Answer:
The minibus traveled 75.66 km
Explanation:
Motion with Constant Speed
An object is said to travel at constant speed if the ratio of the distance traveled by the time taken is constant.
The formula to calculate the speed is:
[tex]\displaystyle v=\frac{d}{t}[/tex]
Where
v = Speed of the object
d = Distance traveled
t = Time taken to travel d.
From the equation above, we solve for d:
d = v . t
The minibus has a constant speed of v=39 km/h and it's required to find the distance it travels in t=1.94 hours.
Calculating the distance:
d = 39 km/h * 1.94 h
d = 75.66 km
The minibus traveled 75.66 km
A 540,000 kg plane is coming in to land at 210 m/s. If it hits a 24,000 kg firetruck that's on the runway, inelastically, what will the final velocity be after the collision?
Answer:
201.06m/s
Explanation:
The solve this problem, we must understand that momentum must be conserved.
This implies that the starting momentum must be the same as the ending momentum. Since we are dealing with an inelastic collision:
Momentum of Plane + Momentum of truck = Momentum of plane + truck
Mp x Vp + Mt x Vt = V(Mp + Mt)
M is the mass
V is the velocity
t is truck
p is the plane
(540000 x 210) + (24000 x 0) = V (540000+24000)
113400000 = 564000V
V = 201.06m/s
Select the correct location on the image.
The lines on the position-time graph show the velocities of different vehicles. Which line represents a vehicle moving at constant velocity?
Answer:
orange line
Explanation:
it is the only straight line that makes sense. curved lines imply velocity changed. and the completly flat green line implies the vehicle didn't move since its position value stayed the same. must be the downward pointing orange line
What is every type of compound
Answer:
There are four types of compounds, depending on how the constituent atoms are held together: molecules held together by covalent bonds. ionic compounds held together by ionic bonds. intermetallic compounds held together by metallic bonds.
Explanation:
A ball is thrown upward from the ground with an initial speed of 21.4 m/s; at the same instant, another ball is dropped from a building 18 m high. After how long will the balls be at the same height above the ground
Answer:
after 0.8411 seconds, the balls are at the same height
Explanation:
given the data in the question;
we take a look at the equation which relates the vertical displacement of the object, time elapsed, initial velocity and acceleration of the object;
Δy = [tex]V_{i}[/tex]t + [tex]\frac{1}{2}[/tex] [tex]a_{y}[/tex]t²
where Δy is the vertical displacement, [tex]V_{i}[/tex] is the initial velocity, [tex]a_{y}[/tex] is acceleration of the object and t is time elapsed.
Now let h represent the height above ground level where they meet.
so the distance moved by the falling ball will be;
18 m - h
The ball is moving freely under the influence of gravity.
so [tex]a_{y}[/tex] = -g
Using kinematic relation a freely falling ball
Δy = [tex]V_{i}[/tex]t + [tex]\frac{1}{2}[/tex] [tex]a_{y}[/tex]t²
-( 18 m - h ) = 0 - [tex]\frac{1}{2}[/tex]gt²
-18 m + h = - [tex]\frac{1}{2}[/tex]gt²
h = 18 m - [tex]\frac{1}{2}[/tex]gt² --------- let this be equ1
Now lets apply kinematic relation to the raising ball
h = (21.4 m/s)t - [tex]\frac{1}{2}[/tex]gt² ----------- let this be equ 2
so equ1 = equ2
18 m - [tex]\frac{1}{2}[/tex]gt² = (21.4 m/s)t - [tex]\frac{1}{2}[/tex]gt²
18 m = (21.4 m/s)t
t = 18 m / 21.4 m/s
t = 0.8411 s
Therefore, after 0.8411 seconds, the balls are at the same height
B. Why does the moon revolve around the earth? Give
reason.[2]
Answer:
The moon revolves around Earth because Earth is larger than the moon, so it is heavier, and has a greater gravitational pull. The plane of the moon's orbit is very close to the plane of Earth's orbit around the Sun. This is why planets revolve around the Sun, because it is larger, so therefore it has a greater gravitational pull.
The ratio of the magnitude of the charge of a proton compared to the magnitude of the charge on an electron is
A 1:1
B 1:3
C 1:1.63 E18
D 1:1.67 E-27
E 1:9.11 E-31
Answer: A 1:1
Explanation:
They both have a magnitude of 1.6 x 10^-19 C.
