what is democratic means in science
Answer:
relating to or supporting democracy or its principles.
Explanation:
Which quantity or quantities is/are increasing for the object represented by line B?
Answer:
C. Velocity and Position
Explanation:
The quantities that are increasing for the object represented by line B are velocity and position. The correct option is b.
What is velocity?The direction of a body or object's movement is defined by its velocity. In its basic form, speed is a scalar quantity. In essence, velocity is a vector quantity. It is the speed at which distance changes. It is the displacement change rate.
Velocity is the pace and direction of an object's movement, whereas speed is the time rate at which an object is traveling along a path. In other words, velocity is a vector, whereas speed is a scalar value.
The graph is given which represents the velocity and time with terms A, B, and C. As opposed to the position-time graph, which describes an object's motion over time, the velocity-time graph reveals an object's speed.
Therefore, the correct option is b. velocity and position.
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The question is incomplete. Your most probably complete question is given below:
Velocity and acceleration
velocity and position
velocity only
velocity, position, and acceleration
Croquet ball A moving at 8.3 m/s makes a head on collision with ball B of equal mass and initially at rest. Immediately after the collision ball B moves forward at 6.4 m/s .
What fraction of the initial kinetic energy is lost in the collision?
Answer:
0.25
Explanation:
That is the right answer.
The fraction of the initial kinetic energy is lost in the collision is 35.3%.
The given parameters:
Initial velocity of ball A = 8.3 m/sInitial velocity of ball B = 0Final velocity of ball B = 6.4 m/sThe initial kinetic energy of the system collision is calculated as follows;
[tex]K.E_i = \frac{1}{2} mv_1_i^2 + \frac{1}{2} mv_2_i^2\\\\K.E_i = \frac{1}{2} (m)(8.3)^2 + \frac{1}{2} (m) (0)^2\\\\K.E_i = 34.445 m[/tex]
The final velocity of ball A after collision is calculated as follows;
[tex]u_1 + v_1 = u_2 + v_2\\\\8.3 + v_1 = 0 + 6.4\\\\v_1 = 6.4 - 8.3\\\\v_1 = -1.9 \ m/s[/tex]
The final kinetic energy of the system after collision is calculated as follows;
[tex]K.E_f = \frac{1}{2} m(-1.9)^2 + \frac{1}{2} m(6.4)^2\\\\K.E_f = 22.285 \ m \[/tex]
The fraction of the initial kinetic energy is lost in the collision is calculated as follows;
[tex]= \frac{K_i - K_f}{K_i} \\\\= \frac{34.445 - 22.285}{34.445} \\\\= 0.353\\\\= 35.3\%[/tex]
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Power can be defined as?
A. The distance over which work has done
B. How much work can be done in a given time
C. All the work in an given area
D. The energy required to do work
( Last question was wrong according to the test I took)
A stone is allowed to fall from the top of the tower 100m high at the same moment another stone is projected vertically upward with a velocity of 25m/s. Where and when will the two cross each other
Answer:
both the stone will meet at a distance of 80 m from the top of tower.
Explanation:
let "t" = time after which both stones meet
"S" = distance travelled by the stone dropped from the top of tower
(100-S) = distance travelled by the projected stone.
◆ i) For stone dropped from the top of tower
-S = 0 + 1/2 (-10) t²
or, S = 5t²
◆ ii) For stone projected upward
(100 - S) = 25t + 1/2 (-10) t²
= 25t - 5t²
Adding i) and ii) , We get
100 = 25t
or t = 4 s
Therefore, Two stones will meet after 4 s.
◆ ¡¡¡) Put value of t = 4 s in Equation i) , we get
S = 5 × 16
= 80 m.
Thus , both the stone will meet at a distance of 80 m from the top of tower.
(Hope this helps can I pls have brainlist (crown)☺️)
A tuning fork of frequency 254 Hz and an open orang pipe of slightly lower frequency are at 15oC. When
sounded together, they produce 4 beats per second. On altering the temperature of the air in the pipes,
it is observed that the number of beats per second first diminishes to zero and then increases again to 4.
