The torque 0,1 acting on the meter stick due to 1 is 3.92Nm.
The torque 0,2 acting on the meter stick due to 1 is 20.94Nm.
The torque 0,stick due to weight of meter stick is 2.45Nm.
The torque 0,pivot acting due to pivot is 0.
What is torque?The rotating equivalent of force is torque. The moment of force is another name for it (also abbreviated to moment). It illustrates how a force can cause a change in the body's rotational motion.
The definition of torque is the product of the force's perpendicular component's magnitude and the force's line of action's separation from the point being measured. Torque can also be understood in terms of the law of conservation of energy.
Using the formula:
τ= r × F
where,
τ is the torque,
F is the force and,
r is the distance of the line of action of a force from the point around which torque is being determined.
Substituting the values and solving for torque:
The torque 0,1 acting on the meter stick due to 1 is 3.92Nm.
The torque 0,2 acting on the meter stick due to 1 is 20.94Nm.
The torque 0,stick due to weight of meter stick is 2.45Nm.
The torque 0,pivot acting due to pivot is 0.
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A 32.9 kg child on a sled slides down a hill, reaching a speed of 9.94 m/s. How high was the hill (in m)?
Answer:
To find the height of the hill, we can use the conservation of energy principle, which states that the initial potential energy of the child and sled at the top of the hill is equal to their final kinetic energy at the bottom of the hill. The potential energy is given by:
PE = mgh
where m is the mass of the child and sled, g is the acceleration due to gravity (9.81 m/s^2), and h is the height of the hill.
The kinetic energy is given by:
KE = (1/2)mv^2
where v is the speed of the child and sled at the bottom of the hill.
Equating the potential and kinetic energies, we have:
mgh = (1/2)mv^2
Canceling the mass, we get:
gh = (1/2)v^2
Solving for h, we have:
h = (1/2) v^2 / g
Substituting the given values, we get:
h = (1/2) (9.94 m/s)^2 / 9.81 m/s^2
h = 5.06 m
Therefore, the height of the hill is 5.06 meters.
a ring of aluminium bronze alloy has internal diameter 300mm and 50mm wide .the coefficient of cubical expansion of alloy is 51×10^-6 /degree celcius .for a temperature rise of 600 degree celsius find the final internal diameter?
According to the given statement The ring's final internal diameter would be 1218 mm.
What is cubical expansion explanation?Cubical expansion is the term for the phenomenon wherein the volume of a solid increases as it is heated. Also called volumetric expansion. The coefficient of volumetric expansion measures how much a material's volume expands as its temperature rises by one degree.
The following formula can be used to get the ring's ultimate interior diameter:
ΔL = αLΔT
where ΔL is the length increase, is the cubical growth coefficient, L is the starting length, and ΔT is the temp change.
In this instance, the change in width, which is double the change in length, is what we are searching for. Thus, the formula may be rewritten as follows:
ΔD = 2αDLΔT
where D represents the starting diameter and ΔD represents the diameter change.
By putting in the indicated values, we get:
ΔD = 2(51×10⁻⁶ /degree celsius)(300 mm)(600 degree celsius)
ΔD = 918 mm
As a result, the ring's final internal diameter would be:
Dfinal = Dinitial + ΔD = 300 mm + 918 mm = 1218 mm
Hence, the ring's final internal diameter would be 1218 mm.
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Question #1. 500 pages of a book has a total mass of 2.5 kg. What is the mass of each page in: (i) kg: (ii) mg: (iii). µg:
Answer:
i) 2.5kg ii)2.5x10^6mg III) 2.5x10^9μgA car is moving at 32 miles per hour. The kinetic energy of that car is 5 × 10^5 J.
How much energy does the same car have when it moves at 101 miles per hour?
Answer in units of J.
Answer:
The car has approximately 1.42 × 10^6 J of energy when it moves at 101 miles per hour.
Explanation:
First, we need to convert the initial velocity and kinetic energy to SI units:
Initial velocity: 32 miles per hour = 14.3 meters per second (rounded to 2 decimal places)
Kinetic energy: 5 × 10^5 J (given)
Next, we can use the formula for kinetic energy:
KE = (1/2)mv^2
where KE is the kinetic energy, m is the mass of the car, and v is the velocity of the car.
