The true statement among the following is that the air is not affected by human movement. Hence, option C is correct.
What is Air?Air relates to the atmosphere of the planet. Several gases and minute dust particles make up the air. Living organisms breathe and thrive in this pure gas. Its shape and volume are ill-defined. Considering that it is matter, it has mass and weight. Atmospheric pressure is generated by air weight. The space vacuum lacks air.
About 78% of the gas within air is nitrogen, 21% of the gas is oxygen, 0.9% of the gas is argon, 0.04% of the gas is co2, and very little other gas is present.
A typical amount of water vapor is around 1%.
Since respiration requires oxygen, animals must breathe it to survive. The lungs transfer back carbon dioxide back into the atmosphere when breathing, putting oxygen into the body.
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Calculate the potential energy of a 2 kg ball that is about to be dropped of a height of 25 m
Given:
The mass of the ball, m=2 kg
The height from which the ball is about to be dropped, h=25 m
To find:
The potential energy of the ball.
Explanation:
The type of potential energy that is stored in this ball is called gravitational potential energy. The gravitational potential energy is the energy that an object possesses due to its height. The gravitational potential energy is directly proportional to the height at which the object is situated.
The potential energy of the ball is given by,
[tex]E=\text{mgh}[/tex]Where g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} E=2\times9.8\times25 \\ =490\text{ J} \end{gathered}[/tex]Final answer:
The potential energy of the ball is 490 J
Which of the following is an appropriate measure of electric power on a toaster label?220 W55 Ω110 V2.0 A
A Watt is the unit of electrical power
How many cubic inches are there in 3.25 yd3?Express the volume in cubic inches to three significant figures.What is the mass in grams of 16.86 mL of acetone?Express your answer to four significant figures and include the appropriate units.What is the volume in milliliters of 7.06 g of acetone?Express your answer to three significant figures and include the appropriate units.
Answer:
[tex]\text{ 3.25 yd}^3=151,632in^3[/tex]Explanation: We need to convert cubic-yards into cubic inches, this can be simply done as follows:
[tex]\frac{46656\text{ Cubic inches}}{1\text{ Cubic Yard}}^{}[/tex]Therefore we have:
[tex]\begin{gathered} 3.25\text{ Cubic yards }\times\text{ }\frac{46656\text{ Cubic Inches}}{1\text{ Cubic Yard}}=151,632\text{ Cubic Inches} \\ \therefore\rightarrow \\ \text{ 3.25 yd}^3=151,632in^3 \end{gathered}[/tex]A car travels at an average speed of 60 km/h for 15 minutes.How far does the car travel in 15 minutes?D900 kmC 240 kmA4.0 km B 15 km
Notice that the speed is written using units of km/h and the time is written using units of minutes.
Convert the time interval to hours. Remember that 1 hour equals 60 minutes:
[tex]15\min =15\min \times\frac{1h}{60\min }=0.25h[/tex]The distance d that a particle travels during a time t if it moves at an average speed v is given by the formula:
[tex]d=v\cdot t[/tex]Replace v=60km/h and t=0.25h to find the distance traveled by the car:
[tex]d=(60\frac{km}{h})\times(0.25h)=15km[/tex]Therefore, the car travels 15km in 15 minutes.
A truck covers 40.0 m in 9.00 s while uniformly slowing down to a final velocity of 2.20 m/s.(a) Find the truck's original speed. m/s(b) Find its acceleration. m/s2
Given:
The distance covered by truck: d = 40.0 m
The time taken to cover the distance is: t = 9.00 s
The final velocity of the truck is: v2 = 2.20 m/s
To find:
a) the speed of the truck.
b) the acceleration
Explanation:
a)
The speed of the truck before it slows down can be calculated as:
[tex]d=\frac{1}{2}(v_2+v_1)t[/tex]Substituting the values in the above equation, we get
[tex]\begin{gathered} 40=\frac{1}{2}(2.20+v_1)\times9 \\ \\ \frac{40\times2}{9}-2.20=v_1 \\ \\ v_1=6.69\text{ m/s} \end{gathered}[/tex]b)
The truck is initially moving at a speed of 6.69 m/s. It then slows down to the final velocity of 2.20 m/s. The acceleration of the truck can be determined as:
[tex]d=v_1t+\frac{1}{2}at^2[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} 40=6.69\times9+\frac{1}{2}\times a\times9^2 \\ \\ 40=60.21+40.5a \\ \\ a=\frac{40-60.21}{40.5} \\ \\ a=-0.499 \\ \\ a\approx-0.5\text{ m/s}^2 \end{gathered}[/tex]Final answer:
a) The original speed of the truck is 6.69 m/s.
