Explanation:
Potential Energy = m g h
400 J = 8 kg * 9.81 m/s^2 * h
400 / (8 *9.81) = h = 5.1 m
When a car is stopped, facing upwards on a hill, which way does friction act?
When a car is stopped, facing upwards on a hill, the friction acts in the opposite direction to the motion that the car would naturally take if it were not stopped.
In this case, the car would roll backwards down the hill due to the force of gravity. The friction between the tires and the road surface acts in the opposite direction to this motion, providing a force that opposes the car's tendency to roll backwards. Therefore, the friction acts in the forward direction, up the hill, to prevent the car from rolling backwards.
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2. Billiard ball A moves with speed VA = 3 ft/s at an angle 0 = 70°. It collides with ball B of equal mass which is initially at rest and moves horizontally after impact. The coefficient of restitution between the two balls is 0.9. Determine the velocity of ball B after impact. Y 6 in. B 10 in. r
The velocity of ball B after impact would be2.7147 i + 2.6987 j ft/s
Conservation of momentumTo solve this problem, we can use the conservation of momentum and energy.
First, let's find the momentum of ball A before the collision. The momentum is given by:
p = mv
The mass of each ball is the same, so we can write:
p_A = mV_A
where V_A is the velocity vector of ball A.
We can break V_A into its x and y components as follows:
V_Ax = V_A cos(θ)
V_Ay = V_A sin(θ)
where θ is the angle between the velocity vector and the x-axis.
Substituting in the given values, we get:
V_Ax = 3 cos(70°) = 0.9063 ft/s
V_Ay = 3 sin(70°) = 2.8830 ft/s
So, the momentum of ball A before the collision is:
p_A = mV_A = m (V_Ax i + V_Ay j) = m (0.9063 i + 2.8830 j) lb·ft/s
Next, we need to find the velocity of ball A after the collision. We can use conservation of momentum and energy to do this.
p_A + p_B = p_A' + p_B'
where p_B is the momentum of ball B before the collision, and p_A', p_B' are their respective momenta after the collision.
Since ball B is initially at rest, its momentum before the collision is zero:
p_B = 0
Conservation of energy tells us that the total kinetic energy of the system before the collision is equal to the total kinetic energy of the system after the collision:
1/2 m V_A^2 = 1/2 m V_A'^2 + 1/2 m V_B'^2
where V_A' and V_B' are the velocities of the balls after the collision.
We can use the coefficient of restitution (e) to relate the velocities of the balls before and after the collision:
e = (V_B' - V_A') / (V_A - V_B)
Substituting in the given values, we get:
e = (V_B' - V_A') / (3 - 0)
Solving for V_B', we get:
V_B' = e (V_A - V_B) + V_A'
Substituting in the known values, we get:
V_A' = (0.9063 i + 2.8830 j) ft/s
e = 0.9
Solving for V_B', we get:
V_B' = e (V_A - V_B) + V_A'
= 0.9 (3 i + 0 j) + (0.9063 i + 2.8830 j)
= 2.7147 i + 2.6987 j ft/s
So, the velocity of ball B after the collision is:
V_B' = 2.7147 i + 2.6987 j ft/s
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The magnetic field through a circular loop of radius r = 15 cm varies with time, as shown. The direction of the field is perpendicular to the loop. What is the magnitude of the induced emf, ε1, in millivolts, at time t1 = 1.0ms?
B=0I2R(at centre of loop), B = 0 I 2 R(at centre of loop), where R is the loop's radius, gives the magnetic field strength at the loop's centre.
How does the magnetic field change as the loop's radius increases?The magnetic field strength diminishes with increasing radius. Radius of the loop has an inverse relationship with magnetic field intensity.
With its plane normal to an external field of magnitude 5.0102T, a circular coil with a radius of 10 cm and 16 turns that is carrying a current of 0.75 A is at rest. The coil is unrestricted in its ability to rotate around an axis in a plane perpendicular to the field direction.
Hence, the integral around any circle's diameter that is centred on a wire.
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Explain the relationship between binary stars, the Chadrasekhar limit, and Type 1a supernovae
if in a binary star system a white dwarf exceeds this limit through mass transfer, it will explode and a Type Ia Supernova will be the end result.
Need help please and thanks In advance
Answer:
Human development option D
A 70.0 kg
ice hockey goalie, originally at rest, has a 0.110 kg
hockey puck slapped at him at a velocity of 43.5 m/s.
Suppose the goalie and the puck have an elastic collision, and the puck is reflected back in the direction from which it came. What would the final velocities goalie
and puck
of the goalie and the puck, respectively, be in this case? Assume that the collision is completely elastic.
