Answers
Answer:
The size will increase.
Explanation:
When you bend a plastic ruler, it's size will increase because it is elastic and will exhibit elastic deformation. When it is been bent, it will continue to stretch until it get to a point where it will not be able to regain it formal shape, it size wound of increase. Therefore when the ruler get to elastic limit and you have bend it to the point it cannot regain it's formal shape back, it will remain bent and if further force is apply on it,it will break.
Related Questions
As a part of a project to construct a Rube Goldberg machine, a student wishes to construct a spring-loaded cube launcher. For examples of Rube Goldberg machines, see the beautiful Cog video or the Mythbusters Happy Holidays Rube Machine. Our student plans to trigger the release of the spring with a falling ball. The small cube to be launched has a mass of 119.0 grams. The available spring is a very stiff but light spring that has a spring constant of 461.0 N/m (the spring is so light that it may be considered massless). The spring will be mounted at the bottom of a wooden ramp which makes an angle of 50.0 degrees with the horizontal; the length of the spring is parallel to the ramp with the lower end of the spring fixed in place. For simplicity, assume that the part of the wooden ramp which is underneath the spring is highly polished and very slick; you may assume no friction on the cube by the ramp when the cube is moving on this portion of the ramp. For the rest of the wooden ramp, the coefficients of friction between the ramp surface and the cube surface are 0.590 for static friction and 0.470 for kinetic friction. Measured from the equilibrium position of the free end of the mounted spring, the distance to the top of the ramp is 17.0 cm (this is measured along the ramp) The plan is to compress the spring and maintain the compression with a simple trigger, which will be released by a falling ball. The cube will be at rest on the free end of the compressed spring. When the trigger is released, the spring will expand, shooting the cube up the ramp. The desired speed for the cube when it reaches the top of the ramp (where the cube will leave the ramp as a projectile to continue the Rube Goldberg sequence) is 45.0 cm/s. Your ultimate task will be to calculate the necessary amount of compression for the spring. The Spring is Compressed by 5.90 cm Before attempting the ultimate task, answer the following questions (Parts A-B) for the case in which the spring is compressed by 5.90 cm and then released, shooting the cube up the ramp and into the air.
A) Find the work done on the cube by the spring during the time that the cube is in contact with the spring.
B) What is the speed of the cube at the the instant just before the sliding cube leaves the ramp?
Answers
Answer:
1. The work done on the cube during the time the cube is in contact with the spring is 0.8023705 J
2. The speed of the cube at the instant just before the sliding cube leaves the ramp is approximately 31.5 cm/s
Explanation:
The given parameters of the Rube Goldberg machine are;
The distance from the free end of the spring to the top of the ramp, d = 17.0 cm = 0.17 m
The mass of the small cube to be launched, m = 119.0 g = 0.119 kg
The spring constant of the spring, k = 461.0 N/m
The angle of elevation of the ramp to the horizontal, θ = 50.0°
The coefficient of static friction of the wood, \(\mu_s\) = 0.590
The coefficient of dynamic friction of the wood, \(\mu_k\) = 0.470
The velocity of the cube at the top of the ramp, v = 45.0 cm/s = 0.45 m/s
The amount by which the cube is compressed, x = 5.90 cm = 0.059 m
The work done on the cube during the time the cube is in contact with the spring = The energy of the spring, E = (1/2)·k·x²
∴ E = (1/2) × 461.0 N/m × (0.059 m)² = 0.8023705 J
The work done on the cube during the time the cube is in contact with the spring= E = 0.8023705 J
2. The frictional force, \(F_f\) = \(\mu_k\)·m·g·cos(θ)
∴ \(F_f\) = 0.470 × 0.119 × 9.8 × cos(50) ≈ 0.35232 N
The work loss to friction, W = \(F_f\) × d
∴ W = 0.35232 N × 0.17 m ≈ 0.05989 J
The work lost to friction, W ≈ 0.05989 J
The potential energy of the cube at the top of the ramp, P.E. = m·g·h
∴ P.E. = 0.119 kg × 9.8 m/s² × 0.17 m × sin(50°) ≈ 0.151871375 J
By conservation of energy principle, the Kinetic Energy of the cube at the top of the ramp, K.E. = E - W - P.E.
