We are given:
the initial height of the object (h) = 100 m
initial velocity (u) = 0 m/s
we will let the value of g = 10 m/s/s
Speed of the object just before hitting the ground:
From the third equation of motion:
v² - u² = 2ah (where v is the final velocity)
replacing the variables, we get:
v² - (0)² = 2(10)(100)
v² = 2000
v = 10√20 = 44.7 m/s
Therefore, the speed of the object just before hitting the ground is 44.7 m/s
1. Which statement is true of culture?
O Culture do not change or evolve
O Culture refers to the degree in which resources are used to address for problems.
O Culture has little affect on a persons life
Food can reflect culture
which two types of information are found in an elements box in the periodic table
Answer:
Each box represents an element and contains its atomic number, symbol, average atomic mass, and (sometimes) name.
Explanation:
Answer:
An element's period and group
If an object is moving with a constant velocity to the right, what direction is the net force.
Group of answer choices
A.To the right
B.To the left
C.Net force is 0
D.Not enough information
Answer:
At constant velocity, his weight equals the force of friction. In other words, there is no net force. If however, he loosens his grip and decreases the friction force, he will accelerate downward.
Explanation:
in the figure shown if angle i increases slightly angle r will
Answer:
we need the image to do so.
Explanation:
sorry
The horizontal surface on which the block slides is frictionless. The speed of the block before it touches the spring is 6.0 m/s. How fast is the block moving at the instant the spring has been compressed 15 cm
Answer:
The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.
Explanation:
The spring constant is 2000 newtons per meter. Let consider the spring-block system, from Principle of Energy Conservation we can represent it by the following model:
[tex]U_{k,1}+K_{1} = U_{k,2}+K_{2}[/tex]
[tex]K_{2} = K_{1}+(U_{k,1}-U_{k,2})[/tex] (Eq. 1)
Where:
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final kinetic energies of the block, measured in joules.
[tex]U_{k,1}[/tex], [tex]U_{k,2}[/tex] - Initial and final elastic potential energy, measured in joules.
And we expand the equation above by definitions of elastic potential energy and kinetic energy:
[tex]\frac{1}{2}\cdot m \cdot v_{2}^{2} = \frac{1}{2}\cdot m\cdot v_{1}^{2} + \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})[/tex]
[tex]v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) }[/tex] (Eq. 1b)
Where:
[tex]m[/tex] - Mass of the block, measured in kilograms.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final velocities of the block, measured in meters per second.
[tex]x_{1}[/tex], [tex]x_{2}[/tex] - Initial and final positions of spring, measured in meters.
If we know that [tex]v_{1} = 6\,\frac{m}{s}[/tex], [tex]k = 2000\,\frac{N}{m}[/tex], [tex]m = 2\,kg[/tex], [tex]x_{1} = 0\,m[/tex] and [tex]x_{2} = 0.15\,m[/tex], the final speed of the block moving at the instant the spring has been compressed is:
[tex]v_{2} = \sqrt{\left(6\,\frac{m}{s} \right)^{2}+\left(\frac{2000\,\frac{N}{m} }{2\,kg} \right)\cdot [(0\,m)^{2}-(0.15\,m)^{2}]}[/tex]
[tex]v_{2}\approx 3.674\,\frac{m}{s}[/tex]
The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.
A person has a mass of 1000g and an acceleration of 20 m/s/s. What is the force on the person
Answer:
20000
Explanation:
Newtons Second law states that the force acting on an object is equal to its mass times its acceleration, f=ma. To solve for force, plug in your values for m and a, and then solve. f = (1000)*(20) = 20000
A car traveling at 27 m/s slams on its brakes to come to a stop. It decelerates at a rate of 8 m/s2 . What is the stopping distance of the car?
v² - u² = 2 a ∆x
where u = initial velocity (27 m/s), v = final velocity (0), a = acceleration (-8 m/s², taken to be negative because we take direction of movement to be positive), and ∆x = stopping distance.
