Answer:
Dense, rocky objects orbiting near the central star.
Cluster of rock loosely held together by gravity
Explanation:
"Stars are huge, glowing balls of gases. "The closest star to Earth is the Sun. Most of the pinpricks of light that shine in the night sky are also stars. Countless more stars are too far from Earth to be seen without a telescope. Most stars are incredibly far away.
What is gravity ?"Gravity is the force by which a planet or other body draws objects toward its center. The force of gravity keeps all of the planets in orbit around the sun."
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On a low-friction track, a 0.66-kg cart initially going at 1.85 m/s to the right collides with a cart of unknown inertia initially going at 2.17 m/s to the left. After the collision, the 0.66-kg cart is going at 1.32 m/s to the left, and the cart of unknown inertia is going at 3.22 m/s to the right. The collision takes 0.010 s.
What is the unknown inertia?
What is the average acceleration of the heavier cart?
What is the average acceleration of the lighter cart?
Answer:
(a) the unknown inertia is 0.388 kg
(b) the average acceleration of the heavier cart is 317 m/s²
(c) the average acceleration of the lighter cart is 539 m/s²
Explanation:
Given;
mass of the first cart, m₁ = 0.66 kg
initial speed of the first cart, u₁ = 1.85 m/s
let the mass of the cart with unknown inertia be m₂
initial velocity of the second cart, u₂ = 2.17 m/s to the left
velocity of the first cart after collision, v₁ = 1.32 m/s to the left
velocity of the second cart after collision, v₂ = 3.22 m/s
time of collision, t = 0.010 s
(a) What is the unknown inertia?
Apply the principle of conservation of linear momentum, to determine the unknown inertia.
let leftward direction be negative direction
let rightward direction be positive direction
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.66(1.85) + m₂(-2.17) = 0.66(-1.32) + m₂(3.22)
1.221 - 2.17m₂ = -0.8712 + 3.22m₂
1.221 + 0.8712 = 3.22m₂ + 2.17m₂
2.0922 = 5.39m₂
m₂ = 2.0922 / 5.39
m₂ = 0.388 kg
The unknown inertia is 0.388 kg
(b) the average acceleration of the heavier cart
the heavier cart has a mass of 0.66 kg
[tex]a = \frac{v_1 - u_1}{t} \\\\a = \frac{-1.32 - 1.85}{0.01} \\\\a = -317 \ m/s^2\\\\|a| = 317 \ m/s^2[/tex]
(c) the average acceleration of the lighter cart;
the lighter cart has a mass of 0.388 kg
[tex]a = \frac{v_2 - u_2}{t} \\\\a = \frac{3.22 - (-2.17)}{0.01} \\\\a =\frac{3.22 \ +\ 2.17}{0.01} \\\\a= 539\ m/s^2[/tex]
Two point charges, A and B, are separated by a distance of 19.0 cm . The magnitude of the charge on A is twice that of the charge on B. If each charge exerts a force of magnitude 45.0 N on the other, find the magnitudes of the charges.
Answer:
QA = 19μC
QB = 9.5 μC
Explanation:
The force that each charge exerts on the other must obey Coulomb's Law, as follows:[tex]F_{AB} = \frac{k*Q_{A} * Q_{B}}{r_{AB}^{2}} (1)[/tex]
We know that the value of the magnitude of FAB is 45.0 N, the distance between QA and QB is 0.19 m, and that QA = 2*QB.Replacing in (1), we can solve for QB, as follows:[tex]Q_{B} = \sqrt{\frac{F_{AB}*r_{AB} ^{2}}{2*k} } = \sqrt{\frac{45.0N*(0.19m) ^{2}}{2*9e9N*m2/C2} } = 9.5e-6 C (2)[/tex]
Since QA = 2*QB⇒ QA = 2* 9.5μC = 19.0 μC⇒ QB = 9.5μC
In the study of personality, the
model includes different traits that are believed to underlie each individual's basic tendencies.
Answer:
Five-Factor
Explanation:
The Five-Factor model is a trait model of personality that includes different traits that are believed to underlie each individual's basic tendencies.
These different traits are:
1. Openness
2. Conscientiousness
3. Extraversion
4. Agreeable Ness
5. Neuroticism.
It is often shortened and referred to as OCEAN or CANOE.
It is a majorly acknowledged personality theory among scientists.