b) The acceleration of the truck is - 0.5 m/s^2.
what is the smallest amount of time in which the person can accelerate the car from rest to 23 m/s and still keep the coffee cup on the roof. The coefficient of the static friction is 0.21. The maximum acceleration of the car that is allowed so that the cup does not fall is 2.1 m/s^2
Given:
The coefficient of the static friction, μ=0.21
The maximum acceleration of the car so that the cup does not fall, a=2.1 m/s²
The initial velocity of the car, u=0 m/s
The final velocity of the car, v=23 m/s
To find:
The smallest amount of the time in which the car can accelerate so that the coffee cup will still be on the roof.
Explanation:
From the equation of motion,
[tex]v=u+at[/tex]Where t is the smallest amount of time in which the person can accelerate and still keep the cup on the car.
On rearranging the above equation,
[tex]t=\frac{v-u}{a}[/tex]On substituting the known values,
[tex]\begin{gathered} t=\frac{23-0}{2.1} \\ =10.95\text{ s} \end{gathered}[/tex]Final answer:
Thus the smallest amount of the time in which the person can accelerate the car at the given rate and still keep the cup on the roof of the car is 10.95 s
The word _____ in contrast,refers to the accumulation of such a borrowing, year after year
Answer:
Budget
Explanation:
Have good day!!!
A cheetah is running at a velocity of 8 m/s when it accelerptes at 1.5 m/s^2 for 6 seconds. If the
cheetah started at a position of 20m what is the final Position of the cheetah?
Answer:
[tex]95[/tex]
Explanation:
[tex]x= x. + vt+\frac{1}{2} at^{2}[/tex]
[tex]s=20 + (8*6)+ \frac{1}{2} (1.5 * 6^{2} )[/tex]
[tex]s= 95[/tex]
10. ABC.Per=1200 NNet Force:Pit=600 NEngen-SONFrid=20 NPry=800 NPrax=800 NF50NWhich situation above would best describe free fall velocity?Which situation above would best describe a crane lifting an object?If situation C had a Fapp of 40N to the right, the net force on the object would beIf situation Chad Fapp of 20N to the right, the forces would be (balanced, unbalanced) and thehorizontal velocity would be (constant, + accelerating, - accelerating). Circle the correct terms
When the body is under free fall its apperaent weight will be zero.
Therefore
If a 4 kg ball is dropped from rest and falls without air resistance, what is its speed after 0.5 seconds?
Answer:
4.9 m/s
Explanation:
The velocity of the ball after 5 seconds can be calculated using the following equation
[tex]v_f=v_i+at[/tex]vi = the initial velocity, in this case, it is equal to 0 because the ball is dropped from the rest
a = acceleration, this is the acceleration due to gravity so it is -9.8 m/s²
t = time, it is equal to 0.5 s
So, replacing the values, we get:
[tex]\begin{gathered} v_f=0-9.8(0.5) \\ v_f=-4.9\text{ m/s} \end{gathered}[/tex]Therefore, the speed after 0.5 seconds is 4.9 m/s
To make peanut butter, a machine grinds peanuts into a paste. Which of the following can be considered as a process of this system?Question 16 options:PeanutsMachine grinds peanutsPeanut butter
A process is an action that involves individual items, not the items themselves.
Answer: Machine grinds peanuts
An air compressor has a volume of 100.L What volume of gas is pumped into the tank if the pressure goes from 750 torr to a pressure of 145 psi?Remember to convert pressure to atm. Refer to picture for conversions if needed.
