Neither technician A nor B. No technician is correct.
What is waste spark Ignition?Two spark plugs are connected to the ends of one coil in a waste spark ignition system (for a two-cylinder engine) ( for 4 cylinder engine two coils are used, each driving two spark plugs ).
As a result, both spark plugs will light up simultaneously. The first (A) cylinder's compression stroke will ignite one spark plug, while the opposing cylinder's exhaust stroke will fire the second ( wasted ). Cylinder "A" will fire during the exhaust stroke when it is time for cylinder "B" to fire during compression.
Due of the opposing current flow, both spark plugs burn simultaneously from the ends of just one coil. The spark "flows" in this manner from the center to the exterior.
Therefore, Neither technician A nor B. No technician is correct.
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Final answer:
Both Technician A and Technician B are correct; the waste spark ignition system fires two spark plugs at the same time and uses ignition coils connected to companion cylinders. So the correct option is c.
Explanation:
The subject of this question is the waste spark ignition system which is used in internal combustion engines. When evaluating the statements made by Technician A and Technician B, we find that Technician A says that a waste spark ignition system fires two spark plugs at the same time. This is correct as in a waste spark system, each ignition coil fires two plugs simultaneously - one on the compression stroke and the other on the exhaust stroke (which is the 'waste' spark). Technician B says that the system uses ignition coils connected to companion cylinders. This is also correct, as the coils are connected to two cylinders that are at the top dead center at the same time, one on the power stroke, and the other on the exhaust stroke.
Therefore, the answer is:
c. Both technicians A and B
A ball rolls from x=-5 to x=32.4 m in 8.9 seconds. What was it’s average velocity
Answer:
4.157 m/s
Explanation:
Average velocity: This can be defined as the ratio change in position to time interval. The S.I unit of average velocity is m/s
The expression for average velocity is given as,
V = Δx/t.............. Equation 1
Where V = average velocity, Δx = change in position on the x- axis, t = time.
But,
Δx = x₂-x₁........... Equation 2
Substitute equation 2 into equation 1
V = (x₂-x₁)/t................ Equation 3
Given: x₂ = 32.4 m, x₁ = -5 m, t = 8.9 s
Substitute into equation 3
V = [32.4-(-5)]/8.9
V = (32.4+5)/8.9
V = 37.4/8.9
V = 4.157 m/s
Hence the average velocity = 4.157 m/s
Answer:
v = 4.20m/s
Average velocity is 4.20m/s
Explanation:
Average velocity is the change in displacement per unit time;
v = ∆x/t
Given;
∆x = 32.4 - (-5) = 37.4 m
t = 8.9 seconds
Substituting the values;
v = 37.4m/8.9s
v = 4.20m/s
Average velocity is 4.20m/s
V LESSON 4 Check
1. Classify What kind of material transmits
some light, making objects behind it appear
blurry?
Answer:
Some objects are wax paper, newspaper, and white pape
Answer:
the correct answer is Translucent
What is the magnitude of the acceleration a of the chair? What is the magnitude of the normal force fn acting on the chair?
The magnitude of the normal force and acceleration is calculated using Newton's second law and free body diagrams, considering all forces acting on the object.
Explanation:The question pertains to finding the magnitude of the acceleration and the normal force acting on a chair, block, or similar object using Newton's second law of motion. To determine these quantities, one must analyze the forces acting on the object in a free body diagram and apply the equations of motion. The normal force is the force exerted by a surface to support the weight of an object resting on it, and it acts perpendicular to the surface. To find the normal force (FN), we typically assume static equilibrium if the object is at rest, which implies that the net force in the vertical direction is zero; thus, FN equals the object's weight (mg), where m is the mass and g is the acceleration due to gravity. The magnitude of acceleration (a) can be calculated using the net force acting on the object (resultant force) divided by its mass, as stated by Newton's second law (F = ma).
For example, if a block is placed on a table, the normal force is equal to the gravitational force acting on it (FN = Fg). If the block is accelerating, you would calculate the net force by including all forces acting on the block, such as tension, friction, or applied forces, and then use F = ma to find the acceleration.