Solving for mass:
m = 2KE/v^2
Substituting the given values:
m = 2(5 × 10^5 J) / (14.3 m/s)^2 ≈ 1569.93 kg (rounded to 2 decimal places)
Now, we can use the same formula to calculate the kinetic energy of the car when it moves at 101 miles per hour (rounded to 2 decimal places):
KE = (1/2)mv^2 = (1/2)(1569.93 kg)(45.06 m/s)^2 ≈ 1.42 × 10^6 J
Therefore, the car has approximately 1.42 × 10^6 J of energy when it moves at 101 miles per hour.
A horizontal pipe of diameter 1.11 m has a
smooth constriction to a section of diameter
0.666 m . The density of oil flowing in the pipe
is 821 kg/m3
.
If the pressure in the pipe is 8130 N/m
2
and in the constricted section is 6097.5 N/m2
,
what is the rate at which oil is flowing
PLEASE ANSWER THISSSSSSSSS!!!!!
The rate at which οil is flοwing is 0.494 m³/s. This means that the prοduct οf the fluid's density (ρ), crοss-sectiοnal area (A), and velοcity (v) is cοnstant.
What is Density?Density is a physical prοperty οf matter that represents hοw much mass is cοntained within a given vοlume οf a substance. It is defined as the amοunt οf mass per unit vοlume and is typically expressed in units οf kilοgrams per cubic meter (kg/m³) οr grams per cubic centimeter (g/cm³).
We can use the principle οf cοntinuity tο sοlve this prοblem. Accοrding tο this principle, the mass flοw rate οf a fluid remains cοnstant as it flοws thrοugh a pipe οf varying diameter.
Therefore, we can write:
[tex]\rho_1A_1v_1 = \rho_2A_2v_2[/tex]
where the subscripts 1 and 2 refer to the sections of the pipe before and after the constriction, respectively.
We can rearrange this equation to solve for the velocity of the oil in the pipe:
[tex]v_2 = (A_1/A_2) \times (v_1 \times (\rho_1/\rho_2))[/tex]
where A1 and A2 are the cross-sectional areas of the pipe before and after the constriction, respectively.
Using the given values, we get:
[tex]v_2 = (\pi/4) \times (1.11 m)^2 \times (8130 N/m^2 / 821 kg/m^3) / [(\pi/4) \times (0.666 m)^2] \times (6097.5 N/m^2 / 821 kg/m^3)[/tex]
v₂ = 1.74 m/s
Finally, we can calculate the rate at which oil is flowing using the formula:
Q = A₂ × v₂
Using the given values, we get:
Q = (π/4) × (0.666 m)² × 1.74 m/s
Q = 0.494 m³/s
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A student heats the same amount of two different liquids over Bunsen burners. Each liquid is at room temperature when the student begins. If Liquid 1 has a higher specific heat than Liquid 2, then Liquid 1 will
Answer: Liquid 1 heats more slowly than 2
Explanation:
Liquid 1 heats more slowly than 2 because it has a higher specific heat.
(Hope this helps!)
A spring is compressed while two steel balls of mass m1 = 2.5 kg and m2 = 1.6 kg. If m1 experiences an acceleration of 2.0 m/s² to the left, then what is the acceleration (in m/s²) of mass m2 to the right?
The acceleration of mass m2 to the right is 3.125 m/s².
Let's denote the compression of the spring by x, the acceleration of m1 by a1 to the left, and the acceleration of m2 by a2 to the right.
According to Newton's second law, the force exerted on an object is equal to its mass times its acceleration:
F = ma. In this case, the forces acting on the two masses are the force of the compressed spring and the force of friction between the masses.
For mass m1, the net force is given by:
F1 = -kx - f
where k is the spring constant, x is the compression of the spring, and f is the force of friction. The negative sign in front of kx indicates that the force of the spring is acting to the left. Using Newton's second law, we have:
m1a1 = -kx - f
For mass m2, the net force is given by:
F2 = kx - f
where the force of the spring is now acting to the right. Using Newton's second law, we have:
m2a2 = kx - f
We want to find the acceleration of mass m2, which is given by a2. To do this, we need to eliminate the force of friction f from the above two equations.
