What is the direction of the centripetal force when applied to an object?
the direction opposite to the object’s velocity
the same direction as the object’s velocity
perpendicular to an object’s motion
parallel with an object’s motion
The direction of the centripetal force when applied to an object is perpendicular to the object's motion.
What is centripetal force?Any motion along a curved road is accelerated, necessitating the application of force to the path's center of curvature. This force is known as the centripetal force, which is a force that seeks its center.
The direction of centripetal force, or even net force, is always that of acceleration. As an illustration, consider how quickly the earth is moving toward the sun. We are not moving toward the sun since the acceleration vector for planet Earth is constantly pointing in that direction and the velocity is perpendicular to it.
When force applied to an item, the centripetal force's direction is perpendicular to the object's motion.
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You push a shopping cart filled with groceries (total mass = 20 kg) by applying a force to the cart 30° from the horizontal. if the force you apply has a magnitude of 86 n, what is the cart's acceleration? assume negligible friction.
A car is traveling with a constant speed when the driver suddenly applies the brakes, causing the car to slow down with a constant acceleration of magnitude 3.50 m/s2. if the car comes to a stop in a distance of 30.0 m, what was the car's original speed?
The car's original speed can be calculated by utilizing the kinematic equation which is a principle in physics. Given that the car came to a stop with constant deceleration in a certain distance, the estimated initial speed was approximately 26.9 m/s.
Explanation:The subject of the question involves concepts in physics specifically related to the equations of motion. To solve this, we can utilize the kinematic equation which relates initial velocity, acceleration, and distance: v2 = u2 + 2as, where 'v' is the final velocity, 'u' is the initial velocity, 'a' is acceleration and 's' is distance. Since the car comes to a stop, our final velocity v = 0 m/s. We know that the distance s = 30m and the acceleration is given as a = -3.50 m/s2 (negative because it's deceleration). From these, we can find the initial velocity 'u'. When we substitute these values into the equation, we can solve for 'u' as √(v2 - 2as), which equals approximately √(0 - 2*-3.50*30) ≈ 26.9 m/s. Therefore, the car's original speed was approximately 26.9 m/s.
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Calculate the efficiency of an engine in a power plant operating between 40 ° c and 320 °
c. remember that the efficiency is a decimal number less than 1.
A vector A is added to B=6i-8j. The resultant vector is in the positive x direction and has a magnitude equal to A . What is the magnitude of A?
a)11
b)7.1
c)5.1
d)8.3
e)12.2
The correct answer would be letter d, 8.3.
Solution for the problem follows:
Given are:
B = 6i - 8j
A is unknown; let A be = mi + nj
A+B is along the x axis (therefore A+B = Ki + 0j, where K is unknown,
but then again the magnitude of A+B is the similar as the magnitude of A,
so mag(A+B)=K=sqrt(m^2+n^2), or K^2 = m^2+n^2.
A+B, from simple vector addition, will be now (m+6)i + (n-8)j.
Ever since we previously
know A+B = Ki + 0j, we now know that:
m+6 = K
n-8 = 0, which implies n=8.
Thus, K^2=m^2+n^2 ====> (m+6)^2 = m^2 +8^2
= m^2 + 12m + 36 = m^2 + 64
= 12m = 28
= m = 2.33333...
Therefore, the magnitude of A is sqrt[(2.333...)^2 + 8^2] = 8.3333
The magnitude of [tex]\vec A[/tex] is [tex]8.3[/tex].
Further explanation:
A vector is a quantity having magnitude and direction both. It is represented as the product of magnitude and direction vector.
Given:
The vector is given as [tex]\vec B = 6\hat i - 8\hat j[/tex].
The direction of resultant vector is in the X- direction.
Concept used:
Consider a vector [tex]\vec A = a\hat i + b\hat j[/tex] which is added to [tex]\vec B[/tex] to get a resultant vector [tex]\vec C[/tex]. The resultant vector is directed in the positive X-direction which means that the y- component of the vector is zero.
The expression for the resultant vector is given as.
[tex]\vec C = \vec A + \vec B[/tex]
Substitute [tex]6\hat i - 8\hat j[/tex] for [tex]\vec B[/tex] in the above expression.
[tex]\begin{gathered}\vec C = \left( {a\hat i + b\hat j} \right) + \left( {6\hat i - 8\hat j} \right) \\= \left( {a + 6} \right)\hat i + \left( {b - 8} \right)\hat j \\ \end{gathered}[/tex]
The resultant vector is represented as.
