The work done by the spring in kilojoules is 13.5 KJ.
Spring constant of the spring = 3 KN/cm
[tex] = 3 \times 10 ^{7} \: N/m[/tex]
Spring has been compressed to 3 cm.
So, the displacement covered by the spring = 3 cm
[tex] = \frac{3}{100} = 0.03 \: m[/tex]
Work done by the spring is,
[tex]Work \: done = \frac{1}{2} Spring \: constant × (Displacement ) ^{2} [/tex]
[tex]W = \frac{1}{2} Kx ^{2} [/tex]
[tex]W = \frac{1}{2} \times 3 \times 10 ^{7} \times (0.03) ^{2} [/tex]
[tex]W = 13500 \:Joules[/tex]
Thus, the work done by the spring is 13500 Joules.
Now, the work by the spring in kilojoules is,
[tex]1 \: Kilojoules =1000 \: Joules[/tex]
[tex]1 \: KJ = 1000 \: J[/tex]
[tex]1 \: J = \frac{1} {1000} \:K J[/tex]
[tex]13500 \: J = \frac{13500}{1000}[/tex]
[tex] = 13.5 \: KJ[/tex]
Therefore, the work done by the spring in kilojoules is 13.5 KJ.
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a body of mass m rests on a horizontal plane with the stattic and kinetic fricion coefficient both being euqal to u
The body of mass is resting on a horizontal plane, then the normal force exerted onto it by stattic and kinetic frictional forces both have an equal magnitude. In addition, the net resultant force acting upon the body of mass will be zero.
What do you mean by statistic friction?Statistic friction is a term that refers to the resistance we experience when trying to measure and monitor our performance. This can be due to a variety of factors, including statistical noise or bias. Statistics can often feel like one big puzzle that we're never able to fully understand or control.Statistical friction can lead us down two possible paths: the path of confusion and frustration, or the path of discovery and mastery.This covers everything from fluids and vapors to tangible objects like tools or automobiles. Through figuring out the forces that are at play when two substances come into contact, you can then understand how those forces will impact the things involved. This is done by understanding stattic friction.To learn more about statistic friction, visit:
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a jet fighter pilot wishes to accelerate from rest at 5 g to reach mach-3 (three times the speed of sound) as quickly as possible. experimental tests reveal that he will black out if this acceleration lasts for more than 4.9 s . use 331 m/s for the speed of sound.
The calculated greatest speed is v=0.725 vs.
The ratio of a moving object's speed to the fundamental speed of sound is known as the Mach number. Therefore, an object travelling at twice the speed of sound has a Mach number of 2. speed of sound, the rate at which sound waves move through various substances. Modern estimates of the speed of sound, in particular, for dry air at a temperature of 0 °C (32 °F), are 331.29 meters (1,086.9 feet) per second.
The sound speed is vs=331 m/s.
The duration is t=4.9s.
A=5g is the acceleration.
u=0 is the beginning speed.
Let v be the fastest possible speed without a blackout.
The initial equation of motion is given by v=u+at
v=0+5gt
v=5(9.81m/s2)(4.9s)
v=240m/s
v=240m/svs/331m/s
v=0.725 vs.
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If you have two convex lenses with different focal lengths, which has the larger magnification?
The larger magnification will be based on Focal length
The narrower the focal length, the smaller the magnification, and the larger the focal length, the larger the magnification.
it is obvious that the angle of the photograph from the objective lens or reflect to the eyepiece could be narrower the higher the focal duration. therefore, the photo may be considered via the eyepiece deeper with out dropping attention the narrower the angle (higher focal period).
The intensity of consciousness is shorter than the focal period. It approach that in comparison to large focal lengths, decrease focal lengths provide you with much less focuser play.
although you could nevertheless acquire the equal magnification with either focal period, the shorter focal duration will lose awareness a great deal earlier than the bigger focal period.
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Given the equation p2 = a3, what is the orbital period, in days, for the planet venus? (venus is located 0.72 au from the sun?) 72 days 225 days 500 days 5,375 days
The correct answer is 223 days.