To eliminate f, we can add the two equations:
m1a1 + m2a2 = -kx + kx - 2f
Simplifying and substituting the given values, we get:
(2.5 kg)(2.0 m/s²) + (1.6 kg)a2 = 0
Solving for a2, we get:
a2 = -(2.5 kg)(2.0 m/s²)/(1.6 kg) = -3.125 m/s²
Therefore, the acceleration of mass m2 to the right is 3.125 m/s².
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A ball rolls along flat ground with a speed of 5.2 m/s when it encounters a hill. What vertical height (in m) above the ground does the ball reach?
A key challenge with renewable energy is that the energy must be transported to the place where it's needed _ or devices that store the energy needed improvement.
A key challenge with renewable energy is that the energy must be transported to the place where it's needed batteries or devices .
Option A is correct.
What is renewable energy?Renewable energy sources such as solar, wind, and hydroelectric power are often located far away from the areas where the energy is needed. This makes it necessary to transport the energy over long distances, which can be expensive and lead to losses due to transmission and distribution. In addition, renewable energy sources are often intermittent, meaning they don't produce a constant supply of energy.
To address this, energy storage devices are needed to store excess energy produced during peak times for use during times of low energy production. However, the current technology for energy storage is not yet efficient or cost-effective enough to meet the growing demand for renewable energy. Developing more effective energy storage devices is therefore an important area of research and development in the field of renewable energy.
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Question incomplete:
A key challenge with renewable energy is that the energy must be transported to the place where it's needed _ or devices that store the energy needed improvement.
A. batteries
B. refineries
C. solar panels
D. wind turbines
Answer:
batteries
Explanation:
in the summer people often wear light colored clothing to stay cool this is a good idea because light colors tend to_____electromagnetic waves
1. absorb
2. refract
3. reflect
7. A student pushed a box 25.0 meters across a smooth, horizontal floor using a constant force of 112 Newtons. If the force was applied for 7.00 seconds, how much power was developed?
The power developed was 1,568 Watts, the calculation is seen in the section below.
Computation of PowerIn science and engineering, power is the rate at which work is done or energy is delivered. It can be expressed as the product of the work done (W) or the energy transferred (E) divided by the time interval (t).
Given data
Distance = 25.0 metersApplied Force = 112 NewtonsTime = 7.00 seconds,We know that expression for Power is given as
Power = (Force x Distance) / Time
Substituting our data into the expression we have
Power = (112 N x 25.0 m) / 7.00 s
Power = 1,568 Watts.
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Select which of the following is (are) the correct unit(s) for energy?
Joule (J). This is the fundamental energy unit of the metric system, or the International System of Units in a later, more thorough version (SI).
What does "energy unit" mean?The SI unit of energy, the joule (J), was created in honour of James Prescott Joule and his research on the mechanical equivalent of heat since energy is defined through labour. In terms of SI base units, 1 joule is equivalent to 1-newton metre and, in slightly more basic words.
What is the formula for the energy SI unit?Potential energy, which can be calculated using the formula P.E. = mgh. Unit is the energy that an object has stored due to its position and height. Joules is the SI unit for energy (J).
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A faulty model rocket moves in the xy-plane (the positive y-direction is vertically upward). The rocket's acceleration has components ax(t)=αt2
and ay(t)=β−γt
, where α
= 2.50 m/s4
, β
= 9.00 m/s2
, and γ
= 1.40 m/s3
. At t=0
the rocket is at the origin and has velocity v⃗ 0=v0xi^+v0yj^
with v0x
= 1.00 m/s
and v0y
= 7.00 m/s
.
The rocket travels a horizontal distance of 57.4 m before hitting the ground.
What is the initial speed of the rocket?The initial speed of the rocket is v0=√(v0x^2+v0y^2)=7.28 m/s.
