The maximum quantity of repetitions that the athlete could complete with their current level of energy was 229 for such a workout.
Energy efficiency is the ratio of input power to energy output. As just a result, a cookie contains 54.0 kcal in calories. An athlete utilizes the 54.0 kcal in just this cookie for input energy.
The following formula could be employed to measure efficiency:
Energy intake minus power generated is efficiency.
Efficiency = output energy / input energy
After rearrangement above equation.
output energy = efficiency × input energy
Put the given data in above equation.
Output energy = 0.25 × 54 kcal = 13.5 kcal.
Again for power capacity, the lifting exercise was already carried out n times. The source of total output energy is mgh within the form of electromagnetic potential.
So, Energy per repetition = mgh
Where, m will represent mass g will represent gravity h is height.
By entering the data provided above.
Energy per repetition = 103 kg × 9.8 m/[tex]s^{2}[/tex] × (2.45 × 0.1m)
Energy per repetition= 247.303 J
Energy per repetition = 0.059 kcal.
So, quantity of repetitions = overall output energy / energy per repetition
quantity of repetitions = 13.5 kcal / 0.059 kcal
quantity of repetitions will be equal to 229 repetitions.
Therefore, the quantity of repetition would be be 229.
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mick took his friend out to dinner the bill was $40 but he applied a coupon then the total price was $33 what was the % off?
ANSWER
17.5 %
EXPLANATION
We have to find the percent change, given that the initial price was $40 and the final price was $33.
[tex]\text{ \% }change=\frac{final.price-initial.price}{initial.price}\times100[/tex][tex]\text{\% }change=\frac{33-40}{40}\times100=\frac{-7}{40}\times100=-0.175\times100=-17.5\text{\%}[/tex]The coupon was for 17.5% off
. A circular loop of wire of area 10 cm2 carries a current of 25 A. At a particular instant, the loop lies in the xy-plane and is subjected to a magnetic field B=(2.0iˆ+6.0jˆ+8.0kˆ)×10−3T. As viewed from above the xy-plane, the current is circulating clockwise. (a) What is the magnetic dipole moment of the current loop? (b) At this instant, what is the magnetic torque on the loop?
The magnetic dipole moment of the current loop is 0.025 Am².
The magnetic torque on the loop is 2.5 x 10⁻⁴ Nm.
What is magnetic dipole moment?The magnetic dipole moment of an object, is the measure of the object's tendency to align with a magnetic field.
Mathematically, magnetic dipole moment is given as;
μ = NIA
where;
N is number of turns of the loopA is the area of the loopI is the current flowing in the loopμ = (1) x (25 A) x (0.001 m²)
μ = 0.025 Am²
The magnetic torque on the loop is calculated as follows;
τ = μB
where;
B is magnetic field strengthB = √(0.002² + 0.006² + 0.008²)
B = 0.01 T
τ = μB
τ = 0.025 Am² x 0.01 T
τ = 2.5 x 10⁻⁴ Nm
Thus, the magnetic dipole moment of the current loop is determined from the current and area of the loop while the magnetic torque on the loop is determined from the magnetic dipole moment.
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Set the cannon to have an initial speed of 20 m/s. For which situation do you think the cannon ball will go father: if it is set at a 60-degree angle, or if it is set at a 70-degree angle?
Question 2 options:
60 degrees
70 degrees
The cannon ball will go farther if the the angle of projection is set at 60 degrees
How to determine which angle will result in farther distance
Case 1:
Initial velocity (u) = 20 m/sAngle of projection (θ) = 60 ° Acceleration due to gravity (g) = 9.8 m/s²Horizontal distance (R) =?R = u²Sine(2θ) / g
R = 20² × Sine (2×60) / 9.8
R = 346.41 / 9.8
R = 35.35 m
Case 2:
Initial velocity (u) = 20 m/sAngle of projection (θ) = 70 ° Acceleration due to gravity (g) = 9.8 m/s²Horizontal distance (R) =?R = u²Sine(2θ) / g
R = 20² × Sine (2×70) / 9.8
R = 257.12 / 9.8
R = 26.24 m
From the above calculations, we can conclude that the ball will go farther, if the angle is 60 °
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If your 1000 kg car ran out of gas and you had to push it into the gas station with a
force of 50 N, what would the acceleration of the car be?
Answer:
a=0.05m/s²
Explanation:
force=mass × acceleration
f=ma
f=50N
m=1000kg
a=?
50=1000×a
50=1000a
a=50/1000
a=0.05m/s²
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A projectile is thrown at an angle 30° from horizontal. Which statement about its vertical component of velocity is true?
