 0
S = 120 m
=> Vf² - 0 = 2a(120)
=> Vf² = 240a
m = 760 kg
Force = F
F = ma
=> F =760 a
=> a = F/760
Vf² = 240F/760
Case 2 :When the bus is towing a 330-kg small car,
m = 760 + 330 = 1090 kg
a = F/1090
Vf² = 2aS
=> 240F/760 = 2 (F/1090) S
=> S = 120 x 1090 /760
=> S = 172.1 m
172.1 m distance needed to reach same Vf
Explanation:
10) A soccer player kicks a soccer ball (m = 0.42 kg) accelerating from rest to 32.5m/s in 0.21s. Determine the force that sends soccer ball towards the goal.
G
U
E
S
S
Formula
11) Small rockets are fired to make small adjustments in the speed of a satellite. A certain small rocket can change the velocity of a 72,000kg satellite from 0m/s to 0.63m/s in 1296s. What force is exerted by the rocket on the satellite?
G
U
E
S
S
Formula
please I need help I don't understand it and I had to deliver it yesterday helpp:(
Answer:
10. 65 N
11. 35 N
Explanation:
10. Determination the force that sends the soccer ball towards the goal.
We'll begin by calculating the acceleration of the ball. This can be obtained as follow:
Initial velocity (u) of the ball = 0 m/s
Final velocity (v) of the ball = 32.5m/s time (t) = 0.21 s
Acceleration (a) of the ball =?
a = (v – u) /t
a = (32.5 – 0) / 0.21
a = 32.5 / 0.21
a = 154.76 m/s²
Finally, we shall determine the force that sends the soccer ball towards the goal. This can be obtained as follow:
Mass (m) of the ball = 0.42 kg
Acceleration (a) of the ball = 154.76 m/s²
Force (F) =?
F = ma
F = 0.42 × 154.76
F = 65 N
Thus, the force that sends soccer ball towards the goal is 65 N
11. Determination of the force exerted by the rocket on the satellite.
We'll begin by calculating the acceleration of the satellite. This can be obtained as follow:
Initial velocity (u) of satellite = 0 m/s
Final velocity (v) of satellite = 0.63 m/s
Time (t) = 1296 s
Acceleration (a) of the satellite =?
a = (v – u) /t
a = (0.63 – 0) / 1296
a = 0.63 / 1296
a = 4.861×10¯⁴ m/s²
Finally, we shall determine the force exerted by the rocket on the satellite. This can be obtained as follow:
Mass (m) of the satellite = 72000 Kg
Acceleration (a) of the satellite = 4.861×10¯⁴ m/s²
Force (F) =?
F = ma
F = 72000 × 4.861×10¯⁴
F = 35 N
Thus, the force exerted by the rocket on the satellite is 35 N
A racecar makes 24 revolutions around a circular track of radius 2 meters in
162 seconds. Find the racecar's frequency
Answer:
[tex]0.15\: \mathrm{Hz}[/tex]
Explanation:
The frequency is of an object is given by [tex]f=\frac{1}{T}[/tex], where [tex]T[/tex] is the orbital period of the object.
Since the racecar makes 24 revolutions around a circular track in 162 seconds, it will take the racecar [tex]\frac{162}{24}=6.75\:\mathrm{s}[/tex] per revolution.
Therefore, the frequency of the racecar is [tex]\frac{1}{6.75}=\fbox{$0.15\:\mathrm{Hz}$}[/tex] (two significant figures).
The radius of the track is irrelevant in this problem.
A 71-kg swimmer dives horizontally off a 500-kg raft. If the diver's speed immediately after leaving the raft is 6m/s, what is the corresponding raft speed?
Answer:
The answer is below
Explanation:
Momentum is used to measure the quantity of motion in an object. Momentum is the product of mass and velocity.
Momentum = mass * velocity
The principle of conservation of momentum states that momentum cannot be created or destroyed but can be transferred. Therefore the momentum before and after an action is equal.
Initial momentum = Final momentum
Let m be the mass of the diver, M be the mass of the raft, u be the initial velocity of the diver, U be the initial velocity of the raft, v be the final velocity of the diver and V be the final velocity of the raft.
m = 71 kg, M = 500 kg, v = 6 m/s
Initial both the raft and diver are at rest, hence u and U is zero, hence:
mu + MU = mv + MV
71(0) + 500(0) = 71(6) + 500(V)
0 = 426 + 500(V)
500(V) = -426
V = -426/500
V = -0.852 m/s
1. The volume of a given mass of gas is 20cm when its
pressure is 400mmHg. Calculate its pressure when the
volume becomes 80cm'at constant temperature,
Answer:
Explanation:
The way to show a cubed substance is either like this³ or like this x^3. The small three is found at the bottom toolbar at the bottom of the question space marked by the Ω symbol.