By how much has the temperature of the air in the pipe been altered?
The temperature of the air in the open orang pipe has been altered by 18.73° C
The frequency of an open orang pipe is estimated by using the formula:
[tex]\mathbf{f = \dfrac{v}{2L}}[/tex]
Then, the combination of the frequency of the tuning fork and the open orang pipe is:
[tex]\mathbf{254 - \dfrac{v}{2L} }[/tex]
These combinations of frequency produce 4 beats per sound.
i.e.
[tex]\mathbf{254 - \dfrac{v}{2L} =4}[/tex]
[tex]\mathbf{ \dfrac{v}{2L} = 254-4 }[/tex]
[tex]\mathbf{ \dfrac{v}{2L} = 250 ----(1)}[/tex]
When it is altered, the beats first diminish and increase again by 4.
i.e.
[tex]\mathbf{ \dfrac{v'}{2L} = 254+4 }[/tex]
[tex]\mathbf{ \dfrac{v'}{2L} = 258 --- (2) }[/tex]
If we equate both equations (1) and (2) together, we have:
[tex]\mathbf{\dfrac{v'}{v}= \dfrac{258}{250}}[/tex]
However, from our previous knowledge, we understand that the velocity of an object varies directly proportional to the square root of its temperature.
Hence;
when the temperature of the pipe = unknown ???the temperature of the open orang pipe = 15∴
[tex]\implies \mathbf{\sqrt{\Big(\dfrac{273 + T}{273 + 15}\Big)}= \dfrac{258}{250}}[/tex]
By squaring both sides, we have:
[tex]\implies \mathbf{\Big(\dfrac{273 + T}{273 + 15}\Big)}= \Big (\dfrac{258}{250}\Big )^2}[/tex]
[tex]\implies \mathbf{\Big(\dfrac{273 + T}{273 + 15}\Big)= \Big (\dfrac{66564}{62500}\Big )}[/tex]
[tex]\implies \mathbf{\Big(\dfrac{273 + T}{288}\Big)= \Big (1.065024\Big )}[/tex]
[tex]\implies \mathbf{273 +T =306.726912 }[/tex]
T = 306.726912 - 273
T ≅ 33.73 ° C
∴
The change in temperature ΔT = 33.73° C - 15° C
The change in temperature ΔT = 18.73° C
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What was different about the molecules you needed to make protein 3 compared to the molecules you used to make protein 2?
Answer:
the different about the molecules we needed to make protein 3 compared to the molecules we used to make protein 2 is that if we used 2 molecules than it will be smaller than using protein 3.
In an oscillating LC circuit, when 81.0% of the total energy is stored in the inductor's magnetic field, (a) what multiple of the maximum charge is on the capacitor and (b) what multiple of the maximum current is in the inductor
Answer:
21
Explanation:
9+10=21
when a torque is acting on a fly wheel the angular velocity of the fly wheel changes from 10rad/sec to 25rad/sec in 5sec.what will be the magnitudes of the angular acceleration of the fly wheel?
Hello!
We can use the following angular kinematic equation to solve:
α = Δω/Δt, or (ωf-ωi)/t
Plug in the given values:
α = (25 - 10)/5 = 15/5 = 3 rad/sec²
Which of the following statements are true?
(a) An object can move even when no force acts on it.
(b) If an object isn't moving, no external forces act on it.
(c) If a single force acts on an object, the object accelerates.
(d) If an object accelerates, a force is acting on it.
(e) If an object isn't accelerating, no external force is acting on it.
(f) If the net force acting on an object is in the positive x-direction, the object moves only in the positive x-direction.
I can't understand how A is true.
Answer: a, c, d
Explanation: a is true because the object will continue to move even without any force because of inertia (so yh thats why a is true). c is true because an object can accelerate if a single force acts on it. to accelerate (not move), it needs a force to act on it
Would appreciate brainly <3
Explanation:
inertia ...
you push or throw something, and you apply some force to it at that moment, but then it moves and keeps moving even long after you have no more connection to it, and no more force is applied to it.
please consider : we are only talking about moving. not about acceleration.
so, yes, (a) is true.
