Answer:
c < a<=b
Explanation:
Tensile stress = (force) /Area
for A:
Tensile stress = 200/1 =200N/mm²
For B:
Tensile stress = 200/1 =200N/mm2
For C:
Tensile stress = 100/2 =50N/mm²
Ranking from smallest to largest we have;
C<A<=B which is option 4
HELP!!! how does gravity affect how objects fall to the ground
Answer:
c
Explanation:
Answer:
When objects fall to the ground, gravity causes them to accelerate. Acceleration is a change in velocity, and velocity, in turn, is a measure of the speed and direction of motion. Gravity causes an object to fall toward the ground at a faster and faster velocity the longer the object falls.
Explanation:
If matter cannot be created or destroyed, then how do you end up with
rust? Below is the equation for rust.
4Fe + 302 → 2Fe203
oxygen from the air
water in the atmosphere
oxygen from in the metal
there shouldn't be any oxygen
Your question is a "non sequitur", which means "it doesn't follow".
Your "then" doesn't contradict your "If", so no mystery is implied.
Maybe you're trying to say that matter is somehow not conserved in the equation . . . 4Fe + 302 → 2Fe203 . But it is. There are 4 Irons and 6 Oxygens on each side, so conservation is not violated here.
I looked up "rust" on Floogle, and got slapped with pages and pages of chemistry that I don't completely understand. But what it's saying is that rusting is a very complex chemical process, AND it doesn't happen unless there's some water involved.
So the bottom line is that there's a lot more going on than simply
4Fe + 302 → 2Fe203 ,
there's water going in and out of the process at every stage, and when it's all over, you have rusty iron, and mass has been conserved.
if an input of 100 j in pulley system increases potential energy of load 60 J, what efficency of the system?
Answer:
Efficiency of the system = 100%
Explanation:
Given:
Input energy = 100 J
Potential energy load = 60 J
Find:
Efficiency of the system
Computation:
Efficiency of the system = [Potential energy load/Input energy]100
Efficiency of the system = [60/100]100
Efficiency of the system = 100%
Interactive Solution 8.29 offers a model for this problem. The drive propeller of a ship starts from rest and accelerates at 2.38 x 10-3 rad/s2 for 2.04 x 103 s. For the next 1.48 x 103 s the propeller rotates at a constant angular speed. Then it decelerates at 2.63 x 10-3 rad/s2 until it slows (without reversing direction) to an angular speed of 2.42 rad/s. Find the total angular displacement of the propeller.
Answer:
Δθ = 15747.37 rad.
Explanation:
The total angular displacement is the sum of three partial displacements: one while accelerating from rest to a certain angular speed, a second one rotating at this same angular speed, and a third one while decelerating to a final angular speed.Applying the definition of angular acceleration, we can find the final angular speed for this first part as follows:[tex]\omega_{f1} = \alpha * \Delta t = 2.38*e-3rad/s2*2.04e3s = 4.9 rad/sec (1)[/tex]
Since the angular acceleration is constant, and the propeller starts from rest, we can use the following kinematic equation in order to find the first angular displacement θ₁:[tex]\omega_{f1}^{2} = 2* \alpha *\Delta\theta (2)[/tex]
Solving for Δθ in (2):[tex]\theta_{1} = \frac{\omega_{f1}^{2}}{2*\alpha } = \frac{(4.9rad/sec)^{2}}{2*2.38*e-3rad/sec2} = 5044.12 rad (3)[/tex]
The second displacement θ₂, (since along it the propeller rotates at a constant angular speed equal to (1), can be found just applying the definition of average angular velocity, as follows:[tex]\theta_{2} =\omega_{f1} * \Delta_{t2} = 4.9 rad/s * 1.48*e3 s = 7252 rad (4)[/tex]
Finally we can find the third displacement θ₃, applying the same kinematic equation as in (2), taking into account that the angular initial speed is not zero anymore:[tex]\omega_{f2}^{2} - \omega_{o2}^{2} = 2* \alpha *\Delta\theta (5)[/tex]
Replacing by the givens (α, ωf₂) and ω₀₂ from (1) we can solve for Δθ as follows:[tex]\theta_{3} = \frac{(\omega_{f2})^{2}- (\omega_{f1}) ^{2} }{2*\alpha } = \frac{(2.42rad/s^{2}) -(4.9rad/sec)^{2}}{2*(-2.63*e-3rad/sec2)} = 3451.25 rad (6)[/tex]
The total angular displacement is just the sum of (3), (4) and (6):Δθ = θ₁ + θ₂ + θ₃ = 5044.12 rad + 7252 rad + 3451.25 rad ⇒ Δθ = 15747.37 rad.A projectile is fired with an initial velocity of 120.0 m/s at an angle, θ, above the horizontal. If the projectile’s initial horizontal speed is 55 meters per second, then angle θ measures approximately
Answer:
algm sabe tô precisando muito
An 5 kg object moving at 10 m/s will have a momentum equaling ____________.
