The magnitude of the car's acceleration is 1.1 m/[tex]s^{2}[/tex].
Use this formula of acceleration, [tex]v^{2}[/tex] = [tex]u^{2}[/tex] + 2as
[tex]v^{2}[/tex] = [tex]u^{2}[/tex] + 2as, Here, initial velocity is zero.
a = [tex]v^{2}[/tex] / 2s
v = velocity
s = distance
a = acceleration
Put the values in this formula, a = [tex]v^{2}[/tex] / 2s
a = 15 × 15 / 2 × 100
a = 1.125 m/[tex]s^{2}[/tex]
Hence, the magnitude of the car's acceleration is 1.125 m/[tex]s^{2}[/tex].
Acceleration is the rate of change of velocity per unit time. It is vector quantity and its S.I. unit is m/[tex]s^{2}[/tex].
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A trumpet plays middle C (262 Hz). How fast would it have to be moving to raise the pitch to C sharp (277 Hz)? Use 343 m/s as the speed of sound.
The trumpet would have to be moving at a speed of approximately 345.8 m/s to raise the pitch to C sharp.
What is speed?
In everyday language and kinematics, an object's speed (typically abbreviated as "v") is defined as the size of the change in position that occurs over time or the amount of change that occurs per period of time; it therefore a vector quantity. The instantaneous speed is the upper limit of the average speed as the time interval's duration gets closer to zero; it is calculated by dividing the distance travelled by the time interval's duration. Speed and velocity are distinct concepts.
Speed is measured as distance divided by time.
To raise the pitch of a note by one semitone, the speed of sound must be increased by approximately 6%. In this case, the trumpet would have to be moving at approximately 364 m/s to raise the pitch to C sharp.
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3. Hakeem was participating in a psychology experiment. For the first ten minutes Hakeem would hear a click then a puff of air would blow in Hakeem's eye. His eyes would blink and tear up. For the second ten minutes Hakeem would hear a click but the puff of air would only blow in his eye every third time. Hakeem knows this but his eyes would blink and tear up after every click.
UCS: UCR:
CS: CR:
NS:
In the psychology experiment in which Hakeem participates, UCS is click, UCR is blink and tear up, CS is blowing a puff of air, CR is thought of the puff of air and NS is the click before puff of air was introduced.
In psychology the abbreviations of the given terms are as follows:
UCS - Unconditioned Stimulus UCR - Unconditioned ResponseCS - Conditioned Stimulus CR - Conditioned ResponseNS - Neutral StimulusHere the Unconditioned Response is the blinking and tearing up because he does that even though there is no puff of air. Since clicking provokes the Unconditioned Response, clicking is the Unconditioned Stimulus. The clicking is paired with puff of air for multiple times, that is why Unconditioned Response is stimulated. So the puff of air is the Conditioned Stimulus.
Therefore,
UCS - Clicking UCR - Blinking and tearing upCS - Blowing of puff of air in the eyeCR - Thought of puff of air being blown into the eye NS - Clicking before any puff of air was blownTo know more about stimulus and response
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describe how two objects can have the same speed but different velocities
Two objects can have the same speed but different velocities. The detailed description is given below.
We can take two escalators to explain the concept. The two escalators are moving up at a speed of 10 m/s. In this case, the two escalators have the same speed and they are moving in the same direction. So, they have the same velocity also.
Now one escalator is moving up at a speed of 10 m/s and the other escalator moving down at a speed of 10 m/s. In this case, both escalators move at 10 m/s, so their speed is the same. But they are not moving in the same direction. Hence their velocity is not the same. One has 10 m/s downwards and the other one has 10 m/s upwards velocity.
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An orange dropped from a tree and had a velocity of 8 m/s just before it hits the ground. How far is the ground from orange's starting position?
The orange was at a height of 3.26 m above the ground.
State third equation of motion.The third equation of motion is
v² - u² = 2aS
Given is an orange dropped from a tree and had a velocity of 8 m/s just before it hits the ground.