ANSWER:
1000 L
STEP-BY-STEP EXPLANATION:
The first thing is to convert the unit of both pressures to atm, with the help of the conversion equivalences shown, just like this:
[tex]\begin{gathered} P_1=750\text{ torr }\cdot\frac{1\text{ atm}}{760\text{ torr}}=0.987\text{ atm} \\ P_2=145\text{ psi }\cdot\frac{1\text{ atm}}{14.7\text{ psi}}=9.87\text{ atm} \end{gathered}[/tex]Now, applying Boyle's Law, we calculate the value of the volume, like this:
[tex]\begin{gathered} P_1\cdot V_1=P_2\cdot V_2 \\ V_1=\frac{P_2\cdot V_2}{P_1} \\ \text{ replacing} \\ V_1=\frac{9.87\cdot100}{0.987}_{} \\ V_1=1000\text{ L} \end{gathered}[/tex]The volume of gas is 1000 L
a) your one friend and their bumper boat (m = 210 kg) are traveling to the left at 3.5 m/s and your other friend and their bumper boat (m = 221 kg) are traveling in the opposite direction at 1.8 m/s. The two boats then collide in a perfectly elastic one-dimensional collision. How fast is your first friend and their boat moving after the collision?b) after the collision, a constant drag force between the water and the boat causes your first friends boat to come to a stop. If the boat travels 2.7 m before stopping, what is the magnitude of the constant drag force?
ANSWER:
a) 1.757 m/s
b) 119.91 N
STEP-BY-STEP EXPLANATION:
Given:
Mass 1 (m1) = 210 kg
Initial speed 1 (u1) = 3.5 m/s
Mass 2 (m2) = 221 kg
Initial speed (u2) = 1.8 m/s
We make a sketch of the situation:
a)
We make a momentum balance by taking into account the conservation of momentum:
[tex]\begin{gathered} m_1u_1-m_2u_2=m_1v_1+m_2v_2 \\ \\ v_2=\frac{m_1u_1-m_2u_2-m_1v_1}{m_2}\rightarrow(1) \end{gathered}[/tex]Now an energy balance taking into account the conservation of energy, as follows:
[tex]\begin{gathered} \frac{1}{2}m_1(u_1)^2+\frac{1}{2}m_2(u_2)^2=\frac{1}{2}m_1(v_1)^2+\frac{1}{2}m_2(v_2)^2 \\ \\ m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+m_2\lparen v_2)^2\rightarrow(2) \end{gathered}[/tex]Now, we substitute equation (1) in (2) and we get the following:
[tex]\begin{gathered} m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+m_2\left(\frac{m_1u_1-m_2u_2-m_1v_1}{m_2}\right)^2 \\ \\ m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+m_2\left(\frac{\left(m_1u_1\right)^2-2\left(m_1u_1\right)\left(m_2u_2\right)-2\left(m_1u_1\right)\left(m_1v_1\right)+\left(m_2u_2\right)^2+2\left(m_2u_2\right)\left(m_1v_1\right)+\left(m_1v_1\right)^2}{(m_2)^2}\right) \\ \\ m_1\left(u_1\right)^2+m_2\left(u_2\right)^2=m_1\left(v_1\right)^2+\left(\frac{\left(m_1u_1\right)^2-2\left(m_1u_1\right)\left(m_2u_2\right)-2\left(m_1u_1\right)\left(m_1v_1\right)+\left(m_2u_2\right)^2+2\left(m_2u_2\right)\left(m_1v_1\right)+\left(m_1v_1\right)^2}{m_2}\right) \\ \\ v_1=u_1\frac{m_1-m_2}{m_1+m_2}+u_2\frac{2m_2}{m_1+m_2} \\ \\ \text{ Now, we substitute each value, like so:} \\ \\ v_1=3.5\cdot\frac{210-221}{210+221}+1.8\cdot\frac{2\cdot221}{210+221} \\ \\ v_1=1.757\text{ m/s} \end{gathered}[/tex]b)
We use the following formula to determine the force:
[tex]\begin{gathered} v^2=u^2+2as \\ \\ \text{ Wee replacing:} \\ \\ (1.756)^2=0^2+2\cdot a\cdot2.7 \\ \\ a=0.003375\text{ m/s}^2 \\ \\ \text{ Therfore:} \\ \\ F=m\cdot a=210\cdot0.571=119.91\text{ N} \end{gathered}[/tex]A worker is holding a filled gas cylinder still. Which two sentences are true about the energy of the filled gas cylinder?
A man in a blue dress holding a red color cylinder
It has no energy because it’s being held still.
It has gravitational potential energy because of its height.
Its atoms and molecules have thermal energy.
It has motion energy because it will fall if let go.
Its kinetic energy is being converted to potential energy.
The two sentences that are true about the energy of the filled gas cylinder are;
(b) It has gravitational potential energy because of its height
(d) It has motion energy because it will fall if let go.
What is the principle of conservation of conservation of energy?The principle or law of conservation of energy states that energy can neither be created nor destroyed but can be converted from one form to another.