[tex]\vec C = x\hat i + 0\hat j[/tex]
Compare the above two expression of resultant vector.
[tex]x = a + 6[/tex] …… (1)
[tex]\begin{aligned}0&=b-8\\b&=8\\\end{aligned}[/tex]
The magnitude of resultant vector is equal to the magnitude of [tex]\vec A[/tex].
The expression for the magnitude of [tex]\vec A[/tex] is given as.
[tex]\left| {\vec A} \right| = \sqrt {\left( {{a^2}} \right) + \left( {{b^2}} \right)}[/tex] …… (2)
The expression for the magnitude of [tex]\vec C[/tex] is given as.
[tex]\left| {\vec C} \right| = \sqrt {{{\left( {a + 6} \right)}^2}}[/tex]
Compare the above two expressions.
[tex]\sqrt {\left( {{a^2}} \right) + \left( {{b^2}} \right)}= \sqrt {{{\left( {a + 6} \right)}^2}}[/tex]
Substitute 8 for [tex]b[/tex] in above expression.
[tex]\begin{aligned}\sqrt {\left( {{a^2}} \right) + \left( {{8^2}} \right)}&=\sqrt {{{\left( {a + 6} \right)}^2}}\\{a^2} + 64&={a^2} + 36 + 12a \\12a&=28 \\a&=2.33 \\ \end{aligned}[/tex]
Substitute 2.33 for a and 8 for b in equation (2).
[tex]\begin{aligned}\left| {\vec A} \right|&=\sqrt {{{\left( {2.33} \right)}^2} + {{\left( 8 \right)}^2}}\\&=8.33 \\ \end{aligned}[/tex]
Thus, the magnitude of vector A is [tex]8.33[/tex].
Learn more:
1. Motion under friction https://brainly.com/question/7031524.
2. Conservation of momentum https://brainly.com/question/9484203.
3. Circular motion https://brainly.com/question/9575487.
Answer Details:
Grade: College
Subject: Physics
Chapter: Vectors
Keywords:
Vectors, product, magnitude, direction, resultant vector, adding vector, subtraction of vector, 2.33, 8, 8.33, 8.3, 8.33.
A 30-kg chandelier hangs from a ceiling on a vertical 4.4-m-long wire. what horizontal force would be necessary to displace its position 0.16 m to one side?
The horizontal force would be necessary to displace its position 0.16 m to one side will be 10.7 N.
What is horizontal force?Horizontal force is defined as a force applied in a direction parallel to the horizon. A force exerted on a body has two components: a horizontal component and a vertical component.
Given that,
m = 30 kg,
W = mg = (30 kg)*(9.8 m/s²)
W = 294 N, the weight of the chandelier.
Let T = tension in the wire.
Let F = the horizontal force required to displace the chandelier by 0.16 m.
The angle that the wire makes with the vertical is given by:
sin(θ) = 0.16/4.4 = 0.0364
θ = sin⁻¹ 0.0364
θ = 2.084°
For equilibrium,
T cos(θ) = W
0.9993T = 294
T = 294.195 N
We know that,
F = Tsin(θ)
F = 294.195*0.0364
F = 10.709 N
Thus, The horizontal force would be necessary to displace its position 0.16 m to one side will be 10.7 N.
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____ is the rate of change in velocity.
Answer:
Acceleration is the rate of change in velocity.
Explanation:
Velocity,v-
"It is the change in a body's displacement or change in speed,d over the time,t."
v=s/sec,Units: meter/second.Acceleration,a-
"When the body has a varying velocity,v across a given time frame,t is called as the acceleration,a."
a=v/t,Unit: meter/second².A flat surface of area 3.20 m2 is rotated in a uniform electric field of magnitude e = 6.20 x 105 n/c. determine the electric flux through this area (a) when the electric field is perpendicular to the surface and (b) when the electric field is parallel to the surface.
The electric flux through this area:
When the electric field is perpendicular to the surface= 1.98 x 10⁶ Nm²/CWhen the electric field is parallel to the surface= 0What is Electric flux?
This is defined as the number of electric field lines that intersect a given area.
Electric flux(φ)= EA cos θ
where θ = angle between the directions of E and A is the surface area.
When the electric field is perpendicular to the surface, θ = 0°.