The relationship between the duration of revolution and the separation between the sun is shown by Kepler's third law. Using the notions of circular motion and the gravitational and centripetal forces, we may obtain this equation.
According to Kepler's third rule, the semi-major axis of an orbit is linked to the orbital period of a planet around the sun as follows:
p² = a³
where an is the semi-major axis/distance to the star and p is the orbital period in years.
It is said that a = 0.72 AU for Venus.
P= √(0.72 AU)^3 = 0.61 years.
365 days in a year = 222.9 ≈ 223 days.
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9. A girl standing upright exerts a pressure of 15000 N/m 2 on the floor. Given
that the total area of contact of shoes is 0.02m ².
a) Determine the mass of the girl.
The mass of the girl is 30.6 Kg.
Inertia, a fundamental property of all matter, can be quantified as mass. In essence, it's the resistance of a mass of matter to changing its direction or speed in reaction to the application of a force. The more mass a body possesses, the smaller the change that is brought about by an applied force. The dimensionless quantity mass is used to express the amount of matter contained in a particle or object (symbolized m).
Given,
The pressure exerted by the girl = 15000 [tex]\frac{N}{m^{2}}[/tex]
Total area of contact = 0.02[tex]m^{2}[/tex]
As we know,
Pressure = [tex]\frac{Force}{area}[/tex]
Force = Area x Pressure
Force = 0.02 x 15000
Force = 300N
For the conversion of weight into mass, use the formula "w = m x g." The gravitational pull on an object is referred to as its weight. The equation for the statement is written as w = m x g, or w = mg, by scientists.
Weight of the girl = Mass x Gravitational constant
300N = Mass of the girl x 9.8
Mass of the girl = 300 / 9.8
Mass of the girl = 30.6 Kg
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What is the potential difference between the plates of a 3.7-f capacitor that stores sufficient energy to operate a 75.0-w light bulb for one minute?
The potential difference between the plates will be 1552 Volts.
What is a potential difference?Voltage, or the difference in electric potential between two points, is defined as the amount of labor per unit of charge needed to move a test charge between the two points.
Given that a 3.7-f capacitor that stores sufficient energy to operate a 75.0-w
The potential difference will be calculated by the formula below:-
Q = I t
Where I = charge / time
Q = V * C
V C = I t
V = I t / C
V = 75 C/s x 60 sec / 2.9 faraday
V = 1552 Volts
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there’s nothing like the freedom of the open road. and there’s nowhere on the planet more thrilling to traverse on
The lines there’s nothing like the freedom of the open road are taken from the poem Song of the open road. The road in the poem signifies the symbol of freedom.
These lines are from the poem "Song of the Open Road” written by Walt Whitman. In the poem, a journey characterized by freedom, and independence is described. The road in the poem is a symbol of self-awareness and freedom as the author travels along the road.
Whitman describes that the journey includes a test of the wisdom of the soul, revealed through provoking questions and experiences.
The poem also conveys the narrator's joy and happiness in being on the open road, which converts into a love of freedom and the open spaces that are the outdoors.
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You whirl a 2-kg body attached to a 1-meter cord around your head in a nearly horizontal circle with a speed of 4 m/s. the tension in the cord is:________
The tension in the cord is equal to centripetal force which is 32 newton.
We need to know about centripetal force to solve this problem. When an object moves in a circular motion, the object is experiencing centripetal force. The magnitude of centripetal force is
Fc = m . v²/R
where Fc is the centripetal force, m is mass, v is velocity and R is the radius.
From the question, we know that
m = 2 kg
v = 4 m/s
R = 1 m
The tension in the cord is equal to centripetal force
T = Fc
T = m . v²/R
T = 2 . 4² / 1
T = 32 N
Hence, the tension in the cord is 32 newton.