∴ K.E. = 0.8023705 J - 0.05989 J - 0.151871375 J ≈ 0.590609125 J
K.E. = (1/2)·m·v²
Where;
v = The speed of the cube at the instant just before the sliding cube leaves the ramp
∴ K.E. = (1/2) × 0.119 kg × v² ≈ 0.590609125 J
v² ≈ 0.590609125 J/((1/2) × 0.119 kg) ≈ 9.92620378 m²/s²
v = √(9.92620378 m²/s²) ≈ 3.15058785 m/s ≈ 31.5 cm/s
The speed of the cube at the instant just before the sliding cube leaves the ramp, v ≈ 31.5 cm/s.
A rock, a book and a can of soda all have the same mass. Which of the following best describes their relationship?
A. They are all the same size
B. They each contain the same amount of matter.
C. They are each made of the same element.
D. They each occupy the same amount of space.
Answers
Answer:
Explanation:
d
A rock, a book and a can of soda all have the same mass. They each contain the same amount of matter.
Hence, the correct option is D.
Since the rock, book, and can of soda all have the same mass, it means that they contain the same amount of matter, regardless of their size or the material they are made of.
Mass is a measure of the amount of matter in an object, so if their masses are equal, it implies that the quantity of matter is the same in each of them.
The size, shape, and material composition can be different for each object, but their masses remain the same in this scenario.
Hence, A rock, a book and a can of soda all have the same mass. They each contain the same amount of matter.
Hence, the correct option is D.
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(2) Put 5kg mass at left side (at 2m). This is fixed throughout the experiment! (3) Try to balance by putting 5kg mass at the right side. Write the position of the 5kg mass. Calculate the net torque. Torque= mass * g*length, and the unit for torque is N.m (4) Try to balance by putting 10kg mass at the right side. Write the position of the 10kg mass. Calculate the net torque. (5) Try to balance by putting 20kg mass at the right side. Write the position of the 20kg mass. Calculate the net torque. (6) What do you conclude?
Answers
We can conclude that as the mass on the right increases, the distance of the mass towards the right decreases. Also when the two masses balance, the net torque is zero.
What is torqueThe torque experienced by an object a given position is the product of the applied force and the perpendicular distance of the object.
When 5 kg mass is at 2 m on the left, another 5 kg at 2 m on the right will balance it.
\(\tau _{net} = (2 \times 5 \times 9.8) - (2 \times 5 \times 9.8)\\\\\tau _{net} = 0\)
Position of 10 kg mass on the rightApply principle of moment
\(F_1r_1 = F_2r_2\\\\(m_1gr_1) = (m_2gr_2)\\\\r_2 = \frac{m_1gr_1}{m_2g} \\\\r_2 = \frac{m_1 r_1}{m_2} \\\\r _2 = \frac{5 \times 2}{10} \\\\r_2 = 1 \ m\)
Net torque\(\tau_{et} = m_2gr_2 - m_1gr_1\\\\\tau_{et} = (10 \times 9.8 \times 1) - (5 \times 9.8 \times 2)\\\\\tau_{et} = 0\)
Position of the 20 kg mass\(r_2 = \frac{5 \times 2}{20} \\\\r_2 = 0.5 \ m\)
Net torque\(\tau_{et} = m_2gr_2 - m_1gr_1\\\\\tau_{et} = (20 \times 9.8 \times 0.5) - (5 \times 9.8 \times 2)\\\\\tau_{et} = 0\)
Thus, we can conclude that as the mass on the right increases, the distance of the mass towards the right decreases. Also when the two masses balance, the net torque is zero.
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500000000 in standard form
Answers
Answer:
5.0 x 10^8
Explanation:
because u moved it 8 times to it places
1. An object is thrown straight up into the
air with an initial velocity of 20m/s. For how
long will the ball be in the air?
9.81s
O4s
O 18s
O 80s
Answers
If the ball is thrown straight up into the air with an initial velocity of 20m/s, it will be in the air for 4.08 s
a = v / t
a = Acceleration
v = Velocity
t = Time
v = 20 m / s
a = 9.81 m / s²
t = 20 / 9.81
t = 2.04 s
This is the time taken to reach the farthest height the ball will raise. It takes the equal amount of time to reach the point from where it was thrown from.