We can write -
[v] = 8 m/s
[a] = 9.8 m/s²
[u] = 0 m/s
Using third equation of motion, we get -
v² - u² = 2aS
64 - 0 = 2 x 9.8 x S
64 = 19.6 x S
S = 3.26 m
Therefore, the orange was at a height of 3.26 m above the ground.
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For questions 7 through 9, Albert and friends are stranded in a clearing on top of a hill at an
elevation of 350m above sea level. You are on a rescue plane flying in supplies to tide them over
until help arrives. Your plane is flying at a constant speed of 260 km/h from west to east at an
elevation of 810m.
PLEASE HELP WOTH THESE QUESTIONS!!!!!!
Answer:
Assume that the air resistance on the supplies is negligible, and that [tex]g = 9.8 \; {\rm m\cdot s^{-2}}[/tex].
The plane need to drop the supplies when it is horizontally approximately [tex]700\; {\rm m}[/tex] away from the hill.
The supplies will hit the tree.
Explanation:
Let [tex]u_{y}[/tex] and [tex]v_{y}[/tex] denote the initial and final vertical velocity of the supply; [tex]u_{y} = 0\; {\rm m\cdot s^{-1}}[/tex] since the plane was flying horizontally.
Let [tex]x_{y}[/tex] denote the vertical displacement of the supply; [tex]x_{y} = 350\; {\rm m} - 810\; {\rm m} = (-460)\; {\rm m}[/tex].
Let [tex]a_{y}[/tex] denote the vertical acceleration of the supply; [tex]a = (-g) = (-9.8)\; {\rm m\cdot s^{-2}}[/tex].
Make use of the SUVAT equation [tex]{v_{y}}^{2} - {u_{y}}^{2} = 2\, a_{y}\, x_{y}[/tex] to find [tex]v_{y}[/tex], the final vertical velocity of the supply:
[tex]\begin{aligned} {v_{y}}^{2} &= {u_{y}}^{2} + 2\, a_{y}\, x_{y} \end{aligned}[/tex].
[tex]\begin{aligned} v_{y} &= -\sqrt{{u_{y}}^{2} + 2\, a_{y}\, x_{y}} \\ &= -\sqrt{0^{2} + 2\, (-9.8)\, (-460)}\; {\rm m\cdot s^{-1}} \\ &\approx (-94.953)\; {\rm m\cdot s^{-1}} \end{aligned}[/tex].
(Negative since the supply would be travelling downwards.)
Let [tex]t[/tex] denote time it takes for the supply to land on the hill after being dropped from the plane. Make use of the SUVAT equation [tex]t = (v_{y} - u_{y}) / (a)[/tex] to find the value of [tex]t\![/tex]:
[tex]\begin{aligned} t &= \frac{v_{y} - u_{y}}{a} \\ &\approx \frac{(-94.953) - 0}{(-9.8)} \; {\rm s}\\ &\approx 9.6890 \; {\rm s} \end{aligned}[/tex].
Apply unit conversion and ensure that [tex]v_{x}[/tex], the horizontal speed of the plane is in the standard unit [tex]{\rm m\cdot s^{-1}}[/tex]:
[tex]\begin{aligned} v_{x} &= \frac{260\; {\rm km}}{1\; {\rm h}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \\ &\approx 72.222\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Under the assumptions, the horizontal speed of the supply will be the same as that of the plane- [tex]v_{x} \approx 72.222\; {\rm m\cdot s^{-1}}[/tex]- until it lands.
While in the air, the supply will travel a horizontal distance of:
[tex]\begin{aligned}x_{x} &= v_{x}\, t \\ &\approx 72.222\; {\rm m\cdot s^{-1}} \times 9.6890\; {\rm s} \\ &\approx 699.76\; {\rm m}\end{aligned}[/tex].
Hence, for the supply to land exactly at the top of the hill, the plane need to drop the supply while at a horizontal distance of approximately [tex]700\; {\rm m}[/tex] away from the hill.