15 kg m/s/s
15 kg m/s
Answer:
50Kgm/s
Explanation:
Momentum=Mass*Velocity
P=mv
Given Mass=5Kg. Given Velocity=10m/s
Momentum=5*10=50Kgm/s
__________ and __________ both heavily relied on dream analysis in their treatment of patients. A. Alfred Adler . . . Albert Ellis B. Alfred Adler . . . Carl Jung C. Sigmund Freud . . . Carl Jung D. Sigmund Freud . . . Alfred Adler
Answer:
C. Sigmund Freud . . . Carl Jung
Explanation:
edge 2021
C.Sigmund Freud and Carl Jung both heavily relied on dream analysis in their treatment of patients.
What is Freud most famous for?Freud is well-known for inventing and developing the approach of psychoanalysis; for articulating the psychoanalytic idea of motivation, intellectual infection, and the structure of the unconscious; and for influencing medical and popular conceptions of human nature by using positing both everyday and strange thought.
Sigmund Freud was an Austrian neurologist who's perhaps maximum known as the founding father of psychoanalysis. Freud advanced a fixed of therapeutic strategies centered on communication therapy that worried the use of techniques that include transference, loose affiliation, and dream interpretation.
Learn more about Sigmund Freud here: https://brainly.com/question/2706543
#SPJ2
Discuss how the pressure cooker is designed to achieve temperatures above 100°C.
With rising heat, the steam pressure inside the pot builds up beyond atmospheric pressure, allowing the temperatures to rise well above boiling point. This design enables to save time, energy, and resources. The temperature inside a pressure cooker can well go beyond 110° C, which reduces the time needed to cook food.
2. A uniform wire of resistance R is stretched until its length doubles. Assuming its density and resistivity remain constant, what is its new resistance
Answer:
Resistance is quadrupled.
Explanation:
Solving this requires us to use the formula of resistivity.
Resistivity is usually said to be the measure of the resistance of a particular size of any given material to the electrical conduction. It is mathematically represented as
ρ = RA/L, where
ρ = the resistivity of the given material
R = the resistance of the material
A = the area of the material
L = length of the material.
From the question, we're told that the length is doubled with the resistivity and density remaining constant. If the density is constant, this makes the volume constant as well.
Volume, V = A * L. We're then told that the length is doubled. If the length is doubled, for the volume to remain constant, then the area must be halved.
Volume, V = A/2 * 2L
Making, Resistance R, subject of the formula, we have
R = ρL/A.
Since resistivity is constant and the area is halved, we then have
R = 2L / (1/2A)
R = 4L / A
If the length is doubled, we have the resistance to be quadrupled
A scientist adds different amounts of salt to 5 bottles of water. She then measures how long it takes for the water to boil. What is the responding variable in this experiment
Answer:
the responding variable is the water boiling
Explanation:
a responding variable is the same thing as a dependent variable and an independent variable you change the independent variable is the amount of salt, the control group is how long water takes to boil without adding salt, and a constant is the same amount of water
Rhodium is in period 5 of the periodic table. What does this tell you about this element
Answer:
. It is an extraordinarily rare, silvery-white, hard, corrosion-resistant, and chemically inert transition metal. It is a noble metal and a member of the platinum group.