You can determine the index of refraction of a substance by measuring its critical angle for total internal reflection. True or False
Answer:
True. Hope you have a good day, internet stranger. :)
A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop is 17.6 m, with what minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top? Assume the rider is not strapped to the car.
Answer:
The minimum speed must the car must be 13.13 m/s.
Explanation:
The radius of the loop is 17.6 m. We need to find the minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top.
We know that, mg be the weight of car and rider, which is equal to the centripetal force.
[tex]mg = \dfrac{mv^2}{r}\\v = \sqrt{rg}\\v = \sqrt{17.6\times 9.8}\\v = 13.13\ m/s\\[/tex]
So, the minimum speed must the car must be 13.13 m/s.
Which type of field is present near a moving electric charge
Answer:
magnetic field is the correct answer.
Explanation:
The magnetic field is produced by the movement of a charge or by current flow and through a magnetized material.magnetic fields are the areas where the objects display a magnetic force and the magnetic fields also affect the neighboring objects.The magnetic field is used as a tool to explain how the magnetic force is distributed inside thing magnetic.The range of wavelengths for visible light is from 3.9*10^-7m. What is the corresponding range of frequencies for visible light
Answer:
from 7.69×10¹⁵ Hz
Explanation:
Frequency: This can be defined as the number of complete cycle a wave complete in one seconds, the S.I unit of frequency is Hertz (Hz).
From the question,
v = λf............. Equation 1
Where v = speed of visible light in air, λ = wave length of visible light, f = frequency of visible light.
make f the subject of the equation
f = v/λ.................... Equation 2
Note: all electromagnetic wave have the same speed = 3×10⁸ m/s
Given: v = 3×10⁸ m/s, λ = from 3.9×10⁻⁷ m
Substitute into equation 2
f = 3×10⁸/ 3.9×10⁻⁷
f = 7.69×10¹⁵ Hz
Hence the range of the frequencies for visible light = from 7.69×10¹⁵ Hz
Final answer:
The range of frequencies for visible light, corresponding to wavelengths from about 400 nm to 700 nm, extends from approximately 4.28 × 10¹⁴ Hz to 7.50 × 10¹⁴ Hz.
Explanation:
The range of wavelengths for visible light is commonly accepted to be from about 400 nm to 700 nm. To find the corresponding range of frequencies for visible light, we use the formula c = λf, where c is the speed of light (≈ 3.00 × 10¸ m/s), λ is the wavelength in meters, and f is the frequency in hertz (Hz). For the visible light range, converting nanometers to meters (× 10⁻⁹), we calculate the frequencies at both ends of the spectrum: for 400 nm (4.00 × 10⁻⁷ m), the frequency is approximately 7.50 × 10¹⁴ Hz, and for 700 nm (7.00 × 10⁻⁷ m), it's approximately 4.28 × 10¹⁴ Hz. Therefore, the range of frequencies for visible light extends approximately from 4.28 × 10¹⁴ Hz to 7.50 × 10¹⁴ Hz.
When an object moves in a circular path, it accelerates toward the center of the circle as a result of ______.
A. terminal velocity B. momentum
C. centripetal force D. friction
Answer:
Centripetal acceleration
Explanation:
An object moving around a xirxular path maintains its route as a result of centripetal force. However, its acceleration is caused by centripetal acceleration. Despite centripetal acceleration not being among the choices, it is the right answer.
Centripetal acceleration helps an object that navigates around a circular path to accelerate while centripetal force enables the movement of an object around a circular path to move inwards. Momentum, given as one of the choices is product of mass and velocity while friction is the force opposing movement of an object around a surface.
oung's experiment is performed with light of wavelength 502 nm from excited helium atoms. Fringes are measured carefully on a screen 1.50 m away from the double slit, and the center of the 20th fringe (not counting the central bright fringe) is found to be 9.60 mm from the center of the central bright fringe. What is the separation of the two slits?