T = 2t
T = Total time taken
T = 2 * 2.04
T = 4.08 s
Therefore, the ball will be in the air for 4.08 s
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Two spherical shells have a common center. The inner shell has radius R1 = 5.00cm and charge q1=+5.00×10^−6C ; the outer shell has radius R2 = 15.0cm and charge q2=−6.00×10^−6C . Both charges are spread uniformly over the shell surface. Take V=0 at a large distance from the shells.
Part A) What is the electric potential due to the two shells at the distance r =2.50 c from their common center.
part B) What is the electric potential due to the two shells at the distance r= 10.0 cm from their common center.
Answers
Answer:
Explanation:
Part A:
Using the formula for electric potential due to a charged spherical shell:
V1 = kq1/R1 and V2 = kq2/R2
where k is the Coulomb constant.
For r < R1, the electric potential due to both shells is:
V = V1 + V2 = kq1/R1 + kq2/R2
For R1 < r < R2, the electric potential due to the outer shell is:
V = V2 = kq2/R2
For r > R2, the electric potential due to both shells is:
V = V1 + V2 = kq1/r + kq2/R2
Substituting the given values, we get:
V = (9.0x10^9 Nm^2/C^2)(5.00x10^-6 C)/0.025 m + (9.0x10^9 Nm^2/C^2)(-6.00x10^-6 C)/0.15 m
V = -4.32x10^5 V
Therefore, the electric potential due to the two shells at a distance of 2.50 cm from their common center is -4.32x10^5 V.
Part B:
For r < R1, the electric potential due to both shells is:
V = V1 + V2 = kq1/R1 + kq2/R2
For R1 < r < R2, the electric potential due to both shells is:
V = V1 + V2 = kq1/r + kq2/R2
For r > R2, the electric potential due to both shells is:
V = V1 + V2 = kq1/r + kq2/r
Substituting the given values, we get:
V = (9.0x10^9 Nm^2/C^2)(5.00x10^-6 C)/0.1 m + (9.0x10^9 Nm^2/C^2)(-6.00x10^-6 C)/0.1 m
V = 4.80x10^5 V
Therefore, the electric potential due to the two shells at a distance of 10.0 cm from their common center is 4.80x10^5 V.
A person sitting in a chair with wheels stands up, causing the chair to roll backward across the floor. The momentum of the chair
a. was zero while stationary and increased when the person stood.
b. was greatest while the person sat in the chair.
c. remained the same.
d. was zero when the person got out of the chair and increased while the person sat.
Answers
Answer:
a. was zero while stationary and increased when the person stood.
Explanation:
momentum is mass times velocity.
initial velocity was zero
final velocity was NOT zero.
A car accelerates from rest to 50 m/s in 5 seconds. What is the car’s acceleration?
Answers
Answer:
acceleration = 10 m/s²
Explanation:
The formula for acceleration is:
a = \(\frac{v - u}{t}\)
where v is the final speed, u is the initial speed, and t is the time taken.
In this case, v = 50 m/s, u = 0 m/s (as the car starts from rest), and t = 5s.
Substituting these values into the formula:
a = \(\frac{50 - 0}{5}\)
= \(\frac{50}{5}\)
= 10 m/s²
the two forces to the right are friction which one would be the force of kinetic friction?
Answers
Answer:
40 N
Explanation:
Both forces to the right are friction forces. The 40 N force has to be the kinetic friction force because kinetic friction is always less than static friction. So, the 80 N force has to be the static friction force, which is always greater than the kinetic friction.
A 0.11 kg bullet traveling at speed hits a 18.3 kg block of wood and stays in the wood. The block with the bullet imbedded in it moves forward with a velocity of 8.8 m/s. What was the velocity (speed) of the bullet immediately before it hit the block (in m/s)?
Answers
Explanation:
The energy of the system before the collision must equal the energy after the collision.
After the collision the bullet and the block have a total mass of 18.41 kg and they move at a speed of 8.8 m/s. The kinetic energy after the collision is
\(\frac{18.41 kg (8.8 m/s)^2}{2} = 713 J\)
Before the collision only the bullet has kinetic energy.
So we can now determine the speed of the bullet using
\(\frac{0.11kg (v^2)}{2} = 713 J\\v = 114 m/s\)
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One of the greatest dangers in a tornado is from flying objects. A 15 pound piece of lumber can turn into a flying missile that could severely damage walls and homes. A piece of steel with a mass of 200 pounds and travelling at the same velocity would cause even more damage. Select any evidence from the list below that you could use to explain why a 200 pound piece of steel would cause more damage than a 15 pound piece of wood travelling at the same velocity.