The horizontal distance between the trees and the location where the plane dropped the supply would be approximately [tex](700\; {\rm m} - 30\; {\rm m}) = 670\; {\rm m}[/tex]. The time required for the the supply to reach that horizontal position would be:
[tex]\begin{aligned} t &= \frac{x_{x}}{v_{x}} \approx \frac{669.76\; {\rm m}}{72.222\; {\rm m\cdot s^{-1}}} \approx 9.2736\; {\rm s}\end{aligned}[/tex].
Let [tex]h_{0}[/tex] denote the initial height of the supply (relative to the sea level.) In this question, [tex]h_{0} = 810\; {\rm m}[/tex].
Let [tex]h(t)[/tex] denote the height of the supply (relative to the sea level) after being dropped from the plane for time [tex]t[/tex].
The SUVAT equation [tex]h(t) = (1/2)\, a\, t^{2} + u_{y}\, t + h_{0}[/tex] gives an expression for [tex]h(t)[/tex]. Make use of this equation to find the height of the supply (relative to the sea level) when the supply reach the horizontal position of the trees at [tex]t \approx 9.2736\; {\rm s}[/tex]:
[tex]\begin{aligned} h(t) &= \frac{1}{2}\, a\, t^{2} + u_{y}\, t + h_{0} \\ &= \frac{1}{2}\times (-9.8)\, (9.2736)^{2} + 0\times 9.2736 + 810 \\ &\approx 388.60\; {\rm m} \end{aligned}[/tex].
Note that the altitude of the top of the trees is [tex]350\; {\rm m} + 40\; {\rm m} = 390\; {\rm m}[/tex] relative to the sea level. Since [tex]388.90\; {\rm m} < 390\; {\rm m}[/tex], the supplies will run into the trees.
What is the distance from the moon to the sun?
Answer:
About 150 million kilometers
Explanation:
have good day♡
Mr. Red Herring was found shot dead in his backyard. He was about a foot away from his back porch, lying next to his personal handgun. A bloody footprint was also found on the porch. Analysis of a bullet found nearby suggests that, based on the striation marks, the bullet that killed Herring came from a gun discarded in a nearby trash can. He apparently had a date that night with Mrs. Scarlet. Based on various evidence, the forensics team also created a digital rendering of the crime.
An example of demonstrative evidence is the digital rendering of the crime scene that shows where the gun was fired from. That is option B.
What is demonstrative evidence?A demonstrative evidence is the type of evidence that is being represented using visuals to help enhance the facts of a claim made against an opponent in the law court.
The components of demonstrative evidence include the following:
photos, x-rays, videotapes, movies, sound recordings, diagrams, forensic animation,maps and drawings.Based on the various evidence presented concerning the sudden death of Mr. Red Herring who was short dead in his backyard, the demonstrative evidence is when the forensics team also created a digital rendering of the crime showing where the gun was shot.
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Options of question:
In this story, which is an example of demonstrative evidence?
The fingerprint that showed that Mrs. Scarlet handled the gun
The digital rendering of the crime scene that shows where the gun was fired from
The report about bullet striations that prove the shot came from the discarded weapon
The DNA evidence gathered from the blood splatters
An object moves 60.0 m on a bearing (angle from North) of 60.0°. If the object then moves 30.0 m North, how far is it from the start point? You may use a scale diagram or trigonometry to answer this question.
Answer: x(t2)−x(t1) over the time interval [t1,t2]
Explanation: hope this helps
A force of -4.4 x 103 N exists between a positive charge of 8.0 x 10-4 C and a negative charge of -3.0 x 10-4 C. What is the distance that separates the charges?
The distance that separates the charges is 0.7 m.
What is the distance between the charges?The distance between the charges is determined by applying Coulomb's law of electrostatic force as shown below.
F = kq₁q₂/r²
where;
r is the distance between the chargesk is coulomb's constantF is the force between the chargesr² = kq₁q₂/F
r² = (9 x 10⁹ x 8 x 10⁻⁴ x 3 x 10⁻⁴) / (4.4 x 10³)
r² = 0.49
r = √0.49
r = 0.7 m
Thus, the distance that separates the charges is 0.7 m.