The final velocity of the goalie is -0.613 m/s (indicating that he moves backwards), and the final velocity of the puck is 30.2 m/s (indicating that it moves forwards).
What does mechanics' collision mean?When particles, collections of particles, or solid bodies move in the same direction and get close enough to each other, they collide and generate mutual force.
This issue can be resolved by applying the laws of motion and kinetic energy conservation.
momentum conservation
Before the collision:
Total momentum = 0 (since the goalie is at rest)
After the collision:
Total momentum = m_goalie * v_goalie + m_puck * v_puck
Conservation of kinetic energy:
Before the collision:
Total kinetic energy = 0
After the collision:
Total kinetic energy = (1/2) * m_goalie * v_goalie² + (1/2) * m_puck * v_puck^2
We can use these two equations to solve for the final velocities of the goalie and the puck.
First, let's use the conservation of momentum equation to solve for v_goalie:
0 = m_goalie * v_goalie + m_puck * v_puck
v_goalie = - m_puck * v_puck / m_goalie
Now, we can substitute this expression for v_goalie into the conservation of kinetic energy equation:
(1/2) * m_puck * v_puck² = (1/2) * m_goalie * (- m_puck * v_puck / m_goalie)² + (1/2) * m_puck * 43.5²
Simplifying this equation and solving for v_puck, we get:
v_puck = 2 * (m_oalie / (m_goalie + m_puck)) * 43.5
v_puck = 30.2 m/s
Finally, we can substitute this value for v_puck back into the equation for v_goalie
v_goalie = - m_puck * v_puck / m_goalie
v_goalie = -0.613 m/s
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A refrigerator removes 37.0 kcal of heat from the freezer and releases 82.5 kcal through the condenser on the back. How much work was done by the compressor?
Answer:
work = (37.0 kcal) - (82.5 kcal) = -45.5 kcal
Explanation:
The first law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. In this case, the refrigerator is removing heat from the freezer and releasing it through the condenser on the back. Therefore, the work done by the compressor is:
work = heat removed - heat released
The negative sign indicates that work was done on the refrigerator by an external agent (e.g., an electric motor) to remove heat from the freezer and release it through the condenser.
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Identify the constellation, label any of the names stars or asterisms that were covered in class
Answer: I believe this constellation is cygnus, the swan. The brightest star is Deneb at the tail. The bright double star at the head of the swan is Albireo.
Explanation:
A straight piece of conducting wire with mass M and length L is placed on a frictionless incline tilted at an angle θ from the horizontal (the figure (Figure 1)). There is a uniform, vertical magnetic field B⃗ at all points (produced by an arrangement of magnets not shown in the figure). To keep the wire from sliding down the incline, a voltage source is attached to the ends of the wire. When just the right amount of current flows through the wire, the wire remains at rest. Using the given variables and appropriate constants, determine the magnitude of the current in the wire that will cause the wire to remain at rest.
I = M g sin / (L B), where M, L,, and B are the provided variables and g is the acceleration due to gravity (9.81 m/s2), determines the amount of current in the wire that will keep it at rest.
Is the strength of the magnetic field surrounding a wire inversely proportional to the current flowing through it?The magnitude of the magnetic field is inversely related to the perpendicular distance from the wire and proportionate to the current.
We need to use the equation for the force on a current-carrying wire in a magnetic field to estimate the size of the current in the wire that will keep it at rest:
F = I L × B
The weight of the wire's component moving down the incline must be balanced by the force acting on the wire:
F = M g sinθ
For the wire to remain at rest, these two forces must be equal:
I L × B = M g sinθ
Solving for I, we get:
I = M g sinθ / (L × B)
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Find the heat energy per unit time per unit area (power/area, or 'flux') given off by a body having an emissivity € = 0.500 and temperature of 20.0° C. [Convert to Kelvin temperature.]
Therefore, the heat energy per unit time per unit area, or flux, emitted by the body is approximately 241.7 W/m².
What is the thermal energy equation?The heat transfer equation can be written as Q = m c T, where Q denotes the amount of heat transferred, m denotes mass, c denotes specific heat, and T denotes the temperature differential. Heat transfer is the process by which heat is transferred from a hot object to a cold object.
The Stefan-Boltzmann Law can be used to determine the thermal energy per unit time per unit area, or flux, released by a body:
F = σ * € * T⁴
To convert 20.0°C to Kelvin, we add 273.15 K to get 293.15 K.
Now we can plug in the values:
F = 5.67 x 10⁻⁸ W/m² K⁴ * 0.500 * (293.15 K)⁴
F ≈ 241.7 W/m²
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What is the mechanical advantage of the wedge?
400
0.25
4
50
A ball is dropped from the height of 10m.at the same time, another ball is thrown vertically upward at an initial speed of 10m/s. How high above the ground will thr two balls collide?