Hence, in the study of personality, the FIVE FACTOR model includes different traits that are believed to underlie each individual's basic tendencies.
Choose all the answers that apply.
Which of the following will cause an increase in the acceleration of an object?
increase force
decrease force
increase mass
decrease mass
Whats the answer!?
plzzzzzzzzzzzzz help SERIOUSLY
what type of waves can only travel through a medium?
Answer:
Mechanical waves
Explanation:
Mechanical waves are the waves that can travel only through a medium. Mechanical waves are disturbance of matter and require medium to transfer the energy. There are three types of mechanical waves that include transverse wave, longitudinal wave and surface wave.
Some of the examples of mechanical waves are sound waves and seismic waves etcetera.
Hence, the correct answer is "Mechanical waves".
How do you calculate area when pressure and force are given to you
Answer:
This is my answer
Explanation:
First convert 150 kPa to Pa:
150 × 1,000 = 150,000.
Next substitute the values into the equation:
force normal to a surface area = pressure × area of that surface.
force = 150,000 × 180.
force = 27,000,000 N.
1.First convert 150kPato Pa:
2.150 x 1,000 + 150,000
3.next substitute the values into the equations:
4.force normal to a surface area =pressure x area of that surface.
5.force=150,000 x 180.
6.force = 27,000,000N.
can i have brainliest please
This cat is A tired B lazy or C Dreaming
the answer is obviously c because cats cant have nightmares
Magnification of lens is 1. What does it mean?
Answer:
It means when you look into the lens your vision magnifies by x1
Explanation:
A stretched string has a mass per unit length of 5.00 g/cm and a tension of 10.0 N. A sinusoidal wave on this string has an amplitude of 0.12 mm and a frequency of 100 Hz and is traveling in the negative direction of an x axis. If the wave equation is of the form y(x, t) = ym sin(kx ± ωt), what are (a) ym, (b) k, (c) ω, and (d) the correct choice of sign in front of ω?
Answer:
0.12 mm ; 140.50 rad/m ; 628.32 rad/sec ; +
Explanation:
Given the wave equation of the form :
y(x, t) = ym sin(kx ± ωt)
Mas per unit length (u) = 5 g/cm = (5÷1000)kg / 0.01m) = 0.005kg/0.01m = 0.5kg/m
Tension, T = 10 N
Amplitude, A = 0.12 mm
Frequency, F = 100 Hz
Comparing with the general wave equation :
y = Asin(kx ± ωt)
A = amplitude = ym = 0.12 mm
2.) k = 2π / λ
Recall :
v = fλ
v = sqrt(T/u) = sqrt(10/0.5) = sqrt(20) = 4.472
λ = v/ f = 4.472 / 100 = 0.04472
Hence,
k = (2 * π) / 0.04472
k = 140.50 rad/m
3.) Angular frequency, ω
ω = 2πf = 2 * 3.14 * 100 = 628.32 rad/sec
4.) sign is +ve
Direction of wave propagation as given is in the negative x axis
12. A bag weighing 20 N CARRIED horizontally a distance of 35 m, How much
work is done on the bag in Joules? (Do not put units with your answer.) W=Fd *
Your answer
13. A child performs 10J of work in lifting a box 1 m in 2 seconds. How much
power did the child apply to the box? (Do not include units with your answer.)
P=W/t *
Your answer
Answer:
Explanation:
Well they told you the exact formula to use. Work is the force multiplied by the distance through which its applied.
W = (20N)(35m)
= 700 Joules
13.) Power is the amount of work done over the time through which the work is being done.
P = W/t
= 10J/2s
= 5J/s
One component of a metal sculpture consists of a solid cube with an edge of length 24.5 cm . The alloy used to make the cube has a density of 8050 kg/m3 . Find the cube's mass.
Answer:
118kg
Explanation:
answered
Given
density of the cube= 8050 kg/m3 .
length of the sizes of the cube=24.5 cm
We can convert the length to cm for unit consistency.
It's length =24.5 cm =0.245m
✓ the length of sizes of the cube is the same, then the volume can be calculated as
Volume= L^3
= (0.245m)^3
=0.01470625 m^3
✓ but we know that
Density = mass/ volume
Then,
Mass= (Volume × density)
= (0.01470625)(8050)
= 118 kg
Hence, the mass of the cube is 118 kg
Why does it rain more in West Ferris than in East Ferris? Explain your answer.