100 mmHg
Givens
V1 = 20 cm^3
V2 = 80 cm^3
P1 = 400 mmHg
P2 = ?
Formula
V1 * P1 = V2 * P2
Solution
20 * 400 = 80 * P2 Divide by 80
20 * 400/80 = P2
P2 = 8000 / 80
P2 = 100 mmHg
A 6.00-kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the end of the rope, and the height of the crate above its initial position is given by y(t)=(2.80m/s)t+(0.610m/s^3)t^3. What is the magnitude of F when t = 4.00 s?
Answer:
The magnitude of F when t=4 s=146.64 N
Explanation:
We are given that
Mass of crate,m=6.00 kg
Height of crate above its initial position is given by
[tex]y(t)=(2.80 m/s)t+(0.610 m/s^3)t^3[/tex]
We have to find the magnitude of F when t=4.00 s
Differentiate w.r.t t
[tex]\frac{dy}{dt]=2.8+3(0.61)t^2[/tex]
[tex]\frac{d^2y}{dt^2}=6(0.61)t[/tex]
[tex]a(t)=\frac{d^2y}{dt^2}=6(0.61)t m/s[/tex]
[tex]a(4)=6(0.61)(4)=14.64 m/s^2[/tex]
Now, magnitude of force
[tex]F=m(a+g)=6(14.64+9.8)=146.64N[/tex]
Hence, the magnitude of F when t=4 s=146.64 N
You are holding two balloons of the same shape and size. One is filled with helium, and the other is filled with ordinary air. On which balloon the buoyant force is greater?
a. The helium filled balloon experiences the greater buoyant force.
b. The air filled balloon experiences the greater buoyant force.
c. Both balloons experiences same buoyant force.
Answer:
Both balloons experiences same buoyant force.
Explanation:
Buoyant force is defined as the upward force that is exerted on an object which is wholly or partly immersed in a fluid. We can also define it as the upward force exerted by any fluid upon a body immersed in it. We may also refer to this buoyant force as the upthrust.
This buoyant force is the push of air on the balloon and it is independent of the contents of the balloons. Hence, both balloons experiences same buoyant force.
A bar magnet was placed underneath a sheet of paper where a pile of iron filings sits. In the presence of the energy stored in the magnetic field, the iron filings arranged themselves, creating lines of force. How do the energy and the lines of force change when a stronger bar magnet is used?
The answer is "The energy increases, and the lines of force are denser"
Answer:
The energy increase, and the lines of force are denser
Explanation:
The rest of the answers
1.) The field energy will increase.
3.) It points toward the field of earths magnetic poles.
4.) l, ll, and lll only
5.) ll, lV, l, lll
Based on the information given, the energy and the lines of force change as the energy increases, and the lines of force are dense.
Energy
In physics, it should be noted that energy is the quantitative property which must be transferred to a body or physical system to perform work on the body.
From the information, the bar magnet was placed underneath a sheet of paper where a pile of iron filings sits and in the presence of the energy stored in the magnetic field, the iron filings arranged themselves, creating lines of force. Therefore, when a stronger bar magnet is used, the energy increases and the lines of force are denser.
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in the periodic table, the properties repeat in what direction?
Answer:
Left to right and top to bottom
Explanation:
On the periodic table, the properties repeat from left to right and from top to bottom.
Periodic properties have a pattern from the top to the bottom or down a group or family.
Also, across the period from left to right, they also show a repeating pattern.
Certain properties increase from left to right and decreases from top to bottom. E.g. electronegativity. Also, some properties decreases from left to right and increases from top to bottom e.g. atomic radius.In the picture shown below A represents a characteristic of only geocentric model, B represents a characteristic common to both geocentric and heliocentric models, C represents a characteristic of only heliocentric model, and D represents a characteristic which the geocentric and heliocentric models do not have.
Under which label will the characteristic, "The sun and planets revolve around a central moon in the solar system" fall?
A
B
C
D
A car of weight 6,400 N has four wheels. Each wheel has an area of
80 cm-touching the road. Find the pressure the car puts on the
ground. I
Answer:
80N/cm^2
Explanation:
Given data
A car of weight 6,400N
Area= 80cm^2
The weight is distributed evenly on the 4 wheels
hence 6400/4
=1600 N
We know that
Pressure = force/Area
Pressure= 1600/80
Pressure= 20 N/cm^2
For the four wheels, the pressure is
=20*4
=80N/cm^2
Two 2.1-cm-diameter electrodes with a 0.20-mm-thick sheet of Teflon between them are attached to a 9.0 V battery. Without disconnecting the battery, the Teflon is removed.
Required:
a. What is the charge before the Teflon is removed?
b. What is the potential difference before the Teflon is removed?
c. What is the electric field before the Teflon is removed?
d. What is the charge after the Teflon is removed?
e. What is the potential difference after the Teflon is removed?
f. What are the electric field after the Teflon is removed?