Explanation:
A girl weighing 45kg is standing on the floor, exerting a downward force of 200N on the floor. The force exerted on her by the floor is ..............
Select one:
a.
No force exerted
b.
Less than 2000N
c.
Equal to 200 N
d.
Greater than 200 N
Answer:
c.
Equal to 200 N..........
Which time interval has the greatest speed?
Answer:
es la 2
Explanation:
epero que te curva
a pendulum clock having Copper keeps time at 20 degree Celsius it gains 15 second per day if cooled to 0°C celsius calculate the coefficient of linear expansion of copper.
?.............................
A potter's wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/srev/s . The wheel can be considered a uniform disk of mass 4.5 kgkg and diameter 0.30 mm . The potter then throws a 2.8-kgkg chunk of clay, approximately shaped as a flat disk of radius 8.0 cmcm , onto the center of the rotating wheel. Part A What is the frequency of the wheel after the clay sticks to it
Answer:1.7 rev/s
Explanation:
Given
Frequency of wheel [tex]N_1=2\ rev/s[/tex]
angular speed [tex]\omega_1=2\pi N_1=4\pi\ rad/s[/tex]
mass of wheel [tex]m_1=4.5\ kg[/tex]
diameter of wheel [tex]d_1=0.30\ m=30\ cm[/tex]
radius of wheel [tex]r_1=\frac{d_1}{2}=\frac{30}{2}=15\ cm[/tex]
mass of clay [tex]m_2=2.8\ kg[/tex]
the radius of the chunk of clay [tex]r_2=8\ cm[/tex]
Moment of inertia of Wheel
[tex]I_1=\dfrac{m_1r_1^2}{2}=\dfrac{4.5\times 15^2}{2}\ kg-cm^2[/tex]
Combined moment of inertia of wheel and clay chunk
[tex]I_2=\dfrac{m_1r_1^2}{2}+\dfrac{m_2r_2^2}{2}=\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2}\ kg-cm^2[/tex]
Conserving angular momentum
[tex]\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow \dfrac{4.5\times 15^2}{2}\cdot 4\pi=(\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2})\omega_2\\\\\Rightarrow \omega _2=\dfrac{4\pi }{1+\dfrac{2.8}{4.5}\times (\dfrac{8}{15})^2}=\dfrac{4\pi}{1+0.1769}=0.849\times 4\pi[/tex]
Common frequency of wheel and chunk of clay is
[tex]\Rightarrow N_2=\dfrac{4\pi \times 0.849}{2\pi}=1.698\approx 1.7\ rev/s[/tex]
A hook in boxing primarily involves horizontal flexion of the shoulder while maintaining a constant angle at the elbow. During this punch, the horizontal flexor muscles of the shoulder contract and shorten at an average speed of 75 cm/s. They move through an arc length of 5 cm during the hook, while the fist moves through an arc length of 100 cm. What is the average speed of the fist during the hook
Answer:
average speed of the fist during the hook = 15 m/s or 1500 cm/s
Explanation:
We are given;
Speed of shoulder contraction, v_s = 75 cm/s = 0.75 m/s
Distance moved through the arc length by shoulder, d_s = 5 cm = 0.05 m
Distance moved by the fist, d_f = 100 cm = 1 m
Now, we are to find the average speed of the fist during the hook; v_f
Thus can be gotten from proportion;.
d_f/d_s = v_f/ v_s
Making V_f the subject, we have;
v_f = (d_f × v_s)/d_s
Thus;
v_f = (1 × 0.75)/0.05
v_f = 0.75/0.05
v_f = 15 m/s
Carl works hard to get a grades on his report card because his mother pays him 25 dollars for each semester he earns straight as Carl’s behavior is being influenced by
. Assume that the batter does hit the ball. If the bat's instantaneous angular velocity is 30 rad/s at the instant of contact, and the distance from the sweet spot on the bat to the axis of rotation is 1.25 m, what is the instantaneous linear velocity of the sweet spot at the instant of ball contact
Answer:
37.5 m/s
Explanation:
Using,
Formula
v = ωr....................... Equation 1
Where ω = instantaneous angular velocity, v = instantaneous linear velocity, r = radius or distance from the sweet spot of the bat to the axis of rotation.