Answer:
0.00156875 m
Explanation:
y = Distance from the 20th fringe = 9.6 mm
l = Distance to screen = 1.5 m
[tex]\lambda[/tex] = Wavelength = [tex]502\ nm[/tex]
m = Number of fringe = 20
We have the relation
[tex]\dfrac{y}{l}=\dfrac{m\lambda}{d}\\\Rightarrow d=\dfrac{m\lambda l}{y}\\\Rightarrow d=\dfrac{20\times 502\times 10^{-9}\times 1.5}{9.6\times 10^{-3}}\\\Rightarrow d=0.00156875\ m[/tex]
The seperation of the two slits is 0.00156875 m
Our local piano tuner is making a house call to tune our piano (... yes, she's maintaining social distance!) She strikes a middle C tuning fork (about 262 Hz frequency), and strikes that key on your piano. You both hear a beat frequency of about 4 Hz. What is the actual frequency of your middle C piano string
Answer:
258 Hz or 266 Hz
Explanation:
When two sounds of similar frequency occur at the same time, a phenomenon called "beat" occurs. The beat is an interference patter between the two sounds; the frequency of this beats (called beat frequency) is given by the difference between the frequencies of the two sounds:
[tex]f_B = |f_1 -f_2|[/tex]
where
[tex]f_B[/tex] is the beat frequency
[tex]f_1[/tex] is the frequency of the 1st sound
[tex]f_2[/tex] is the frequency of the 2nd sound
In this problem,
[tex]f_B=4 Hz[/tex] is the beat frequency
[tex]f_1=262 Hz[/tex] is the frequency of the middle C tuning fork
[tex]f_2[/tex] is the frequency of the middle C piano string
Solving for f2, we find two solutions:
[tex]f_2=f_1-f_B = 262-4 = 258 Hz\\f_2 = f_1+f_B = 262+4=266 Hz[/tex]
state laws of refraction
Answer:
The law of refraction states that the incident ray, the refracted ray, and the normal to the interface, all lie in the same plane.
HOPE IT HELPS^_^
Answer:
There are two laws of refraction. The behavior of light as it travels through different media is described by the rules of refraction, sometimes referred to as Snell's laws. These principles control how light is affected by a barrier between two transparent materials in terms of its direction and speed.
Explanation:
First Law of refraction:- The incident ray, the refracted ray, and the normal—a line perpendicular to the boundary—all reside on the same plane according to the first law of refraction. This indicates that when light travels through a barrier between two media, its orientation changes.
Second Law of refraction:- The second law of refraction states that there is a continuous relationship between the sines of the angles of incidence and refraction (i.e., the angle between the incident ray and the normal and the angle between the refracted ray and the normal). The refractive index is a ratio that is unique to the two media in question. This may be written mathematically as:
n₁ * sin(θ₁) = n₂ * sin(θ₂)
where n₁ and n₂ are the refractive indices of the two media, θ₁ is the angle of incidence, and θ₂ is the angle of refraction.
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What is the angular acceleration if the weight is 24.0 cm from the elbow joint, her forearm has a moment of inertia of 0.240 kg⋅m2, and the net force she exerts is 759 N at an effective perpendicular lever arm of 2.00 cm?
The complete question is;
To develop muscle tone, a woman lifts a 2.00-kg weight held in her hand. She uses her biceps muscle to flex the lower arm through an angle of 60.0º . What is the angular acceleration if the weight is 24.0 cm from the elbow joint, her forearm has a moment of inertia of 0.240 kg.m², and the net force she exerts is 759 N at an effective perpendicular lever arm of 2.00 cm?
Answer:
α = 42.76 rad/s²
Explanation:
We are given;
Mass; m = 2 kg
Distance; r = 24cm = 0.24m
Moment of inertia of the forearm; I = 0.24 kg.m²
Muscle Force; F = 759N
Perpendicular distance to lever arm; r' = 2cm = 0.02m
Angle at which she flexes arm; θ =60°
Let's assume that the arm starts extended vertically downwards. and we take the elbow joint as the point about which we calculate the torque.
Thus, we know that torque is given by the formula ;
τ = Force x Perpendicular distance
Thus, the toque exerted by force in the muscle is;
τ_muscle = 759 x 0.02 = 15.18 N.m
Also, torque exerted by the lifted weight is given as 0 because perpendicular distance is zero.
Thus, the net torque on the lower arm is;
τ_net = τ_muscle - τ_weight
τ_net = 15.18 - 0 = 15.18 N.m
Now, let's calculate moment of inertia of lifted weight.