Answer:
This idea helps students explain why more rain forms over West Ferris than East Ferris. ... Therefore, when students explain that water vapor condenses higher in the atmosphere, they are actually explaining that water vapor condenses high in the troposphere, which is relatively low in the atmosphere.
Explanation:
Plz mark me brainliest thank u> have a good day
A sprinter starts from rest and accelerated at a rate of 0.16 m/s over a distance of 50.0 meters. How fast is the athletes traveling at the end of the 50.0 meters?
Answer:
40m/s
Explanation:
v²=u²+2as
v²=0²+2(16)(50)
v²=160v=40m/s
You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 92.2 kg hop on board for a ride through the woods and the springs (one for each wheel) each compress by 5.97 cm. When you pull the trailer over a tree root in the trail, it oscillates with a period of 1.14 s. Determine the force constant of the springs in N/m.
Answer:
k = 1400.4 N / m
Explanation:
When the springs are oscillating a simple harmonic motion is created where the angular velocity is
w² = k / m
w = [tex]\sqrt{ \frac{k}{m} }[/tex]
where angular velocity, frequency and period are related
w = 2π f = 2π / T
we substitute
2π / T = \sqrt{ \frac{k}{m} }
T² = 4π² [tex]\frac{m}{k}[/tex]
k = π² [tex]\frac{m}{T^{2} }[/tex]
in this case the period is T = 1.14s, the combined mass of the children is
m = 92.2 kg and the constant of the two springs is
k = 4π² 92.2 / 1.14²
k = 2800.8 N / m
to find the constant of each spring let's use the equilibrium condition
F₁ + F₂ - W = 0
k x + k x = W
indicate that the compression of the two springs is the same, so we could replace these subtraction by another with an equivalent cosecant
(k + k) x = W
2k x = W
k_eq = 2k
k = k_eq / 2
k = 2800.8 / 2
k = 1400.4 N / m
Pete applies a 10.9-Newton force to a 1.32-kg mug of root beer in order to accelerate it from rest over a distance of 1.25-m? How much work does Pete do on the mug of root beer?
Answer: 4 J
explanation:
A 5.7 kg particle starts from rest at x = 0 and moves under the influence of a single force Fx = 4.5 + 13.7 x − 1.5 x 2 , where Fx is in Newtons and x is in meters. Find the work done by this force on the particle as the particle moves from x = 0 m to x = 1.9 m. Answer in units of J.
Answer:
The work done by the force on the particle is 29.85 J.
Explanation:
The work is given by:
[tex] W = ^{x_{2}}_{x_{1}}\int F_{x} dx [/tex]
Where:
x₁: is the lower limit = 0 m
x₂: is the upper limit = 1.9 m
Fₓ: is the force in the horizontal direction = (4.5 + 13.7x - 1.5x²)N
[tex]W = ^{1.9}_{0}\int (4.5 + 13.7x - 1.5x^{2}) dx[/tex]
[tex] W = 4.5x|^{1.9}_{0} + \frac{13.7}{2}x^{2}|^{1.9}_{0} - \frac{1.5}{3}x^{3}|^{1.9}_{0} [/tex]
[tex] W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3} [/tex]
[tex] W = 4.5N(1.9 m) + \frac{13.7N}{2}(1.9 m)^{2} - \frac{1.5N}{3}(1.9 m)^{3} [/tex]
[tex]W = 29.85 J[/tex]
Therefore, the work done by the force on the particle is 29.85 J.
I hope it helps you!
A sinusoidal wave is traveling on a string with speed 19.3 cm/s. The displacement of the particles of the string at x = 6.0 cm is found to vary with time according to the equation y = (2.6 cm) sin[1.8 - (5.8 s-1)t]. The linear density of the string is 5.0 g/cm. What are (a) the frequency and (b) the wavelength of the wave? If the wave equation is of the form
Answer:
Explanation:
equation of wave is given by the following equation
y = (2.6 cm) sin[1.8 - (5.8 s-1)t].
Comparing it with standard form of wave
y = A sin ( ωt - kx )
we get
ω = 5.8
2πn = 5.8
n = .92 per second
kx = 1.8
k x 6 = 1.8
k = 0.3
[tex]\frac{2\pi}{\lambda}[/tex] = 0.3
λ = 20.9 cm
A child uses one hand to charge a balloon by rubbing it against her shirt. She then holds a rod in her other hand and finds that the balloon and rod, when brought close to one another, repel. Which one of the following is true?
a. the rod must be a conductor
b. the rod must be an insulator
c. it could be either
Answer:
B. The rod must be an insulator
Explanation:
We have that this rod must be an insulator. After this child rubbed the balloon, the balloon acquired static charge. So holding a rid against it is going to cause it to repel, this is to say it is repelling because the rod also is carrying some static charges. If this rid was a conductor, there would be no charge in its surface. The charge would have passed through her hand as it comes in contact.