The true statement is the vertical component of the velocity is half of the magnitude of the velocity.
What is vertical component of velocity?
The vertical component of a velocity is the velocity of the object acting along vertical direction.
The vertical component of the velocity acting along the vertical direction is calculated as follows;
Vy = V sinθ
where;
Vy is the vertical component of the velocityθ is the angle of projectionV is the magnitude of the velocitySubstitute the value of the angle of projection and evaluate the vertical component of the velocity.
Vy = V sin(30)
Vy = V(0.5)
Vy = ¹/₂V.
Thus, the vertical component of the velocity is half of the magnitude of the velocity.
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The complete question is below
A projectile is thrown at an angle 30° from horizontal. Which statement about its vertical component of velocity is true?
the vertical component of the velocity is half of the magnitude of the velocitythe vertical component of the velocity is one-third of the velocitythe vertical component of the velocity is the same as the magnitude of the velocity.When is a secondary source more helpful than a primary source?
A. When you want to confirm the conclusion with your own experiment
B. When you are not an expert in the field being studied in the experiment
C. When you do not need to know the conclusion of the experiment
D. When you want to know the exact data values
Option A is the correct answer: A secondary source is more helpful than a primary source when you want to confirm the conclusion with your own experiment.
Primary and secondary sources work well together to support the argument you are trying to make. Although secondary sources demonstrate how your work connects to earlier research, primary sources are more reliable as evidence. When you want to confirm the conclusion of your experiment, secondary sources prove to be more useful.
Original research is built on primary sources. You are able to discover novel information, offer solid justification for your claims, and provide reliable facts on your subject.
Secondary sources are useful for getting a comprehensive perspective of your subject and learning other researchers' methods. They frequently combine numerous primary materials that would take a lot of time and effort to obtain independently.
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What is the magnification of a curved mirror if a 10.0 cm tall object is placed 8.00 cmfrom the mirror and produces an image 12.00 cm in front of the mirror?-0.6671.50-1.500.667
We will have that the magnification will be given by:
[tex]m=-\frac{v}{u}[/tex]That is:
[tex]\begin{gathered} m=-\frac{8cm}{12cm}\Rightarrow m=-\frac{2}{3} \\ \\ \Rightarrow m\approx-0.667 \end{gathered}[/tex]So, the magnification is approximately -0.667.
5. Was you hypothesis supported by the data collected for question 2? Why or why not ?
Answer:
Your hypothesis was not supported by the data collected
Explanation:
On question 2 you said that the point with the greatest kinetic energy would be point 5. However, when you analyze the data, the point with the greatest velocity was point 2 which means that this is the point with the greatest kinetic energy.
Therefore, your hypothesis was not supported by the data collected because based on the picture, it is hard to say where is the lowest point of the roller coaster.
I just started a new lesson and have this study guide, but the material is unfamiliar to me and
2.79 eV
Explanation:Given
The wavelength, λ = 445 nm
The planck's constant, h = 6.626 x 10^-34 m² kg/s
The speed of light, c = 3 x 10^8 m/s
The energy of the photon is calculated as:
[tex]\begin{gathered} E=\frac{hc}{\lambda} \\ \\ E=\frac{6.626\times10^{-34}\times3\times10^8}{445\times10^{-9}} \\ \\ E=\frac{6.626\times3}{445}\times10^{-34+8+9} \\ \\ E=0.0447\times10^{-17} \\ \\ E=4.47\times10^{-2}\times10^{-17} \\ \\ E=4.47\times10^{-19}J \end{gathered}[/tex]But, 1 eV = 1.6 x 10^-19J
[tex]\begin{gathered} E=\frac{4.47\times10^{-19}}{1.6\times10^{-19}} \\ \\ E=2.79eV \end{gathered}[/tex]
A tensile load of 190 kN is applied to a round metal bar with a diameter of 16mm and a gage length Of 50mm. Under this load the bar elastically deforms so that the gage length increases to 50.1349 mm and the diameter decreases to 15.99 mm. Determine the modulus of elasticity and Poisson s ratio for this metal.
The elasticity and Poisson's ratio for this metal is 0.232.
What is ratio?
The realation between two numbers which shows how much bigger one quantity is than another.