(c) is true.
(d) is true.
calculate 18% of 2758 correct to 4 significant figure
Answer:
......the answer is 496.4
Which statement is true?
O A ball with more kinetic energy rolls farther because it is harder to stop.
A ball at its highest speed demonstrates its greatest potential energy.
When a ball comes to a stop, its kinetic energy is greatest.
A ball with less kinetic energy rolls farther because it is harder to stop.
Answer:
c when a ball stop it's kinetic enegy is the greatest
When a ball comes to a stop, its kinetic energy is greatest is true.
What are the types of energy ?The energy is the ability to do work or produce action and / or movement and manifests itself in many different ways, such as body movement, heat, electricity, etc.
The various types of energy include Kinetic energy which is associated with the movement of bodies and the Potential energy is stored by virtue of a body's position also called gravitational potential energy.
Thermal Energy can be referred as the energy associated with the kinetic energy of the molecules that make up an element, it can be manifested if there is a temperature difference between two bodies.
Chemical energy released or formed from chemical reactions, Solar energy from sunlight. This form of energy is used to generate electricity through photovoltaic plates, for example.
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Figure 1 shows that the two vectors in different direction respectively. Determine the resultant for the two vectors
Answer:
15-45/2 = -7.5 downwards
Explanation:
cos(60) x 45 = z
z = 45/2
sin (30) x 70 = y
y = 15
15-45/ 2 = -7.5 downwards
PLZ HURRY
When you hold a racquet and swing your arm toward the ball, there are two kinds of resistance working against your muscles—the ______ and the ______.
A.
racket, air
B.
air, ball
C.
ball, net
D.
air, racket
Answer:
It should just be A
Explanation:
I dont see the difference between these 2 but ill choose A and update you
Answer:
A: racket, air
Explanation:
Calculate the torque produced by a 50.0 N perpendicular force at the end of a 0.300 m long wrench.
Answer:
Torque = 50N x 0.3m = 15Nm
Explanation:
Torque = Force x length of lever arm. To obtain the torque simply multiply the two given values.
Students are asked to create roller coasters for marbles. Their goal is to design a coaster with the tallest possible hill that a marble released from a height of 1.5 m (meters) can clear. The marbles will experience some air resistance and friction as
they move.
What should the students keep in mind as they build their designs?
a)The hill can be taller than 1.5 m (meters), because the marble will be moving faster than its initial velocity allowing it to travel higher than its release height.
b)The hill can be taller than 1.5 m (meters), because the marble will gain mechanical energy as it moves allowing it to travel higher than its release height.
c)The hill should be a little less than 1.5 m (meters) high, because the marble will lose mechanical energy as it moves, preventing it from reaching its release
height.
d)The hill can be exactly 1.5 m (meters) high, because mechanical energy is always
conserved allowing the marble to travel to its release height.
Answer:
Kinetic Energy.
Explanation:
The movement of a roller coaster is accomplished by the conversion of potential energy to kinetic energy. The roller coaster cars gain potential energy as they are pulled to the top of the first hill. As the cars descend the potential energy is converted to kinetic energy.
Since the marble will loose mechanical energy, the hill should be a little less than 1.5 m (meters) high.
A roller coaster is used to demonstrate the conversion of mechanical energy. In a roller coaster, potential energy is converted to a kinetic energy hence it conveniently serves as a device for demonstrating energy conversions.
As the students make their design, they must bear in mind that the hill should be a little less than 1.5 m (meters) high, because the marble will lose mechanical energy as it moves, preventing it from reaching its release
height.
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what are people words in english
Answer: Plenty
Explanation:
some words are Hi, Banana, Dude and many more
in vacuum , the shorter the wavelength of an electromagnetic wave is , the:
A. lower its frequency
B. higher its energy
C. longer its period
D. slower its speed
Answer:
Higher its Energy
Explanation:
A weather forecaster uses a computational model on a Monday to predict the weather on Friday. Why might that forecast change? (1 point)
A. The forecaster may have done the calculations a second time because he made a mistake in the calculation.
B. Someone may have reported the weather incorrectly before the first computation.
C. The forecaster may have had the computer model do the calculations a second time, and he found that the prediction changed.