It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energy and store it until it is needed during peak load times, such as the middle of the day. One suggestion has been to store the energy in large flywheels spinning on nearly frictionless ball-bearings. Consider a flywheel made of iron, with a density of 7800 kg/m^3 , in the shape of a uniform disk with a thickness of 11.3 cm.
Required:
a. What would the diameter of such a disk need to be if it is to store an amount of kinetic energy of 14.1 MJ when spinning at an angular velocity of 93.0 rpm about an axis perpendicular to the disk at its center?
b. What would be the centripetal acceleration of a point on its rim when spinning at this rate?
Answer:
Explanation:
kinetic energy = 14.1 MJ = 14.1 x 10⁶ J
Let radius of flywheel be r .
volume of flywheel = π r² x t where t is thickness
= 3.14 x r² x .113 m³
= .04 r² m³
mass = volume x density
= .04 r² x 7800 = 312.73 r²kg
moment of inertia I = 1 / 2 mass x radius²
= .5 x 312.73 r² x r²
= 156.37 r⁴ kg m²
angular velocity ω = 2π x 93/60
= 9.734 rad /s
kinetic energy = 1/2 Iω² where ω is angular velocity
= .5 x 156.37 r⁴ x 9.734²
= 7408.08 r⁴
Given
7408.08 r⁴ = 14.1 x 10⁶
r⁴ = .19 x 10⁴
r = .66 x 10
= 6.60 m .
Diameter = 13.2 m
b )
centripetal acceleration of a point on its rim = ω² r
= 9.734² x 6.6
= 625.35 m /s²
take a picture of an object in your house, describe the
energy stores and transfers that happen with it. You can be as imaginative as you wish
with the object (choose something unusual), but the stores you identify and transfers
that happen must be real.
pls give me ideas of what to take a photo of for this I'm really stuck :(
A ball of mass m makes a head-on elastic collision with a second ball (at rest) and rebounds in the opposite direction with a speed equal to one-fourth its original speed. what is the mass of the second ball?
When a ball of mass m makes a head-on elastic collision with a second ball (at rest) and rebounds in the opposite direction with a speed equal to one-fourth its original speed, then mass of the second ball having v/3 is velocity after collision is 9m/4.
What is momentum ?Momentum is defined as mass times velocity of body. it is denoted by p and its SI unit is Kg.m/s. It has both magnitude and direction. it is a vector quantity. it tells about the moment of the body. it is denoted by p and expressed in kg.m/s. mathematically it is written as p = mv. A body having zero velocity or zero mass has zero momentum. its dimensions is [M¹ L¹ T⁻¹]. Momentum is conserved throughout the motion.
initial momentum = final momentum
Given,
mass of first body m₁ = m
initial velocity of first body = v₁' = v
final velocity of first body = v₁'' =v/4
mass of second body m₂ = ?
initial velocity of second body = v₂' = 0
final velocity of second body = v₂'' = v/3
According to conservation of momentum,
initial momentum = final momentum
m₁v₁' + m₂v₂' = m₁v₁'' + m₂v₂''
putting al above values
m₁v + 0 = m₁v/4 + m₂v/3
m₁v - m₁v/4 = m₂v/3
m (1 - 1/4)v = m₂v/3
3m/4 = m₂/3
m₂ = 9m/4
Hence mass of the second body is 9m/4.
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On the periodic table , the vertical columns that extend down the periodic table are called ?
Answer:
groups
Explanation:
Answer: Groups
Explanation: They are in the same group! Like the Alkaline Metals are all the group. They all lose an electron. :)
a particle with a charge of 5.5 x 10^-8 c is 3.5 cm from a particle with a charge of -2.3 x10^-8 c. the potential energy of this two particle system relative to the potential energy at infinite separation is:
Answer:
-32.5 * 10^-5 J
Explanation:
The potential energy of this system of charges is;
Ue = kq1q2/r
Where;
k is the Coulumb's constant
q1 and q2 are the magnitudes of the charges
r is the distance of separation between the charges
Substituting values;
Ue = 9.0×10^9 N⋅m2/C2 * 5.5 x 10^-8 C *( -2.3 x10^-8) C/(3.5 * 10^-2)
Ue= -32.5 * 10^-5 J
The potential energy of this two particle system relative to the potential energy at infinite separation is [tex]\bold {-32.5x 10^-^5\ J}[/tex].