Answer: H=5.1m
Explanation:
Given:
Ball 1 height= 10m
Ball 2 initial velocity=10m/s
use the kinematic equation:
S=(vi)t+12at2
I choose my sign convention to be up=positive, down=negative, so a=−9.81ms2 (a is taken as the value of gravity)
Ball 1 dropped from 10m :
−(10−H)=0+12(−9.81)t2
Note that (10-S) is negative because that displacement is *below* the starting point.
12(9.81)t2=10−H
——- equation (1)
Ball 2 thrown upward at 10 m/s :
H=(10)t+12(−9.81)t2
or
12(9.81)t2=10t−H
——- equation (2)
equation (1) minus equation (2):
0=(10−H)−(10t−H)
t=1 equation (1):
12(9.81)12=10−H
H=5.1m
Given Information:
Ball One:
[tex]\vec y_{0} = 10 \ m[/tex] (Indicating the initial position)
We also know the ball was dropped from rest. So, [tex]\vec v_{0_{1} } = 0 \ m/s[/tex].
Ball Two:
[tex]\vec v_{0} = 10 \ m/s[/tex] (Indicating the initial velocity)
We also know the ball was throw from the ground. So, [tex]\vec y_{0_{1} } = 0 \ m[/tex].
The Information we want to Find:
[tex]\vec y_{c} = ?? \ m[/tex] (Indicating the position the two projectiles collide)
Using the Following Kinematic Equation to Solve:
[tex]\Delta \vec x = \vec v_{0}t + \frac{1}{2} \vec at[/tex]
For ball one...
[tex]\Delta \vec y = \vec v_{0}t + \frac{1}{2} \vec a_{y} t[/tex]
[tex]\Longrightarrow \vec y_{c}- \vec y_{0} = \vec v_{0_{1} }t + \frac{1}{2} \vec a_{y}t[/tex]
[tex]\Longrightarrow \vec y_{c}- \vec y_{0} = (0)t + \frac{1}{2} \vec a_{y}t[/tex]
[tex]\Longrightarrow \vec y_{c}- \vec y_{0} = \frac{1}{2} \vec a_{y}t[/tex]
[tex]\Longrightarrow \vec y_{c} = \frac{1}{2} \vec a_{y}t +\vec y_{0}[/tex] => Equation 1
For ball two...
[tex]\Delta \vec y = \vec v_{0}t + \frac{1}{2} \vec a_{y}t[/tex]
[tex]\Longrightarrow \vec y_{c}- \vec y_{0_{1} } = \vec v_{0}t + \frac{1}{2} \vec a_{y}t[/tex]
[tex]\Longrightarrow \vec y_{c} = \vec v_{0}t + \frac{1}{2} \vec a_{y}t +\vec y_{0_{1} }[/tex]
[tex]\Longrightarrow \vec y_{c} = \vec v_{0}t + \frac{1}{2} \vec a_{y}t + 0[/tex]
[tex]\Longrightarrow \vec y_{c} = \vec v_{0}t + \frac{1}{2} \vec a_{y}t[/tex] => Equation 2
Set equations 1 and 2 equal to each other and solve for the time that they collide.
[tex]\left \{ {{\vec y_{c} = \frac{1}{2} \vec a_{y}t +\vec y_{0}} \atop { \vec y_{c} = \vec v_{0}t + \frac{1}{2} \vec a_{y}t } \right.[/tex]
[tex]\Longrightarrow \frac{1}{2} \vec at +\vec y_{0}= \vec v_{0}t + \frac{1}{2} \vec at[/tex]
[tex]\Longrightarrow \vec y_{0}= \vec v_{0}t[/tex]
[tex]\Longrightarrow t=\frac{\vec y_{0}}{\vec v_{0}}[/tex]
[tex]\Longrightarrow t=\frac{10}{10}[/tex]
[tex]\Longrightarrow t=1 \ s[/tex]
Thus, the balls collide at time, t=1 s. We can now use this time to plug into equation 1 or 2 to find the height at which they collide. I will use equation 1.
[tex]\Longrightarrow \vec y_{c} = \frac{1}{2} \vec a_{y}t +\vec y_{0}[/tex]
[tex]\Longrightarrow \vec y_{c} = \frac{1}{2} (-9.8)(1) +10[/tex]
[tex]\Longrightarrow \vec y_{c} = 5.1 \ m \ \therefore \ Sol.[/tex]
*Note* [tex]\vec a_{y}[/tex] is the acceleration of gravity ([tex]-9.8 \ m/s^2 \ or \ -32 \ ft/s^2[/tex])
Final Answer: The balls collide at the height 5.1 m.