What is the rocket's velocity at the maximum height?The rocket's velocity at the maximum height is zero, as it momentarily stops moving vertically and starts falling back down.
To solve this problem, we can use the kinematic equations of motion. Let's first find the velocity and position as a function of time:
vx(t) = v0x + ∫ax(t) dt = v0x + (1/3)αt^3
vy(t) = v0y + ∫ay(t) dt = v0y + βt - (1/2)γt^2
x(t) = ∫vx(t) dt = v0x t + (1/12)αt^4
y(t) = ∫vy(t) dt = v0y t + (1/2)βt^2 - (1/6)γt^3
Now, let's find the time t1 when the rocket reaches its maximum height:
ay(t1) = 0
β - γt1 = 0
t1 = β/γ = 6.43 s
At t1, the rocket's height is:
y(t1) = v0y t1 + (1/2)βt1^2 - (1/6)γt1^3
y(t1) = 7.00 m/s × 6.43 s + (1/2) × 9.00 m/s2 × (6.43 s)^2 - (1/6) × 1.40 m/s3 × (6.43 s)^3
y(t1) = 92.5 m
Now, let's find the time t2 when the rocket hits the ground. We can do this by solving for the positive root of the quadratic equation:
y(t) = 0
(1/2)γt^2 - βt - v0y = 0
Using the quadratic formula, we get:
t2 = (β + √(β^2 + 2γv0y))/γ = 8.01 s
Finally, let's find the horizontal distance traveled by the rocket:
x(t2) = v0x t2 + (1/12)αt2^4
x(t2) = 1.00 m/s × 8.01 s + (1/12) × 2.50 m/s4 × (8.01 s)^4
x(t2) = 57.4 m
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A mini copper of mass 1350kg is travelling at speed of 100km/hr and travel at a distance of 45 m before coming to rest . Determine the magnitude of the net force required to bring the machine to rest and state the direction with regards to initial velocity
Answer:
The net force required to bring the car to rest is 11,900 N. Since the car is slowing down, the direction of the net force must be opposite to the initial velocity. Therefore, the direction of the net force is opposite to the initial direction of the car's motion.
Explanation:
First, we need to convert the speed from km/h to m/s:
100 km/h * (1000 m/km) / (3600 s/h) = 27.78 m/s
The initial kinetic energy of the car is:
KE = (1/2) * m * v^2
KE = (1/2) * 1350 kg * (27.78 m/s)^2
KE = 535,500 J
To bring the car to a stop, the net force applied over the distance traveled must equal the initial kinetic energy. So:
work done = force * distance = KE
Rearranging this equation, we get:
force = KE / distance
Substituting the values we have, we get:
force = 535,500 J / 45 m
force = 11,900 N
The net force required to bring the car to rest is 11,900 N. Since the car is slowing down, the direction of the net force must be opposite to the initial velocity. Therefore, the direction of the net force is opposite to the initial direction of the car's motion.
PLEASE HELPPP!!!! i really need help with this report if anyone can!!!!
The U.S. Army is planning to drop supplies from a plane at a refugee camp. The supplies are divided into 700-kilogram parcels, and the parachutes have an area of 100 square meters. The only problem is that the parcels cannot hit the ground at a velocity of more than 5 meters per second without damaging the contents. Are these parachutes suitable for this task?
For the purposes of this exercise, assume that the for the drag coefficient of the parachute is 1.5 and that the air density is 1.22 kilograms per cubic meter. Write a report detailing why these parachutes are or are not suitable and determining the minimum size parachute that can be used in this situation.
Answer:
search it up
Explanation:
5. While driving a car Rahul saw one sign board and suddenly, he reduced the speed. Which of
the following signboard he has seen?
The signboard Rahul saw was likely a speed limit sign. This sign is used to inform drivers of the maximum speed they should be travelling at in that area.
What is limit?Limit is a mathematical concept that describes the highest or lowest value that a given function or expression can reach. It is used as a tool to measure how a function or expression behaves as its variables approach a certain value. For example, the limit of a function as x approaches infinity is the highest value that the function can take on. It is often used to calculate the area under a curve, the rate of change of a function, and the slope of a line at a given point. Limits are also used to compare the relative sizes of different functions, and to determine the continuity of a function.