The potential energy of this system of charges,
[tex]\bold {Ue = k\dfrac{q1q2}{r}}[/tex]
Where;
k - Coulumb's constant
q1 and q2 - magnitudes of the charges
r - distance between the charges
Put the values in the equation,
[tex]\bold {Ue = 9.0x10^9\times \dfrac {5.5 x 10^{-8} C \times -2.3 x10^{-8} C}{3.5 \times 10^{-2}}}\\\\\bold {Ue= -32.5 x 10^-^5\ J}[/tex]
Therefore, the the potential energy of this two particle system relative to the potential energy at infinite separation is [tex]\bold {-32.5x 10^-^5\ J}[/tex].
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If a person visits an exercise facility, buys a new piece of fitness/sporting equipment,
or just starts planning to be active, which of the five stages of change for physical
activity are they at?
Planning
Maintenance
Precontemplation
Contemplation
Answer:planning
Explanation:
The person is in the stage of planning due to its action of planning to be active.
What is planning stage?The person is in the planning stage among the five stages of change for physical activity because the person just started planning to be active not yet started the activity. If a person is in the state of looking thoughtfully at something for a very long time then it is said to be Contemplation.
While on the other hand, if a person is in a stage in which there is no intention to change behavior in the foreseeable future then it is called precontemplation so we can conclude that the person is in the stage of planning due to its action of planning to be active.
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Billiard ball A (0.35 kg) is struck such that it moves at 10 m/s toward a
second identical ball (Ball B). After the collision Ball A continues to move
in the same direction at 2 m/s. What is the magnitude of the velocity for
Ball B after the collision?
Before Collision:
10 m/s
A
After Collision:
2 m/s
O
Answer:
6m/s
Explanation:
Using the law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the momentum after collision.
Using the expression
m1u1 + m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the final velocity after collision
Substitute the given values in the formula
0.35(10)+0.35(2) = (0.35+0.35)v
3.5+0.7 = 0.7v
4.2 = 0.7v
v = 4.2/0.7
v = 6m/s
Hence the magnitude of the velocity for Ball B after the collision is 6m/s
A mass m is gently placed on the end of a freely hanging spring. The mass then falls 33 cm before it stops and begins to rise. What is the frequency of the oscillation
Answer:
Explanation:
The mass falls by .33 m before it begins to rise . At that point loss of potential energy is equal to gain of elastic energy .
1/2 k x² = mgx
.5 x k x .33² = m x 9.8 x .33
k / m = 59.4
frequency of oscillation = [tex]\frac{1}{2\pi} \times\sqrt{\frac{k}{m} }[/tex]
= [tex]\frac{1}{2\pi} \times\sqrt{59.4}[/tex]
= 1.22 per second .
A 75 kg window cleaner uses a 10 kg ladder that is 5.0 m long. He places one end on the ground 2.5 m from a wall, rests the upper end against a cracked Window, and climbs the ladder. He is 3.0 m up along the ladder when the window breaks. Neglect friction between the ladder and window and assume that the base of the ladder does not slip. When the window is on the verge of breaking, what are (a) the magnitude of the force on the window from the ladder, (b) the magnitude of the force on the ladder from the ground, and (c) the angle (relative to the horizontal) of that force on the ladder
Solution :
Given : Mass of ladder = 10 kg
Length of ladder = 5 m
Weight of window cleaner = 75 kg
a). Now equate the torque about the lowermost point of the ladder is given by :
[tex]$=10 \times 9.8 \times \frac{2.5}{2} + 75 \times 9.8 \times \frac{3}{5} \times 2.5 = N \times \sqrt{5^2 - 2.5^2}$[/tex]
Here, N = normal force that the glass exerts on the ladder
Therefore, [tex]$N = 282.9 \ N$[/tex]
= 280 N (in 2 significant figures)
b). Equate the forces along horizontal direction,
The horizontal component of the friction, [tex]$F_x = N = 282.9 \ N $[/tex]
The vertical component of the friction, [tex]$F_y = (10+75) \times 9.8$[/tex]
= 833 N
Therefore, the net frictional force, [tex]$F = \sqrt{F_x^2+F_y^2}$[/tex]
[tex]$F = \sqrt{(282.9)^2+(833)^2}$[/tex]
= 879.7 N
= 880 N (in 2 significant figures)
c). The angle the forces makes [tex]$= \tan \frac{833}{282.9} $[/tex]
[tex]$= 71.2 ^\circ $[/tex]
Therefore in 2 significant figures = [tex]$71 ^\circ$[/tex]
An unmarked police car traveling a constant 38.6 m/s is passed by a speeder traveling 53.4 m/s. Precisely 2.2 seconds after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is a constant 1.6 m/s2, how much time passes before the police car overtakes the speeder after the speeder passes (assumed moving at constant speed).
