Answer:
A) The maximum height above the roof reached by the rock can be found using the formula:
h = (v₀²sin²θ)/(2g)
where v₀ is the initial velocity (32.0 m/s), θ is the angle of the initial velocity (29.0°), and g is the acceleration due to gravity (9.81 m/s²).
Plugging in the values, we get:
h = (32.0²sin²29.0)/(2(9.81)) = 31.1 m
Therefore, the maximum height above the roof reached by the rock is 31.1 meters.
B) The vertical component of the velocity just before the rock strikes the ground is:
vᵥ = v₀sinθ - gt
where t is the time it takes for the rock to reach the ground.
We can find t by using the formula:
h = v₀sinθt - (1/2)gt²
where h is the height of the building (14.0 m). Rearranging this formula and solving for t, we get:
t = (v₀sinθ + sqrt((v₀sinθ)² + 2gh))/g
Plugging in the values, we get:
t = (32.0sin29.0 + sqrt((32.0sin29.0)² + 2(9.81)(14.0)))/9.81 = 4.01 s
Therefore, the vertical component of the velocity just before the rock strikes the ground is:
vᵥ = 32.0sin29.0 - 9.81(4.01) = -14.3 m/s
Note that the negative sign indicates that the velocity is directed downwards.
C) The horizontal distance from the base of the building to the point where the rock strikes the ground can be found using the formula:
d = v₀cosθt
Plugging in the values, we get:
d = 32.0cos29.0(4.01) = 96.4 m
Therefore, the horizontal distance from the base of the building to the point where the rock strikes the ground is 96.4 meters.
Find I1, V1, and V2 for the circuit shown in Fig. 4
From the given circuit, the value of I1 = -4 Amp, V1 = -4 V, and the V2 = 4V. Ohm's law has three different variations that are related to voltage, current, and resistance. The relationship between the voltage across the terminals and the resistor's resistance determines the current for a constant flow of current.
What is the tenet of Ohm's law?According to Ohm's law, the voltage across a conductor determines how much current flows through it. For many materials, including metals, this is true so long as the temperature (as well as other physical parameters) stay constant.
What does "nodal analysis" mean?Any electrical network may be solved using nodal analysis, which is what it is called. the formula used to determine how much voltage is shared across circuit nodes. Since the node voltages are with respect to the ground, this method is also referred to as the node-voltage method.
Applying KCL,
I1 = -1 -3
I1 = -4 A
Applying concept of super-node,
V2 = 4V
[tex]\frac{4-V1}{2} -1-3=0[/tex]
[tex]2-\frac{V1}{2} =4[/tex]
[tex]\frac{-V1}{2} =4-2[/tex]
[tex]V1=-4V[/tex]
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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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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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Use the table to answer the question.
Wave Wavelength (meters) Frequency (hertz)
W 5 200
X 3 300
The table shows information about two waves. Based on the given information, which conclusion can be made?
O Wave W has a faster speed.
O Wave W has a greater amplitude.
O Wave X has a greater amplitude.
O Wave X has a faster speed.
We can see from the table that we have in the question that Wave W has a faster speed.
What is the relationship of wavelength and frequency?Wavelength and frequency are two fundamental properties of waves, including electromagnetic waves and sound waves. The relationship between wavelength and frequency can be described by the following equation:
c = λν
where c is the speed of light (or speed of sound), λ is the wavelength, and ν is the frequency.
According to this equation, the wavelength and frequency are inversely proportional to each other.
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When landing after a spectacular somersault, a 35.0 kg gymnast decelerates by pushing straight down on the mat. Calculate the force (in N) she must exert if her deceleration is 7.00 times the acceleration of gravity. (Enter a number.)
The force (in N) she must exert if her deceleration is 7.00 times the acceleration of gravity is 245 N
Calculation of ForceGiven data
Deceleration = 7.00 times the acceleration of gravity (9.81 m/s2)Mass = 35.0 kgForce = Mass x Acceleration
Force = 35.0 kg x (7.00 x 9.81 m/s2)
Force = 245 N
The force required for the 35.0 kg gymnast to decelerate after a somersault is 245 N. This force is calculated by multiplying the mass of the gymnast by the deceleration of 7.00 times the acceleration of gravity.
The amount of force is necessary to slow down the gymnast and reduce the impact of the landing. Without this force, the gymnast could be injured from the sudden stop.
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16. The density difference between warm, moist air and cold air causes the moist
air to rise. This is key to forming
A. lightning.
B. clouds.
C. stars.
D. snow.
Warm, moist air rises because of the disparity in density between warm, moist air and cold air. This is crucial for clouds to develop.