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if we want to find the size of the force necessary to just barely overcome static friction (in which case fs
Force necessary to overcome static friction= (μs mg ) / (cos Ф + μs sinФ)
We have to find Force(F) in terms of m, g, Ф ,and μs
To get the value of F first we have to find the value of N by using the formula of friction,
F sin Ф + N - mg =0
N = mg - F sinФ
As the body does not move so the net force is zero, therefore we have,
FcosФ -μs N = 0
Now we have to put the value of N in the equation. To know the force necessary to overcome the static friction we have,
F cosФ - μs ( mg - FsinФ) = 0
F cosФ - μs mg + μs FsinФ = 0
F( cos Ф + μs sinФ ) = μs mg
F = μs mg/ (cosФ +μs sin Ф)
Therefore we get the value of F i.e., force in terms of m, g, Ф, and μs.
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The complete question is:
If we want to find the size of the force necessary to just barely overcome static friction (in which case fs=μsN), we use the condition that the sum of the forces in both directions must be 0. Using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as Fcosθ−μsN=0 Fsinθ+N−mg=0 To find the magnitude of force F, we have to solve a system of two equations with both F and the normal force N unknown. Use the methods we have learned to find an expression for F in terms of m, g, θ, and μs (no N).
A mass weighing 4 pounds is attached to a spring whose spring constant is 25 lb/ft. what is the period of simple harmonic motion?
The basic harmonic motion of the mass-spring system lasts for 5.024 seconds.
We have a 16-pound mass that is fastened to a spring with a 25-pound-per-foot spring constant. This entire spring and mass system is moving harmonically.
What is simple harmonic motion?Simple harmonic motion (sometimes referred to as SHM) is a particular type of periodic motion in mechanics and physics in which the restoring force on the moving object is inversely proportional to the magnitude of the object's displacement and acts in the direction of the object's equilibrium position.
The particle's acceleration in a simple harmonic motion is geared toward its mean location and directly proportionate to its displacement. Simple harmonic motion results in the conservation of the particle's total energy. SHM is a cyclical motion.
Simple harmonic motion is demonstrated by bungee jumping. Due to the flexibility of the bungee cord, the jumper is experiencing SHM as it oscillates up and down.
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A water main is to be constructed with a % grade in the north direction and a % grade in the east direction. Determine the angle required in the water main for the turn from north to east.
The angle between the two pipes is 88.88°
The objective is to find the angle between the vectors u and v.
First, construct the vectors u and v,
The water main with 20% grade in the north direction is,
20% = rise/ run
rise/run = 20/100 = 1/5
That is rise = 1 and run = 5
Hence, rise corresponds to the z- component, and run corresponds to the y-component to the vector u. There is no x-component in this direction since the vector does not go to the east at all.
Therefore, the vector
v = xi + yj + zk
Here x = 0, y = 5, z = 1
v = 0i +5j +k is parallel to the pipe in the north direction.
The water main with 10% grade in the east direction is,
10% = rise/run
rise/run = 1/10
Thus, rise =1 and run = 10
Hence, rise corresponds to the z-components, and run corresponds to the x-components to the vector v. There is no y-component in this direction since the vector does not go to the north at all.
Therefore, the vector
v = xi + yj + zk
Here x = 10, y = 0, z = 1
v =10i +0j + k is parallel to the pipe in the east direction.
The angle between two vectors is,
Ф= [tex]cos^{-1} \frac{u-v}{|u| |v|}[/tex]
The dot product of the two vector u and v is,
u.v = (0.5,1).(10,0,1)
= 0.10+5.0+1.1
=1
The magnitude of these two vectors is,
|u| =[tex]\sqrt{x^{2} + y^{2}+z^{2} }[/tex]
|u| = [tex]\sqrt{0^{2} + 5^{2} +1^{2} }[/tex]
=[tex]\sqrt{0+25+1}[/tex]
= [tex]\sqrt{26}[/tex]
|v| = [tex]\sqrt{x^{2} + y^{2} +z^{2} }[/tex]
|v| = [tex]\sqrt{0^{2}+5^{2} +1^{2} }[/tex]
= [tex]\sqrt{100+0+1}[/tex]
=[tex]\sqrt{101}[/tex]
Therefore, the angle between the two pipes is,
Ф = [tex]cos^{-1} \frac{u-v}{|u||v|}[/tex]
= [tex]cos^{-1}[/tex][tex]\frac{1}{\sqrt{26}\sqrt{101} }[/tex]
≈ 1.55rad
= 88.88°
Hence, the angle between the two pipes is 88.88°
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The complete question is:
A water main is to be constructed with a 20% grade in the north direction and a 10% grade in the east direction. Determine the angle required in the water main for the turn from north to east.