Answer:
a. Q = 1881.73 x [tex]10^{-13}[/tex] C
b. As battery is not removed so, potential difference will remain same.
c. E = 21.42 x [tex]10^{3}[/tex] V/m
d. Q = 895.5 x [tex]10^{-13}[/tex] C
e. Again the potential difference will not change it will remain same as 9 V
f. E = 45 x [tex]10^{3}[/tex] V/m
Explanation:
Solution:
Here, Teflon is used so, the dielectric constant of the Teflon K = 2.1
Diameter = 2.1 cm
Radius = 2.1/2 cm
Radius = 1.05 cm
Radius = 0.015 m
Now, we need to find the area of each plate:
A = [tex]\pi r^{2}[/tex]
A = (3.14) ([tex]0.015^{2}[/tex])
A = 0.000225 [tex]m^{2}[/tex]
A = 2.25 x [tex]10^{-4}[/tex] [tex]m^{2}[/tex]
We are given the thickness of the plate which equal to the distance between the two plates.
d = 0.20 mm = 0.2 x [tex]10^{-3}[/tex] m
d = 0.2 x [tex]10^{-3}[/tex] m = distance between two plates.
Hence, the capacitance of the dielectric without the dielectric
C = [tex]\frac{E.A}{d}[/tex]
Putting up the values we get,
E = 8.85 x [tex]10^{-12}[/tex]
C = [tex]\frac{8.85 . 10^{-12} x 2.25 . 10^{-4} }{0.002}[/tex]
C = 99.5 [tex]10^{-13}[/tex]
If dielectric is included then,
[tex]C^{'}[/tex] = K C
[tex]C^{'}[/tex] = (2.1) ( 99.5 x [tex]10^{-13}[/tex])
[tex]C^{'}[/tex] = 209.08 x [tex]10^{-13}[/tex] F
As we know the voltage of the battery V = 9V So,
a) Charge before the Teflon is removed:
Q = CV
Q = [tex]C^{'}[/tex]V
Q = (209.08 x [tex]10^{-13}[/tex] F) (9V)
Q = 1881.73 x [tex]10^{-13}[/tex] C
b) Potential Difference before the Teflon is removed = ?
As battery is not removed so, potential difference will remain same.
c) Electric Field =?
As we know,
E = V/(K.d)
E = 9V/(2.1 x 0.2 x [tex]10^{-3}[/tex])
E = 21.42 x [tex]10^{3}[/tex] V/m
d) After the Teflon is removed
Q = CV
Q = (99.5 [tex]10^{-13}[/tex] ) ( 9)
Q = 895.5 x [tex]10^{-13}[/tex] C
e) Again the potential difference will not change it will remain same as 9 V
f) Electric Field = ?
E = [tex]\frac{V}{d}[/tex] (Teflon is removed)
E = 9/0.2 x [tex]10^{-3}[/tex]
E = 45 x [tex]10^{3}[/tex] V/m
____ is factual information not subject to bias.
Interpretation
Analysis
Data
Opinion
Answer:
Data
Explanation:
Data is factual information not subject to bias.
This ultimately implies that, data connotes fact, thus, it is an information that is credible, accurate, a statement of truth, evidential and proven.
In Computer programming, a data dictionary can be defined as a centralized collection of information on a specific data such as attributes, names, fields and definitions that are being used in a computer database system.
In a data dictionary, data elements are combined into records, which are meaningful combinations of data elements that are included in data flows or retained in data stores.
This ultimately implies that, a data dictionary found in a computer database system typically contains the records about all the data elements (objects) such as data relationships with other elements, ownership, type, size, primary keys etc. This records are stored and communicated to other data when required or needed.
A .2g sphere is suspended by a thread in an electreic field of 5000 N/C that is directed straight up. The tension in the string is 2.3 x 10^-3 N. Determine the charge on the shpere.
Answer:
6.8*10^-8C
Explanation:
The formula for calculating the Electric field on the sphere is expressed as;
E = Sum of force on the sphere/charge q
q is the charge on the sphere
E is the electric field
Given
E = 5000N/C
T = 2.3 x 10^-3 N.
q = ?
W = 0.0002 * 9.8
W = 0.00196N
W = 1.96*10^-3N
From the formula;
q = T-W/E
W is the weight of the sphere
T is the tension in the string
q = (2.3*10^-3 - 1.96 x 10^-3)/5000
q = 0.00034/5000
q = 6.8*10^-8C
Hence the charge on the sphere is 6.8*10^-8C
On a slope where does a marble have to most kinetic energy?
a) it is always the same
b) at the initial position
c) at the final position
d) somewhere between the initial and the final position
Answer:
C
Explanation:
Kinetic energy is the energy of motion. It has the most potential energy at the top but the most kinetic at the bottom after it's accelerated fully down the slope.