D. An area of low pressure might move more quickly on Tuesday and Wednesday than expected.
Answer:
(d) is your answer for your question
Students perform a set of experiments by placing a block of mass m against a spring, compressing the spring a distance x along a horizontal surface of negligible friction, releasing the block, and measuring the velocity v of the block as it leaves the spring, as shown in Figure 1. The experiments indicate that as x increases, so does v in a linear relationship. The surface is now lifted so that the surface is at an angle θ above the horizontal. Which of the following indicates how the relationship between v and x changes?
Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.
The statement that indicates how the relationship between v and x changes is; As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x.Reasons:
The energy given to the block by the spring = [tex]\mathbf{0.5 \cdot k \cdot x^2}[/tex]
According to the principle of conservation of energy, we have;
On a flat plane, energy given to the block = [tex]0.5 \cdot k \cdot x^2[/tex] = kinetic energy of
block = [tex]0.5 \cdot m \cdot v^2[/tex]
Therefore;
0.5·k·x² = 0.5·m·v²
Which gives;
x² ∝ v²
x ∝ v
On a plane inclined at an angle θ, we have;
The energy of the spring = [tex]\mathbf{0.5 \cdot k \cdot x^2}[/tex]
The force of the weight of the block on the string, [tex]F = m \cdot g \cdot sin(\theta)[/tex]
The energy given to the block = [tex]0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta)[/tex] = The kinetic energy of block as it leaves the spring = [tex]\mathbf{0.5 \cdot m \cdot v^2}[/tex]
Which gives;
[tex]0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta) = 0.5 \cdot m \cdot v^2[/tex]
Which is of the form;
a·x² - b = c·v²
a·x² + c·v² = b
Where;
a, b, and c are constants
The graph of the equation a·x² + c·v² = b is an ellipse
Therefore;
As x increases, v increases, however, the value of v obtained will be lesser than the same value of x as when the block is on a flat plane.Please find attached a drawing related to the question obtained from a similar question online
The possible question options are;
As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of xThe relationship is no longer linear and v will be more for the same value of xThe relationship is still linear, with lesser value of vThe relationship is still linear, with higher value of vThe relationship is still linear, but vary inversely, such that as x increases, v decreasesLearn more here:
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The coefficient of kinetic friction for a 22 kg bobsled on a track is 0.10. What force is required to push it down a 5.0 degree incline and achieve a speed of 62 km/h at the end of 75 m
Hi there!
We must begin by converting km/h to m/s using dimensional analysis:
[tex]\frac{62km}{1hr} * \frac{1hr}{3600sec}*\frac{1000m}{1km} = 17.22 m/s[/tex]
Now, we can use the kinematic equation below to find the required acceleration:
vf² = vi² + 2ad
We can assume the object starts from rest, so:
vf² = 2ad
(17.22)²/(2 · 75) = a
a = 1.978 m/s²
Now, we can begin looking at forces.
For an object moving down a ramp experiencing friction and an applied force, we have the forces:
Fκ = μMgcosθ = Force due to kinetic friction
Mgsinθ = Force due to gravity
A = Applied Force
We can write out the summation. Let down the incline be positive.
ΣF = A + Mgsinθ - μMgcosθ
Or:
ma = A + Mgsinθ - μMgcosθ
We can plug in the given values:
22(1.978) = A + 22(9.8sin(5)) - 0.10(22 · 9.8cos(5))
A = 46.203 N
A solenoid 91.0 cm long has a radius of 1.50 cm and a winding of 1300 turns; it carries a current of 3.60 A. Calculate the magnitude of the magnetic field inside the solenoid.
The magnitude of the magnetic field inside the solenoid is [tex]6.46 \times 10^{-3} \ T[/tex].