What is the acceleration of a car moving along a straight road that increases its speed from 0 to 100 km/h in 10 s?.
The acceleration of a car moving along straight road that increases its speed from 0 to 100 km/h in 10 s is 10 km/h-s.
What is acceleration?The rate at which an object's velocity with the minimum time changes is called an accelerator in mechanics. They were the vector quantities and accelerations.
Acceleration has dimensions of velocity (L/T) divided by time, i.e. L T−2. SI unit of acceleration is the meter per second squared (m s−2); or "meter per second per second", as the velocity in meters per second changes by the acceleration value, every second
The direction of net load acting on an item affects the direction of its acceleration.
Average acceleration = (change in velocity)/(time it takes)
The acceleration of a car moving along straight road that increases its speed from 0 to 100 km/h is:
car's acceleration = (100 km/h)/(10 s) = 10 km/h/s
Thus, acceleration of a car moving along a straight road that increases its speed from 0 to 100 km/h in 10 s is 10 km/h-s.
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1. S1 shows a change as a result of price. Is this a change in supply or change in quantity
supplied? Change in quantity supplied
2. D1 to D2 shows a change. Is this a change in demand or change in quantity demanded?
Change in demand
3. What is the D1 equilibrium price? $2500
4. What is the D2 market clearing price? $4000
5. What is the quantity supplied and demanded at the D1 and D2 equilibrium price?
6. What is the quantity demanded (D1 & D2) and supplied (S1) at $10,000? 13,000
7. What is the quantity supplied and demanded at $1000?
8. What could have caused the shift from D2 to D3 (non-price factors)
Answer: yes
Explanation:
so basically i need points
If a fisherman applies a horizontal force with magnitude 43.5 nn to the box and produces an acceleration of magnitude 3.30 m/s2m/s2, what is the mass of the box?
The mass of the box which the fisherman applies a horizontal force is 13.18 Kg
To solve this exercise, the formula and procedure to be applied are:
F = m * a
Where:
m = massF = Forcea = accelerationInformation about the problem:
a= 3.30 m/s2F= 43.5 N1 N = kg * m/s²m = ?Applying the force formula and isolating the mass, we get:
F = m * a
m = F/a
m = 43.5 Kg * m/s² / 3.30 m/s2
m = 13.18 Kg
What is force?It is the physical magnitude that expresses the effort necessary to move the mass of one kilogram with an acceleration of one meter per second squared, this is expressed in units of the international system in Newton.
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Find the velocity and position vectors of a particle that has the given acceleration and the given initial velocity and position. a(t) = 3 i 4 j, v(0) = k, r(0) = i
The velocity vector is (v = 3ti + 4tj + k) and the position vector is (x = (3t²/2 + 1) i + 2t²j + tk)
We need to know about vector to solve this problem. The velocity is depend on position vector and acceleration. It can be written as
v = dx / dt
where v is velocity, dx is displacement and dt is time interval.
By differentiating the velocity function, we can also get the acceleration
a = dv / dt
where a is acceleration.
From the question above, we know that
a(t) = 3i + 4j
Hence, the velocity is
a = dv / dt
dv = a . dt
v = ∫a . dt
v = ∫(3i + 4j) . dt
v = (3i + 4j)t + v(0)
because v(0) is k, thus
v = 3ti + 4tj + k
Find the position vector
v = dx / dt
dx = v . dt
x = ∫v . dt
x = ∫(3ti + 4tj + k) . dt
x = (3t²/2 i + 2t²j + tk) + x(0)
because x(0) is i, thus
x = 3t²/2 i + 2t²j + tk + i
x = (3t²/2 + 1) i + 2t²j + tk
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Because of the ______________________, we can predict that gravity works the same way on earth as it does on mars.
Because of the cosmological principle, we can predict gravity works similarly on Earth as on Mars.