From the question,
Given: ω = 30 rad/s, r = 1.25 m
Substitute these values into equation 1
v = 30(1.25)
v = 37.5 m/s.
Hence the instantaneous linear velocity of the sweet spot at the instant of ball contact is 37.5 m/s
(will give brainliest to whoever is first and gives reason) A mass is spun in a circle with a frequency of 40Hz. What is the period of its rotation?
Answer:
[tex]\huge\boxed{T = 0.025\ seconds}[/tex]
Explanation:
Given:
Frequency = f = 40 Hz
Required:
Time period = T = ?
Formula:
[tex]\sf T = 1 / f[/tex]
Solution:
T = 1 / 40
T = 0.025 seconds
[tex]\rule[225]{225}{2}[/tex]
Hope this helped!
~AH1807Highest density of electrostatic charges in a metal is found where
I don't know the answer but I just want points sorry
A cylindrical body has 6 m height and its radius is 2 metre calculate its volume. Ans :75.428m3
Answer:
75.4
Explanation:
r= 2
h= 6
v= 22/7 *r*r*h
v= 75.42
1. Alexandra and Rachel are on a train that sounds a whistle at a constant frequency as
it leaves the train station. Compared to the sound emitted by the whistle, the sound that
the passengers standing on the platform hear has a frequency that is
a. lower, because the sound-wave fronts reach the platform at a frequency
lower than the frequency at which they are produced
b. lower, because the sound waves travel more slowly in the still air above the
platform than in the rushing air near the train
c. higher, because the sound-wave fronts reach the platform at a frequency
higher than the frequency at which they are produced
d. higher, because the sound waves travel faster in the still air above the
platform than in the rushing air near the train
Answer: the answer would be C trust me i took the test if its not that its b
hope that helps
Explanation: i took the test
answer:
a) lower because the sound-wave fronts reach the platform at a frequency lower than the frequency at which they are produced
explanation :3
If the train is leaving the train station, then the people who are standing on the platform would hear a sound with a lower frequency since the train is moving further away. ^^
A class is learning about states of matter. The students set up the investigation in the diagram.
Which kinds of energy are needed in this investigation to change the state of matter of the owl made of wax?
A 416 kg merry-go-round in the shape of a horizontal disk with a radius of 1.7 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. How large a torque would have to be exerted to bring the merry-go-round from rest to an angular speed of 3.7 rad/s in 2.9 s
Answer:
The torque exerted on the merry-go-round is 766.95 Nm
Explanation:
Given;
mass of the merry-go-round, m = 416 kg
radius of the disk, r = 1.7 m
angular speed of the merry-go-round, ω = 3.7 rad/s
time of motion, t = 2.9 s
The torque exerted on the merry-go-round is calculated as;
[tex]\tau = Fr= I\alpha\\\\\tau = (\frac{1}{2} m r^2)(\frac{\omega }{t} )\\\\\tau = (\frac{1}{2} \times 416 \times 1.7^2)( \frac{3.7}{2.9} )\\\\\tau = 766.95 \ Nm[/tex]
Therefore, the torque exerted on the merry-go-round is 766.95 Nm
Magnetism is a type of
A. force
B. energy
C. gravity
D. electricity
Answer:
A. force
Explanation:
Magnetism is a type of force.
Answer:
A. force
Hope this helps
A force of 12 N changes the momentum of a toy car from 3kgm/s t0 10kgm/s. Calculate the time the force took to produce this change in momentum.