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a child pushes a merry-go-round that has a diameter of 4.00 m and goes from rest to an angular speed of 17.0 rpm in a time of 45.0 s. 1) calculate the average angular acceleration of the merry-go-round. (express your answer to three significant figures.)
The total tangential speed = 3.56 m/s
, r =2 m ,
t =44sw
=17rpm = 17*2*3.14/60
= 1.7793 rad/swo =
01) from rotational kinematic equation
w =wo+t1.7793 =0 + *44 =0.0404 rad/s^2
(2) from rotationla kinematic equation
w^2 -wo^2 = 21.7793*1.7793
= (2*0.0404*)=39.18 rad
3) v =rw =2*1.7793
tangential speed = 3.56 m/s
What is average angular acceleration?
A spinning object's change in angular velocity per unit of time is expressed quantitatively as angular acceleration, also known as rotational acceleration. It is a vector quantity with either one of two predetermined directions or senses and a magnitude component.
The formula for average angular acceleration is: a = change in velocity divided by change in time. a=(w2−w1)/(t2−t1) a = ( w 2 − w 1 ) / ( t 2 − t 1 ) . w2 represents the final velocity measured in radians per second. w1 represents the initial velocity measured in radians per second.
Therefore, tangential speed = 3.56 m/s
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as the pressure of an enclosed gas decreases to half its original value, what happens to the volume of the gas if temperature is held constant?(1 point)
Answer:
See below
Explanation:
P1V1/T1 = P2V2/T2 IF T1 = T2
Then
P1V1 = P2V2 P2 = 1/2 P1
P1V1 = 1/2 P1 V2 DIVIDE BOTH SIDES BY P1
V1 = 1/2 V2 MULTIPLY BOTH SIDES BY 2
2V1 = V2 OR V2 = 2 V1 THE NEW VOLUME IS TWICE ORIGINAL
according to a simplified model of a mammalian heart, at each pulse approximately 20 g of blood is accelerated from 0.25 m/s to 0.35 m/s during a period of 0.10 s. what is the magnitude of the force exerted by the heart muscle?
The magnitude of the force exerted by the heart muscle is 0.02N
since
F = m dv/dt
= 0.02(0.35-0.25)/0.10
F =0.02N.
A force is an influence in physics that can change the motion of an object. A force can cause a mass object to change its velocity, or accelerate. Intuitively, force can be described as a push or a pull. A force is a vector quantity because it has both magnitude and direction.
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A horse pulls a cart with force F. As a result of this force the cart accelerates with constant acceleration. The
magnitude of the force that the cart exerts on the horse
A) is zero newtons. B) greater than the magnitude of F. C) equal to the magnitude of F. D).
less than the magnitude of F.
Answer:
c. equal equal to the magnitude of f
1. on the planet arrakis a male ornithoid is flying toward his mate at 25.0 m/s while singing at a frequency of 1200 hz. if the stationary female hears a tone of 1240 hz, what is the speed of sound in the atmosphere of arrakis?
If the stationary female hears a tone of 1240 hertz, then the speed of the sound in the atmosphere of arrakis will be 775 meter per second.
The term "Ducler shift" refers to the apparent change in frequency caused by the source moving closer to the observer or by the relative motion between the source and the object. The formula: true frequency over 1 minus v s over v gives the observed frequency. Let's call this equation 1, shall we? The real frequency in our case is stated as 1200 hertz, but the perceived frequency is reported as 1240 hertz. It is stated that the source's velocity, vs, is 25.0 meters per second. Rearranging equation 1 yields the result that the real frequency over the absorbed is equal to 1 minus b s over v.
As a result, we can be represented as v s over 1 minus f, actual over f absorbed now 1 substituting the values we get 25.0 meter per and over 1 minus 1200 hertz over 1240 hertz. Frequency fur, the rearranging we get 1 minus f real over f absorbed, is equal to v s over v. The sound travels at a speed of 775 meters per second after further calculation.
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what diameter must a copper wire have if its resistance is to be the same as that of an equal length of aluminum wire with diameter 1.94 mm ?