The moment of inertia is given by;
I = mr²
Thus, moment of Inertia of lifted weight is; I_weight = 2 x 0.24² = 0.115 kg.m²
Thus,total moment of inertia is;
I_total = I_arm + I_weight
Thus, I_total = 0.24 + 0.115 = 0.355 kg.m²
Now, we can calculate the angular acceleration.
It's gotten from the equation;
τ_net = I_total•α
Where α is angular acceleration.
Thus making α the subject, we have ; α = τ_net/I_total
α = 15.18/0.355 = 42.76 rad/s²
The average intensity of light emerging from a polarizing sheet is 0.708 W/m2, and that of the horizontally polarized light incident on the sheet is 0.960 W/m2. Determine the angle that the transmission axis of the polarizing sheet makes with the horizontal.
Answer:
Angle θ = 30.82°
Explanation:
From Malus’s law, since the intensity of a wave is proportional to its amplitude squared, the intensity I of the transmitted wave is related to the incident wave by; I = I_o cos²θ
where;
I_o is the intensity of the polarized wave before passing through the filter.
In this question,
I is 0.708 W/m²
While I_o is 0.960 W/m²
Thus, plugging in these values into the equation, we have;
0.708 W/m² = 0.960 W/m² •cos²θ
Thus, cos²θ = 0.708 W/m²/0.960 W/m²
cos²θ = 0.7375
Cos θ = √0.7375
Cos θ = 0.8588
θ = Cos^(-1)0.8588
θ = 30.82°
Is the magnetic field created by a current loop uniform?
The magnetic field produced by a current loop is not uniform. It is strongest at the center of the loop and decreases as you move away from the center. This change in strength across the space surrounding the loop makes it non-uniform.
Explanation:No, the magnetic field created by a current loop is not uniform. A current loop means a closed electrical circuit or a current flowing in a loop of conducting wire. When a current flows through the loop, a magnetic field is generated surrounding the loop. This magnetic field is strongest at the center of the loop and diminishes as you move away from the center. This variance in strength implies that the magnetic field is not uniform.
Example:
If you place a compass around a loop with flowing current, you'd notice that the compass needle's direction (which aligns with the magnetic field) changes as you move the compass around. This indicates that the magnetic field strength (and direction) changes across the space surrounding the current loop, which makes it non-uniform.
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What is the measure of how difficult it is for current to flow in a circult?
O A
voltage
B.
resistance
C.
power
D.
energy
Answer:
Resistance
Explanation:
Resistance "resists" the flow of electricity and makes it more difficult to travel. The higher the resistance the less current there is in a circuit. Example being, open circuit (infinity ohms) means there is no current flowing with ohms law.
A bird is flying directly toward a stationary bird-watcher and emits a frequency of 1420 Hz. The bird-watcher, however, hears a frequency of 1456 Hz. What is the speed of the bird, expressed as a percentage of the speed of sound
Answer:
2.47 %
Explanation:
We are given;
Frequency emitted by source(bird); f_s = 1420 Hz
Frequency heard by observer; f_o = 1456 Hz
From doppler shift frequency, we know that;
f_o = f_s [c/(c - c_s)]
Where c_s is speed of source which is the bird and c is speed of sound.
Thus;
Rearranging the equation, we have;
f_s/f_o = (c - c_s)/c
f_s/f_o = 1 - (c_s/c)
Plugging in the relevant values to get ;
1420/1456 = 1 - (c_s/c)
0.9753 = 1 - (c_s/c)
1 - 0.9753 = (c_s/c)
(c_s/c) = 0.0247
Since we want it expressed im percentage,
Thus, (c_s/c) % = 0.0247 x 100 = 2.47 %
A spaceship negotiates a circular turn of radius 2925 km at a speed of 29960 km/h. (a) What is the magnitude of the angular speed? 0.0028 Correct: Your answer is correct. rad/s (b) What is the magnitude of the radial acceleration? 23.68 Correct: Your answer is correct. m/s2 (c) What is the magnitude of the tangential acceleration? m/s2
a) 0.0028 rad/s
b) [tex]23.68 m/s^2[/tex]
c) [tex]0 m/s^2[/tex]
Explanation:
a)
When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by
[tex]\omega = \frac{\theta}{t}[/tex]
where
[tex]\theta[/tex] is the angular displacement
t is the time interval
The angular speed of an object in circular motion can also be written as
[tex]\omega = \frac{v}{r}[/tex] (1)
where
v is the linear speed of the object
r is the radius of the orbit
For the spaceship in this problem we have:
[tex]v=29,960 km/h[/tex] is the linear speed, converted into m/s,
[tex]v=8322 m/s[/tex]
[tex]r=2925 km = 2.925\cdot 10^6 m[/tex] is the radius of the orbit
Subsituting into eq(1), we find the angular speed of the spaceship:
[tex]\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s[/tex]
b)
When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the centre of the orbit: this is called centripetal acceleration, or radial acceleration.