What happens to artificial satellites sent into the Earth’s orbit?
Artificial satellites are launched into Earth's orbit for various purposes, including communication, navigation, weather monitoring, scientific research, and military surveillance.
What exactly happens when artificial satellites are sent into the Earth’s orbit?Artificial satellites sent into Earth's orbit the Earth, typically at an altitude between 200 and 22,000 miles, depending on its intended purpose. Satellites in low Earth orbit (LEO) travel at a speed of about 17,500 miles per hour, completing one orbit in about 90 minutes, while satellites in geostationary orbit (GEO) remain stationary above the equator at an altitude of about 22,236 miles.
Satellites can remain in orbit for many years, but eventually, they can fall out of orbit due to atmospheric drag or collisions with space debris. When a satellite falls out of orbit, it typically burns up in Earth's atmosphere, although larger satellites may leave debris that can pose a risk to other spacecraft.
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A block is attached to two identical ideal springs. It is initially held so the whole setup is horizontal and the springs are not extended. The unextended length of each spring is 0.9 m with a spring constant of 473 N/m. The block is slowly lowered until the mass is in static equillibrium. The springs now make an angle θ = 20o with the horizontal as shown. What is the mass of the block? You may assume the block is a point mass (no volume).
Answer:
Approximately [tex]1.9\; {\rm kg}[/tex] (assuming that [tex]g = 9.81\; {\rm N\cdot kg^{-1}}[/tex].)
Explanation:
Let [tex]L = 0.9\; {\rm m}[/tex] denote the unextended length of each spring.
The length of each spring is now [tex](L / \cos(\theta))[/tex]. The displacement of each spring would be [tex]x = L - (L / \cos(\theta)) = (1 - (1 / \cos(\theta)))\, L[/tex].
The tension in each spring would be [tex]T = k\, x[/tex], where [tex]k[/tex] is the spring constant.
Decompose the tension that each spring exerts on the block into two components:
Horizontal: [tex]T\, \cos(\theta)[/tex].Vertical: [tex]T\, \sin(\theta)[/tex].The two horizontal components balance each other since they are equal in magnitude. The two vertical components add on to each other to exert a total upward force of [tex]2\, T\, \sin(\theta)[/tex] on the block.
Since the block is in equilibrium, the resultant force on the block will be [tex]0[/tex]. The sum of these two (upward) vertical components of tension should balance the (downward) weight of the block:
[tex]2\, T\, \sin(\theta) = m\, g[/tex], where [tex]m[/tex] is the mass of the block.
Rearrange this equation to find the mass of the block:
[tex]\begin{aligned} m &= \frac{2\, T\, \sin(\theta)}{g} \\ &= \frac{2\, k\, x\, \sin(\theta)}{g} \\ &= \frac{2\, k\, L\, (1 - (1 / \cos(\theta))\, \sin(\theta)}{g} \\ &= \frac{2\, (473)\, (0.9)\, (1 - (1 / \cos(20^{\circ})))\, \sin(20^{\circ})}{(9.81)}\; {\rm kg} \\ &\approx 1.9\; {\rm kg}\end{aligned}[/tex].
[tex]\blue{\huge {\mathrm{MASS \; OF \; THE \; BLOCK}}}[/tex]
[tex]\\[/tex]
[tex]{===========================================}[/tex]
[tex]{\underline{\huge \mathbb{Q} {\large \mathrm {UESTION : }}}}[/tex]
A block is attached to two identical ideal springs. It is initially held so the whole setup is horizontal and the springs are not extended. The unextended length of each spring is 0.9 m with a spring constant of 473 N/m. The block is slowly lowered until the mass is in static equillibrium. The springs now make an angle θ = 20° with the horizontal as shown. What is the mass of the block? You may assume the block is a point mass (no volume).[tex]{===========================================}[/tex]
[tex] {\underline{\huge \mathbb{A} {\large \mathrm {NSWER : }}}} [/tex]
The mass of the block is 1.9 kg.[tex]{===========================================}[/tex]
[tex]{\underline{\huge \mathbb{S} {\large \mathrm {OLUTION : }}}}[/tex]
To solve for the mass of the block, we can use the forces acting on the block at equilibrium. We know that the force of gravity pulling down on the block is equal to the force of the springs pulling up.