The diameter of the copper wire if its resistance is to be the same as that of an equal length of aluminum wire with diameter 1.94 mm is 1.568mm
what is resistance?The obstruction to current flow in an electrical circuit is measured by resistance. The Greek letter omega Ω represents resistance, which is measured in ohms.
what is resistivity?The term "resistivity" refers to a quality that quantifies how much an object resists having current flow through it. It is a feature of the material itself, independent of the size or form of the sample, and is typically temperature dependent, though it could also be pressure-dependent.
by formula we have ;
R=ρL/A =ρL/πd²/4
where
R = resistance
ρ= resistivity
L = length ; and
A = area of cross section = πd²/4
we know that,
ρ₁ of aluminum = 2.63×10⁻⁸Ωm ; d₁ is diameter of aluminum
ρ₂ of copper=1.72×10⁻⁸Ωm ; d₂ is diameter of copper
given: L of copper= L of aluminum
also, R of copper= R of aluminum
on cancelling the equal terms we get,
ρ₁/d₁²=ρ₂/d₂²
here d₂=d₁[tex]\sqrt\frac{resistivity of copper}{resistivity of aluminum}[/tex]
d₂=1.94[tex]\sqrt \frac{1.72}{2.63}[/tex]
d₂=1.568mm
Hence the diameter of copper wire is 1.568mm
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Light travels at a speed of about 3.0 108 m/s.(a) How many miles does a pulse of light travel in a time interval of 0.1 s, which is about the blink of an eye?Δx = mi(b) Compare this distance to the diameter of Earth. (Use 6.38 106 m for the radius of the Earth.)ΔxDE =
Given:
Speed of light = 3 x 10⁸ m/s
Let's solve for the following:
• (a). How many miles does a pulse of light travel in a time interval of 0.1 s, which is about the blink of an eye?
Apply the formula:
[tex]\Delta x=v*t[/tex]Where:
v is the speed of light
t is the time.
Thus, we have:
[tex]\begin{gathered} \Delta x=3.0\times10^8*0.1 \\ \\ \Delta x=3.0\operatorname{\times}10^7\text{ m} \end{gathered}[/tex]Now let's convert the answer from meters to miles.
Where:
1 mile = 1609.34 meters
[tex]\begin{gathered} 3.0\times10^7=\frac{3.0\times10^7}{1609.34} \\ \\ =18641.14\text{ mi} \end{gathered}[/tex]Δx = 18641.14 mi
• (b). Compare this distance to the diameter of Earth.
Apply the formula:
[tex]\frac{\Delta x}{D_E}=\frac{\Delta x}{2*r}[/tex]Where:
r = 6.38 x 10⁶ m.
Thus, we have:
[tex]\frac{\Delta x}{D_E}=\frac{3.0\times10^7}{2*6.38\times10^6}=2.35[/tex]ANSWER:
• (a). 18641.14 mi
,• (b). 2.35
what are the 4 types of graphs and when do we use each type?
A wire of a certain length (α = 0.0065 1/°C) has a resistance of 15 Ω at 20°C.
Calculate the temperature at which the resistance will be 22.8 Ω
The temperature at which the resistance will be 22.8 Ω is 100 °C
How to determine the temperatureThe following data were obtained from the question:
Coefficient of epansion (α) = 0.0065 °C¯¹ Original resistance (R₁) = 15 Ω Original temperature (T₁) = 20 °C New resistance (R₂) = 22.8 ΩNew temperature (T₂) =?The new temperature can be obtained as illustrated below:
α = R₂ – R₁ / R₁(T₂ – T₁)
0.0065 = 22.8 – 15 / 15(T₂ – 20)
0.0065 = 7.8 / 15(T₂ – 20)
Cross multiply
0.0065 × 15 (T₂ – 20) = 7.8
0.0975 (T₂ – 20) = 7.8
Divide both side by 0.0975
T₂ – 20 = 7.8 / 0.0975
T₂ – 20 = 80
Collect like terms
T₂ = 80 + 20
T₂ = 100 °C
Thus, the temperature is 100 °C
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for the following trajectory, find the speed associated with the trajectory and then find the length of the trajectory on the given interval. r(t), for 0
The Fundamental Theorem of Calculus is used to evaluate the definite integral. A important tool for simulating the dynamics of moving objects and other physical phenomena is the parametric form.