Sol-
As per the given question
P=190KN
d=16 mm
Lo=50mm
X=50.1349-50=0.1349mm
Y=15.99-16=-0.01mm
The formula-
E=ó/€
Ó=P/A
A=r/4 d^2 =π/4(16)^2=201.062 mm
ó={190(1000)}201.062=944.982 Mpa
E=944.982/0.002698=350.253 GPa
€y=-0.000625
v=0.232(answer)
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make G the subject of the formula
F = GMM²/1²
HENCE WRITE THE DIMENSION FOR G
The value of G is FR²/M₁M₂. and the dimension of G is [M⁻¹L³T⁻²]..
The provided formula is of gravitational force F between two objects,
F = GM₁M₂/R²
Where M₁ is the mass of first object and M₂ is the mas of the other object while R is the distance between there centers and G is the universal gravitation constant.
To find the dimension of G, making G the subject of formula,
G = FR²/M₁M₂.
As we know, unit of mass is Kilogram (Kg), unit of force is Newton (N) and unit of distance is Meter (M).
Putting all the values, Units in the place of quantities,
G = N.R²/Kg.Kg
Now, using Dimensional analysis, and writing the dimensions of all the other units,
G = [MLT⁻²][L²]/[M][M]
G = [ML³T⁻²]/[M²]
G = [M⁻¹L³T⁻²]
The dimensions of G are [M⁻¹L³T⁻²].
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A resistor uses energy at a rate of 2.50W when there is a current of 4.00A passing through it. What must be the potential difference across the resistor?1.25V1.50V0.625V10.0V
As we know
[tex]P=\text{ V}\times I;[/tex]Where,
P= electric power= 2.50 W
V= potential difference= ?
I= current = 4.00A
Using above formula we get,
[tex]\begin{gathered} P=V\times I; \\ \therefore2.50=\text{ V}\times4; \\ V=\text{ }\frac{2.50}{4}=\text{ 0.625V} \end{gathered}[/tex]Final answer is :- 0.625 V
What is the relationship of the force on the spring and stretch of the spring.(This is the spring we are looking at weights can be attached to it)
We will have the following:
We know that for a simple spring the following is true:
[tex]\begin{cases}F=-kx \\ \\ F=ma\end{cases}[/tex]So:
[tex]-kx=ma[/tex]So, the stretch of the spring is directly proportional to the force.
Students were experimenting with objects in a collision. Ball A moves with a constant acceleration by sliding down a frictionless incline plane before colliding with Ball B. Students used motion detectors to measure the velocity of Ball A at various points. Ball A (mass of 1.00 kg) began from rest, at position 1, on a ramp at a height of 1.25 m. At position 2, ball A was moving at 5.00 m/s. Ball A continues to roll at a constant 5.00 m/s when it collides with Ball B (mass of 1.00 kg) at position 3. Ball A comes to a complete stop. Ball B moves at a constant velocity after the collision.FORMULAS: PE = m•g•h (g=9.8m/s²) KE = 1/2•m•v² momentum = m•v
Given data:
* The mass of ball A is m_1 = 1 kg.
* The mass of ball B is m_2 = 1 kg.
* The initial velocity of ball A is u_1 = 5 m/s.
* The final velocity of ball A is v_1 = 0 m/s.
* The initial velocity of the ball B is u_2 = 0 m/s.
Solution:
According to the law of conservation of momentum, the net momentum of the system before the collision is equal to the net momentum of the system after the collision.
Thus,
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]Substituting the known values,
[tex]\begin{gathered} 1\times5+1\times0=1\times0+p_2 \\ 5+0=0+p_2 \\ p_2=5\text{ kgm/s} \end{gathered}[/tex]where p_2 is the momentum of the ball B,
Thus, the momentum of ball B after the collision is 5 kgm/s.
Hence, the third option is the correct answer.
3. A boulder rolls with speed of 3.5 m/s off a cliff. It hits the ground 2.25 m from the base ofthe ledge. A) How high is the ledge? B) How long did it take the boulder to fall to the bottomof the cliff?DrawingVerticalHorizontal
Given data
*The given distance from the base of the ledge is R = 2.25 m
*The given speed is v = 3.5 m/s
The diagram is given below
(a)
Let (h) be the height of the edge
The formula for the distance from the base of the ledge is given as
[tex]\begin{gathered} R=v\times t \\ R=v\times\sqrt[]{\frac{2h}{g}} \\ h=\frac{R^2\times g}{2v^2} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} h=\frac{(2.25)^2\times9.8}{2\times(3.5)^2} \\ =2.025\text{ m} \end{gathered}[/tex]Hence, the height of the ledge is h = 2.025 m
(b)
The formula for the time taken by the boulder to fall to the bottom of the cliff is given as
[tex]t=\sqrt[]{\frac{2h}{g}}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} t=\sqrt[]{\frac{2\times2.025}{9.8}} \\ =0.642\text{ s} \end{gathered}[/tex]Hence, the time taken by the boulder to fall to the bottom of the cliff is t = 0.642 s
Each set of protons and electrons represents a different atom. Place the atoms in order of their overall charge. Order them from most positive to most negative.swap_vert22 protons, 18 electronsswap_vert12 protons, 10 electronsswap_vert17 protons, 10 electrons
Protons have positive charge
Electrons have negative charge
• 22 protons (+22) + 18 electrons (-18 )
22 - 18 = 4 (positive)
• 12 -10 = 2
• 17 - 10 = 7
• 2 - 1 = 1
From most positive to most negative overall charges.