The magnitude of the radial acceleration is given by
[tex]a_r=\omega^2 r[/tex]
where
[tex]\omega[/tex] is the angular speed
[tex]r[/tex] is the radius of the orbit
For the spaceship in the problem, we have
[tex]\omega=0.0028 rad/s[/tex] is the angular speed
[tex]r=2925 km = 2.925\cdot 10^6 m[/tex] is the radius of the orbit
Substittuing into the equation above, we find the radial acceleration:
[tex]a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2[/tex]
c)
When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.
The tangential acceleration is given by
[tex]a_t=\frac{\Delta v}{\Delta t}[/tex]
where
[tex]\Delta v[/tex] is the change in the linear speed
[tex]\Delta t[/tex] is the time interval
In this problem, the spaceship is moving with constant linear speed equal to
[tex]v=8322 m/s[/tex]
Therefore, its linear speed is not changing, so the change in linear speed is zero:
[tex]\Delta v=0[/tex]
And therefore, the tangential acceleration is zero as well:
[tex]a_t=\frac{0}{\Delta t}=0 m/s^2[/tex]
The 3 stages of the water cycle are
Answer:
Evaporation, condensation, precipitation
Explanation:
Answer:
Evaporation, condensation, precipitation & collection
Explanation:
In a parallel circuit, each resistor has:
A: no connection to the voltage source
B: a shared connection to the voltage source
C: its own connection to the voltage source
Answer:
C
Explanation:
Answer:
In a parallel circuit, each resistor has: its own connection to the voltage source.
Explanation:
A cross-country skier is skiing along at a zippy 8.0 mis. She stops pushing and simply glides along, slowing to a reduced speed of 6.0 mis after gliding for 5.0 m. What is the magnitude of her acceleration as she slows
Answer:
Acceleration, a = [tex]2.8\ m/s^2[/tex]
Explanation:
Given that,
Initial speed of the skier, u = 8 m/s
Final speed of the skier, v = 6 m/s
Distance, d = 5 m
We need to find the magnitude of her acceleration. It is equal to the rate of change of speed. It is given by :
[tex]v^2-u^2=2as\\\\a=\dfrac{v^2-u^2}{2s}\\\\a=\dfrac{6^2-8^2}{2\times 5}\\\\a=-2.8\ m/s^2[/tex]
So, the magnitude of her acceleration is [tex]2.8\ m/s^2[/tex].
The magnitude of the skier's acceleration as she slows down is 0.4 m/s^2.
Explanation:To find the magnitude of the skier's acceleration, we can use the formula:
Acceleration = (Final Velocity - Initial Velocity) / Time
In this case, the initial velocity is 8.0 m/s, the final velocity is 6.0 m/s, and the time is 5.0 seconds.
Plugging these values into the formula, we get:
Acceleration = (6.0 m/s - 8.0 m/s) / 5.0 s = -0.4 m/s^2
So the magnitude of the skier's acceleration as she slows down is 0.4 m/s^2.
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Twin space probes have a mass of 722 kg each. If the gravitational force between the two space probes is 8.61
10-16 N. what is the distance between them?