For a parameter t, the dynamics or trajectory of a moving object is given in parametric form. We determine the speed of the object over the range of its parameter using derivatives and a definite integral. The Fundamental Theorem of Calculus is used to evaluate the definite integral. A important tool for simulating the dynamics of moving objects and other physical phenomena is the parametric form.
The velocity function or vector's absolute value or norm can be used to calculate an object's speed.
We obtain the velocity v(t) as the first derivative of the position trajectory given as follows since the velocity is the rate of change of the distance traveled:
v ( t ) = r ′ ( t ) = ⟨ ( 2 t 3 ) ′ , ( − t 3 ) ′ , ( 3 t 3 ) ′ ⟩ = ⟨ 6 t 2 , − 3 t 2 , 9 t 2 ⟩
The speed is given as either its norm or absolute value in the velocity function above as follows:
Speed(t)=∥v(t)∥=∥⟨6t2,−3t2,9t2⟩∥
=√(6t2)2+(−3t2)2(9t2)2=√36t4+9t4+81t4=√126t2(2)
The speed at t = 7 will be (2) from (2).
Speed(7)=√126×72=49√126.
Using (2) and the Calculus Fundamental Theorem, the trajectory's length L can be determined as follows:
L = ∫ 7 0 ∥ r ′ ( t )∥ d t = ∫ 7 0 ∥ v ( t ) ∥ d t = ∫ 7 0 √ 126 t 2 d t = √ 126 [ t 3 3 ] 7 0 = √ 126 [ 7 3 3 − 0 3 3 ] = 7 3 √ 126 3. \s.
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Since gravity is a conservative force, the work done against gravity can be replaced by gravitational potential energy. When an object is thrown vertically upward, its gravitational potential energy true or false?.
A ball's gravitational potential energy will rise as it is launched vertically upward.
An object's gravitational potential energy is the power it has as a result of being in a gravitational field. The gravitational potential energy is most frequently used for an object close to the Earth's surface, when the gravitational acceleration is believed to remain constant at around 9.8 m/s2. Since any point can be chosen as the gravitational potential energy zero (just like any coordinate system zero), the potential energy at a height h above that point is equal to the work that would be needed to lift the item to that height with no net change in kinetic energy. Its weight must be lifted with the same amount of force, so the gravitational potential energy must also be equal. The gravitational potential energy is most frequently used for an object close to the Earth's surface, when the gravitational acceleration is believed to remain constant at around 9.8 m/s2. Multiple factors affect the gravitational potential energy of an object: its mass, the gravitational acceleration it experiences from the earth, and its distance from the ground.
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inner parts of the flattening cloud begin to fall freely inward, raining down on growing object at the center. gravitational potential energy of collapsing gas cloud is converted into heat and radiative energy.
These 2 statements are 2 steps of solar planet formation 1st statement is 3rd step 2nd statement is 4th step.
In the cosmos, there are a lot of planetary systems with solar planets around a host star, similar to our own. Because we refer to things that are connected to our star as "solar," our planetary system is sometimes known as "the solar system," after the Latin word for the Sun, "solis." The Sun, our star, and everything gravitationally connected to it, including the moons, planets, and dwarf planets like Pluto, Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune, make up our solar system. Thousands of planetary systems orbiting other stars in the Milky Way have been found, in addition to our solar system. The Milky Way galaxy's outer spiral arm is where our solar system is situated. In our solar system, there is only the Sun.
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two cars are diriving down the road. car a has a mass of 1,100 kg and is moving at 20 m/s. car b has a mass of 1000 kg and is moving at 30 m/s. which car has more kinetic energy and why
Answer:
Hope the pictures will help you
Two cars are moving along a stright line in the same direction with a velocity of 25 km/h and 30 km/h respectively. find the velocity of car a relative to car b
Answer:
Explanation:
V = Vb - Va = 30 - 25 = 5 km/h
Describe how well you think your modeled position matches the observed position for the man.