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Two identical blocks are connected by a lightweight string that passes over a lightweight pulley that can rotate about its axle with negligible friction. The two-block system is released from rest and the blocks accelerate. Which of the following correctly relates the potential energy gained by the block 1-Earth system |∆U1| to the potential energy lost by the block 2-Earth system |∆U2| and provides correct evidence?
An astronaut floating in space is motionless. The astronaut throws a wrench in one direction, propelling her in the opposite direction. Which of the following statements are true? (Choose all that apply.)
The statement that are true on the astronaut throwing the wrench in space are:
(A) The wrench will have a greater velocity than the astronaut.(C) The wrench will have greater kinetic energy than the astronaut.How would the astronaut be affected ?When the astronaut throws the wrench, the wrench exerts a force on the astronaut, pushing her in the opposite direction. The velocity of the wrench will be greater because it has less mass than the astronaut and therefore can be propelled with greater velocity.
Kinetic energy is proportional to the mass and velocity of an object. Since the wrench has less mass but greater velocity than the astronaut, it will have greater kinetic energy.
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A bear walks 9 meters north, stops to munch on some berries, then walks another 12 meters north. What is the bear's total distance travelled AND his displacement?
The bear's total distance traveled is 21 meters and its displacement is 3 meters north.
How to determine the bear's of total distance travelled and his displacement ?
The bear's total distance traveled is the sum of the distances covered in each leg of the journey, which is:
Total distance traveled = 9 meters + 12 meters = 21 meters
The bear's displacement is the straight-line distance between the starting point and the ending point of the journey, regardless of any intermediate stops.
Since the bear is moving only north, the displacement is simply the difference between the final and initial positions in the north direction, which is:
Displacement = 12 meters - 9 meters = 3 meters north
Therefore, the bear's total distance traveled is 21 meters and its displacement is 3 meters north.
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200g of olive oil is at 80°C in a container. If another 50g of olive oil at 20°C is added to the container, what will the final temperature be?
(please with solution)
The resultant final temperature is roughly around 61.7°C.
What is Principle of conservation of energy?According to the principle of conservation of energy, a system's overall energy level is constant.
According to the question,
Let Q1 = heat that the heated olive oil has absorbed.
Let Q2= heat that the heated olive oil has released.
Q1 + Q2 = 0 {according to the energy conservation principle}
What is Heat?For absorption of heat by the hot olive oil
Q1 = m1 × c × ΔT1 {Q is the heat absorbed or emitted, m is the sample's mass, and c is olive oil's specific heat capacity = 4.18 J/g °C ΔT stands for temperature change.}
Given- M1= 200g, c= 4.18g, T=80°C
Q1 = 200 g × 4.18 J/g °C × ([tex]T_{f}[/tex] - 80°C)
For the heat released by the the warmth that the cold olive oil emits:
Q2 = m2 × c × ΔT2
Q2 = 50 g × 4.18 J/g°C × ( - 20°C)
Q1 + Q2 = 0 {on equating the equation}
200 g × 4.18 J/g °C × ([tex]T_{f}[/tex] - 80°C) + 50 g × 4.18 J/g °C × ([tex]T_{f}[/tex] - 20°C) = 0
[tex]T_{f}[/tex] = (200 g × 4.18 J/g °C × 80°C + 50 g × 4.18 J/g °C × 20°C) / (200 g × 4.18 J/g °C + 50 g × 4.18 J/g °C)
[tex]T_{f}[/tex] = 61.7°C
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4. Radiation safety and protection, including: A. Radiation safety and emergency measures in radiotherapy B. Compliance with local legislative and licensing requirements, code of practice and local rules C. Room shielding design and calculation for radiotherapy equipment and facilities D. Optimization
Radiation safety and protection, including: all the given options.
What is Radiation safety and protection?Radiation safety and protection includes various measures taken to minimize radiation exposure to individuals, the environment, and property.