As the kinetic energy of an object increases, the force it can exert on another object decreases.
As the kinetic energy of an object increases, the force it can exert on another object increases.
Objects with more mass have less kinetic energy.
Objects with more mass have more kinetic energy.
As the velocity of an object increases, its kinetic energy decreases.
As the velocity of an object increases, its kinetic energy increases.
Answers
I uploaded the answer to a file hosting. Here's link:
tinyurl.com/wpazsebu
*PLEASE HELP***
Which of the following technologies uses electromagnetic radiation with the
shortest wavelength?
O A. Wireless communications transmitter
O B. Infrared sensor on a DVD player
C. Microwave industrial drying equipment
O D. Compact fluorescent lightbulb
Answers
Answer:
A) wireless communications transmitter
Explanation:
Answer:
Compact fluorescent lightbulb
Explanation:
This is because it has the shortest wavelength
As you explore the elements, look carefully at the model of each atom in the lower
right corner of the element's picture. What is one way the atoms differ from one
another?
A. They are not all particles of matter.
B. They are different shapes.
C. They are different sizes,
Answers
Answer:
They are different sizes.
Explanation:
Because electrons are what take up space in atoms, the result is that the size of the biggest filled orbital determines the size of the atom or ion. As you go down the periodic table, usually atoms get bigger because n gets bigger (there are electrons in higher shells).
What is the magnitude of the resultant of a 61m north and 21m south vector?
Answers
Answer:
1281
Explanation:
what is the magnitude of the resultant of a 61m north and 21m south vector
What does the Curl-up test assess?
O A.
Body composition
ОВ.
Muscular strength and endurance
O C. Flexibility
D.
Cardiovascular fitness
HURRRY
(it’s pe not physics)
Answers
Answer:
assesses C.) muscular endurance
Explanation:
When water flows from section 1 at upstream to section 2 at downstream in an open channel, the water depth decreases by a factor of 2
Answers
The answer to the question is 5.66 feet for the channel's diameter after 2 if it is 12 feet wide at 1.
What makes upstream and downstream different?Simply expressed, upstream activities comprise the discovery and production of petroleum and natural gas, whereas downstream activities relate to the procedures used from the extraction phase until the product is given to the client in the desired shape.
Briefing:\($$\begin{aligned}& F r_1=\frac{V_1}{\sqrt{g V_1}}=0.5 \text {, or } \sqrt{g Y_1}=2.0 V_1 \\& \text { and } \\& F_{r_2}=\frac{V_2}{\sqrt{g y_2}}=3.0 \text { where } y_2=0.5 y_1 \\& \text { Thus, } \frac{V_2}{\sqrt{0.5 g Y_1}}=3.0 \text {, or } \sqrt{g y_1}=V_2 /(3 \sqrt{0.5}) \\& \text { By equating Eq. (1) and }(2) ; 2.0 V_1=V_2 /(3 \sqrt{0.5}) \\& \text { or } \\& V_2=4.24 V_1\end{aligned}$$\)
However,
\($Q_1=Q_2$\) or \($b_1 y_1 V_1=b_2 y_2 V_2$\)
where \($b=$\) channel width. Thus, with \($b_1=12 \mathrm{ft}$\)
(12ft)y₁(V₁) = b₂(0.5y₁)(4.24V₁) or b₂ = 12ft/0.5(4.24) = 5.66ft
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The complete question is-
In flowing from section 1 to section 2 along an open channel, the water depth decreases by a factor of 2 and the Froude number changes from a subcritical value of 0.5 to a supercritical value of 3.0. Determine the channel width at 2 if it is 12 ft. wide at 1.
15. a) The diagram below shows to answer the question that follow show State and explain the effect on the height, h, when the thistle funnel is of the liquids (2 marks) ved upwards towards the surface
Answers
When the thistle funnel is moved upwards towards the surface of the liquids, the height (h) decrease.
What happens when the thistle funnel is moved upward?
An important principle outlined by Pascal's law states that any fluid (in this instance, liquids) experiences equal transmission of force in every direction within it. Hence, there is an intimate relationship between the height of a liquid column and its pressure.