What does rising moist, warm air in the atmosphere form?Puffy cumulus clouds can form in the atmosphere as warm, humid air rises in an updraft. As it rises, the moisture in the air condenses into water droplets. As long as warm air rising from below persists, the cloud will expand.
Why do lightning and sound happen?The air in the lightning channel may reach temperatures of 50,000 degrees Fahrenheit, which is five times hotter than the surface of the sun. After the flash, the air swiftly cools and contracts.
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A scientist makes a model of Earth's water by drawing 100 drops of water, all the same size. How many of the 100 drops represent ocean water?
A.3
B.50
C.75
D.97
Answer:
D
Explanation:
If the water represents the oceans water then you'd would need to calculate how much of earth is water (96.5)
Sort the following items from most to least inertia.
Answer: In order of most inertia to least.
car stopped
motorcycle driving
dog laying
large hot coffee
small mouse
Explanation:
Honestly, I'm not 100% sure about this, but inertia is the want to remain the way it currently is, So the larger the object, the more inertia it should have. So I basically arranged the things by size, I think the moving and not is a red herring to throw you off. Also given that the dog laying in the middle option makes sense because a dog would be in the middle of the size options.
if 100g of iron at 100°c is dropped into 390g of water at 20°c what will the final temperature be?
which one defines force?
Answer:
a
Explanation:
a push or a pull that occurs when an object interacts with another object or field.
pls mrk me brainliest
I need guidance in this one
The clay ball that could stick to the door would be more effective when you want to shut the door from a distance be throwing a n object.
Will a rubber ball or a clay ball do better at helping you shut a door when thrown at the door?The effective way to shut the door could be through the use of the clay ball that could stick to the door and push it to the place where it could jam rather than the rubber ball that would bound back.
The best way to shut a door is to use the door handle or knob to pull the door towards you and guide it into the door frame until it clicks shut.
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At the moment when a shot putter releases a 5.00kg shot, the shot is 3.00m above the ground and traveling at 15.0m/s. It reaches a maximum height of 14.5m above the ground and then falls to the ground. If air resistance is negligible, what was the potential energy of the shot as it left the hand relative to the ground?
The potential energy of the shot as it left the hand relative to the ground would be -524 J.
Potential energy calculationAt the moment the shot putter releases the shot, the total energy of the system is:
E = KE + PE
where KE is the kinetic energy of the shot and PE is its potential energy relative to the ground. We can assume that the kinetic energy is entirely due to the motion of the shot in the horizontal direction, so we can write:
KE = (1/2)mv^2
where m is the mass of the shot, and v is its horizontal velocity. Substituting the known values, we get:
KE = (1/2)(5.00 kg)(15.0 m/s)^2 = 1125 J
At the maximum height of 14.5 m, the shot has zero kinetic energy, so all its energy is potential energy:
PE = mgh
where g is the acceleration due to gravity and h is the height above the ground. Substituting the known values, we get:
PE = (5.00 kg)(9.81 m/s^2)(14.5 m - 3.00 m) = 601 J
Therefore, the potential energy of the shot as it left the hand relative to the ground was:
PE = E - KE = 601 J - 1125 J = -524 J
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A 7g bullet is fired into a 345g block that is initially at rest at the edge of a frictionless table of height 1.9m. The bullet remains in the block and after impact the block lands 2m from the bottom of the table. Find the initial speed of the bullet. The acceleration due to gravity is 9.8m/s^2. Answer in units of m/s
Answer:
≈ 192.7 m/s
Explanation:
Let's start by finding the velocity of the block just before it hits the ground. We can use the conservation of energy:
The initial potential energy of the block is mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height of the table.The final kinetic energy of the block is (1/2)mv^2, where v is the velocity of the block just before it hits the ground.Conservation of energy tells us that these two energies are equal:
mgh = (1/2)mv^2
Solving for v, we get:
v = sqrt(2gh)
Plugging in the given values, we get:
v = sqrt(2 * 9.8 m/s^2 * 1.9 m) = 6.06 m/s
Now, let's use conservation of momentum to find the initial speed of the bullet. We know that the total momentum of the system (bullet + block) is conserved before and after the collision. Before the collision, the momentum is:
p = mb * vb
where mb is the mass of the block and vb is its initial velocity, which is 0 since it is at rest.
After the collision, the bullet and block move together with a common velocity v. The total momentum is:
p = (mb + m) * v
where m is the mass of the bullet. Since momentum is conserved:
mb * vb = (mb + m) * v
Solving for vb, we get:
vb = (mb + m) * v / mb
Plugging in the given values, we get:
vb = (345 g + 7 g) / 7 g * 6.06 m/s = 192.7 m/s
Therefore, the initial speed of the bullet was approximately 192.7 m/s.
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.