So
0² - (27 m/s)² = 2 (-8 m/s²) ∆x
∆x = (27 m/s)² / (16 m/s²)
∆x ≈ 45.6 m
The stopping distance of car achieved during the braking is of 45.56 m.
Given data:
The initial speed of car is, u = 27 m/s.
The final speed of car is, v = 0 m/s. (Because car comes to stop finally)
The magnitude of deacceleration is, [tex]a = 8\;\rm m/s^{2}[/tex].
In order to find the stopping distance of the car, we need to use the third kinematic equation of motion. Third kinematic equation of motion is the relation between the initial speed, final speed, acceleration and distance covered.
Therefore,
[tex]v^{2}=u^{2}+2(-a)s[/tex]
Here, s is the stopping distance.
Solving as,
[tex]0^{2}=27^{2}+2(-8)s\\\\s = 45.56 \;\rm m[/tex]
Thus, we can conclude that the stopping distance of car achieved during the braking is of 45.56 m.
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Converting compound units
You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfortunately, you have the density of mercury in units of kilogram/meter3 and the density of silicon in other units: 2.33 gram/centimeter3. You decide to convert the density of silicon into units of kilogram/meter3 to perform the comparison. By which combination of conversion factors will you multiply 2.33 gram/centimeter3 to perform the unit conversion?
Answer:
Dividing the silicon density by 1000 and then multiply it by 1000000.
Explanation:
A kilogram equals 1000 grams and a cubic meter equals 1000000 cubic centimeters. Hence, we must divide the silicon density by 1000 and then multiply itby 1000000 to convert the value into kilograms per cubic centimeter. That is:
[tex]x = 2.33\,\frac{g}{cm^{3}}\times \frac{1\,kg}{1000\,g}\times \frac{1000000\,cm^{3}}{1\,m^{3}}[/tex]
[tex]x = 2330\,\frac{kg}{m^{3}}[/tex]
In a nutshell, we must multiply the density of silicon by 1000 to obtains its value in kilograms per cubic meter.
An object is dropped from a high building. It reaches the ground in two seconds.
a) Find the distance travelled by the object.
b) Find the speed of the object, when it hits the ground.
Answer:
a: 19.6 meters b: 9.8meters per second
Explanation:
speed and object falls is 9.8 meters per second. 9.8 meter times 2 is 19.8 meters.
A car traveling at 32.4 m/s skids to a stop in 4.55 s. Determine the skidding distance of the car (assume uniform acceleration).
Answer:
x = 73.71 [m]
Explanation:
In order to solve this problem we must use two formulas of kinematics. It is important to make it clear that these formulate are for uniformly accelerated motion, i.e. with constant acceleration.
[tex]v_{f }= v_{i}-(a*t)[/tex]
where:
Vf = fnal velocity = 0
Vi = initial velocity = 32.4 [m/s]
t = time = 4,55 [s]
a = acceleration or desacceleration [m/s^2]
0 = 32.4 - (a*4.55)
a = 7.12 [m/s^2]
Note: it is important to clarify that the negative sign in the above equation is because the car stops and decreases its speed to zero, thus its final velocity is equal to zero.
Now using the following equation:
[tex]x = x_{o} + (v_{i}*t)-(\frac{1}{2} )*a*t^{2}[/tex]
where:
xo = initial distance = 0
x = final distance [m]
Therefore we have:
x = 0 + (32.4*4.55) - (0.5*7.12*4.55^2)
x = 73.71 [m]
Why do you feel that you are being thrown upward out of your seat when going over an arced hump on a roller coaster
Answer: The options are not given.
Here are the options.
a) There is an additional force lifting up on you.
(b) At the top you continue going straight and the seat moves out from under you.
(c) You press on the seat less than when the coaster is at rest.Thus the seat presses less on you. (
d) Both b and c are correct.
(e) a, b, and c are correct.
The correct option Is D.