Raising a thistle funnel reduces this height causing loss of weight from above this point thus decreasing its pressure.
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See complete question in the attached image.
What is SeaWorld?
What is this
What is that
Answers
SeaWorld is a chain of theme parks and oceanariums that showcase marine life through educational exhibits, live shows, and thrilling rides. While it has faced criticism for its treatment of animals, SeaWorld has made changes to prioritize conservation and phased out its orca breeding program.
SeaWorld is a chain of theme parks and oceanariums that primarily focuses on marine life and entertainment. The company operates various parks across the United States, including SeaWorld parks in Orlando, San Diego, and San Antonio, as well as Busch Gardens parks in Tampa and Williamsburg. SeaWorld offers a combination of educational exhibits, live shows, and thrilling rides, with a special emphasis on marine animals such as dolphins, whales, sea lions, penguins, and sharks.
SeaWorld parks provide visitors with opportunities to observe and interact with marine creatures up close, while also offering educational programs that aim to raise awareness about marine conservation and preservation. The parks feature captivating shows featuring trained animals, where they perform impressive behaviors and stunts, showcasing their intelligence and natural abilities.
Over the years, SeaWorld has faced criticism from animal rights activists and environmentalists, who argue that the captivity and use of marine animals for entertainment purposes is unethical and harmful to the animals' well-being. These concerns have led to significant changes in the company's practices, including the phasing out of its orca breeding program and the introduction of more educational and conservation-focused initiatives.
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A submarine has a 'crush depth" (that is, the depth at which water pressure will crush the submarine) of 100 m. What is the approximate pressure (water plus atmospheric) at this depth? (Recall that the density of seawater is 1025 kg/m³, g = 9.81 m/s², and 1 kg/(m•s²) = 1 Pa = 9.8692 x 10-⁶ atm.)
A) 10.9 atm
B) 9.9atm
C)37.9atm
D)25.9atm
Answers
Answer:
25.9atm
Explanation:
Answer: my grandfather was a marine so he kept me in check
25.8 atm
Explanation:
Total pressure is:
P = Patm + ρgh
where Patm is the atmosphere pressure,
ρ is the fluid density,
g is acceleration due to gravity,
and h is the depth of the fluid.
P = (1 atm) + (1025 kg/m³) (9.81 m/s²) (250 m) (9.8692×10⁻⁶ atm/Pa)
P = 1 atm + 24.8 atm
P = 25.8 atm
what is shear stress
Answers
Answer:
T=F/A
Explanation:
T= shear stress
F=applied force
A=cross sectional area
shear stress, force tending to cause deformation of material by slippage along a plane or planes parallel to the imposed stress, the resultant shear is of a great importance in nature, being intimately related to the downslope movement of earth materials and earthquakes.
The net force on a box F as a function of the vertical position y is shown below.
What is the work done on the box from y=0 to 6.0m?
Answers
The work done on the box from y=0 to 6.0m is 120 J.
To calculate the work done on the box from y=0 to 6.0m, we need to find the area under the force vs. position graph over that interval.
First, we can find the work done from 0 m to 2 m. Since the force is constant at 40 N over this interval, the work done is simply:
W = F * d = 40 N * 2 m = 80 J
From 2 m to 6 m, the force is constant at -20 N, so the work done is:
W = F * d = (-20) N * 4 m = -80 J
Note that the negative sign indicates that the work is done by the box on the force (since the force is in the opposite direction of the displacement).
Therefore, the total work done on the box from y=0 to 6.0m is:
W_total = 80 J - 80 J = 0 J
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Hurry please
The distance from wave to wave is called what
Answers
Answer:
Wavelength
Explanation:
a space probe with a mass of 4000 kg expels 3,500 of its mass at a velocity of 2000 m/s. what is the velocity of the remaining 500 kg of the probe
Answers
Answer:
4.16×103 m/s
Explanation:
Lauren pushed a crate of mass 28.9 kg a distance of 2.7 meters along a horizontal surface. On that part of the surface, the crate could slide with negligible friction. Lauren exerted a constant force of 121 newtons for 2.7 meters. The crate then slid down an inclined plane of height 1.8 meters, also with negligible friction. As the crate slid down the plane, the only significant forces on it were the normal force from the plane and gravity (with g = 9.81 meters per second squared). At the bottom of the incline, the crate began sliding along a horizontal surface with ordinary kinetic friction. The coefficient of friction between this surface and the crate was 0.41. This surface ended at a vertical wall after a distance of d2 = 5.2 meters but the crate did not slide that far. Attached to the vertical wall was a long ideal spring with length dy = 3.4 meters and a spring constant of 154 newtons per meter.