What is the acceleration of gravity from the Earth when you are one full Earth radius height above the surface of the earth?
Select the answer below in terms of the acceleration at the surface, g.
A.2g
B.g
C. g/2
D. g/4
Answer:
Explanation:The acceleration due to gravity at a distance equal to one full Earth radius above the surface of the Earth is given by:
g' = g/(1 + R/E)^2
Where g is the acceleration due to gravity at the surface of the Earth, R is the radius of the Earth, and E is the distance of the object from the center of the Earth.
Substituting R for E+R, we get:
g' = g/[1 + (E+R)/R]^2
g' = g/[1 + (1+E/R)]^2
g' = g/[1 + (1+1)]^2 (Since E/R is very small compared to 1)
g' = g/16
Therefore, the acceleration due to gravity at a distance equal to one full Earth radius above the surface of the Earth is g/16. Answer: D. g/4.
Answer:
The acceleration of gravity decreases as you move away from the surface of the Earth. The acceleration of gravity at a height of one Earth radius above the surface can be found using the formula:
g' = (R/(R+h))^2 * g
where R is the radius of the Earth, h is the height above the surface, and g is the acceleration due to gravity at the Earth's surface.
If we plug in R = 6,371 km (the radius of the Earth) and h = 6,371 km (one Earth radius above the surface), we get:
g' = ((6,371 km)/(2*6,371 km))^2 * g
g' = (1/2)^2 * g
g' = g/4
Therefore, the answer is D. g/4.
Compare the patterns of iron filings to the spiral arms in Interacting galaxies.
The primary locations for the birth of new stars are in galaxies' spiral arms. The proportion of the galaxy that can participate in star formation increases as more gas and dust become available.
What distinguishes the numerous varieties of spiral galaxies?We refer to some spiral galaxies as "barred spirals" because the centre bulge seems extended, like a bar. The spiral arms of the galaxy seem to emerge from the ends of the bar in barred spirals. Elliptical galaxies are round or oval in shape, as their name implies, and have a rather uniform distribution of stars.
The four spiral arms are what?The Norma and Cygnus arm, Sagittarius, Scutum-Crux, and Perseus arms are the four principal spiral arms of the Milky Way.
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A machine has velocity ratio 6 and is 80% efficient. what effort would be needed to lift a load of 300N with the aid of this machine?
The effort needed to lift the load of 300 N, given that the velocity ratio is 6 and the machine has an efficiency of 80% is 62.5 N
How do i determine the effort needed to lift the load?First, we shall determine the mechanical advantage of the machine. Details below:
Velocity ratio (VR) = 6Efficiency = 80%Mechanical advantage (MA) = ?Efficiency = MA / VR
80% = MA / 6
Cross multiply
MA = 80% × 6
MA = 4.8
Finally, we shall determine the effort of the machine. Details below:
Mechanical advantage (MA) = 4.8Load (L) = 300 NEffort (E) = ?Mechanical advantage (MA) = Load (L) / Effort (E)
4.8 = 300 / Effort
Cross multiply
4.8 × Effort = 300
Divide both sides by 4.8
Efoort = 300 / 4.8
Effort = 62.5 N
Thus, we can conclude that the effort of the machine is 62.5 N
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Two identical billiard balls are rolling toward each other at the same speed. What will be true after they collide head–on?
They will both stop rolling immediately after the collision.
One ball will stop rolling as the other bounces backward at a slower speed.
They will both bounce back at a faster speed after the collision.
They will both bounce back at the same speed they had before the collision.
Answer:
After the head-on collision, both identical billiard balls will bounce back at the same speed they had before the collision. This is because of the conservation of momentum, which states that the total momentum of a closed system remains constant if no external forces act on it. In this case, the two billiard balls are a closed system, and their total momentum before the collision is equal and opposite to their total momentum after the collision. Therefore, they will both bounce back with the same speed they had before the collision.
Explanation:
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 car is moving along a road at 13.0 m/s with an engine that exerts a force of 1,775.0 N on the car to balance the drag and friction so that the car maintains a constant speed. What is the power output of the engine?
Answer:
The power output of the engine can be calculated using the formula:
Power = Force x Velocity
where force is the net force acting on the car, and velocity is the speed of the car. In this case, the net force is equal to the force exerted by the engine, which is 1,775.0 N, since the car is moving at a constant speed and there is no acceleration. The velocity of the car is 13.0 m/s. Thus, the power output of the engine can be calculated as:
Power = 1,775.0 N x 13.0 m/s = 23,075 W
Therefore, the power output of the engine is 23,075 watts.