The given parameters;
length of the solenoid, L = 91 cm = 0.91 mradius of the solenoid, r = 1.5 cm = 0.015 mnumber of turns of the solenoid, N = 1300 current in the solenoid, I = 3.6 AThe magnitude of the magnetic field inside the solenoid is calculated as;
[tex]B = \mu_0 nI\\\\B = \mu_o(\frac{ N}{L} )I\\\\[/tex]
where;
[tex]\mu_o[/tex] is the permeability of frees space = 4π x 10⁻⁷ T.m/A
[tex]B = (4\pi \times 10^{-7}) \times (\frac{1300}{0.91} ) \times 3.6\\\\B = 6.46 \times 10^{-3} \ T[/tex]
Thus, the magnitude of the magnetic field inside the solenoid is [tex]6.46 \times 10^{-3} \ T[/tex].
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What is the potential energy when a 100 kg object is raised 4.00 m straight up?
Answer:
100kg x 4 x 10 = 4000J
Explanation:
9. The acceleration (a)-time (t) graph of a particle moving in a straight line is as shown in figure. At time t = 0, the velocity of particle is 10 m/s. What is the velocity at t = 8 s?
(1) 2 m/s
(2) 4 m/s
(3) 10 m/s
(4) 12 m/s
Answer:Acceleration - time graph for a particle moving in a straight line is as shown in figure. Change in velocity of the particle from t = 0 to t = 6s is:-.
1 answer
·
Top answer:
Change in velocity = (sum of area of graph) = ( 12 × 4 × 4 ) + ( 12 × ( + 2) ( - 1) ) - 4 = 8 - 4 = 4 x
Explanation:
At what speed must the electron revolve round the nucleus of
the hydrogen in its ground state in order that it may not be pulled into the
nucleus by electrostatic attraction
Explanation:
I think this is it, give it a try
of the following which is the largest body?
a. the moon
b. Pluto
c. Mercury
d. Ganymede
Answer:
Ganymede is the largest body
Explanation:
it is the satellite of jupiter
Jimmy walks to school by traveling 2.0 miles east and 3.0 miles north from his starting point. What are the magnitude and direction of Jimmy's displacement with respect to his original position?
Answer:
3.6 miles at 56.3 degrees north of east.
Explanation:
If you draw a diagram to calculate Jimmy's displacement, you end up with a triangle.
To solve for the magnitude of Jimmy's displacement with respect to his original position, we need to use the Pythagorean Theorem formula.
Pythagorean Theorem Formula = a^2 + b^2 = c^2
Step by Step to solve for the magnitude of Jimmy's displacement:
1) a^2 + b^2 = c^2
2) 2.0^2 + 3^2 = c^2
3) 4 + 9 = c^2
4) 13 = c^2
5) To find what "c" equals to we need to find the square root of 13.
6) √13 = 3.6 | c = 3.6 miles
To solve for the direction of Jimmy's displacement with respect to his original position, we need to use the following formula:
tan^-1 ∅ (3/2)
tan^-1 ∅ (1.5)
∅ = 56.309° north of east
Therefore, your answer is 3.6 miles at 56.3 degrees north of east.
A fat wire, radius a, carries a constant current I, uniformly distributed over its cross section. A narrow gap in the wire, of width w << a, forms a parallel-plate capacitor. Find the magnetic field in the gap, at a distance s < a from the axis.
The magnetic field in the gap is [tex]\mathbf{B = \dfrac{\mu_oI s}{2 \pi a^2}}[/tex]
In a circuit, a magnetic flux is circulated or followed via a confined environment or passage called a magnetic field gap. The narrow air gap is a non-magnetic component of a magnetic circuit that is normally connected to the remainder of the circuit magnetically in series. This enables a significant amount of magnetic flux to pass via the gap.