φ = EA cos(0°) = (6.20 x 10⁵ N/C)*(3.2 m²) × 1 = 1.984 x 10⁶ Nm²/C
When the electric field is parallel to the area, then θ = 90°, and
φ = EA cos(90°) = 0 as result of cos 90° being zero.
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During an episode of turbulence in an airplane you feel 170 n heavier than usual. if your mass is 73 kg, what are the magnitude and direction of the airplane's acceleration?
Final answer:
The airplane's acceleration that caused the passenger to feel 170 N heavier is 2.33 m/s² upwards.
Explanation:
When a person experiences an increase in weight due to airplane turbulence, it indicates that there's an additional force acting on them because of the airplane's acceleration. To find the magnitude and direction of the airplane's acceleration, we can use Newton's second law of motion (F=ma), where F is the force, m is the mass, and a is the acceleration.
The student feels 170 N heavier, implying that the upward acceleration of the airplane is causing this additional force. Since the student's mass is 73 kg, we can calculate the acceleration using the formula a = F/m, where F is the additional force (170 N) and m is the mass of the student (73 kg). Hence, a = 170 N / 73 kg = 2.33 m/s². This is the magnitude of the acceleration.
The direction of the acceleration is upwards, as it increases the normal force which is felt as an increase in apparent weight.
"on a snowy day, max (mass = 15 kg) pulls his little sister maya in a sled (combined mass = 20 kg) through the slippery snow. when max pulls on the sled with 12 n of force, directed at an angle of 15° above the ground, how much work does max do on the sled as he pulls his sister 25 m in the snow?"
The index of refraction of water is 1.36. what is the speed of light in water
When opening a door, you push on it perpendicularly with a force of 57.0 n at a distance of 0.470 m from the hinges. what torque (in n·m) are you exerting relative to the hinges? (enter the magnitude.)?
At a given point on a horizontal streamline in flowing air, the static pressure is â2.0 psi (i.e., a vacuum) and the velocity is 150 ft/s. determine the pressure at a stagnation
The pressure at the stagnation point is 149.854 lb/ft².
Explanation:According to Bernoulli's equation, the total pressure at a given point is the sum of the static pressure and the dynamic pressure. The dynamic pressure is given by 0.5 * ρ * v^2, where ρ is the density of the fluid and v is the velocity of the fluid. In this case, the static pressure is -2.0 psi (a vacuum) and the velocity is 150 ft/s. The density of air at sea level is approximately 1.14 kg/m³, which is equivalent to 0.075 lb/ft³. Converting the units, we have -2.0 psi = -18.896 lb/ft². Plugging in the values into Bernoulli's equation, we can calculate the dynamic pressure:
Dynamic pressure = 0.5 * (0.075 lb/ft³) * (150 ft/s)^2 = 168.75 lb/ft².
The total pressure at the stagnation point is equal to the sum of the static pressure and the dynamic pressure. Therefore, the pressure at the stagnation point is:
Total pressure = -18.896 lb/ft² + 168.75 lb/ft² = 149.854 lb/ft².
Johannes Kepler used math to show that the planets move in perfect circles around the sun.
true
false
Make a general statement concerning how large bodies of water affect the climate of nearby coastal communities.
Final answer:
Large bodies of water like oceans contribute to more moderate climates in coastal areas due to their thermal properties. Global warming is leading to sea level rise through glacial meltwater and thermal expansion, which affects coastal communities. Oceans also impact global weather patterns, including precipitation and climate, due to heat transport and storage.
Explanation:
Large bodies of water, like oceans and large lakes, have a significant impact on the climate of coastal communities. The thermal properties of water, which heats and cools more slowly than land, lead to moderate climates in coastal areas. These regions typically experience smaller temperature fluctuations both daily and seasonally, in comparison to interior landmasses. Additionally, global warming is causing sea levels to rise due to glacial meltwater and thermal expansion, further complicating the climate effects on coastal communities. The warmth of oceanic currents is transported across vast distances, affecting the weather patterns far inland as well.
As the planet warms, the rise in sea levels can lead to the inundation of shorelines, which poses challenges for coastal cities. This rising sea level can increase the impact of storm surges, putting infrastructure at risk. The warming of oceans also contributes to the continued melting of polar ice, which can disrupt the supply of freshwater and bring about long-term changes to global precipitation and climate patterns. Hence, the oceans play a crucial role in moderating global climate and the long-term implications of climate change.