Some of the key components of radiation safety and protection are:
Radiation safety and emergency measures in radiotherapy: This involves ensuring that appropriate measures are in place to minimize radiation exposure to patients, healthcare workers, and the public during radiotherapy procedures.
Compliance with local legislative and licensing requirements, code of practice and local rules: To ensure radiation safety, there are specific laws, codes of practice, and licensing requirements that must be followed by all organizations and individuals working with radiation.
Room shielding design and calculation for radiotherapy equipment and facilities: Shielding is an essential component of radiation safety and involves designing and constructing radiation-shielded rooms and facilities.
Optimization: Optimization involves using the lowest possible radiation dose necessary to achieve the desired outcome in medical procedures.
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You want to hang a 9.0-kg
sign that advertises your new business. To do this, you use a pivot to attach the base of a 5.0-kg
beam to a wall (Figure 1). You then attach a cable to the beam and to the wall in such a way that the cable and beam are perpendicular to each other. The beam is 1.5 m
long and makes an angle of 37∘
with the vertical. You hang the sign from the end of the beam to which the cable is attached.
A.)What must be the minimum tensile strength of the cable (the minimum amount of tension it can sustain) if it is not to snap?
B.) Determine the horizontal and vertical components of the force exerted by the pivot on the beam.
I have a picture of the system below and I have solved part A to be 68N, but I cannot seem to figure out part B.
Finding unknown responses for statics beam and truss issues can be done by solving for the total of pressures acting in the x- and y-directions. When in balance, the total of the forces acting in each direction on each beam and truss will be equal to zero.
What is a beam in statics?A beam is a long, slender structural component that can support twisting stresses by deforming transverse to its long axis. It should be noted that bending loads are imparted across the long plane.
Steel beams can have cross-sections in a variety of forms, including square, rectangle, circular, I, T, H, C, and tubular.The normal or axial force, the shearing force, and the bending moment are the three potential internal forces that may be generated when a beam or frame is exposed to transverse loadings, as shown in section k of the cantilever.
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If two balls have the same volume, but ball A has twice as much mass as ball B, which one will have the greater density? If ball C is 3 times the volume of ball D and ball D has 1/3 the mass of ball C, which has the greater density? If two balls have the same mass, but ball P is twice as large as ball Q, which one will have the greater density? If ball X is twice as big as ball Y and weighs only half as much as ball Y, then which one will have the greater densitv? Previous Activity
Ball A will have the greater density because it has twice as much mass as ball B for the same volume. Ball C will have the greater density because it has 3 times the volume of ball D and only 1/3 the mass.
What is volume?Volume is the quantity of three-dimensional space occupied by an object or a substance. It is measured in cubic units, such as liters or gallons. Volume is an important concept in mathematics, physics, chemistry, and engineering, and is often used to calculate the amount of material needed for a certain project. For example, in architecture, engineers may use volume to determine the amount of concrete needed to build a bridge. In cooking, cooks use volume to measure the amount of ingredients needed for a recipe. In physics, volume is used to measure the amount of space an object occupies, or the amount of space within an object, such as a liquid or gas.
Ball P will have the greater density because it is twice as large as ball Q for the same mass. Ball X will have the greater density because it is twice as big as ball Y but weighs only half as much.
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We measure the intensity of a sound source in open air to be 0.3 W/m2 when we are located a distance of 15m away from the source. If we were to move to a distance of 25 m, what would be the intensity of the sound? How about at a distance of 45 m?
The intensity of sound are- For distance of 25 m: I = 0.108 W/m² and For distance of 45 m: I = 0.034 W/m².
Explain about the intensity of a sound?The sound becomes softer the further you get from the sound source, particularly when you are outside. This is not shocking at all. Rather like light, sound expands out as it moves away from where it originated. The strength of the sound decreases as you move further away from the source if there aren't surfaces for it to reflect from.Your comprehension of the inverse square law, which states that a sound's intensity varies inversely to the square of its distance from its source, will be put to the test in this challenge.
I = k • (1/R²)
when, distance R is 15m , intensity of a sound source in open air to be 0.3 W/m².