Answer:
Time = 0.58 seconds
Explanation:
Given the following data;
Initial momentum = 3 kgm/s
Final momentum = 10 kgm/s
Force = 12 N
To find the time required for the change in momentum;
First of all, we would determine the change in momentum.
[tex] Change \; in \; momentum = final \; momentum - initial \; momentum [/tex]
[tex] Change \; in \; momentum = 10 - 3 [/tex]
Change in momentum = 7 kgm/s
Now, we can find the time required;
Note: the impulse of an object is equal to the change in momentum experienced by the object.
Mathematically, impulse (change in momentum) is given by the formula;
[tex] Impulse = force * time [/tex]
Making "time" the subject of formula, we have;
[tex] Time = \frac {impulse}{force} [/tex]
Substituting into the formula, we have;
[tex] Time = \frac {7}{12} [/tex]
Time = 0.58 seconds
During which phase is the moon not visible?
A) Full Moon
B) First quarter
C) New moon
D) Waxing crescent
Answer:
they are right it is a new moon
Explanation:
took the test
As an admirer of Thomas Young, you perform a double-slit experiment in his honor. You set your slits 1.03 mm apart and position your screen 3.19 m from the slits. Although Young had to struggle to achieve a monochromatic light beam of sufficient intensity, you simply turn on a laser with a wavelength of 641 nm . How far on the screen are the first bright fringe and the second dark fringe from the central bright fringe
Answer:
Δx = distance between central bright fringe and first bright fringe = 19.85 x 10⁻⁴ m = 1.985 mm
distance between central bright fringe and second dark fringe = 2.978 mm
Explanation:
We have the following data:
λ = wavelength of light = 641 nm = 6.41 x 10⁷ m
L = Distance of Screen from slits = 3.19 m
d = slit separation = 1.03 mm = 1.03 x 10⁻³ m
Δx = distance between consecutive bright fringes = fringe spacing = ?
Using formula:
[tex]\Delta x = \frac{\lambda L}{d}\\\\\Delta x = \frac{(6.41\ x\ 10^{-7}\ m)(3.19\ m)}{1.03\ x\ 10^{-3}\ m}[/tex]
Δx = distance between central bright fringe and first bright fringe = 19.85 x 10⁻⁴ m = 1.985 mm
distance between central bright fringe and second dark fringe = 1.5Δx
distance between central bright fringe and second dark fringe = (1.5)(1.985 mm)
distance between central bright fringe and second dark fringe = 2.978 mm
what dose nuclear reactions produce?
Answer:
a new chromebook for you and you will get to know the other one that
A falling 0.60 kg object experiences a frictional force due to air resistance of 1.5 N. What is the object's acceleration?
Answer:
7.5 m/s².
Explanation:
From the question given above, the following data were:
Mass (m) of object = 0.6 Kg
Force of friction (Fբ) = 1.5 N
Acceleration (a) =?
Next, we shall determine the force of gravity on the object. This can be obtained as follow:
Mass (m) of object = 0.6 Kg
Acceleration due to gravity (g) = 10 m/s²
Force of gravity (F₉) =?
F₉ = mg
F₉ = 0.6 × 10
F₉ = 6 N
Next, we shall determine the net force acting on the object. This can be obtained as follow:
Force of friction (Fբ) = 1.5 N
Force of gravity (F₉) = 6 N
Net force (Fₙ) =?
Fₙ = F₉ – Fբ
Fₙ = 6 – 1.5
Fₙ = 4.5 N
Finally, we shall determine the acceleration of the object. This can be obtained as follow:
Mass (m) of object = 0.6 Kg
Net force (Fₙ) = 4.5 N
Acceleration (a) of object =?
Fₙ = ma
4.5 = 0.6 × a
Divide both side by 0.6
a = 4.5 / 0.6
a = 7.5 m/s²
Therefore, the acceleration of the object is 7.5 m/s²