Based on this law, the kinetic energy of an object can be converted into potential energy and vice versa.
A filled gas cylinder held above the ground possesses gravitational potential energy and if the gas cylinder is held still, the kinetic energy is zero.
Thus, we can conclude that the following statements are true;
It has gravitational potential energy because of its height.It has motion energy because it will fall if let go.Learn more about conservation of energy here: https://brainly.com/question/166559
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How much kinetic energy does Usain Bolt (m=94kg) have when he hits his top
speed of 12 m/s?
Answer:
6768 Joules (J)
Explanation:
kinetic energy = 1/2mv^2
1/2 (94x12^2) = 6768
What is the image distance if a 5.00 cm tall object is placed 2.33 cm from a converging lens with a focal length of 5.75 cm?0.603cm1.66cm-0.255cm-3.92cm
We will have the following:
First, we will recall that:
[tex]\frac{1}{f}=\frac{1}{v}+\frac{1}{u}[/tex]That is:
[tex]\begin{gathered} \frac{1}{5.75}=\frac{1}{2.33}+\frac{1}{u}\Rightarrow\frac{1}{u}=-\frac{1368}{5359} \\ \\ \Rightarrow u=-\frac{5359}{1368}\Rightarrow u\approx-3.92 \end{gathered}[/tex]So, the image distance is approximately -3.92 cm.
Explain why clothes stick together when they are removed from a drier. What is static electricity?
ANSWER:
What happens in clothes is a phenomenon called static cling is a phenomenon caused by static electricity. When dry materials rub against each other, they can exchange electrons, creating an electrical charge. This charge can build up in the form of static electricity and cause two objects, in this case clothing, to stick or stick together.
When the substance that loses electrons becomes positively charged and the substance that gains electrons becomes negatively charged. These charges are stationary and remain on the surface of the material. Since there is no flow of electrons, this is called static electricity.
A 3000-kg satellite orbits the Earth in a circular orbit 11797 km above the Earth's surface (Earth radius = 6380 km, Earth Mass = 5.97x10^24 kg). Reminders:Distance should be in meters, not kilometers. 1000 m = 1 km.The total radius needed for the problem is r=r earth + hightWhat is the gravitational force (in newtons, N) between the satellite and the Earth?Hint: The radius of the Earth + the height of the orbit = the center-to-center distance needed for the equation. You also need the universal gravitational constant (G), which is not 9.81 m/s^2. Be careful.Fg=Gm1m2/r2Answer: __________ N
We have:
m1 = mass 1 = 3000 kg
h = height = 11797 km = 11797000 m
r2 = 6380 km = 6380000
m2 = mass 2 = 5.97x10^24 kg
G = gravitational constant = 6.6743 × 10-11 Nm^2 /kg^2
r= distance = h + r2 = 11797000 m + 6390000 m = 18,177,000 m
Apply:
Fg = G m1m2/ r^2
Replacing:
Fg = 6.6743 × 10-11 Nm^2/kg^2 ( 3000 kg * 5.97x10^24 kg ) / (18,177,000 m)^2
Fg= 3,617.9 N
A student on skateboard pushes off from the top of small hill with a apees of 2.0m/s, and then geos down the hill with a constant acceleration of 0.5 m/s2
After traveling a distancie 12.0m, how fast is the student going?
The final velocity of the student after travelling 12 m is 4 m/s.
What is the final velocity of the student?
The final velocity of the student is determined by applying the following Kinematic equation.
v² = u² + 2as
where;
u is the initial velocity of the studentv is the final velocity of the studenta is the acceleration of the students is the distance travelled by the studentv² = (2)² + 2(0.5)(12)
v² = 16
v = √16
v = 4 m/s
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Which of the following circuits can be used to measure the resistance of the heating element, shown as a resistor in the diagrams below?
In order to measure the resistance in the circuit, we need to know the voltage V and the current I in the circuit, this way we can calculate the resistance using the formula:
[tex]R=\frac{V}{I}[/tex]In order to calculate the current, we can use an amperemeter that must be in series with the circuit, this way it will not affect the circuit.
And in order to calculate the voltage, we can use a voltmeter that must be in parallel with the resistance, this way it will not affect the circuit.
The correct option that shows an amperemeter in series and a voltmeter in parallel is the fourth option.
A boy walks 30m [E25°S] then 60m [E40°N]. Determine his net displacement.