7 - 4 - 2 - 1
The tortoise and the hare A tortoise and a hare run to the East. The hare knows that it is faster, so it gives the tortoise a 30-meter head start. The tortoise is moving east at 1 m/s and the hare is moving east at 4 m/s.
The time taken for the hare to catch up with the tortoise is 10 seconds.
What is the time taken for the hare to catch the tortoise?
The time taken for the hare to catch up with the tortoise is calculated by applying the principle of relative velocity as shown below.
Distance travelled by hare + Distance travelled by tortoise = Total distance
V₁t + V₂t = d
where;
V₁ is the velocity of hareV₂ is the velocity of tortoiset is the time taken for them to meetd is the distance between themSince they are moving in the same direction, the relative velocity becomes
V₁t - V₂t = d
(V₁ - V₂)t = d
(4 - 1)t = 30
3t = 30
t = 30/3
t = 10 seconds
Thus, the time taken for the hare to catch up with the tortoise is 10 seconds.
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The complete question is below:
A tortoise and a hare run to the East. The hare knows that it is faster, so it gives the tortoise a 30-meter head start. The tortoise is moving east at 1 m/s and the hare is moving east at 4 m/s. At what time does the hare catch up with the tortoise?
The figure shows a standing wave oscillating at 100 Hz on a string. What is the wave speed ?
Answer:
The wavelength is 60 cm.
Explanation:
Speed = frequency x wavelength = 100 x 60 = 6000 cm/s
or if you wanted the answer in m/s
Speed = frequency x wavelength= 100 x 0.60 = 60 m/s
009 (part 1 of 2) 10.0 points When a water gun is fired while being held horizontally at a height of 1.19 m above ground level, the water travels a horizontal distance of 1.98 m. Find the initial velocity of the water. The acceleration of gravity is 9.81 m/s^2
Answer in units of m/s.
010 (part 2 of 2) 10.0 points A child, holding the same gun in a horizontal
position, slides down a 33.0◦incline at a constant speed of 1.40 m/s. The child fires the gun when it is 4.36 m above the ground and the water takes 0.868 s to reach the ground. How far will the water travel horizontally?
Answer in units of m.
Answer:
Speed = 4 m/s
Explanation:
Given:
009 (part 1 of 2)
H =1 .19 m
L = 1.98 m
g = 9.81 m/c²
____________
V₀ - ?
Equation of motion horizontally:
L = V₀*t (1)
Equation of vertical motion:
H = g*t² / 2 (2)
From equation (2):
Time:
t = √ (2*H / g) = √ ( 2*1.19 / 9.81) ≈ 0,49 s
From equation (1):
Horizontal speed:
V₀ = L / t = 1.98 / 0.49 ≈ 4 m/s
It takes 225 kJ of work to accelerate a car from 20.1 m/s to 28.1 m/s. What is the car's mass?
Answer:
THE REMAINIG WILL BE 75
Explanation:
HOPE IT HELPS YOU
A dog running of constant speed of 3m/s increases it's speed to 7m/s upon seeing a lion. if the mass of the dog is 20kg, the work it does in achieving the new speed is.......
A dog running of constant speed of 3m/s increases it's speed to 7m/s upon seeing a lion. if the mass of the dog is 20kg, the work it does in achieving the new speed is....... 400 J.
What is Speed?