The ratio of the magnitude of the charge of a proton compared to the magnitude of the charge on an electron is 1:1 option (A) is correct.
What is an atom?It is a basic unit of matter everything is made from the atom. It consists of three small particles in it named electron, proton, and neutron. The electron has a negative charge, a proton has a positive charge, and a neutron has no charge.
We know the charge on the proton:
[tex]\rm Q_p = 1.6 x 10^{-19} \ C[/tex]
And charge on the electron:
[tex]\rm Q_e = -1.6 x 10^{-19} \ C[/tex]
The absolute value of charge on the electron:
[tex]\rm |Q_e| = 1.6 x 10^{-19} \ C[/tex]
[tex]\rm \dfrac{\rm Q_p} {\rm |Q_e| }= \dfrac{1.6 x 10^{-19}}{ 1.6 x 10^{-19}}[/tex]
= 1:1
Thus, the ratio of the magnitude of the charge of a proton compared to the magnitude of the charge on an electron is 1:1 option (A) is correct.
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What happens when air from the balloon has exited the balloon?
How will motion change?
Answer:
When you release the opening of the balloon, gas quickly escapes to equalize the pressure inside with the air pressure outside of the balloon. The escaping air exerts a force on the balloon itself. ... That opposing force—called thrust, in this case—propels the rocket forward.
Which particles are located in the nucleus of the atom? (2 points)
Neutron and electron
Neutron and proton
Proton and electron
Only neutron
Answer:
Neutrons and protons are located in the nucleus of the atom.
Explanation:
And electrons are in the electron cloud.
In scenario X, a device that can measure the volume of sound is placed on the ground in vacuum. An object, also in vacuum, is released from rest above the ground and falls so that it strikes the ground next to the device. In scenario Y, the experiment is repeated in the same way, except the device and object are not in vacuum. In terms of energy transfer, which of the following claims is correct about the volume of sound produced by the object when it hits the ground in scenario X and scenario Y?
a. The measured volume in scenario X is greater than in scenario Y, because sound waves are not impeded by particles in the air as they travel through vacuum. Thus, the sound waves have more energy in scenario X than in scenario Y.
b. The measured volume in scenario X is less than in scenario Y, because sound waves must have a medium to transfer energy. Thus, the sound waves have more energy in scenario X than in scenario Y.
c. The measured volume in scenario X is the same as in scenario Y, because the experiment is identical in both scenarios except for the amount of air where the experiments are conducted. Thus, the sound waves have equal energies.
d. The answer cannot be determined, because the recorded volume of sound is not related to the energy contained within the sound wave.
Answer:
the correct answer is B
Explanation:
Mechanical waves need a material medium that supports them for their transmission, sound is a material wave that is transmitted by the longitudinal vibrations of material media, one of the media through which sound is transmitted is air, but if it is in contact with other media, water, wood, metal, etc., it can also be transmitted by these means.
In this experiment for case X, sound is transmitted through the floor, but since there is no air, it cannot be transmitted
In experiment Y, sound is transmitted through the floor and through the air, therefore, having two transmission media, the energy captured by the sensor must be greater in this experiment.
when examining the answers the correct one is B
A nonconducting spherical shell, with an inner radius of 4.0 cm and an outer radius of 6.0 cm, has charge spread nonuniformly through its volume between its inner and outer surfaces. The volume charge density rho is the charge per unit volume, with the unit coulomb per cubic meter. For this shell rho = b/r, where r is the distance in meters from the center of the shell and b=3.0μC/m². What is the net charge in the shell?
Answer:
the net charge in the shell is 3.7699 × 10⁻⁸ C or 3.7699 × 10⁻² μC
Explanation:
Given that;
Inner radius r = 4.0 cm = 0.04 m
Outer radius R = 6.0 cm = 0.06 m
b = 3.0μC/m² = 3.0 × 10⁻⁶ C/m²
we know that volume density = b/r
Now consider spherical shell inside the given shell with radius r and with of dr
the volume of the spherical shell will be;
dv = 4πr².dr
Now total change in this shell will be;
dQ = P.dv
so
dQ = b/r . 4πr².dr
dQ = b × 4πr.dr
Q = 4πb [tex]\int\limits^R_r {r} \,dx[/tex]
Q = 4πb[ r²/2 ][tex]^R_r[/tex]
Q = 4πb[ R²/2 - r²/2 ]
Q = 2πb[ R² - r² ]
so we substitute our values
Q = 2π×3.0 × 10⁻⁶[ 0.06² - 0.04² ]
Q = 2π×3.0 × 10⁻⁶ × 0.002
Q = 3.7699 × 10⁻⁸ C or 3.7699 × 10⁻² μC
Therefore, the net charge in the shell is 3.7699 × 10⁻⁸ C or 3.7699 × 10⁻² μC
Compare the gravity between these pairs, each consisting of an Earth-like planet and its star. You are given the mass of the planet in Earth masses, the mass of the star in Sun masses, and the distance in AUs.