The cosmological principle says that an observer's view of the universe is independent of both the direction he looks and his location. This principle only applies to the universe's large-scale properties, but it does suggest that the world has no edge, implying that the big-bang origin took place not at a specific point in space, but instead throughout space at the exact time.
These two assumptions allow us to determine the history of the universe after a specific epoch known as the Planck time. Scientists are yet to discover what existed prior to Planck time.
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Just a quick joke but…. Why are apartments called A-P-A-R-T-M-E-N-T-S if they are so close together??????
Answer:
why?
Explanation:
Do you need any help on anything else?
Who would like to Join A Bible Club
Here's the link For You to Sign Up If Interested
Requirements:
Be older than 10 and younger than 20
https://forms.gle/kjbkMRV9u1fLJnGu8
This is for You and For God !!!
“For the wages of sin is death, but the gift of God is eternal life in Christ Jesus our Lord” Romans 3:23
Answer:
alright
Explanation:
that's so cool I'll join
the net force on a box is in the positive x direction. which of the following statements best describes the motion of the box?
Its acceleration is parallel to the x-axis
Using Newton Second Law of motion
Where M is mass of the object, [tex]\overrightarrow{F}[/tex] = Net force,[tex]\overrightarrow{a}[/tex] = acceleration.
Thus net force and acceleration has always the same direction.
Now net force on the box is in the positive x-direction then acceleration on the box is in the positive x-direction.
Acceleration is given by
[tex]\overrightarrow{a}[/tex] = [tex]\frac{d\overrightarrow{v}}{dt}[/tex]
Where [tex]\overrightarrow{v}[/tex] = velocity of the box and t = time
thus acceleration is parallel to change in velocity but not velocity. Thus it is not necessary that velocity of the box is parallel to the x-axis. Because if initially box is moving in arbitary direction then only its x-component of velocity changes only but another component of velocity remain same.
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[15 points] a laboratory sample of sand is formed inside a mold of 0.4 cubic meters. it took 711.2 kg of wet sand (dry mass of sand
The following are the values:-
(a)Water content (W%) = 13.99%
(b) Total unit weight (Υ[tex]_{t}[/tex]) = 1778 kg/m³
(c) Dry unit weight (Y[tex]_{d}[/tex]) = 1559.75 kg/m³
(d) Void ratio (e) = 0.718
(e) Porosity (n) = 0.417
Volume of mold (V[tex]_{t}[/tex]) = 0.4 m³
Dry mass of sand (W[tex]_{s}[/tex]) = 623.9 kg
Wet mass of sand (W[tex]_{t}[/tex]) = 711.2 kg
Specific Gravity (G[tex]_{s}[/tex]) = 2.68
1) Water content (W%)Water content (W%) = [tex]\frac{W_{w} }{W_{s} }[/tex] × [tex]100[/tex]
Water content (W%) = [tex]\frac{711.2 - 623.9}{623.9}[/tex] × [tex]100[/tex]
Water content (W%) = 13.99%
2) Total unit weight (Υ[tex]_{t}[/tex])Total unit weight (Υ[tex]_{t}[/tex]) = [tex]\frac{W_{t} }{V_{t} }=\frac{711.2}{0.4}[/tex]
Total unit weight (Υ[tex]_{t}[/tex]) = 1778 kg/m³
3) Dry unit weight (Y[tex]_{d}[/tex])Dry unit weight (Y[tex]_{d}[/tex]) = [tex]\frac{W_{s} }{V_{t} }=\frac{623.9}{0.4}[/tex]
Dry unit weight (Y[tex]_{d}[/tex]) = 1559.75 kg/m³
4) Void ratio(e)Solid unit weight (Y[tex]_{s}[/tex]) = [tex]\frac{W_{s} }{V_{s} }[/tex]
Or
G[tex]_{s}[/tex] = [tex]\frac{Y_{s} }{Y_{w} }[/tex] ⇒ Y[tex]_{s}[/tex] = G[tex]_{s}[/tex]Y[tex]_{w}[/tex]
Y[tex]_{s}[/tex] = 2.68 × [tex]\frac{10^{-3} }{(10^{-2})^{3} }\frac{kg}{m}[/tex] = 2680 kg/m³
Now,
Void Ratio (e) = [tex]\frac{Y_{s} }{Yd}-1[/tex]
Void Ratio (e) = [tex]\frac{2680}{1559.72}-1[/tex]
Void Ratio (e) = 0.718
5) Porosity (n)Porosity (n) = [tex]\frac{e}{1+e}[/tex]
Porosity (n) = [tex]\frac{0.718}{1+0.718}[/tex]
Porosity (n) = 0.417
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A laboratory sample of sand is formed inside a mold of 0.4 cubic meters. It took 711.2 kg of wet sand (dry mass of sand = 623.9 kg) to fill the mold. Assuming the specific gravity of the solid is 2.68, compute the:
(a)Water content (W%)
(b) Total unit weight (Υ[tex]_{t}[/tex])
(c) Dry unit weight (Y[tex]_{d}[/tex])
(d) Void ratio (e)
(e) Porosity (n)
The correct value for the rocket speed measured by an observer on earth would be the correct value for the rocket speed measured by an observer on earth would be:_______
Therefore, the missile's individual speed would be 14c + 14c = 28c.