7.48 x 103 meters
Answer:200×10^5 meters
Explanation:
Answer:
200×10^5 meters
Explanation:
EDGE
Match the following
1. unit an object that orbits the Sun, is spherical, and has cleared its orbit of other large objects
2. an object that orbits the sun and is large enough for its gravity to make it spherical but that is too small to have cleared its orbit of all other large objects
3. the four inner planets; named for their rocky crusts
4.the four outer planets of the solar system; named for their high concentrations of hydrogen and helium
5. planets a unit of measurement based on the average distance of the earth from the sun; one unit equals 149.6 million kilometers
a. astronomical
b. dwarf planet
c. gas giants
d. planet
e. terrestrial
The definitions correctly match to Planet, Dwarf planet, Terrestrial, Gas giants, and Astronomical unit, respectively. These terms refer to objects and measurements relevant to astronomy and the solar system.
Explanation:
The correct matches for the given definitions are:
1. Planet - unit an object that orbits the Sun, is spherical, and has cleared its orbit of other large objects.
2. Dwarf planet - an object that orbits the sun and is large enough for its gravity to make it spherical but that is too small to have cleared its orbit of all other large objects.
3. Terrestrial - the four inner planets; named for their rocky crusts.
4. Gas giants - the four outer planets of the solar system; named for their high concentrations of hydrogen and helium.
5. Astronomical unit - a unit of measurement based on the average distance of the earth from the sun; one unit equals 149.6 million kilometers.
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The electrical resistance of a metallic conductor is inversely proportional to its
Answer:
Cross sectional area
Explanation:
Resistance of a metallic conductor is the measure of its opposition to the flow of electric current. Resistance is known to be directly proportional to the temperature which means as the temperature is increased, its resistance increases too and vice versa.
The resistance however has an inverse relationship with the cross sectional area of the metallic conductor. This means as the resistance increases the cross sectional area decreases and vice versa.
A 0.200 kg air-track glider moving at 1.20 m/s bumps into a 0.600 kg glider at rest.
a.) Find the total kinetic energy after collision if the collision is elastic.
b.) Find the total kinetic energy after collision if the collision is completely inelastic.
Answer:
a) 0.144J
b) 0.12J
Explanation:
A 0.200 kg air-track glider moving at 1.20 m/s bumps into a 0.600 kg glider at rest. a.) Find the total kinetic energy after collision if the collision is elastic. b.) Find the total kinetic energy after collision if the collision is completely inelastic.
Given that
M1 = 0.2kg
M2 = 0.6kg
U1 = 1.2 m/s
Since both momentum and energy are conserved in elastic collisions, the total kinetic energy after collision will be
1/2M1U^2 + 1/2M2U^2
1/2 × 0.2 × 1.2^2 + 0
K.E = 0.144J
B) In elastic collision, only momentum is conserved
M1U1 + M2U2 = (M1 + M2)V
U2 = 0 since the object is at rest
0.2×1.2 + 0 = (0.2 + 0.6)V
0.24 = 0.8V
V = 0.24/0.8
V = 0.3 m/s
K.E = 1/2(M1+M2)V
K.E = 1/2 (0.2 + 0.6) × 0.3
K.E = 0.4 × 0.3
K.E = 0.12J
Answer:
The other person is almost right, but I see where the numbers got fudged
Explanation:
Answer:a) 0.144Jb) 0.036J
Explanation: A 0.200 kg air-track glider moving at 1.20 m/s bumps into a 0.600 kg glider at rest. a.) Find the total kinetic energy after collision if the collision is elastic. b.) Find the total kinetic energy after collision if the collision is completely inelastic. Given that M1 = 0.2kgM2 = 0.6kgU1 = 1.2 m/s Since both momentum and energy are conserved in elastic collisions, the total kinetic energy after collision will be1/2M1U^2 + 1/2M2U^21/2 × 0.2 × 1.2^2 + 0K.E = 0.144J
B) This is where the other person's wrong:
m1u1+m2u2=(m1+m2)V
V=(0.200kg*1.20m/s)/(0.200kg+0.600kg)=0.3m/s
KE=1/2m*v^2
KE=1/2(0.200kg+0.600kg)*(0.3m/s)^2=1/2(0.8x0.3)^2=0.036 J
How many electrons are there in a diatomic oxygen molecule
Answer:
8 electrons
Explanation:
Brainly i can you please help me
Answer:
bet
Explanation:
slader A baseball player slides to a stop on level ground. Using energy consideration, calculate the distance the 65 kg baseball player slides, given his initial speed is 6 m/s and the force of friction against him is a constant 450N.