Answer:
The unmarked police car needs approximately 71.082 seconds to overtake the speeder.
Explanation:
Let suppose that speeder travels at constant velocity, whereas the unmarked police car accelerates at constant rate. In this case, we need to determine the instant when the police car overtakes the speeder. First, we construct a system of equations:
Unmarked police car
[tex]s = s_{o}+v_{o,P}\cdot (t-t') + \frac{1}{2}\cdot a\cdot (t-t')^{2}[/tex] (1)
Speeder
[tex]s = s_{o} + v_{o,S}\cdot t[/tex] (2)
Where:
[tex]s_{o}[/tex] - Initial position, measured in meters.
[tex]s[/tex] - Final position, measured in meters.
[tex]v_{o,P}[/tex], [tex]v_{o,S}[/tex] - Initial velocities of the unmarked police car and the speeder, measured in meters per second.
[tex]a[/tex] - Acceleration of the unmarked police car, measured in meters per square second.
[tex]t[/tex] - Time, measured in seconds.
[tex]t'[/tex] - Initial instant for the unmarked police car, measured in seconds.
By equalizing (1) and (2), we expand and simplify the resulting expression:
[tex]v_{o,P}\cdot (t-t')+\frac{1}{2}\cdot a\cdot (t-t')^{2} = v_{o,S}\cdot t[/tex]
[tex]v_{o,P}\cdot t -v_{o,P}\cdot t' +\frac{1}{2}\cdot a\cdot t^{2}-a\cdot t'\cdot t+\frac{1}{2}\cdot a\cdot t'^{2} = v_{o,S}\cdot t[/tex]
[tex]\frac{1}{2}\cdot a\cdot t^{2}+[(v_{o,P}-v_{o,S})-a\cdot t']\cdot t -\left(v_{o,P}\cdot t'-\frac{1}{2}\cdot a\cdot t'^{2}\right) = 0[/tex]
If we know that [tex]a = 1.6\,\frac{m}{s^{2}}[/tex], [tex]v_{o,P} = 0\,\frac{m}{s}[/tex], [tex]v_{o,S} = 53.4\,\frac{m}{s}[/tex] and [tex]t' = 2.2\,s[/tex], then we solve the resulting second order polynomial:
[tex]0.8\cdot t^{2}-56.92\cdot t +3.872 = 0[/tex] (3)
[tex]t_{1} \approx 71.082\,s[/tex], [tex]t_{2}\approx 0.068[/tex]
Please notice that second root is due to error margin for approximations in coefficients. The required solution is the first root.
The unmarked police car needs approximately 71.082 seconds to overtake the speeder.
An empty elevator has a mass of 722 kg. It moves between floors at a maximum speed of 6.00 m/s. The elevator is stopped on the 20th floor of the building when someone pushes the call button in the lobby (the first floor). Assuming that it takes 15.25 meters for the elevator to reach its maximum speed, and assuming constant acceleration, calculate the tension in the cable as the elevator car begins to descend (during the period of constant acceleration). Take the acceleration due to gravity to be 9.81 m/s2. Show your work.
Answer:
Explanation:
The lift is going down with acceleration
Initial speed u = 0
Final speed v = 6 m/s
distance s = 15.25 m
acceleration a = ?
v² = u² + 2 a s
6² = 0 + 2 x a x 15.25
a = 1.18 m /s²
Elevator is going down with acceleration .
mg - T = ma where T is tension in the cable .
722 x 9.8 - T = 722 x 1.18
7075.6 - T = 851.96
T = 6223.64 N .