So,
0.3 = k • (1/15²)
k = 225 * 0.3
k = 67.5
For distance of 25 m:
I = k • (1/R²)
I = 67.5 • (1/25²)
I = 0.108 W/m².
For distance of 45 m:
I = k • (1/R²)
I = 67.5 • (1/45²)
I = 0.034 W/m².
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Two young people are discussing going to a party. They are fully aware that drugs and alcohol will be there, also. The two youngsters rationalize that the party will bring great happiness since it will be the first party allowed since the pandemic shutdown. Every youngster within a 40-mile radius is planning to be there. Youngster-A is starting to consider the consequences of the party and whether it will bring happiness, particularly days later. Youngster-B says that at least it will bring happiness when it happens. Youngster-A responds, yes, but what if parents find out days later? The student continues, what if underage minors are there, the police find out, and the whole thing gets reported back to the school? There could be some terrible results. Besides, Youngster-A says, think back to when you have gone to parties with drugs and alcohol; things never go well. Drugs and alcohol are always harmful at parties. What two (2) moral theories are identified as best for this situation
Answer:
Explanation:
There are several moral theories that could be applicable to this situation, but two that stand out are consequentialism and deontology.
Consequentialism is a moral theory that focuses on the consequences of actions. According to consequentialism, an action is morally right if it leads to the best overall outcome or consequences. In this case, Youngster-A seems to be considering the potential consequences of going to the party, particularly in terms of how it could affect their lives days or weeks later. Youngster-B, on the other hand, seems to be more focused on the immediate pleasure and excitement of attending the party. From a consequentialist perspective, it would be important to consider all of the potential consequences of attending the party before making a decision.
Deontology is a moral theory that emphasizes the duty or obligation to follow certain moral rules or principles. According to deontology, some actions are inherently right or wrong, regardless of their consequences. In this case, there are several moral rules and principles that could be relevant, such as the duty to obey the law, the obligation to avoid harm to oneself or others, and the responsibility to act in a way that aligns with one's values and beliefs. From a deontological perspective, it would be important to consider how attending the party aligns with these moral rules and principles.
Both consequentialism and deontology offer different ways of approaching moral decision-making in this situation. Ultimately, the decision of whether or not to attend the party will depend on the individual's values, beliefs, and priorities, as well as the specific circumstances of the party and potential consequences.
Your go-cart breaks down right before the end of a race, so you have to push it over the finish line. The go-cart has a mass of 85 kg.
a. What is the weight of your go-cart?
(I need to show my work)
Explanation:
Weight =mass x force of gravity
Weight = 85kg x 9.8m/s²
Weight = 833N
a conducting wire with conductance of 0.9s what is the conductivity of another wire of the same material and of the same length but the radius of its cross section is 3 times the radius of the cross section of the first wire
originlal wire :
conductance = 0.9
conductivity = n
length = l
area = A
New wire -
conductance = ?
conductivity = n
length = l
area = 3A
Conductance of original wire :
C = (nA)/l = 0.9 s
new conductance :
C' = (n3A)/l = 3× (nA)/l = 3 × 0.9 = 2.7 s
A test rocket starting from rest at point A is launched by accelerating it along a 200.0 m incline at 3.50 m/s2 (Figure 1). The incline rises at 35.0∘ above the horizontal, and at the instant the rocket leaves it, the engines turn off and the rocket is subject to gravity only (ignore air resistance). a)Find the maximum height above the ground that the rocket reaches. b)Find the rocket's greatest horizontal range beyond point A
Answer: a) 123.1m
b) 279.1m
Explanation:
The maximum height reached by the rocket is 0 meters. The rocket's greatest horizontal range beyond point A is also 0 meters.
Explanation:To find the maximum height, we can use the kinematic equation for vertical motion. The rocket starts from rest, so its initial vertical velocity is 0 m/s. We can use the equation:
[tex]h = vi^2 / (2 * g)[/tex]
where h is the maximum height, vi is the initial vertical velocity, and g is the acceleration due to gravity. We can calculate h by substituting the given values:
h = (0 m/s)2 / (2 * 9.8 m/s2)
= 0 m
Therefore, the maximum height reached by the rocket is 0 meters above the ground.