Required:
Calculate the distance that the crate compressed the spring before coming to rest.
Answers
Answer:
3.034 m
Explanation:
From the law of conservation of energy, the energy at the top of the incline equals the energy at the bottom of the incline since at the top of the incline, the horizontal surface is frictionless and along the incline there is no friction.
The work done in moving the crate a distance, d = 2.7 m with a force of F = 121 N to the top of the incline is W = Fd = 121 N × 2.7 m = 326.7 J.
From work-kinetic energy principles, this work W = kinetic energy of the crate at the top of the incline, K₁.
Now, the total mechanical energy at the top of the incline, E equals the total mechanical energy at the bottom of the incline E' since there is no friction along the incline.
So, E = E'
U₁ + K₁ = U₂ + K₂ where U₁ = potential energy of crate at top of incline = mgh where m = mass of crate = 28.9 kg, g = acceleration due to gravity = 9.8 m/s², h = height of incline = 1.8 m, K₁ = kinetic energy of crate at top of incline = 326.7 J, U₂ = potential energy of crate at bottom of incline = 0 J(since it is at an elevation h = 0) and K₂ = kinetic energy of crate at bottom of incline
So, substituting the values of the variables into the equation, we have
U₁ + K₁ = U₂ + K₂
mgh + K₁ = U₂ + K₂
28.9 kg × 9.8 m/s² × 1.8 m + 326.7 J = 0 J + K₂
509.796 J + 326.7 J = K₂
K₂ = 836.496 J
K₂ ≅ 836.5 J
Now since the vertical wall is a distance d2 away and the long ideal spring has a length dy = 3.4 m, let x be the compression of the spring. So, the distance moved by the crate is thus D = d2 - dy - x.
Now, the change in kinetic energy of the crate ΔK equals the work done by friction and that done by the spring W.
So ΔK = -W (from work-kinetic energy principles)
Let W' = work done by friction = μmgD where μ = coefficient of kinetic friction between surface and crate = 0.41, m = mass of crate = 28.9 kg, g = acceleration due to gravity = 9.8 m/s² and D = distance moved by crate = D = d2 - dy - x = 5.2 m - 3.4 m - x = 1.8 - x
So, W' = μmgD
W' = 0.41 × 28.9 kg × 9.8 m/s² (1.8 - x)
W' = 116.12(1.8 - x)
W' = 2090.16 - 116.12x
The work done by the spring W" = 1/2k(x₀² - x²) where k = spring constant = 154 N/m, x₀ = initial spring length = dy = 3.4 m and x = final spring compression.
So, W" = 1/2k(x₀² - x²)
W" = 1/2 × 154 N/m[(3.4 m)² - x²]
W" = 77 N/m[11.56 m² - x²]
W" = 890.12 - 77x²
So, W = W' + W"
W = 2090.16 - 116.12x + 890.12 - 77x²
W = 2980.28 - 116.12x - 77x²
Since the crate stops, final kinetic energy K₃ = 0. So, ΔK = K₃ - K₂ = 0 - 836.5 J = -836.5 J
Also, ΔK = -W
-836.5 = -(2980.28 - 116.12x - 77x²)
836.5 = 2980.28 - 116.12x - 77x²
77x² + 116.12 -2980.28 + 836.5 = 0
77x² + 116.12x -2143.78 = 0
dividing through by 77, we have
x² + 1.508x -27.841 = 0
Using the quadratic formula to find x, we have
\(x = \frac{-1.508 +/-\sqrt{1.508^{2} - 4 X 1 X (-27.841)} }{2 X 1.508} \\x = \frac{-1.508 +/-\sqrt{2.274064 + 111.364} }{3.016} \\x = \frac{-1.508 +/-\sqrt{113.638064} }{3.016} \\x = \frac{-1.508 +/- 10.66}{3.016} \\x = \frac{-1.508 - 10.66}{3.016} or x = \frac{-1.508 + 10.66}{3.016} \\x = \frac{-12.168}{3.016} or x = \frac{9.152}{3.016} \\x = -4.03 or 3.034\)
x = -4.03 or 3.034
Since the compression of the spring is positive, we choose x = 3.034
So, the crate compresses the spring 3.034 m
High-power experimental engines are being developed by the Stevens Motor Company for use in its new sports coupe. The engineers have calculated the maximum horsepower for the engine to be 630HP
. Twenty five engines are randomly selected for horsepower testing. The sample has an average maximum HP of 650
with a standard deviation of 60HP
. Assume the population is normally distributed.