1 answer=10,543 points
Answer:
its upto your mind and magic
Explanation:
How long does it take a car to travels 40 m/s to 80 m/s with an acceleration of 20 m/s2 on a highway?
how would i find acceleration? im really confused rn. i can solve it, i just cant remember the right equation. help
Answer:
See below
Explanation:
acceleration = change in velocity / change in time
Or F = m * a
So the first one Accel = 40 m/s / 20 s = 2 m/s^2
Or 100 N = 50 * a A = 2 m/s^2 Same answer two different ways...
The others are similar ......
A car travels south at 30 m/s for 5 minutes. How many seconds does it travel
for?
A. 350 s
B. 250 s
C. 200 s
D. 300 s
Answer:
The Answer is D. 300 s because in 5 minutes there are 300 seconds
two conductors made of the same material are connected across the same potential difference. conductor a has eight times the diameter and eight times the length of conductor b. what is the ratio of the power delivered to a to the power delivered to b?
The power given to A is 7 times more powerful than the power delivered to B. So, the ratio of the power delivered to a to the power delivered to b is 7:1.
A wire's cross sectional area is computed as follows:
A= πd²/4
A wire's resistance is calculated as;
R= pL/A
R= 4pL/πd²
opposition in wire A;
R= 4pALA/πd²A
opposition in wire B;
P = V²/R
wired power delivery;
P= V²A/RA
energy provided through cable A;
P= V²b/Rb
energy transferred through wire B;
Replace R's value in the power delivered through wire A;
PA= V²A/RA = V²Aπd²/4pALA
PA/PB = d²A/LA x LB/V²B/d²B
The diameter and length of wire A are both seven times greater than those of wire B;
PA : PB = 7 : 1
Consequently, the power given to A divided by the power transferred to B equals 7 : 1
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on a two lane highway, a car is following a pickup truck. at one instant, the car has a speed of 32 m/s and is 184 m behind the truck. at the same time, the truck has a speed of 28 m/s. if neither vehicle accelerates, how long will it take the car to catch up to the truck?
The vehicle traveling in the opposing direction travels the following distance in the same amount of time: 200/9 (2.5) equals 99.38 meters. The safe separation between the two vehicles is therefore: 129.38 + 99.38 = 228.76 meters.
To obtain the velocity function, we must calculate the integral of the acceleration function:
V ( t ) = ∫ a ( t ) d t V ( t ) = ∫ 3 d t = 3 t + c 1
To find the integral's constant, we use our initial conditions as a guide. As of now
t = 0
The speed is 80 km per hour.
80 km/h is equal to (80)(1000)/3600, or 200/9 meters per second.
V ( 0 ) = 3 ( 0 ) + c 1
200/ 9 = 0 + c 1
c 1 = 200/ 9
The distance the vehicle travels in
the number of seconds required to travel the same distance as the truck plus 30 meters (15 meters in front of and 15 meters behind the truck).
The distance covered by the vehicle is
In the same number of seconds, 200/9 t.
This is the amount of time it takes for the car to overtake and pass the truck.
15 metres.
To do this, the automobile travels a distance of:
S ( t ) = 3 /2 ( 20 ) + 200/ 9 ( 2 5 ) = 129.38 meters
The vehicle traveling in the opposing direction travels the following distance in the same amount of time:
99.38 meters is equal to 200/9 ( 2 5).
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a ball is thrown in the air at 5m/s. If the ball was thrown at a 4 degrees angle. How long will it take to return to its original height?
Answer:
0.5seconds
Explanation:
where
initial velocity u=5m/s
gravity g=9.8m/s²
final velocity v=0m/s (since velocity at maximum height is 0)
time t=?
using
v=u-gt
0=5-9.8(t)
0=5-9.8t
9.8t=5
t=5/9.8
t=0.5102040816
t=0.5sec
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