Two solutions, initially at 24.60°C, are mixed in a coffee cup calorimeter (Ccal = 15.5 J/°
c. When a100.0 mL volume of 0.100 M AgNO3 solution is mixed with a 100.0 mL sample of 0.200 M NaClsolution, the temperature in the calorimeter rises to 25.30°
c. Determine the ?H°rxn for thereaction as written below. Assume that the density and heat capacity of the solutions is the sameas that of water.NaCl (aq) + AgNO3(aq) ? AgCl(s) + NaNO3(aq) ?H°rxn = ?
A 1.60-m-long barbell has a 25.0 kg weight on its left end and a 37.0 kg weight on its right end. you may want to review ( pages 204 - 205) . part a if you ignore the weight of the bar itself, how far from the left end of the barbell is the center of gravity
Rank the nonmetals in each set from most reactive (1) to least reactive (3). Bromine: Chlorine: Iodine:
The most reactive element of this list is Chlorine, the next most reactive is bromine, and the least reactive is iodine.
All of these three elements are group 7 elements in the periodic table. It is known that the reactivity of group 7 elements decreases down the group. The most reactive element in this group is Flourine with reactivity decreasing down the group.
The reason for this decrease in reactivity is that as you go down the group, the distance between the positive nucleus that attracts valence electrons increases, decreasing the electrostatic attraction between the nucleus and the outer electrons. The other reason is that the electrons in lower energy levels closer to the nucleus repel and shield the electrons in the outermost shell or energy level of the atom.
Chlorine>Bromine>Iodine.
Answer: The order of reactivity of non-metals from most reactive to least reactive is [tex]\text{Chlorine}>\text{Bromine}>\text{Iodine}[/tex]
Explanation:
Reactivity of a non-metal is defined as the tendency of an element to gain electrons. The reactivity increases as we move across a period and it decreases as we move down the group.
When the size of an element increases, the valence electrons gets away from the nucleus and the tendency of an element to gain electrons decreases.
In a group, the size of an element increases because there is an addition of new shell and electron is added in that shell.
The given elements belong to the same group which is Group 17.
Chlorine has the smallest size, then bromine and then iodine.
Hence, the order of reactivity of non-metals from most reactive to least reactive is [tex]\text{Chlorine}>\text{Bromine}>\text{Iodine}[/tex]
You are pushing on a heavy door, trying to slide it open. If your friend stands behind you and helps you push, how have the forces changed?
Question options:
There will be less friction.
The applied force does not change.
The net force will decrease.
The net force applied will increase.
Answer: Option (d) is the correct answer.
Explanation:
Net force is defined as the sum of total number of forces acting on a substance.
For example, when you and your friend are pushing a door in order to slide it open then it means that the net force is sum of force applied by you and force applied by your friend.
Therefore, the force applied earlier was less but it increase when your friend also started applying force.
Thus, we can conclude that if your friend stands behind you and helps you push, then the net force applied will increase.
Two friends, barbara and neil, are out rollerblading. with respect to the ground, barbara is skating due south at a speed of 6.3 m/s. neil is in front of her. with respect to the ground, neil is skating due west at a speed of 2.4 m/s. find neil's velocity ((a) magnitude and (b) direction relative to due west, as seen by barbara.
Final answer:
To find Neil's velocity relative to Barbara, their individual velocities are considered as vectors and subtracted. Neil's relative velocity is calculated to be approximately 6.74 m/s at 69° north of west.
Explanation:
To determine Neil's velocity relative to Barbara, we need to consider both Neil’s and Barbara's velocities as vectors and subtract the velocity of Barbara from Neil's velocity. Barbara is moving due south at 6.3 m/s, and Neil is moving due west at 2.4 m/s. By defining south as the negative y-direction and west as the negative x-direction, we can represent Barbara’s velocity vector as (0 m/s, -6.3 m/s) and Neil's as (-2.4 m/s, 0 m/s).
To find Neil’s velocity relative to Barbara, we calculate the vector difference: Neil’s velocity minus Barbara's velocity:
VNB = VN - VB = (-2.4 m/s, 0 m/s) - (0 m/s, -6.3 m/s) = (-2.4 m/s, 6.3 m/s)
To find the magnitude of Neil's relative velocity (VNB), we use the Pythagorean theorem: |VNB| = √((-2.4 [tex]m/s^2[/tex] + (6.3 [tex]m/s^2[/tex]), which gives:
|VNB| = √(5.76 + 39.69) = √(45.45) ≈ 6.74 m/s
The direction relative to due west can be found using the arctangent of the y-component over the x-component:
θ = arctan(6.3 / 2.4) ≈ arctan(2.625) ≈ 69°
Therefore, Neil's velocity relative to Barbara is approximately 6.74 m/s at an angle of 69° north of west.