Speed is the time rate at which an object is travelling along a path, whereas velocity is the pace and direction of an object's movement. In other words, velocity is a vector, whereas speed is a scalar valu
Work = Change in Kinetic energy
= ½m(v² - u²)
= ½ × 20 kg × [(7 m/s)² - (3 m/s)²]
= 10 kg × 40 m²/s²
= 400 J
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Calculate the depth in the ocean at which the pressure is three times the atmospheric pressure
ANSWER:
20.17 meters
STEP-BY-STEP EXPLANATION:
Given:
Pressure = Po = 1.013x10^5 Pa
Pressure at depth = P = 3Po
Density of sea water = 1025 kg/m^3
We can calculate the depth as follows:
[tex]\begin{gathered} P=P_o+d\cdot g\cdot h \\ \text{ we solve for h} \\ dgh=P-P_o \\ h=\frac{P-P_o}{d\cdot g} \end{gathered}[/tex]We replacing and calculate the depth:
[tex]\begin{gathered} h=\frac{3\cdot P_o-P_o}{1025\cdot9.8}=\frac{2P_o}{10045}=\frac{2\cdot1.013\cdot10^5}{10045} \\ h=20.17\text{ m} \end{gathered}[/tex]Therefore, the depth is equal to 20.17 meters
Information givenknown: mass of Christine=60 kgmass of cart= 22 kg mass of hailey=69The two girls on the cart to the left pushing off of Conner take .3833 s to reach a distance of 0.3m. Conner reaches the same distance in 0.2333s. What is the mass of Conner?
We will have the following:
We use conservation of momentum to solve, that is:
[tex]\begin{gathered} (60kg+22kg+69kg)(0.3m/0.3833s)=(m+22kg)(0.3m/0.233s) \\ \\ \Rightarrow118.1841899kg\ast m/s=(m+22kg)(\frac{300}{233}m/s) \\ \\ \Rightarrow m+22kg=91.78968976kg\Rightarrow m=69.78968976... \\ \\ \Rightarrow m\approx69.8 \end{gathered}[/tex]So, Conner's mass is approximately 69.8 kg.
29. [-/10 Points]DETAILSSERCPWA11 4.WA.010.An object of mass 0.77 kg is initially at rest. When a force acts on It for 2.9 ms It acquires a speed of 15.5 m/s. Find the magnitude (In N) of the average force acting on the object during the2.9 ms time interval.
We have:
m = mass = 0.77 kg
t = time = 2.9 ms = 0.0029 s
vf = final speed = 15.5 m/s
vi= initial speed = 0 m/s
Apply:
• F = m (vf-vi) / t
Replacing:
F = 0.77 (15.5 - 0 )/0.0029
F= 4,155.5 N
A 12 N force acts at a 25 degrees and an 8 N force acts at 65 degrees. Determine the magnitude and direction (include angle) of the resultant . Scale is 1cm = 1N
NEEDDD HELLPP ASAPPPPP
A 12 N force acts at a 25-degree angle, while an 8 N force acts at a 65-degree angle. The magnitude and direction (including angle) of the resultant force are 21.43.
Consider the formula for a force, F=M×Cos∝
As M is a Mass of an object and, ∝ Is the angle at which force is acting on an object.
[tex]F_{1}[/tex]=M × Cos ∝
M=12N, and Cos∝= Cos25 = 0.906
[tex]F_{1}[/tex] = 12× 0.906
∴ [tex]F_{1}[/tex]= 10.87N
For [tex]F_{2}[/tex] = 25×Cos65
= 10.56N
According to Superposition resulting force is the Sum of total forces.
[tex]F_{RESULTANT} = F_{1} + F_{2}[/tex]
= 10.87+ 10.56
=21.43N
∴The resultant force is 21.43.N
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Diffraction is:A.the difference in density of the compression and rarefaction parts of a sound wave.B.the change of frequency heard by an observer when sound waves come from a moving source.C.when waves suddenly appear in a medium without a source.D.the apparent bending of sound waves around obstacles.
Diffraction is a phenomenon that occurs when a wave hits an object or it passes through a small gap.
The propagation of the wave will have a circular pattern after the diffraction effect:
Therefore the correct option is D.
Show that entropy change due to heat transfer by conduction is given by
∆S=mC (4 marks)
ln
2
T
T
2
The change in the reservoir, the system or device, and the surroundings are added to determine the overall entropy change. The reservoir's entropy change is. Since entropy is a function of state and we are contemplating a complete cycle (return to initial state), the device's entropy change is zero.
For finite variations at constant T, entropy changes (S) are calculated using the relation G=ΔH - TΔS.
The generation, consumption, conversion, and exchange of thermal energy across physical systems is the focus of the thermal engineering field of study known as heat transfer. Different heat transmission techniques, including thermal conduction, thermal convection, thermal radiation, and energy transfer by phase changes, are categorized.
Conduction is the process through which heat is transported from an object's hotter end to its colder end. Heat naturally transfers from a hotter body to a colder body.