The magnetic field in the gap at a distance s < a can be computed by using the formula:
[tex]\mathbf{ \oint Bdl = \mu_oI_{enclosed}}[/tex]
where;
Magnetic flux density = Bdistance = d[tex]\mathbf{B( 2 \pi d) = \mu _o \oint _s J_d da }[/tex]
where;
[tex]\mathbf{J_d}[/tex] = drift current density[tex]\mathbf{B( 2 \pi d) = \mu _o J_d \oint _sda }[/tex]
[tex]\mathbf{B( 2 \pi d) = \mu _o J_d (\pi d^2) }[/tex]
Making the magnetic flux density the subject, we have:
[tex]\mathbf{B =\dfrac{ \mu _o J_d (\pi d^2) }{( 2 \pi d)}}[/tex]
[tex]\mathbf{B =\dfrac{ \mu _o J_dd}{ 2 }}[/tex]
Recall that, the drift current density [tex]\mathbf{J_d = \dfrac{I}{\pi a^2}}[/tex]
[tex]\mathbf{B = \dfrac{\mu_o d}{2}(\dfrac{I}{\pi a^2})}[/tex]
Recall that distance in question is said to be (s);
∴
[tex]\mathbf{B = \dfrac{\mu_oI s}{2 \pi a^2}}[/tex]
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A rod of length 0.82 m, rotating with an angular speed, 4.2 rad/s, about axes that pass perpendicularly through one end, has a mass of 0.63 kg, which is distributed uniformly along its length. Find the kinetic energies of the rod.
Hi there!
[tex]\large\boxed{KE =1.245 J}}[/tex]
The equation of angular kinetic energy is:
[tex]KE = \frac{1}{2}Iw^2[/tex]
Where:
I = moment of inertia (kgm²/s)
ω = angular speed (rad/sec)
A rod rotated about one of its endpoints has a standard moment of inertia of 1/3mR², so:
[tex]KE = \frac{1}{2}(\frac{1}{3}mL^2w^2)[/tex]
[tex]KE = \frac{1}{6}mL^2w^2[/tex]
Plug in the given values:
[tex]KE = \frac{1}{6}(0.63)(0.82^2)(4.2^2) = 1.245 J[/tex]
Entra vapor a una tobera adiabática con un flujo másico de 250 kg/h. Al entrar el vapor
tiene una energía interna específica de 2510 kJ/kg, una presión de 1378 kPa, un
volumen específico de 0.147 m3
/kg y una velocidad de 5 m/s. Las condiciones de salida
son P= 138.7 kPa, volumen específico de 1.099 m3
/kg y energía interna específica de
2263 kJ/kg. Determine la velocidad de salida.
Este problema describe el funcionamiento de una tobera adiabática (sin flujo de calor), la cual tiene una corriente de entrada que difiere de la de salida espacial y energéticamente, pero que conservan el mismo flujo másico de 250 kg/h. De este modo, usamos un balance de energía con el fin the determinar la velocidad a la que sale el fluido, según es requerido en el problema:
[tex]mu_1+mP_1v_1+mgh_1+\frac{1}{2} mv_1^2=mu_2+mP_2v_2+mgh_2+\frac{1}{2} mv_2^2[/tex]
En la que se tiene que la incógnita es la velocidad de salida, [tex]v_2[/tex], y es posible simplificar el flujo másico, [tex]m[/tex], al estar como factor común en ambos lados y despreciar la energía potencial (mgh), ya que no hay diferencia de altura significativa entre la entrada y salida de la tobera.
De este modo, es posible reemplazar los valores dados para obtener la siguiente expresión:
[tex]2510\frac{kJ}{kg} +1378kPa*0.147\frac{m^3}{kg} +\frac{1}{2} (5\frac{m}{s} )^2=2263\frac{kJ}{kg}+138.7kPa*1.099\frac{m^3}{kg} +\frac{1}{2} v_2^2[/tex]
Y así, hallar la velocidad de salida como sigue:
[tex]2725.066\frac{kJ}{kg}=2415.431\frac{kJ}{kg} +\frac{1}{2} v_2^2\\\\v_2=\sqrt{2(2725.066\frac{kJ}{kg}-2415.431\frac{kJ}{kg} )} \\\\v_2=24.9\frac{m}{s}[/tex]
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