4 MEarth / 2 MSolar / 3 AU
1 MEarth / 1 MSolar / 1 AU
1 MEarth / 2 MSolar / 2 AU
Answer:
The answer is below
Explanation:
Newton's law of gravity states that the force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The law is expressed by the formula:
[tex]F=G\frac{m_1m_2}{r^2} \\\\Where\ F=force,G=gravitational\ constant, m_1\ and\ m_1=mass\ of\ objects,r\ =distance\ between \ the\ two\ objects.[/tex]
The masses and distances for this question is in common units, Therefore the result would be in ratios
a) 4 MEarth / 2 MSolar / 3 AU
The force (F) = (4 * 3) / 3² = 4/3
b) 1 MEarth / 1 MSolar / 1 AU
The force (F) = (1 * 1) / 1² = 1
c) 1 MEarth / 2 MSolar / 2 AU
The force (F) = (1 * 2) / 2² = 1/2
who will it benefit if they do butterfly groin stretch
Answer:
ummmmm mmmmm butterflies :)
Explanation:
Answer:
It is a great stretch for athletes who play field or court sports, runners, and anyone who has tight hips or a history of groin injuries.
Its a great flexibility exercise
Explanation:
Hope that helps
write a letter to your elder brother tell him about future plan
Answer:
I don't have friendly brother
Explanation:
So no matter if I write he is gonna throw it out
A tennis ball is shot straight up with an initial velocity of 34 m/s. What is its velocity two seconds after launch?
Answer:
The speed after 2 seconds is 14.4 m/s
Explanation:
Vertical Launch Upwards
In a vertical launch upwards, an object is launched vertically up from a height H without taking into consideration any kind of friction with the air.
The speed changes from an initial value vo to a final value vf, given by:
[tex]v_f=v_o-g.t[/tex]
Where g is the gravitational acceleration ([tex]9.8\ m/s^2[/tex])
The tennis ball is shot up with an initial speed of vo=34 m/s. We are required to find the final speed at t=2 s:
[tex]v_f=34-9.8*2[/tex]
[tex]v_f=14.4\ m/s[/tex]
The speed after 2 seconds is 14.4 m/s
the velocity after 2 seconds of launch is 14.4 m/s
the initial velocity(a) is 34m/s
the time(t) is 2 seconds
acceleration(a) due to gravity is 9.8g
the formula is
v= u - at
v= 34 -(9.8)(2)
= 34 - 19.6
= 14.4 m/s
hence the final velocity is 14.4 m/s
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Traveling waves are generated on a string fixed at both ends. The string has a length L, a linear mass density m, and a tension T. Which of the following will cause the wave speed to increase?
I. Using the same string but increasing the tension.
II Using a longer string with the same μ and T.
III. Using a string with the same L and T but a smaller μ.
a. I only
b. II only
c. III only
d. I or II
d. I or III
Answer: d. I or II
Explanation: A traveling wave has speed that depends on characteristics of a medium. Characteristics like linear density (μ), which is defined as mass per length.
Tension or Force ([tex]F_{T}[/tex]) is also related to the speed of a moving wave.
The relationship between tension and linear density and speed is ginve by the formula:
[tex]|v|=\sqrt{\frac{F_{T}}{\mu} }[/tex]
So, for the traveling waves generated on a string fixed at both ends described above, ways to increase wave speed would be:
1) Increase Tension and maintaining mass and length constant;
2) Longer string will decrease linear density, which will increase wave speed, due to their inversely proportional relationship;
Then, ways to increase the wave speed is
I. Using the same string but increasing tension
II. Using a longer string with the same μ and T.
What do ocean waves and sound waves have in common? A. Both are transverse waves. B. both are longitudinal waves C. Both are electromagnetic waves. D. Both exhibit the same particle-to-particle interaction. E. Both are mechanical waves.