B.At the top you continue going straight and the seat moves out from under you. C.At the same time, you press on the seat less than when the coaster is at rest because the normal force expirienced will be less.
Explanation:
At the top you continue going straight and the seat moves out from under you.At the same time, you press on the seat less than when the coaster is at rest because the normal force expirienced will be less because it is as a result of a phenomenon called Weightlessness. This occur when there is no force or little force is acting on your body. At the top you continue going straight and the seat moves out from under you because there is no force acting on your body and when the body is in free fall i.e acceleration due to gravity , the person is not supported by any thing at.
That is the scenarion that occur...
How long would it take you to walk 3,962 km from New York to Los
Angeles?
Answer:
913 hours ur welcome :)
A certain superconducting magnet in the form of a solenoid of length 0.300 m can generate a magnetic field of 8.90 T in its core when its coils carry a current of 95 A. Find the number of turns in the solenoid.
Answer:
The number of turns in the solenoid is 22366.
Explanation:
The number of turns in the solenoid can be found using the following equation:
[tex] B = \mu_{0} I\frac{N}{L} [/tex]
Where:
B: is the magnetic field = 8.90 T
L: is the solenoid's length = 0.300 m
N: is the number of turns =?
I: is the current = 95 A
μ₀: is the magnetic constant = 4π×10⁻⁷ H/m
By solving equation (1) for N we have:
[tex] N = \frac{BL}{\mu_{0} I} = \frac{8.90 T*0.300 m}{4\pi \cdot 10^{-7} H/m*95 A} = 22366 turns [/tex]
Therefore, the number of turns in the solenoid is 22366.
I hope it helps you!
Calculate the work WC done by the gas during the isothermal expansion. Express WC in terms of p0, V0, and Rv.
Complete Question
The complete question is shown on the first and second uploaded image
Answer:
The expression is [tex]W_c = P_o V_o ln (R_v)[/tex]
Explanation:
Generally smallest workdone done by a gas is mathematically represented as
[tex]dW = PdV[/tex]
Generally for an isothermal process
[tex]PV = nRT = constant [/tex]
=> [tex]P = \frac{nRT}{V}[/tex]
Generally the total workdone is mathematically represented as
[tex]W_c = \int\limits^{v_f}_{V_o} {\frac{nRT}{V} } \, dV[/tex]
=> [tex]W_c = nRT \int\limits^{V_f}_{V_o} {\frac{1}{V} } \, dV[/tex]
=> [tex]nRT [lnV] | \left \ {V_f}} \atop {V_o}} \right.[/tex]
=> [tex]W_c = nRT [ln(V_f) - ln(V_o)][/tex]
=> [tex]W_c = nRT ln \frac{V_f}{V_o}[/tex]
From the question [tex]\frac{V_f}{V_o } = R_v[/tex]
=> [tex]W_c = P Vln (R_v)[/tex]
at initial state
[tex]W_c = P_o V_o ln (R_v)[/tex]
Which pair of objects would be most strongly attracted to each other?
A. A positively charged particle and a negatively charged particle
B. Two positively charged particles
C. Two negatively charged particles
D. A negatively charged particle and a neutral particle
Answer:
Its A
Explanation:
Just did the quiz
The pair of objects that should be strongly attracted is option A. A positively charged particle and a negatively charged particle.
Pair of objects that are most strongly attracted:When there is a positively charged particle & the negatively charged particle so due to this it should be strongly attracted. Also, when there are two positively charged particles, or a negative one or a include negative one or neutral one so it should not be strongly attracted. Therefore, the option A is correct.
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PLS HELP 3. Which graph best represents the relationship between acceleration due to gravity and mass for
objects near the surface of Earth? [Neglect air resistance.)
Acceleration
This question involves the concepts of mass, acceleration due to gravity, weight, Newton's Gravitational Law, gravitational force, and graphs.
Graph "A" re[resents the best relationship between acceleration due to gravity, and mass for objects.