The force of each spring can be found using Hooke's Law:
[tex]\sf F = kx[/tex]where:
F is the force,k is the spring constant, andx is the displacement from the equilibrium position.In this case, the displacement is equal to the extension of the spring, which is given by:
[tex]\sf x = L(1-\cos\theta)[/tex]where:
L is the unextended length of the spring, andθ is the angle it makes with the horizontal.So the force of each spring is:
[tex]\sf F_{spring} = kx = kL(1-\cos\theta)[/tex]At equilibrium, the forces in the vertical direction must balance, so we have:
[tex]\sf 2F_{spring} = mg[/tex]where
m is the mass of the block andg is the acceleration due to gravity.Substituting in the expression for [tex]\sf F_{spring}[/tex] and simplifying, we get:
[tex]\sf\qquad\implies 2kL(1-\cos\theta) = mg[/tex]
Solving for m, we obtain:
[tex]\sf\qquad\implies m = \dfrac{2kL(1-\cos\theta)}{g}[/tex]
Plugging in the given values, we get: [tex]\\\begin{aligned}\sf m&=\sf \dfrac{2(473\: N/m)(0.9\: m)[1-\cos(20^{\circ})][\sin(20^{\circ})]}{(9.81 m/s^2)}\\&=\boxed{\bold{\:1.9\: kg\:}}\end{aligned}[/tex]
Therefore, the mass of the block is 1.9 kg.
[tex]{===========================================}[/tex]
[tex]- \large\sf\copyright \: \large\tt{AriesLaveau}\large\qquad\qquad\qquad\qquad\qquad\qquad\tt 04/02/2023[/tex]
How many nodes are on this standing wave? *
The number of nodes present in this standing wave is 6
How many nodes are on this standing wave?A standing wave is a type of wave that remains in a constant position and does not propagate through a medium. Instead, it oscillates in place between two fixed points, creating a pattern of constructive and destructive interference.
In a standing wave, nodes are the points along the medium that remain stationary, with no displacement or movement of the medium. These nodes occur at locations where the displacement of the medium is always zero, meaning that the amplitude of the wave is zero at those points.
In other words, nodes are the points of minimum energy in the standing wave. They are the points where the crest of the wave meets the trough of the wave, resulting in the cancellation of the wave's amplitude. The distance between two adjacent nodes is half of the wavelength of the standing wave.
In this standing wave, the number of nodes present is 6
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A spring is holding a door open. What type of energy is in the compressed spring?
Group of answer choices
Chemical energy
Spring potential energy
Kinetic energy
Elastic potential energy
250g of glass at 60°C is placed in a container with 700g of alcohol. The specific heat capacities of glass and alcohol are 8401/kg °C and 2 500J/kg°C respectively. At thermal equilibrium, the final temperature of the mixture is 20°C. Determine the initial temperature of the alcohol, assuming that no heat is lost while both substances are mixed.
To solve the problem, we can use the principle of conservation of heat, which states that the heat lost by the glass is gained by the alcohol.Let T be the initial temperature of the alcohol in degrees Celsius.The heat lost by the glass is given by:Q1 = mcΔT1where m is the mass of the glass, c is the specific heat capacity of glass, and ΔT1 is the change in temperature of the glass.Substituting the given values, we get:Q1 = (250 g) × (840 J/kg°C) × (60°C - T)The heat gained by the alcohol is given by:Q2 = mcΔT2where m is the mass of the alcohol, c is the specific heat capacity of alcohol, and ΔT2 is the change in temperature of the alcohol.Substituting the given values, we get:Q2 = (700 g) × (2500 J/kg°C) × (20°C - T)Since the total heat lost by the glass is equal to the total heat gained by the alcohol, we can set Q1 equal to Q2 and solve for T:Q1 = Q2(250 g) × (840 J/kg°C) × (60°C - T) = (700 g) × (2500 J/kg°C) × (20°C - T)Simplifying and solving for T, we get:T = 32.5°CTherefore, the initial temperature of the alcohol was 32.5°C.
Explanation:
There are 30 data in the following distribution: 35, 50, 53, 57, 59, 63, 63, 67, 67, 67, 72, 72, 72,72, 72, 78, 78, 83, 84, 84, 85, 85, 86, 90, 90, 90, 95, 95, 100, 100. Calculate the min, max, mean, standard deviation for the sample (SX), mode, and median for the distribution. Hint: STAT-->Edit to enter data in L1; then STAT-->CALC statistics for 1 variable. Round to two decimals.
Answer: Min: 35
Max: 100
Mean: 73.53
Standard deviation (SX): 20.17
Mode: 72
Median: 72.5
Explanation:
4. (Hard) A coal power station generates electricity by first burning coal to heat water into steam, then passing that steam through a turbine to make it spin. The turbine drives a generator which gives out electrical energy A. Read the paragraph above. Identify the energy stores and transfers for the coal power station. B. The coal has a chemical potential energy store of 20,000J. 90% of the energy is transferred into the heat store of the water. Then, 30% of the energy is dissipated in the turbine. Of the remaining energy, 85% is transferred as electricity by the generator. Calculate how much electrical energy is produced by the generator. C. If the power station is provided with 10,000J of energy per second from the coal, calculate the power output of the power station. D. Give three specific improvements that could be made to the power station to make it more efficient
A coal power station generates electricity by first burning coal to heat water into steam, then passing that steam through a turbine to make it spin.