Given,
The distances; a=30 m
b=60 m
Angles; θ=E25°S
α=E40°N
From the diagram, ∠A is given by,
[tex]\angle A=180\degree-\theta-\alpha[/tex]On substituting the known values,
[tex]\begin{gathered} \angle A=180\degree-25\degree-40\degree \\ =115\degree^{} \end{gathered}[/tex]From the cos rule,
[tex]d^2=a^2+b^2-2ab\cos A[/tex]On substituting the known values,
[tex]\begin{gathered} d^2=30^2+60^2-2\times30\times60\times\cos 115\degree \\ \Rightarrow d=\sqrt[]{30^2+60^2-2\times30\times60\times\cos 115\degree} \\ =77.6\text{ m} \end{gathered}[/tex]Thus the total displacement of the boy is 77.6 m
critical mass depends on ___. Check all that apply.A. the polarityB. the purityC. the densityD. the shape
Related to the amount of a fissionable material's critical mass depends on different factors.
These factors are:
- the shape of the material
- the density
- the purity
all last factors are related to the critical mass, becasue of all of them change the efficiency at which neutrons continue the fission procedure.
Scientists might make a computer model of volcanic eruptions. What is the
biggest benefit of this model?
Answer:
Computer Model May Help to More Accurately Predict Volcano Eruptions. Scientists at the GFZ German Research Center in Potsdam, Germany, have developed a computer model which they say boosts the accuracy of volcanic eruption prediction.
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According to Newton’s second law of motion,how much force will be required to accelerate an object at the same rate if it mass is reduced by half ?
According to Newton's second law of motion,
[tex]\text{Force = mass}\times acceleration[/tex]Let the initial force be F, acceleration be a and the initial mass be m.
The acceleration is the same but now the mass is reduced by half.
So, the force will be
[tex]\begin{gathered} F^{\prime}=\frac{m}{2}\times a \\ =\frac{F}{2} \end{gathered}[/tex]Thus, the force will also be half of the initial force if the mass is reduced by half.
Please help with Question(ii). I don't understand the shown step of calculating the momentum of ball B. Especially after the third line 12+Pb=15.
Given:
m1 = mass 1 = 1kg
v1= initial velocity 1 = 12 m/s
m2= mass 2 = 3 kg
P after = momentum after collision = 15 kgm/s
(i)
Momentum of Ball A before collision
Momentum = mass x velocity
Pa = m1 v1
Replacing with the values given:
Pa = (1 kg) (12 m/s) = 12 kg m/s
(ii)
Momentum before = momentum after
Pa + Pb = P after
12 + Pb = 15
Since The ball B is travelling North, the distances travelled form a right triangle:
Apply pythagorean theorem:
c^2 = a^2 + b^2
Where c is the hypotenuse= P after = 15
a & b are the other 2 legs of the triangle = Pa and Pb
Replacing:
15^2 = 12^2 + Pb^2
Solve for Pb
15^2 - 12^2 = pb^2
√15^2 -12^2 = Pb
pb= 9 kgms^2
A car starts from rest and travels for 9.0 s with a uniform acceleration of +2.4 m/s²?. The driver then applies the brakes, causing a uniform acceleration of -2.5 m/s². If the brakes areapplied for 2.0 s, determine each of the following(a) How fast is the car going at the end of the braking period?m/s(b) How far has the car gone?m
a) At the end of the braking period, the car is moving at a speed of 16.6m/s
b) The car has traveled a total distance of 135.4m
Explanations:The car starts from rest
The initial velocity, u = 0 m/s
Uniform acceleration, a = 2.4 m/s²
time, t = 9.0 seconds
Find the final velocity when the car accelerates at 2.4m/s² using the equation
v = u + at
v = 0 + 2.4(9)
v = 21.6 m/s
The distance covered when the car accelerates at 2.4m/s²
s = ut + 0.5at²
s = 0(9) + 0.5(2.4)(9²)
s = 97.2 m
The distance covered when the car accelerates at 2.4m/s² is 97.2 m
The driver then applied a brake for 2.0 s and accelerates at -2.5m/s²
The initial velocity now becomes 21.6 m/s
That is, u = 21.6 m/s
t = 2 seconds
a = -2.5m/s²
The final speed v is calculated as:
v = u + at
v = 21.6 + (-2.5)(2)
v = 21.6 - 5
v = 16.6m/s
At the end of the braking period, the car is moving at a speed of 16.6m/s
The distance covered during the braking period is calculated as:
[tex]\begin{gathered} s\text{ = (}\frac{u+v}{2})t \\ s\text{ = }\frac{21.6+16.6}{2}\times2 \\ s\text{ = }\frac{38.2}{2}\times2 \\ s\text{ = 38.2 m} \end{gathered}[/tex]The car traveled a distance of 38.2 m during the braking period
Total distance covered = 97.2m + 38.2m
Total distance covered = 135.4m
The car has gone a distance of 135.4m
Changes of state occur at segment _____________ and segment_______________.1st blankA-BB-CC-D2nd blank C-DD-EE-F
The change of phase takes place when the amount of heat added is changing the state and the temperature of the system remains constant.