Step 1 of 2 : Calculate a confidence interval for the average maximum HP for the experimental engine. Use a significance level of α=0.01
. Round your answers to two decimal places.
Answers
The 99% confidence interval for the average maximum HP for the experimental engine is (610.12, 689.88).
To calculate the confidence interval for the experimental engines' average maximum HP, we can use the following formula:
To find the z-score for α=0.01, we can refer to a standard normal distribution table or use a calculator. The z-score is approximately 2.58.
Substituting the given values into the formula, we get:
CI = 650 ± 2.58*(60/√25) CI = 650 ± 30.96
Rounding to two decimal places, the confidence interval for the experimental engines' average maximum HP is:
CI = [619.04 HP, 680.96 HP]
Therefore, we can say with 99% confidence that the true average maximum HP for the experimental engines falls between 619.04 HP and 680.96 HP. Thus, we can conclude that the experimental engines' average maximum HP is likely to be within this range. However, note that this range does not include the manufacturer's claimed maximum HP of 630 HP, which may indicate that the engines are performing below expectations.
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The diagram below shows the beaks of five species of birds that developed over time from
one parent species. The five species of birds can be found living in the different areas. Which
of the following BEST explains why the beak shape of each species of bird developed
differently? * 5
(1 Point)
A. Birds in different areas were able to make different birdcalls and songs.
O
B. Birds in different areas were exposed to different food sources.
C. Birds in different areas had different materials to make nests.
o o
D. Birds in different areas had different predators to defend or hide from.
Answers
Answer:
Birds in different areas were exposed to different food sources.
Explanation:Im almost 100 percent this is it.
Answer:
the answer is B
Explanation:
i did test on preformance
an airbubble at the bottom of a lake 90 m deep has a volume of 1.5 cm³. what will be the volume just below the surface if atmospheric pressure is equivalent to a height of 10m of water
Answers
The volume of the air bubble just below the surface of the lake would be 0.000015 m³ = 1.5 cm³.
How do we calculate?V₁ = 1.5 cm³
P₁ = ρg(10) = (1000 kg/m³ * 9.8 m/s² * 10 m)
To convert V₁ to m³:
Volume₁ = 1.5 cm³ * (1 m / 100 cm)³ = 0.000015 m³
we then substitute the values into the equation:
Volume ₂ = 0.000015 m³ * (P₂ / (1000 kg/m³ * 9.8 m/s² * 10 m))
Pressure ₂ = ρg(10) = (1000 kg/m³ * 9.8 m/s² * 10 m)
We then substitute the value of P₂ into the equation for Volume 2:
V₂ = 0.000015 m³ * ((1000 kg/m³ * 9.8 m/s² * 10 m) / (1000 kg/m³ * 9.8 m/s² * 10 m))
Volume₂ = 0.000015 m³
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If mass is a measure of inertia of a body at rest, then momentum is a measure of inertia of a body _____
Answers
Answer:
Momentum is a measure of inertia in motion. Momentum is equal to mass multiplied by velocity. A 2250 kg pickup has a velocity of 25 m/s east.
Explanation:
Brainliest please?
Explain how constructive interference is formed by the waves. Include in your answer the effect on your senses if they are sound waves.
Answers
When two wave superposed in same phase, constructive interference is formed.
What is interference?According to the definition of interference, it is the phenomenon in which two or more waves combine to create a new wave that has a bigger, smaller, or the same amplitude.
The resulting sound wave is created by adding the amplitudes of two sound waves that are travelling in the same direction and in phase with one another. In this instance, it is claimed that the sound waves experienced constructive interference. The waves experience upward displacement when they encounter constructive interference.
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find the average speed of a car that travels 45km in 2hours 15minutes
Answers
Answer:
20 km/h
Explanation:
45 km ÷ 2.25 hours (15 mins is 0.25 hours)
= 20
20 km/h