A. Energy stores:
Chemical potential energy in the coal
Thermal energy in the water
Kinetic energy in the steam
Kinetic energy in the turbine
Electrical energy in the generator
Energy transfers:
Chemical potential energy in the coal to thermal energy in the water (by burning the coal)
Thermal energy in the water to kinetic energy in the steam (by boiling the water)
Kinetic energy in the steam to kinetic energy in the turbine (by passing through the turbine)
Kinetic energy in the turbine to electrical energy in the generator (by driving the generator)
B. Energy transferred into heat store of water: 20,000 J x 0.9 = 18,000 J
Energy dissipated in the turbine: 18,000 J x 0.3 = 5,400 J
Remaining energy after turbine: 18,000 J - 5,400 J = 12,600 J
Energy transferred as electrical energy: 12,600 J x 0.85 = 10,710 J
C. Power output = energy input per second = 10,000 J/s
So, the power output of the power station is 10,710 J/s (since 85% of the remaining energy is transferred as electrical energy).
D. Three specific improvements that could be made to the power station to make it more efficient are:
Implementing better combustion techniques to increase the efficiency of burning coal.
Using better insulation materials to minimize heat loss in the power station.
Using more efficient turbines and generators to convert kinetic energy to electrical energy.
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The diagram shows a partial model for respiration in the human body.
Cellular respiration
?
WATER
water loss through breathing, sweating, and waste removal
?
Food
Energy
a. identify the two missing parts of this model
The two parts of the model that are missing are as shown in the diagram:
Oxygen (O2): Breathing typically provides oxygen, which is necessary for cellular respiration.Carbon dioxide (CO2) is a byproduct of cellular respiration that is eliminated from the body through sweating, breathing, and other methods of waste removal.How important Oxygen is for cellular respiration?Oxygen is essential for cellular respiration, which is the process by which cells convert food molecules (such as glucose) into energy in the form of ATP (adenosine triphosphate). Cellular respiration occurs in the mitochondria of cells and is a complex series of metabolic reactions that requires oxygen as the final electron acceptor in the electron transport chain.
Without oxygen, cellular respiration cannot proceed beyond glycolysis, which is the first step in the process. Glycolysis breaks down glucose into two molecules of pyruvate and produces a small amount of ATP. However, this process is not very efficient, and it cannot sustain cellular activity for very long.
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a 13.1-g object (object 1) moving to the right with a speed of 20.1 cm/s collides head-on collision with a 32.4-g object (object 2) initially at rest. the collision is perfectly elastic. find the velocity (in cm/s) of the 32.4-g object (object 2) after the collision.
The collision is perfectly elastic the velocity of object 2 after the collision is 44.0 cm/s to the right.
Substituting the given values, we get:
KE = (1/2)(0.0131 kg)(0.201 m/s)^2 + (1/2)(0.0324 kg)(0 m/s)^2
KE = 0.000132 J
Since the collision is perfectly elastic, the total kinetic energy of the system after the collision is also equal to KE. Therefore:
KE' = [tex](1/2)m_1v_1'^2 + (1/2)m_2v_2'^2[/tex]
where KE' is the kinetic energy of the system after the collision.
We can arrange this equation to solve for v2':
v2' = [tex]\sqrt{((2/m2)(KE' - (1/2)m1v1'^2))}[/tex]
Substituting the given values , we get:
v2' = [tex]\sqrt{((2/0.0324 kg)(0.000132 J - (1/2)(0.0131 kg)(20.1 m/s)^2))}\\[/tex]
v2' = 0.440 m/s
Finally, we need to convert the velocity back to cm/s:
v2' = 44.0 cm/s (rounded to three significant figures)
A collision refers to an event where two or more objects come into contact with each other, often resulting in damage or a change in the objects' trajectories. Collisions can occur in various contexts, including physics, engineering, and traffic accidents. In physics, a collision involves the transfer of momentum and energy between objects. The types of collisions include elastic collisions, where there is no loss of kinetic energy, and inelastic collisions, where some of the kinetic energy is lost.
In engineering, collisions can occur when designing structures or machines to prevent them from failing under extreme loads. For example, cars are designed with crumple zones to absorb the impact of a collision and protect the passengers.
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A railroad car of mass 2.00×104 moving at a speed of 3.00 collides and couples with another railroad car of mass 4.00×104 and moving toward it at 1.20 . What is the speed of the coupled cars after the collision?