Answer:
E. Both are mechanical waves
About how much gravitational potential energy is stored in an object with a mass of 4 kg that is at rest at a height of 8 meters above the ground? (Use 10 m/s/s as the gravitational acceleration).
Question 3 options:
128 J
320 J
32 J
0 J
Answer:
it should be C I have has this question before
Answer: Gravitational potential energy = mgh
m=mass of the body
g= acceleration due to gravity
h=height
P.E=4*8*10=320
P.E=320J
Explanation:
Calculate the acceleration of an object slowing from 9.8 m/s to 6.4 m/s over the course of 17s
What is the term for when the
body loses too many fluids? *
A) bulimia
B) frostbite
C) dehydration
D) all of the above
Answer:
C. Dehydration is the word describing too little body fluids
Choose the answer that best describes hydrostatic equilibrium.
1. when gravity equalizes the level of liquid on a spherical celestial object
2. when equilibrium is achieved between two celestial bodies orbiting one another 3. when liquid such as water becomes spheroid under weightless conditions 4. when a body has enough mass and gravity to create a celestial body with a round shape
Answer:
the answer should be the third statement
Answer:
when a body has enough mass and gravity to create a celestial body with a round shape
PLEASE HELP WORTH 30 POINTS!!!
Answer:because the circuit will be fully complete
Explanation:
A 1000 kg car moving a 10 m/s collides with a stationary 2000 kg truck. The two vehicles interlock as a result of the collision. What is the final velocity of the two combined vehicles?
Answer:
3.33m/s
Explanation:
Given parameters:
M1 = 1000kg
V 1 = 10m/s
M2 = 2000kg
V2 = 0m/s
Unknown:
Final velocity of the two combined vehicles = ?
Solution:
This is a typical example of an inelastic collision.
Here, it is pertinent that momentum must be conserved;
So;
M1 V1 + M2 V2 = V (M1 + M2)
1000 x 10 + 2000 x 0 = V (1000 + 2000)
10000 = 3000V
V = 3.33m/s
If the cruising speed of a typical passenger jet is 500 mi/hr, calculate the time required, in years, of reaching the Sun from Earth in a passenger jet. Assume the jet could actually make the journey!
Answer: 21.22 years.
Explanation:
Here we can simply use the relation:
Time = Distance/Speed.
We know that the speed of the passenger jet is 500mi/hr.
And the average distance between Earth and the Sun in miles is:
Distance = 92,955,807 miles
Then the time needed to travel that distance will be:
Time = (92,955,807 miles)/500mi/hr = 185,911.614 hours.
But we want this in years.
We know that in one day, we have 24 hours, then this time in days is:
Time = (185,911.614/24) days = 7,746.3 days.
And in one year we have 365 days, then this time in years is:
Time = (7,746.3/365) years = 21.22 years.
A snail travels at 1 metre per hour. How long will it take to travel 8 metres
Answer:
I think it's 8 hours
Explanation:
1 meter: 1 hour
8 meters : 8 hours
Part 1. Define the following terminologies
1. Element
2. Compound
3. Metalloid
4. Physical property
5. Chemical property
6. Alloys
7. insulator
8. Corrosion
9. Rust
Answer:
1: Elements are substances that cannot be broken down into more substances.
2: Compounds are a group of elements in a specific ratio.
3: A metalloid is between a metal and a non-metal.
4: A physical property is a property you can see or touch.
5: A chemical property is a property seen in chemical reactions.
6: Alloys are the combination of two or more metals to make a better material.
7: An insulator doesn't let heat and sound escape.
8: Corrosion is when a metal reacts to oxygen, water, or sulfur.
9: Rust is when a metal is exposed to oxygen
Explanation:
Each of the following statements is related to conductors in electrostatic equilibrium. Choose the words that make each statement correct. HINT (a) The net charge is always zero ---Select--- the surface of an isolated conductor. (b) The electric field is always zero ---Select--- a perfect conductor. (c) The charge density on the surface of an isolated, charged conductor is highest where the surface is ---Select---
Answer:
a) interior, b) inside, c) minor
Explanation:
In this exercise you are asked to select the correct words so that the statements have been correct
Electric charges always repel each other when they are of the same sign, in conductors this has the consequence that charges accumulate on the surface and the interior remains without electric charges. with this we analyze the statements
a) interior
b) inside
c) minor
therefore the phrase would be:
(a) The net charge is always zero ---INTERIOR--- the surface of an isolated conductor.