To find the rocket's greatest horizontal range, we can use the horizontal motion equations. Since the rocket is subject to gravity only after leaving the incline, we need to find the time it takes for the rocket to reach the highest point on the incline. We can use the equation:
t = vf / a
where t is the time, vf is the final velocity, and a is the acceleration. We can calculate t by substituting the given values:
t = 3.50/ 3.50
= 1s
Now we can use the equation for horizontal motion to find the horizontal range:
[tex]R = v0 * t[/tex]
where R is the horizontal range and v0 is the initial horizontal velocity. Since the rocket starts from rest, v0 = 0 m/s. Therefore, the rocket's greatest horizontal range beyond point A is 0 meters.
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A horizontal ruler from 0 to 100. An image of an upward pointing arrow in a frame, at 0. A double convex lens at 100. X marked in the center of the ruler.
What does the “X” on the horizontal line represent?
Draw the ray diagram in your notes, showing only the principal rays. Explain why you don’t need to draw more rays.
Where will the image appear? On the left or on the right of the lens? At which mark on the ruler?
How will the image look? Upright or inverted? Same size, larger, or smaller?
Horizontal lines are referred to as being parallel to a x-axis in coordinate geometry. A line is referred to as horizontal if two points on the line share the same y-coordinate points.
What does a graph's x-axis / horizontal line represent?The intersection of the vertical and horizontal (X axis) real number lines is shown on the axis graph (Y axis). The Y axis is known as the dependent variable of the data set, whereas the X axis is typically used to refer to the independent variable of the data set.
In the line graph, how does X show up?The horizontal x-axis and vertical y-axis are the two axes of a line graph (vertical). A different type of data is indicated at each of the points where the axes connect, and (0,0). The x-axis is referred to as an independent axis since the numbers it represents are not reliant on any of the variables being assessed.
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- An object in equilibrium has three forces exerted on it. A 33-N force act at 90° from the x-axis and a 46-N force act at 60°. What are the magnitude and direction of the third force
Answer:
Explanation:
The 33 N force is at a 90 degree angle, whereas the 44 N force is at a 60 degree angle with the x-axis.
Assume that the third force makes a theta-angle contact with the x-axis.
Since the object is in balance, the total force acting on it will equal zero.
Find the accumulation of the x-axis forces.
[tex]\begin{aligned} 33\cos 90{}^\circ +44\cos 60{}^\circ +{{F}_{3}}\cos \theta &=0 \\ 0+22\text{ N}+{{F}_{3}}\cos \theta &=0 \\ {{F}_{3}}\cos \theta &=-22\text{ N }......\text{ }\left( 1 \right) \end{aligned}[/tex]
Find accumulation of the y-axis forces.
[tex]\begin{aligned} 33\sin 90{}^\circ +44\sin 60{}^\circ +{{F}_{3}}\sin \theta &=0 \\ 33\text{ N}+38.11\text{ N}+{{F}_{3}}\sin \theta &=0 \\ {{F}_{3}}\sin \theta &=-71.11\text{ N }......\text{ }\left( 2 \right) \end{aligned}[/tex]
Identify the magnitude.
[tex]\begin{aligned} F&=\sqrt{{{\left( {{F}_{3}}\cos \theta \right)}^{2}}+{{\left( {{F}_{3}}\sin \theta \right)}^{2}}} \\ &=\sqrt{{{\left( -22\text{ N} \right)}^{2}}+{{\left( -71.11\text{ N} \right)}^{2}}} \\ &=74.43\text{ N} \end{aligned}[/tex]
Identify the direction.
[tex]\begin{aligned} \tan \theta &=\left( \frac{{{F}_{3}}\sin \theta }{{{F}_{3}}\cos \theta } \right) \\ \theta &={{\tan }^{-1}}\left( \frac{{{F}_{3}}\sin \theta }{{{F}_{3}}\cos \theta } \right) \\ \theta &={{\tan }^{-1}}\left( \frac{-71.11\text{ N}}{-22\text{ N}} \right) \\ \theta &=72.8{}^\circ \end{aligned}[/tex]