Final answer:
The 65 kg baseball player slides a distance of 5.2 meters before coming to a stop, calculated using kinetic energy and the work done by friction.
Explanation:
To calculate the distance the 65 kg baseball player slides using energy considerations, we start by noting that the work done by friction is equal to the loss in kinetic energy as the player comes to a stop. The kinetic energy (KE) of the player can be calculated using the formula KE = (1/2)mv², where m is mass and v is velocity. Since the force of friction (f) is constant, the work (W) done by friction is W = fd, with d being the distance over which the friction acts.
Setting the initial kinetic energy equal to the work done by friction, we get:
(1/2)mv² = fd
Inserting the given values, we have:
(1/2)(65 kg)(6 m/s)² = (450 N)d
Upon calculating, we find:
d = (1/2)(65 kg)(6 m/s)² / (450 N)
d = 5.2 meters
Therefore, the baseball player slides a distance of 5.2 meters before coming to a stop.
After a fall, a 75 kg rock climber finds himself dangling from the end of a rope that had been 18 m long and 11 mm in diameter but has stretched by 2.1 cm. For the rope, calculate (a) the strain, (b) the stress, and (c) the Young's modulus.
Answer:
(A) Strain = 0.0012
(b) Stress [tex]=77.44\times 10^5N/m^2[/tex]
(c) Young's modulus [tex]=6.45\times 10^9N/m^2[/tex]
Explanation:
Mass of the rock m = 75 kg
So weight of the rock [tex]F=mg=75\times 9.8=735N[/tex]
Length of the rope l = 18 m
Diameter of the rope d = 11 mm
Change in length of rope [tex]\Delta l=2.1cm =0.021m[/tex]
So radius r = 5.5 mm = 0.0055 m
Cross sectional area [tex]A=\pi r^2[/tex]
[tex]A=3.14\times 0.0055^2=9.49\times 10^{-5}m^2[/tex]
(a) Strain is equal to ratio change in length to original length
So strain [tex]=\frac{\Delta l}{l}=\frac{0.021}{18}=0.0012[/tex]
(b) Stress [tex]=\frac{Weight}{area}[/tex]
[tex]=\frac{735}{9.49\times 10^{-5}}=77.44\times 10^5N/m^2[/tex]
(c) Young's modulus is equal to ratio of stress and strain
So young's modulus [tex]=\frac{77.44\times 10^5}{0.0012}=6.45\times 10^9N/m^2[/tex]
Sound moves at a speed of 343 meters per second through air, a gas.
Which pairing is most accurate for how sound will travel through
liquid and solid?
Answer:
need more info
Explanation:
Where would you expect to have more touch receptors: on the palm of your hand or on the back of your hand? Explain your reasoning
Explanation:
this is because if your going to touch something dangerouse your hand normally feels it first and its the front not the back so when you pick things up you can tell if its dangerouse hope that helped
The palm of the hand has more touch receptors than the back, due to its glabrous skin type which is thicker and more sensitive. These receptors allow for detection of fine touch and pressure, critical for manual tasks.
Explanation:Considering the density of touch receptors in different parts of our body, you could expect to have more touch receptors on the palm of your hand than on the back of your hand. This is due to the specific type of skin found on the palm, referred to as glabrous skin. This type of skin is typically thicker and more sensitive as compared to hairy skin, hence it contains more touch receptors.
The importance of the high density of touch receptors in areas such as palms and fingertips can be seen in the daily actions, such as the ability to sense fine touch and detailed information, which is necessary for fine motor tasks.
Furthermore, mechanoreceptors such as Merkel's disks, Meissner's corpuscles, Pacinian corpuscles, and Ruffini endings play a vital role in interpreting the touch-related sensory information of these regions. For instance, Merkel's disks and Meissner's corpuscles that are found deeper in the skin and in the upper parts of the skin respectively, detect fine touch. On the other hand, Pacinian corpuscles and Ruffini endings can sense deeper touch like pressure.
Learn more about Touch Receptors here:https://brainly.com/question/32104367
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