The missile is traveling in the opposite direction from the Earth, which has an individual speed of zero.
Both the rocket and the missile are traveling in the same direction relative to one another. The moving item whose speed is being measured and the frame of reference for the point of observation must both be in relative motion.
The difference between their respective speeds is hence the relative speed.
Therefore, the missile's individual speed would be 14c + 14c = 28c.
The missile is traveling in the opposite direction from the Earth, which has an individual speed of zero.
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If the electron just misses the upper plate as it emerges from the field, find the speed of the electron as it emerges from the field?.
The speed of the electron as it emerges from the field is; 388.587 m/s
What is the speed of the electron?
Initial speed; v₀x = 1.1 * 10⁶ m/s
Acceleration in horizontal direction = 0 m/s²
distance; s_x = 2 cm = 0.02 m
Thus, formula to find time here is;
t = s_x/v₀x
t = 0.02/(1.1 * 10⁶)
t = 1.82 * 10⁻⁶ s
Now for the vertical distance; v,y_o = 0 m/s
Thus, the equation of motion becomes;
s_y = ¹/₂at²
0.005 = ¹/₂a(1.82 * 10⁻⁶)²
Solving for a gives;
a = 3.02 * 10¹³ m/s²
Thus the speed of the electron as it emerges from the field is;
v² = u² + 2as
v = √(0² + 2(3.02 * 10¹³ * 0.005))
v = 388.587 m/s
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Complete Question is;
An electron is projected with an initial speed v0 = 1.1 * 10⁶ m/s into the uniform field between the parallel plates. The distance between the plates is 1 cm and the length of the plates is 2 cm. Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates. E = N/C
find the speed of the electron as it emerges from the field?.
a mass of the block is 200 g, the dimensions of the block are 6 cm by 4 cm by 5 cm. what is its density?
Suppose you have a pocket calculator through which 1.25 c of charge passes in 2.75 h?
Complete Question
Suppose you have a pocket calculator through which 1.25 C of charge passes in 3.5 h? What is the average current in milliamperes produced by the solar cells .
It is 0.126mA .
An item has a fine fee if it incorporates more protons than electrons. when a thing loses a few electrons, the difference between the number of protons and electrons in the item rises. As a result, the item acquires a positive rate.
tremendous and negative fees cancel each other out when there are an equal range of high-quality and poor charges, leaving the item neutral.
general rate that passes via the calculator may be denoted by q = 1.25c
total time t = 2.75=9900s
current required is = 1.25/9900
=1.26×10^-4 A
=1.26×10^1mA
=0.126mA
Thus the solar cell must produce an average of 0.126mA of current.
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a billiard ball travels 23 cm in the positive direction, hits the cushion and rebounds in the negative direction, and finally comes to rest 7.5 cm behind its original position.
The billiard ball involves relaxation 80 cm
Displacement refers to the space among the very last and the preliminary role. Hence the displacement of the ball can be the distinction among the preliminary and the very last displacement.