The speed of the car after the collision is determined as 0.2 m/s.
What is the speed of the car after the collision?The speed of the car after the collision is calculated by applying the principle of conservation of linear momentum as follows;
m1u1 + m2u2 = v(m1 + m2)
where;
m1 and m2 are the masses of the carsu1 and u2 are the initial velocity of the carsv is the final velocity of the cars after the collision20000(3) - 40000(1.2) = v (20000 + 40000)
12,000 = 60,000v
v = 12,0000 / 60,000
v = 0.2 m/s
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5. How do liquids and gases differ when pressure is applied to them in a container?
A. Gasses easily compress when pressure is applied, but liquids don't.
B. Gases easily expand when pressure is applied, but liquids don't.
C. Liquids easily expand when pressure is applied, but gases don't.
D. Liquids easily compress when pressure is applied, but gases don't.
Answer:C or B
Explanation:
Gases easily compress when pressure is applied, but liquids don't. So, option A.
What is meant by intermolecular force ?The attracting and repellent forces that develop between the molecules of a substance are known as intermolecular forces.
Here,
The interactions between the individual molecules of a substance are mediated by the intermolecular forces. The majority of the physical and chemical features of matter are caused by intermolecular forces.
Compared to gases, which have relatively far-apart particles, liquids exhibit higher intermolecular forces due of the near proximity of their particles. The greater the intermolecular forces get as they get closer to one another since they are electrostatic in nature.
There is no force between particles in a gas. Particles have no restrictions on their movement. The container's walls colliding with one another provide the only forces that exist. which rely on the quantity of gas (number of collisions) and the momentum of each collision respectively. The pressure is nearly consistent throughout the entire mass of the gas because the molecules are free to move.
There is no compression of liquids. Since their volume is constant, changing pressure has no effect on it.
Hence,
Gases easily compress when pressure is applied, but liquids don't. So, option A.
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The secondary of an induction coil has 12,000 turns . If the flux linking the coil changes from 740 to 40 uWb in 180 us , how great is the induced emf ?
We can use Faraday's Law of Electromagnetic Induction to find the induced emf (electromotive force) in the secondary coil:
emf = -N(dΦ/dt)
where
emf = induced emf
N = number of turns in the coil
dΦ/dt = rate of change of magnetic flux through the coil.
In this problem,
N = 12,000 turns, and the change in magnetic flux is:
dΦ = 740 uWb - 40 uWb = 700 uWb
The time interval for this change =180 us, or 180 x 10^-6 seconds. Therefore, the rate of change of magnetic flux is:
dΦ/dt = (700 uWb) / (180 x 10^-6 s) = 3.89 mWb/s
Now we can find the induced emf:
emf = -N(dΦ/dt) = -(12,000 turns)(3.89 mWb/s) = -46.68 volts
Note that the negative sign indicates that the induced emf will produce a current that opposes the change in magnetic flux that caused it.
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what is the dimension for flow of electric current
Answer:
Explanation:
The dimensions for the flow of electric current is amperes, often denoted by "A" in scientific notation.
In terms of the base SI units, one ampere is defined as the flow of one coulomb of electric charge per second through a conductor. It can also be expressed in milliamperes or macro amperes which is equal to 1/1000 and 1/1000000 of amperes respectively.
On a distance vs. time graph, the slope of the line equals the object’s? 9._____
Answer: speed
Explanation: On a distance vs. time graph, the slop of the line equals how fast an object is going.
In a circular parallel plate capacitor radius of each plate is 5 cm and they are
separated by a distance of 2 mm. Calculate the capacitance and the energy
stored, when it is charged by connecting the battery of 200 V (permittivity of free
space: 8.854 × 10−12 −1
The formula C = (20A) / d can be used to determine the capacitance of a circular parallel plate capacitor. As a result, the capacitor can store 0.000494 J of energy.
How many inches are there in a parallel plate capacitor's plate spacing?A charged parallel plate capacitor has 5 cm between its plates, and its internal electric field is 200 Vcm. The capacitor is totally submerged by a 2 cm wide uncharged metal bar. The metal bar's length is the same as the capacitor's.
C = (2πε₀A) / d
Where,
ε₀ = Permittivity of free space = [tex]8.854 × 10^−12 F/m[/tex]
A = Area of the plate =[tex]πr^2[/tex]
d = Distance between the plates
Therefore, the area of each plate (A) = [tex]πr^2 = π(0.05)^2 = 0.00785 m^2.[/tex]
[tex]C = (2πε₀A) / d = (2π × 8.854 × 10^−12 × 0.00785) / 0.002[/tex]
[tex]= 6.21 × 10^−11 F[/tex]So, the capacitance of the circular parallel plate capacitor is [tex]6.21 × 10^-11 F.[/tex]
[tex]U = 1/2 * C * V^2[/tex]
Substituting the values, we get:
[tex]U = 1/2 * 6.21 × 10^-11 * (200)^2[/tex]
= 0.000494 J
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Help me on this question!