For instance, heat is transferred from the hotplate of an electric stove to the bottom of a saucepan that comes into touch with it.
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A ball is orbits in a circle of radius 10m with a speed 50 m/s. What is its angular velocity?
The angular velocity of the ball orbiting in circle will be 5 rad/s.
What is angular velocity?Angular velocity describes the rate at which an object rotates.
Given is a ball that orbits in a circle of radius 10m with a speed of 50 m/s.
The relation between the linear and angular velocity of a body is given by-
v = rω
where -
v is the linear velocity
r is the radius
ω is the angular velocity
On substituting the values, we get -
50 = 10 x ω
ω = 50/10
ω = 5 rad/s
Therefore, the angular velocity of the ball orbiting in circle will be
5 rad/s.
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For the following questions, refer to the position versus time graph above.a. What is the object's velocity att = 5 s? (2 sig figs)b. What is the object's velocity at t=40s? (2 sig figs)
We are given a graph of position vs time. We are asked to determine the speed at 5 seconds. To do that we need to determine the slope of the line above the time point of 5 seconds. To determine the slope of the line we will use the following formula:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]This means that we need to choose two points in the line. We use the graph to determine these points as follows:
The points we have chosen are:
[tex]\begin{gathered} (x_1,y_1)=(0,0)_{} \\ (x_2,y_2)=(10,5) \end{gathered}[/tex]Replacing in the formula for the slope we get:
[tex]m=\frac{5-0}{10-0}[/tex]Solving the operations:
[tex]m=\frac{5}{10}=\frac{1}{2}[/tex]Therefore, the speed at 5 seconds is 0.5 meters per second.
For part B the following points can be taken:
A rope pulls a 72.5 kg skier upa 21.7°slope with /k = 0.120.The rope is parallel to the slope,and exerts a force of 383 N. What isthe acceleration of the skier?(Unit = m/s?)Enter
A rope pulls a 72.5 kg skier upa 21.7°slope with /k = 0.120.The rope is parallel to the slope,and exerts a force of 383 N. The acceleration of the skier is approximately 2.97 m/s².
To find the acceleration of the skier, we need to use Newton's second law of motion and consider the forces acting on the skier.
Identify the forces acting on the skier:
The forces acting on the skier are the force of gravity (mg) and the force exerted by the rope (T), where m is the mass of the skier (72.5 kg) and g is the acceleration due to gravity (9.8 m/s²). The force exerted by the rope is parallel to the slope and can be calculated using the given value of 383 N.
Resolve the forces:
Since the rope is parallel to the slope, we need to resolve the force of gravity into components parallel and perpendicular to the slope. The component parallel to the slope is m * g * sin(21.7°).
Apply Newton's second law of motion:
Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, Fnet = ma.
Determine the net force:
The net force acting on the skier is the difference between the force exerted by the rope and the component of the force of gravity parallel to the slope. Fnet = T - m * g * sin(21.7°).
Calculate the acceleration:
Using Newton's second law, we can rearrange the equation Fnet = ma to solve for the acceleration (a). a = Fnet / m.
Substitute the values and solve:
Substitute the known values into the equation to find the acceleration.
Therefore, the acceleration of the skier is approximately 2.97 m/s².
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Answer:
about 0.567 m/s²
Explanation:
You want the acceleration of a 72.5 kg skier up a 21.7° slope with µk = 0.120, towed by a rope exerting a force of 383 N.
ForcesThe force up the slope is 383 N.
The forces down the slope will be the sum of the force due to gravity and the friction force.
GravityThe force down the slope due to gravity is ...
F = m·g·sin(θ) = (72.5 kg)(9.8 m/s²)(sin(21.7°) ≈ 262.7 N
FrictionThe force due to friction will be the product of µk and the force normal to the slope:
F = m·g·cos(θ)·µk = (72.5 kg)(9.8 m/s²)cos(21.7°)·0.120 ≈ 79.22 N
Net ForceThe net force up the slope is ...
383 N -262.7 N -79.22 N ≈ 41.08 N
This will accelerate a mass of 72.5 kg in the amount of ...
A = F/m = 41.08 N/(72.5 kg) ≈ 0.567 m/s²
The acceleration of the skier is about 0.567 m/s² up the slope.
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Additional comment
We don't have to figure the forces. Rather we can figure the gross acceleration due to the tow rope, then subtract the accelerations due to gravity and friction. This saves a few math operations as we don't have to multiply, then divide, by 72.5 kg.
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