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A tuba makes a low pitched sound and a flute makes a high pitched sound. Which statement accurately describes the wavelengths of the two sounds?
A) The wavelengths of the two sounds are the same.
B) The wavelength of the high pitched sound is not measureable.
C) The wavelength of the low pitched sound is longer than the wavelength of the high pitched sound.
D) The wavelength of the low pitched sound is shorter than the wavelength of the high pitched sound.
An object is at x = 0 at t = 0 and moves along the x axis according to the velocity–time graph in Figure P2.50.(a) What is the object’s acceleration between 0 and 4.0 s? (b) What is the object’s acceleration between 4.0 s and 9.0 s? (c) What is the object’s acceleration between 13.0 s and 18.0 s? (d) At what time(s) is the object moving with the lowest speed? (e) At what time is the object farthest from x = 0? (f) What is the final position x of the object at t = 18.0 s? (g) Through what total distance has the object moved between t = 0 and t = 18.0 s?
Without the given Figure, precise answers can't be provided. Generally, acceleration is calculated as the slope of the velocity-time graph, and position is provided by the integral (or area under the graph) of the velocity-time graph. The object's speed is lowest when its velocity is minimal, and it is farthest from x= 0 when the accumulated area under the graph is maximum.
Explanation:Unfortunately, without the given Figure P2.50, it's impossible to accurately calculate the object's acceleration, position at different times, or specify when the object is moving with the lowest speed or is farthest from x = 0.
However, I can explain the general method to determine this information. Acceleration is calculated from the slope of the velocity-time graph. The position is generally obtained by calculating the area under the velocity-time graph from the beginning of the interval to the end. The object is moving with the lowest speed when the velocity is lowest (either positively or negatively). The object is farthest from x = 0 when the accumulated area under the velocity-time graph (counting areas below the time axis as negative) is a maximum. The total distance the object has moved is equal to the absolute sum of all the areas (both positive and negative) on the velocity-time graph.
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What are the disadvantages of driverless cars? Check all that apply.
They always drive slower than surrounding traffic to avoid accidents.
They cannot tell the difference between people and other objects.
They will wait for crossing pedestrians even if signaled to keep driving.
They may unexpectedly drive off the road to avoid potential hazards.
They will not allow the human driver to take control while in motion.
Answer:B,C, They cannot tell the difference between people and other objects. & They will wait for crossing pedestrians even if signaled to keep driving.
Explanation: i just did it on edg
Driverless cars can disrupt traffic due to slow speeds and over-caution, struggle to differentiate between people and objects, unexpectedly move to avoid hazards, and may not allow human control in emergency situations.
Explanation:Driverless cars, or autonomous vehicles, have several disadvantages that might affect their performance and usability. Firstly, these vehicles often operate at safer, slower speeds than typical road traffic, which can disrupt normal traffic flow. Secondly, while they use advanced sensors to detect their surroundings, they might struggle to differentiate between people and other objects. This could potentially lead to harmful incidents. Additionally, they might be overly cautious and halt for crossing pedestrians even when it's unnecessary, causing further disruption.
Moreover, unpredictable actions like veering off the road to circumvent hazards could pose risks to passengers or other vehicles. Lastly, a key issue is if these vehicles don't permit the human driver to override control manually in crisis situations, which can decrease the human's ability to avert dangerous situations.
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What is the speed of the roller coaster at the top of the loop if the radius of curvature there is 11.0 m and the downward acceleration of the car is 1.50 g?
Final answer:
The roller coaster's speed at the top of the loop is calculated using the centripetal acceleration formula, with the given downward acceleration of 1.50 g which equals 14.7 m/s² and the radius of curvature of 11.0 m to find the velocity.
Explanation:
The speed of the roller coaster at the top of the loop can be determined by utilizing the concept of centripetal acceleration. Centripetal acceleration is provided by the gravitational force when the roller coaster is at the top of the loop. Given that the downward acceleration is 1.50 g, and knowing that 1 g equals 9.8 m/s2, we can calculate the centripetal acceleration as 1.50 times 9.8 m/s2. The formula for centripetal acceleration (ac) is ac = v²/r, where v is the velocity and r is the radius of curvature. Rearranging the formula to solve for v gives us v = √(ac × r). Plugging in the values, we have v = √((1.50 × 9.8 m/s2) × 11.0 m), which yields the speed of the roller coaster at the top of the loop.