The relationship between the mass of earth and the acceleration due to gravity can be found by equating the weight of the object and the gravitational force, from Newton's Gravitational Law on it:
[tex]Weigth = Gravitational\ Force\\\\mg = \frac{GmM}{r^2}\\\\g = \frac{GM}{r^2} ---------- eqn(1)[/tex]
where,
g = acceleration due to gravity
G = universal gravitational constant
M = mass of Earth
r = radius of Earth
Hence, it is clear from equation (1), that mass of the Earth and the acceleration due to gravity have a direct relationship with each other. Therefore the graph between them will be a straight line, which is Graph A.
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The attached picture illustrates Newton's Law of Gravitation.
One student runs with a velocity of +10 m/s while a second student runs with a velocity of –10 m/s. Which student has the faster velocity? Why?
Answer:
The one with the faster velocity is the one with a velocity of -10m/s
What function do you think a flower that can stay warm at night might have for a plant?
Answer:
it keeps it growing because if it stays warm and yuh put it by a plant that needs sun yk
Explanation:
A flower of a plant that can stay warm at night as it keeps it growing.
What are the function of different parts of plant ?
The roots are the underground part of the plant which plays a major role in pulling the water and minerals, expands within the ground for better water absorption, anchors which helps in creating better stability.
A stem present above the ground bears leaves, fruits plus flowers. as it distributes the nutrients and minerals to the leaves, shields the plant and assists in asexual dissemination.
leaf is one of the most major parts of a plant as it contain chlorophyll pigment which assists the plants in preparation for food, the veins allow the flowing of nutrients plus water, it has the parts like petiole, leaf base and lamina help in photosynthesis.
Flower is the most bright and beautiful part of the plant show an important role in making food, used in fertilization process , the basic parts are petals, sepals, stamens, and pistil.
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Two particles are separated by 0.38 m and have charges of -6.25x 10 C and 2.91 x 10 C. Use Coulomb's law to predict the force between the particles if the distance is doubled. The equation for Coulomb's law is Fe = g, and the constant, k, equals 9.00 x 10° Nm/C A. -1.13 x 10-6 N OB. 1.13x 106N O C. 2.83 x 10-7 N OD.-2.83x 10N sUBMIT
Answer:
I do not understand what you are asking
What do you mean is a variable velocity and uniform velocity
Answer:
Uniformly accelerated rectilinear motion (MRUA), also known as uniformly varied rectilinear motion (MRUV), is one in which a mobile moves along a straight path being subjected to a constant acceleration.
Explanation:
Before digital filmmaking, what tool was used to control the speed of movement on the screen after filming?
Answer:
The appropriate response is "Optical printer ".
Explanation:
A photographic printer used mostly for optical aberrations, comprised simply of either a camera that captures the frame to expand, minimize, deform, respectively. through magnifying lenses. A projector that always, as distinct from some kind of touch printer, transferred the image to something like the printing supply.Compare and contrast the CONFLICT (choose one) in the short story you read with the elements appearing in The Watsons Go to Birmingham—1963. Explain how they are similar or different in a few sentences.
Answer:
they were in two places in flint and Birmingham and in Birmingham it is hot but flint of cold the Simi is they both have Sunday school for Joetta
Explanation:
use in your own words teachers know when your not trust me.
how much min the basketball 1 player play
Answer:
A professional basketball game depends on the association presiding over the game. An NBA game lasts for 48 minutes whereas FIBA games take 40 minutes. The total time taken to play for any specialized game is over 2 hours 15 minutes. The time includes the time disruptions like fouls, timeouts, and breaks.
I hope it helps you...
Peter is running laps around a circular track with a diameter of 100 meters. If it takes Peter 12 minutes to run 4 laps, how quickly is he running (in meters per second)?