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:
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 12.000 milligram particle is sliding across a friction-less one-dimensional path at 55.000 m/s and collides with a 68.000 milligram particle moving at -48.000 m/s in a perfectly inelastic collision. What are the velocities of the particles after the collision?
answer with correct units
Answer:
-3525.000 m/s
Explanation:
In a perfectly inelastic collision, the two particles stick together and move with a common velocity after the collision. We can use the conservation of momentum to solve for this common velocity.
The initial momentum of the system is:
p_initial = m1 * v1 + m2 * v2
= (12.000 mg)(55.000 m/s) + (68.000 mg)(-48.000 m/s)
= -282.000 kg·m/s
Here, we convert the masses to kilograms to match the units of velocity.
Since the particles stick together after the collision, their masses add up:
m_final = m1 + m2
= 12.000 mg + 68.000 mg
= 80.000 mg
= 0.080 g
Now, we can use the conservation of momentum to find the final velocity:
p_final = m_final * v_final
where p_final = p_initial and m_final = 0.080 g.
Therefore:
v_final = p_final / m_final
= -282.000 kg·m/s / 0.080 g
= -3525.000 m/s
The wave in the liquid travels towards the surface at an angle. Fig 9.2 shows the centres of the compressions of the sound wave in liquid. Some compressions shown have reached the liquid-air boundary. The parts of these compressions in the air are not shown on Fig 9.2 These waves are also reflected at the boundary. Draw on the diagram the reflected wavefronts.
The reflected sound wavefronts at the given boundary are waves that have bounced off a surface and changed direction.
The reflected sound wavefront is shown in the attachment.
What are reflected wavefronts?A reflected wavefront is a wavefront that has bounced off a surface and changed direction. When a wave, such as a light wave or sound wave, encounters a surface, some of the wave energy is reflected back in the opposite direction.
An example of reflected sound wavefronts in water can be seen in underwater sonar imaging.
In sonar imaging, a sound wave is emitted from a source and travels through the water. When the sound wave encounters an object, some of the wave energy is reflected back toward the source.
Reflected wavefronts play an important role in many areas of science and engineering, such as optics, acoustics, and electromagnetism. They are used to model the behavior of waves in complex systems and to design and optimize devices such as mirrors, lenses, and antennas.
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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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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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A rock group is playing in a bar. Sound
emerging from the door spreads uniformly in
all directions. The intensity level of the music
is 32.2 dB at a distance of 4.99 m from the
door.
At what distance is the music just barely
audible to a person with a normal threshold
of hearing? Disregard absorption.
Answer in units of m.
Answer:
5292.64 m
Explanation:
004 (part 1 of 2) 10.0 points
In a 95 s interval, 216 hailstones strike a glass
2
window of area 1.156 m² at an angle 67 to the
window surface. Each hailstone has a mass of
5 g and speed of 13.5 m/s.
If the collisions are elastic, find the average
force on the window.
Answer in units of N.
005 (part 2 of 2) 10.0 points
Find the pressure on the window.
Answer in units of N/m².
Average force F = 97.52 N acting on window surface.
Pressure on the window: P = 84.39 N/m²
Explain about the elastic collision?An elastic collision is one in which the system does not experience a net loss of kinetic energy as a result of the collision. In elastic collisions, momentum as well as kinetic energy are both conserved. An example of an elastic collision is when two balls collide at a pool table.Average force F:
F = mg sinФ
Put the values:
F = 0.05*9.81* sin 67°
F = 0.45 N
Number of hailstones = 216
Average force F = 0.45* 216 = 97.52 N
Pressure on the window:
Pressure = Average force / area
P = 97.52 / 1.156
P = 84.39 N/m²
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A particle executes simple harmonic motion with an amplitude of 3.00 cm. At what position does its speed equal half of is maximum speed?
Answer:
2.34
Explanation:
To find the position where the speed is half of its maximum speed, we can set v = v_max/2 and solve for x:
v_max/2 = ω√(A^2 - x^2)
Substituting the given values, we have:
(ωA)/2 = ω√(A^2 - x^2)
Simplifying and rearranging:
A^2 - x^2 = (A/2)^2
x^2 = A^2 - (A/2)^2
x^2 = (3.00 cm)^2 - (1.50 cm)^2
x = √(6.75 cm^2 - 2.25 cm^2)
x = √5.50 cm^2
x ≈ 2.34 cm
Therefore, the position where the speed of the particle is half of its maximum speed is approximately 2.34 cm from the equilibrium position.
Hans figures out one math problem easily and then applies that solution to the rest of the problems on the page; however, he gets the rest of the answers wrong. Hans has encountered a common problem with:
Answer:
Explanation:
overgeneralization or applying a single solution to a variety of problems without considering their unique characteristics. This is a common issue in math and other problem-solving situations where individuals may rely too heavily on past successful solutions without adjusting for the specific details of each new problem.