A car slams on its brakes, coming to a complete stop in 4.0 s. The car was traveling south at 60.0 mph. Calculate the acceleration.
Answer:
Acceleration, [tex]a=-6.705\ m/s^2[/tex]
Explanation:
It is given that,
Velocity of the car, u = 60 mph = 26.82 m/s
Finally it comes to stop, v = 0
Time taken, t = 4 s
We need to find the acceleration of the car. The change in velocity divided by time is called the acceleration of the object. Mathematically, it is given by :
[tex]a=\dfrac{v-u}{t}[/tex]
[tex]a=\dfrac{0-26.82}{4}[/tex]
[tex]a=-6.705\ m/s^2[/tex]
Negative sign shows the car is decelerating. So, the acceleration of the car is [tex]6.705\ m/s^2[/tex]. Hence, this is the required solution.
What is the energy of a photon with the wavelength 150 nm?
A 2kg particle moving along the x-axis experiences the net force shown at right. The particle’s velocity is 3.0m/s at x = 0m.
A) At what position is the particle moving the fastest?
B) What is the particle’s velocity at x=8m? Show your work
The solid aluminum shaft has a diameter of 50mm and an allowable shear stress of 60 mpa. determine the largest torque t1 that can be applied to the shaft if it is also subjected to the other torsional loadings
Final answer:
The largest torque that can be applied to the solid aluminum shaft is 1.5 MPa.
Explanation:
To determine the largest torque that can be applied to the solid aluminum shaft, we need to consider the allowable shear stress and the diameter of the shaft. The formula to calculate torque is t = rF sin 0, where t is the torque, r is the radius, F is the applied force, and 0 is the angle between the force and the radius. In this case, since the force is perpendicular to the radius, the angle is 90 degrees, so sin 0 = 1.
The diameter of the shaft is given as 50mm, which means the radius is 25mm. We need to convert the radius to meters, so the radius is 25/1000 = 0.025m. The allowable shear stress is given as 60MPa.
Using the formula for torque, t = rF = (0.025m)(60MPa) = 1.5MPa.
Therefore, the largest torque that can be applied to the shaft is 1.5MPa.
A field mouse trying to escape a hawk runs east for 3.5 m , darts southeast for 4.5 m , then drops 2.0 m down a hole into its burrow. part a what is the magnitude of the net displacement of the mouse?
A cart for hauling ore out of a gold mine has a mass of 413 kg, including its load. the cart runs along a straight stretch of track that is sloped 4.69° from the horizontal. a donkey, trudging along and to the side of the track, has the unenviable job of pulling the cart up the slope with a 4.10 × 102-n force for a distance of 175 m by means of a rope that is parallel to the ground and makes an angle of 14.3° with the track. the coefficient of friction for the cart\'s wheels on the track is 0.0163. use g = 9.81 m/s2. find the work that the donkey performs on the cart during this process.
The total work done by the donkey to pull the cart out of the mine is calculated by finding the work done against gravity and friction. The work done against gravity is 37012.5 J and against friction is 11637.5 J. The sum, and thus the total work done, is 48650 J.
Explanation:To find the total work done by the donkey, we need to consider the work done against both the gravitational force and the frictional force. The total work done will be equal to the sum of these two works.
Firstly, let's find the work done against the gravitational force. The force of gravity acting on the cart can be found using the equation F = m x g x sin(θ), where m is the mass of the cart, g is the acceleration due to gravity, and θ is the angle of the slope. Therefore, F = 413 kg x 9.81 m/s² x sin(4.69°) = 211.5 N. The work done against gravity is then W = F x d, where d is the distance the cart is hauled, resulting in W = 211.5 N x 175 m = 37012.5 J.
Secondly, let's calculate the work done against friction. The frictional force can be found using the equation F = μ x m x g x cos(θ), where μ is the coefficient of friction. Therefore, F = 0.0163 x 413 kg x 9.81 m/s² x cos(4.69°) = 66.5 N. The work done against friction is again W = F x d, giving W = 66.5 N x 175 m = 11637.5 J.
The total work done by the donkey is then the sum of these two works, giving 37012.5 J + 11637.5 J = 48650 J.
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