(b) The electric field is always zero ---INSIDE--- a perfect conductor.
(c) The charge density on the surface of an isolated, charged conductor is highest where the surface is ---MINOR---to)
Brent is a crime scene investigator. Which two documents must he attach to every item of evidence?
With every item of evidence, Brent fills out a(n)
0, which includes the case number, description of the evidence, names of the suspect(s) and/or
victim(s), date and time and signatures of the investigator and witnesses. He also fills out the
document to ensure that the whereabouts of the
evidence is accounted for at all times.
WE
Reset
Next
Answer:
1. evidence log
2. chain of custody
Explanation:
plato
Evidence log and chain of custody are two documents must he attach to every item of evidence.
What are evidence logs and chain of custody?A chain of custody log is used to demonstrate that the integrity of the evidence was upheld from the time it was seized until it was produced in court. Chain of custody logs provides information on the collection, examination, and preservation of data for production.
Evidence logs keep track of the operations carried out when Event Data was being created. Together with the Evidence Records, they should be read. Evidence logs are compiled every day and consist of a series of entries, each of which describes a specific action.
Therefore, he should attach an evidence log and chain of custody to every item of evidence.
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Two traveling sinusoidal waves are described by the wave functions y1 = 4.85 sin [(4.35x − 1270t)] y2 = 4.85 sin [(4.35x − 1270t − 0.250)] where x, y1, and y2 are in meters and t is in seconds. (a) What is the amplitude of the resultant wave function y1 + y2?
Answer:
Approximately [tex]9.62[/tex].
Explanation:
[tex]y_1 = 4.85\, \sin[(4.35\, x - 1270\, t) + 0][/tex].
[tex]y_2 = 4.85\, \sin[(4.35\, x - 1270\, t) + (-0.250)][/tex].
Notice that sine waves [tex]y_1[/tex] and [tex]y_2[/tex] share the same frequency and wavelength. The only distinction between these two waves is the [tex](-0.250)[/tex] in [tex]y_2\![/tex].
Therefore, the sum [tex](y_1 + y_2)[/tex] would still be a sine wave. The amplitude of [tex](y_1 + y_2)\![/tex] could be found without using calculus.
Consider the sum-of-angle identity for sine:
[tex]\sin(a + b) = \sin(a) \cdot \cos(b) + \cos(a) \cdot \sin(b)[/tex].
Compare the expression [tex]\sin(a + b)[/tex] to [tex]y_2[/tex]. Let [tex]a = (4.35\, x - 1270)[/tex] and [tex]b = (-0.250)[/tex]. Apply the sum-of-angle identity of sine to rewrite [tex]y_2\![/tex].
[tex]\begin{aligned}y_2 &= 4.85\, \sin[(\underbrace{4.35\, x - 1270\, t}_{a}) + (\underbrace{-0.250}_{b})]\\ &= 4.85 \, [\sin(4.35\, x - 1270\, t)\cdot \cos(-0.250) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}[/tex].
Therefore, the sum [tex](y_1 + y_2)[/tex] would become:
[tex]\begin{aligned}& y_1 + y_2\\[0.5em] &= 4.85\, [\sin(4.35\, x - 1270\, t) \\ &\quad \quad \quad\;+\sin(4.35\, x - 1270\, t)\cdot \cos(-0.250) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \\[0.5em] &= 4.85\, [\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}[/tex].
Consider: would it be possible to find [tex]m[/tex] and [tex]c[/tex] that satisfy the following hypothetical equation?
[tex]\begin{aligned}& (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)\\&= 4.85\, [\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}[/tex].
Simplify this hypothetical equation:
[tex]\begin{aligned}& m\cdot \sin((4.35\, x - 1270\, t) + c)\\&=\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)\end{aligned}[/tex].
Apply the sum-of-angle identity of sine to rewrite the left-hand side:
[tex]\begin{aligned}& m\cdot \sin((4.35\, x - 1270\, t) + c)\\[0.5em]&=m\, \sin(4.35\, x - 1270\, t)\cdot \cos(c) \\ &\quad\quad + m\, \cos(4.35\, x - 1270\, t)\cdot \sin(c) \\[0.5em] &=\sin(4.35\, x - 1270\, t)\cdot (m\, \cos(c)) \\ &\quad\quad + \cos(4.35\, x - 1270\, t)\cdot (m\, \sin(c)) \end{aligned}[/tex].