Let the preliminary role be 0
Final role = 8 cm
So the distinction among preliminary role and very last role = 0 - 88 cm
So the billiard ball involves relaxation 80 cm at the back of its orbital role.
What is displacement thickness in boundary layer theory?The displacement thickness for the boundary layer is described as the gap the floor could need to pass withinside the y-course to lessen the glide passing via way of means of a extent equal to the actual impact of the boundary layer.
The distance from the boundary floor where the rate reaches 99% of the mainstream speed is used as an arbitrarily defined measure of the thickness of the boundary layer indicated by the symbol.
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Ben wants to buy a new cell phone. Select the conditions that Ben should consider for this buying decision.
The number of sisters and brothers Ben has
The amount of money Ben has to spend
The price of the cell phone
Ben's age
When you push a shopping cart the friction that is opposing the motion is called _____ friction.
What is the friction used when pushing a shopping cart?
Answer:
Rolling Friction
Explanation:
Friction always acts in the opposite direction to that of the motion of the other object. Meaning, friction slows down the motion of an object by exerting opposing forces on it
The scale of the horizontal axis is 5s per division on the vertical axis 5 m/s per división. The initial position is 57 m.
A) what is the position when t=35s?
Answer in units of m.
B) what is the acceleration is represented by the graph? Answer in units of m/s^2
The position when time, t =35s is 87.59 m.
The acceleration represented by the graph is 0.143 m/s².
Acceleration of the objecta = Δv/Δt
where;
Δv is change in velocityΔt is change in time of motiona = (6 - 5) / (7 - 0)
a = 0.143 m/s²
Position of the object at the given time, 5 secondsThe position of the object is determined from the product of velocity and time of motion.
s = vt + ¹/₂at²
where;
v is initial velocity = 0t is time = 35 secondsa is acceleration = 0.143 m/s²s = 0 + ¹/₂(0.143)(35)²
s = 87.59 m
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A block of mass 5 kg is pulled across a horizontal bench by a horizontal force. the acceleration of the block is 2.5 m/s2 when the friction force is 10 n. calculate the horizontal pulling force.
The horizontal pulling force on the block is 22.5 N.
When a block is pulled by applying a force, friction also acts on it.
The net force is calculated by taking the net value of the two forces.
The free-body diagram(FBD) of the block, is shown in the adjoining diagram.
Thus,
F - f = ma
Here F is the applied horizontal force and f is the frictional force.
Putting the given values in the above equation
F - 10 N = 5 x 2.5
F = 22.5 N
Thus, the horizontal pulling force on the block is 22.5 N.
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The horizontal pulling force on the block is 22.5 N.
When a block is pulled by applying a force, friction also acts on it.
The net force is calculated by taking the net value of the two forces.
The free-body diagram(FBD) of the block, is shown in the adjoining diagram.
Thus,
F - f = ma
Here F is the applied horizontal force and f is the frictional force.
Putting the given values in the above equation
F - 10 N = 5 x 2.5
F = 22.5 N
Thus, the horizontal pulling force on the block is 22.5 N.
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two stones are dropped from a tower at an interval of 2 s. find the relative velocity of 1st stone with respect to 2nd stone.
The relative velocity of the first stone concerning the second stone is 20 m/s.
The time interval at which two stones are dropped from the tower is t.
t = 2 seconds
Let the initial velocity of the first stone be u1.
Let the initial velocity of the second stone be u2.
Let the final velocity of the first stone be v1.
Let the final velocity of the second stone be v2.
The velocity of the first stone is,
[tex]v _{1} = u _{1} + at[/tex]
[tex]v _{1} = 0 + gt[/tex]
[tex]v _{1} = gt[/tex]
The velocity of the second stone is,
[tex]v _{2} = g(t - 2)[/tex]
The relative velocity of the first stone concerning the second stone is,
[tex]v = v _{1} - v _{2}[/tex]
= g - g ( t - 2)
= gt -gt - 2g
= 2 × g
= 2 × 10
= 20 m/s
Therefore, the relative velocity of the first stone concerning the second stone is 20 m/s.
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