In the given graph, the ice is changed into the water state through segment B-C.
Then the water is changed into a water vapor state in the segment D-E.
Hence, changes of state occur at segment B-C and segment D-E.
A spring of length 9.7 meters stretches to 9.8 meters when a 0.4 kg mass is hung vertically from one end. What is the spring constant?
Given,
The initial length of the spring, l=9.7 m
The length of the spring after stretching, L=9.8 m
The mass, m=0.4 kg
The magnitude of the restoring force of the spring due to the stretching from the mass will be equal to the force applied by the mass, which is nothing but the weight of the mass.
Thus,
[tex]\begin{gathered} mg=k\Delta x \\ =k(L-l) \end{gathered}[/tex]Where g is the acceleration due to gravity, k is the spring constant, and Δx is the stretch in the length of the spring.
On substituting the known values,
[tex]\begin{gathered} k=\frac{mg}{(L-l)_{}} \\ =\frac{0.4\times9.8}{9.8-9.7} \\ =\frac{3.92}{0.1} \\ =39.2\text{ N/m} \end{gathered}[/tex]Thus the spring constant is 39.2 m
The acceleration of gravity depends on (click all that apply)
The expression for the acceleration due to gravity can be given as,
[tex]g=\frac{GM}{R^2}[/tex]Here, g is the acceleration due to gravity, G is the gravitational constant, M is the mass of planet and R is the distance from the center of planet.
Therefore, the acceleration due to gravity depends upon the distance from center of planet and the mass of planet.
100 POINTS
A man pulls a crate with a rope. The crate slides along the floor in the horizontal direction (x direction). The man exerts a force of 50 N on the rope, and the rope is at an angle . Describe how the force components change as the angle increases from 0° to 90° and use your graph to explain your answer. Give a detailed explanation of the forces at . Show a sample calculation at one angle for both components.
The exerted 50 N force at an angle on the crate can be resolved into a horizontal and vertical force component in which the horizontal component, Fₓ, decreases, while the vertical force component, [tex]F_y[/tex], increases as the value of the angle formed by the rope, θ, increases from 0° to 90°.
What is a component of a force?The components of a force are the force parts acting in perpendicular directions, horizontal and vertical, that combine to give the specified force.
The direction the crate is sliding = The horizontal, x-direction
The force exerted by the man = 50 N
The angle of the rope = θ
The components of the force are therefore:
Horizontal component, Fₓ = 50 × cos(θ)
Vertical component, [tex]F_y[/tex] = 50 × sin(θ)
The value of cos(θ) and sin(θ) as the angle the rope makes with the horizontal, θ, increases from 0° to 90° are as follows:
[tex]\begin{center} \begin{tabular}{ |c|c |c |} \theta & cos(\theta) & sin(\theta) \\ 0^{\circ} & 1 & 0 \\ 30^{\circ} & \sqrt{3} /2 & 0.5 \\ 45^{\circ}&\sqrt{2}/2 &\sqrt{2}/2 \\ 60^{\circ}&0.5&\sqrt{3}/2 \\ 90^{\circ} &0&1\\ \end{tabular}[/tex]
Therefore, the horizontal component of the force exerted by the man, Fₓ has a maximum value at θ = 0, and decreases to 0, as θ increases from 0° to 90°.
The vertical component of the force exerted, [tex]F_y[/tex], has a minimum value of 0 at θ = 0°, and the value of sin(θ) and therefore [tex]F_y[/tex], increases to a maximum of (sin(90°) = 1) 50 N as increases to 90°.
Please find attached the graph showing the components of the force, Fₓ, and [tex]F_y[/tex], exerted by the man as the angle formed by the rope increases from 0° to 90°
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