Compare this expression with the right-hand side. For this hypothetical equation to hold for all real [tex]x[/tex] and [tex]t[/tex], the following should be satisfied:
[tex]\displaystyle 1 + \cos(-0.250) = m\, \cos(c)[/tex], and
[tex]\displaystyle \sin(-0.250) = m\, \sin(c)[/tex].
Consider the Pythagorean identity. For any real number [tex]a[/tex]:
[tex]{\left(\sin(a)\right)}^{2} + {\left(\cos(a)\right)}^{2} = 1^2[/tex].
Make use of the Pythagorean identity to solve this system of equations for [tex]m[/tex]. Square both sides of both equations:
[tex]\displaystyle 1 + 2\, \cos(-0.250) + {\left(\cos(-0.250)\right)}^2= m^2\, {\left(\cos(c)\right)}^2[/tex].
[tex]\displaystyle {\left(\sin(-0.250)\right)}^{2} = m^2\, {\left(\sin(c)\right)}^2[/tex].
Take the sum of these two equations.
Left-hand side:
[tex]\begin{aligned}& 1 + 2\, \cos(-0.250) + \underbrace{{\left(\cos(-0.250)\right)}^2 + {\left(\sin(-0.250)\right)}^2}_{1}\\ &= 1 + 2\, \cos(-0.250) + 1 \\ &= 2 + 2\, \cos(-0.250) \end{aligned}[/tex].
Right-hand side:
[tex]\begin{aligned} &m^2\, {\left(\cos(c)\right)}^2 + m^2\, {\left(\sin(c)\right)}^2 \\ &= m^2\, \left( {\left(\sin(c)\right)}^2 + {\left(\cos(c)\right)}^2\right)\\ &= m^2\end{aligned}[/tex].
Therefore:
[tex]m^2 = 2 + 2\, \cos(-0.250)[/tex].
[tex]m = \sqrt{2 + 2\, \cos(-0.250)} \approx 1.98[/tex].
Substitute [tex]m = \sqrt{2 + 2\, \cos(-0.250)}[/tex] back to the system to find [tex]c[/tex]. However, notice that the exact value of [tex]c\![/tex] isn't required for finding the amplitude of [tex](y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)[/tex].
(Side note: one possible value of [tex]c[/tex] is [tex]\displaystyle \arccos\left(\frac{1 + \cos(0.250)}{\sqrt{2 \times (1 + \cos(0.250))}}\right) \approx 0.125[/tex] radians.)
As long as [tex]\! c[/tex] is a real number, the amplitude of [tex](y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)[/tex] would be equal to the absolute value of [tex](4.85\, m)[/tex].
Therefore, the amplitude of [tex](y_1 + y_2)[/tex] would be:
[tex]\begin{aligned}|4.85\, m| &= 4.85 \times \sqrt{2 + 2\, \cos(-0.250)} \\&\approx 9.62 \end{aligned}[/tex].
The position of a particle moving along the x axis is given by x = (21 + 22t – 6 t2) m, where t is in s. What is the average velocity during the time interval t = 1 s to t = 4 s?
Answer:
12m/s
Explanation:
x = (21 + 22t – 6 t2) m,
We can express change bin velocity as
dx/dt= -12t + 22
But Velocity is changing
interval t = 1 s
t = 4 s
at t = 1, x = 21 + 22(1) -6(1)^2
=37
at t = 4, x = 21 + 22(4) -6(4)^2
=13
Distance travelled at interval t = 1 s
t = 4 s
X2 - X1=13- 37. = - 24m
Velocity= displacement/ time
= 24/(3-1)=12 m/s
Conservation of energy is explained as a scientific law and not a _______ because it does not explain why energy is conserved
Answer:
Theory
Explanation:
Conservation of energy is explained as a scientific law and not a theory because it does not explain why energy is conserved.
A law is a the statement of a scientific fact. It is a product of repeated experiment and observation through time. Most laws do not explain the reason for the logic behind their premise.
A theory on the other hand provides an explanation for an observed phenomenon. Most theories are no immutable. They are often changed when new finds are reported or made.
Laws are immutable and they stand still.