The other ball will go up to a height of 4.3 m after the collision.
What is the height risen by the ball after the collision?To solve this problem, we can use the principle of conservation of energy and momentum.
Let's first find the initial potential energy of the dropped ball:
E_i = mgh = 5 kg x 9.8 m/s² x 12 m = 588 J
When the ball hits the bar, the momentum is conserved:
m1v1 + m2v2 = (m1 + m2)vf
where;
m1 is the mass of the dropped ball (5 kg), v1 is its initial velocity (0), m2 is the mass of the bar (9 kg), v2 is the velocity of the bar after the collision, and vf is the final velocity of the system (the bar and the dropped ball) after the collision.We can simplify this equation by noting that the bar is initially at rest, so v2 = 0:
m1v1 + m2v2 = (m1 + m2)vf
5 kg x 0 m/s + 9 kg x 0 m/s = (5 kg + 9 kg + 5 kg)vf
vf = 0 m/s
This means that the system (the bar and the dropped ball) comes to a momentary stop just after the collision.
Now, let's find the final potential energy of the system:
E_f = (m1 + m2)gh
E_f = (5 kg + 9 kg) x 9.8 m/s² x h
E_f = 137.2 J x h
We can equate the initial and final energies:
E_i = E_f
588 J = 137.2 J x h
h = 4.3 m
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2. A pencil case is dropped vertically from a height at rest and hits the ground 0.5 seconds later. a) What vertical velocity did the pencil case hit the ground with? b) What horizontal velocity did the pencil case hit the ground with? 3. A stone is projected vertically upwards with a velocity of 12ms¹. Calculate or find: a) How long it took the stone to reach its maximum height. b) Maximum height reached by the stone. 4. A balloon is climbing vertically upwards with a constant velocity of 4.2ms¹. A sandbag is dropped from the balloon and hits the ground 3.6 seconds later. Calculate or find: a) Velocity of the sandbag as it hits the ground.
An individual dropped the sandbag from a height of 63.504 metres.
How does moment of inertia change with angular momentum?The relationship between angular momentum L and moment of inertia I and angular speed, expressed in radians per second, is shown. Moment of inertia is different from mass in that it depends on both the form and location of the axis of rotation in addition to the amount of stuff present.
vf = vi + at
vf = 0 + (9.8 m/s²)(0.5 s)
vf = 4.9 m/s
vf = vi + at
0 = 12 m/s - (9.8 m/s²)t
t = 1.22 s
Δy = vi t + 1/2 a t²
Δy = (12 m/s)(1.22 s) + 1/2 (9.8 m/s²) (1.22 s)²
Δy = 7.33 m
Δy = vi t + 1/2 a t²
Δy = vi t + 1/2 a t²
0 = (4.2 m/s)(3.6 s) + 1/2 (9.8 m/s²) (3.6 s)² + Δy
Δy = -63.504 m
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Block A has a mass of 10 kg and Block B has a mass of 70 Kg.
The Friction coefficient of Block A on the plane is 0.18, and the pulley friction is neglected. The block B is maintained at rest at a position 15 m above the ground. It is then released from rest and falls on the ground.
- Using the Conservation of Energy, what will be the velocity of Block B when it is at half of its path?
Velocity of block B when it is at half of its path is approximately 10.17 m/s.
What is Conservation of Energy?Law of conservation of energy is a fundamental principle in physics that states that energy cannot be created or destroyed, but can only be transformed from one form to another.
Potential energy of block B is : PE = mgh
m is mass of the block, g is acceleration due to gravity, and h is height of the block above the ground.
Initial potential energy of block B is: PEi = mgh = 70 kg × 9.81 m/s² × 15 m = 10290 J
When block B is at half of its path, its height above the ground is: h/2 = 15 m / 2 = 7.5 m
Final potential energy of block B at this height is: PEf = mgh/2 = 70 kg × 9.81 m/s² × 7.5 m = 5143.5 J
Change in potential energy is:
ΔPE = PEf - PEi = 5143.5 J - 10290 J = -5146.5 J
Kinetic energy of block B at half of its path is: KE = -ΔPE = 5146.5 J
Kinetic energy of block B is given by: KE = (1/2)mv²
v = √(2KE/m) = √(2ΔPE/m)
v = √(2 × 5146.5 J / 70 kg) = 10.17 m/s
Therefore, velocity of block B when it is at half of its path is approximately 10.17 m/s.
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