Answer:
v = 1.74 m/s
Explanation:
Given that,
Diameter of a circular track, d = 100 m
Distance covered for the 4 laps,
[tex]D=4\pi d\\\\D=4\pi \times 100\\\\D=1256.63\ m[/tex]
Time, t = 12 minutes = 720 s
We need to find the velocity of the peter. It can be calculated as follows :
[tex]v=\dfrac{D}{t}\\\\v=\dfrac{1256.63\ m}{720\ s}\\\\v=1.74\ m/s[/tex]
So, the speed is running with a velocity of 1.74 m/s.
Peter is running at 1.7453 m/sec.
Given to us,
Diameter of the circular track, D = 100 meters,
Number of laps Peter run, L = 4 laps,
Time taken by Peter, t = 12 minutes,
1 lap = circumference of the circle,
4 laps = 4 x circumference of the circle,
As we know, the circumference of a circle is given by πD.
So, 4 laps = 4 x circumference of the circle,
[tex]\begin{aligned}4 laps &= 4\times \pi \times D\\&= 4 \times \pi \times 100\\& = 1,256.6370\ meters\\\end{aligned}[/tex]
Also, we know that 1 minute has 60 sec.
so, 4 minutes = (4 x 60) seconds
Further, speed is given [tex]\bold{(\dfrac{Distance}{Time} )}[/tex]
Thus,
[tex]\begin{aligned}speed &= \dfrac{Distance\ coverd\ by\ Peter}{Time\ taken\ by\ Peter}\\&=\dfrac{1,256.6370}{12\times 60}\\&=1.7453\ m/sec \end{aligned}[/tex]
Hence, Peter is running at 1.7453 m/sec.
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HELP PLEASE!!!
If we have a sample of silicon (Si) atoms that has 14 protons, 14 electrons, and 18 neutrons
What is the name of this specific silicon isotope?
si-14
si-32
si-46
si-153
Answer:
It is si-32
Explanation:
Answer:
silicon-32
Explanation:
just took the quiz and got it right
A car with a mass of 1500kg is traveling at a speed of 30m/s. What force must be applied to stop the car in 3 seconds?
Answer:
The answer is 15,000 NExplanation:
To find the force given the mass , velocity and time can be found by using the formula
[tex]f = \frac{m \times v}{t} \\ [/tex]
where
m is the mass
v is the velocity
t is the time
From the question
m = 1500 kg
v = 30 m/s
t = 3 s
We have
[tex]f = \frac{1500 \times 30}{3} = \frac{45000}{3} \\ [/tex]
We have the final answer as
15,000 NHope this helps you
what is the volume of an object that has a density of 65g/cm3 and a mass of 130g.
Density ρ is mass m per unit volume v, or
ρ = m / v
Solving for v gives
v = m / ρ
So the given object has a volume of
v = (130 g) / (65 g/cm³) = 2 cm³
Question 4
Which of the following is unique for any given element?
O the mass of a neutron
o the number of neutrons
o the charge on the electons
O the number of protons
A light wave in space is described by the general function: u(r, t) = 1 (2π) 3/2 Z dk A(k) e i(k·r−ωt) . (1) (a) Find the particular solution knowing that at t = 0 the wave is given by
Answer:
hello your question is incomplete attached below is the complete question
Answer : attached below
Explanation:
A) finding the particular solution at t = 0
attached below is the detailed solution of the particular solution knowing that t = 0
A 30.0-kgkg box is being pulled across a carpeted floor by a horizontal force of 230 NN , against a friction force of 210 NN . What is the acceleration of the box?
Answer:
The acceleration of the box is 0.67 m/s²
Explanation:
Given that,
Mass of box = 30.0 kg
Horizontal force = 230 N
Friction force = 210 N
We need to calculate the acceleration of the box
Using balance equation
[tex]F-f_{k}=ma[/tex]
[tex]a=\dfrac{F-f_{k}}{m}[/tex]
Where, F = horizontal force
[tex]f_{k}[/tex] =frictional force
m= mass of box
a = acceleration
Put the value into the formula
[tex]a=\dfrac{230-210}{30}[/tex]
[tex]a=0.67\ m/s^2[/tex]
Hence, The acceleration of the box is 0.67 m/s²