Answer:
If the resistors are in series then each resistor
carries the "same current" - the first option
URGENT!!
A runner completes a 10k race (10 km) at an average speed of 13. km/hr. How long does it take for her to complete the race?
Answer: 46 min
Explanation:
Imagine that you could increase the gravitational force on Earth to 200% its current force. What would life be like?
Be sure to answer these questions in your response:
• How would your weight change?
• What challenges would this increase cause?
• What benefits would this increase bring about?
• Would you choose to keep the gravity increase? Why or why not?
(100 POINTSS, ASAP PLS!)
Answer:
You weight would be doubled because gravitational force is doubled
2. Challenges:
a. Everyone’s body would have to adjust to the new weight and learn to walk again
b. Earth would be closer to the sun, so everything would be hotter on Earth
c. Flying airplanes and driving cars might no longer work at all (talk about how difficult this would make traveling)
3. Benefits:
a. Time would go by much slower so technically everyone would live longer
b. Balloons wouldn’t fly off anymore
4. You have to decide for yourself if you’d like it better or worse
Explanation:
A car coasts into a hill at 16.0 m/s. It slows down with a uniform acceleration of -1.0 m/S/S.
What is the displacement after 5.0s?
What is the displacement after 10.0s?
In the given question, a car coasts into a hill at 16.0 m/s and it slows down with a uniform acceleration of -1.0 m/S/S.
to find,
the displacement after 5.0s, s=60m
the displacement after 10.0s, s=80m
according to the equation of motion,
s=ut+1/2at^2
a) t=5second, u=16m/s
s=ut+1/2at^2
s= 16*5 + 1/2 (-1.6) 5^2
s=80-20
s=60m
b)t= 10second , u=16m/s
s=ut+1/2at^2
displacement s=16*10 + 1/2 (-1.6)*10^2
displacement s=80m
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heat is added to a 3.0 kg piece of ice at a rate of how long will it take for the ice at to melt? (for water and heat is added to a 3.0 kg piece of ice at a rate of how long will it take for the ice at to melt? (for water and 1,000 s 640,000 s 1.6 s 0.0 s
It will take 1.6s for the ice to melt.
First we have to write the expression for the heat balance equation for the system
Q supply = Qgain/
t= mLf/t
Here, m is the ice,
Lf is the latent heat of fusion
and t is the time taken to melt the ice.
Now we have to substitute the required value in the above expression to determine the value of time.
636 kJ/s = (3 kg)(334 kJ/kg) / tt =1.575 s
What is latent heat?
The heat or energy that is absorbed or released during a substance's phase shift is known as latent heat. It could go from a solid to a liquid or from a liquid to a gas, or vice versa. Enthalpy, a characteristic of heat, is connected to latent heat.
Therefore, the closest option is 1.6s
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shows an overhead view of three particles on which external forces act. the magnitudes and directions of the forces on two of the particles are indicated. what are the magnitude and direction of the force acting on the third particle if the center of mass of the three-particle system is (a) sta
The net force's work is described as d=(3.82)(4.00)=15.3J for W=F net, where we made use of the fact that the canister began at rest and traveled horizontally while being affected by horizontal forces; the resulting effect is represented by the symbol F net.
Since all of these cases are in closed systems, the result is in a new preserved quantity.
Linear momentum and this new quantity, angular momentum, are comparable. In this chapter, we first define angular momentum and then examine it from a number of angles.
But first, we look into a single particle's angular momentum. This enables us to compute the force angular momentum of a rigid cylindrically symmetric body and a system of particles.
We used the fact that the canister started at rest and moved horizontally under the influence of horizontal forces; the ensuing effect is denoted by the symbol F (force) net.
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Amber walked to the swimming pool, stopping to talk to Maria on her way there.
Amber walked at a constant speed of 100 m/min for the first four minutes; then she
visited with Maria for three minutes; then walked for two more minutes at a constant
speed of 75 m/min. In words, describe how a distance-versus-time graph of Amber's
progress would look.
Answer:
Amber use 25m for looking
Answer:
25 min's, that's what Amber used
if the temperature of star b is twice the temperature of star a, what can we say about the energy emitted by the surface of star b compared to the energy emitted by star a?
The blackbody radiation energy emitted by star a is 1/16 factor of star b energy.
We need to know about black body radiation to solve this problem. The energy radiated by a black body object is proportional to the area and the fourth power of temperature. It can be determined as
P = A . e . σ . T⁴
where P is power, A is surface area, e is emissivity, σ is Stefan Boltzmann's constant ( 5.67 x 10¯⁸ W/m²K⁴) and T is temperature.
From the question above, we know that
Tb = 2 Ta
By using the black body radiation formula, the ratio of temperature is
Ta⁴ / Tb⁴ = Pa / Pb
Ta⁴ / Tb⁴ = Ea / Eb
Hence,
Ta⁴ / Tb⁴ = Ea / Eb
Ta⁴ / (2Ta)⁴ = Ea / Eb
Ta⁴ / 16Ta⁴ = Ea / Eb
1/16 . Eb = Ea
Thus, the energy of star a will have 1/16 factor of star b energy
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a 1.0 kg is dropped from a height of 6.0m. at what height is the rock's kinetic energy twice its potential energy?
Answer:
2.0m
Explanation:
three uncharged capacitors with equal capacitances are combined in parallel. the combination is connected to a 6.99 v battery, which charges the capac
..............................
The capacitance of each capacitor is 2.1 ×[tex]10^{-5}[/tex]F.
Let
Equivalent capacitance of the circuit = [tex]C_{eq}[/tex]
[tex]C_{eq}[/tex] =Q/V
where Q is the charge and V is the potential difference.
Q = [tex]3.5[/tex]×[tex]10^{-4[/tex]
V = 5.55 V
Putting the values
[tex]C_{eq}[/tex] = [tex]3.5[/tex]×[tex]10^{-4[/tex]/5.5V
[tex]C_{eq}[/tex] = [tex]6.4[/tex]×[tex]10^{-5}[/tex]F
The circuit consists of 3 capacitors in parallel, each one having the same capacitance C.
To find C:
C = [tex]C_{eq}[/tex]/3
C = 2.1 ×[tex]10^{-5}[/tex]F
The capacitance of each capacitor is 2.1 ×[tex]10^{-5}[/tex]F.
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The full Question is :
Three uncharged capacitors with equal capacitances are combined in parallel. the combination is connected to a 5.55-v battery, which charges the capacitors. the charging process involves 3.45 Ã 10-4 c of charge moving through the battery. find the capacitance of each capacitor.
For the velocity-time graph shown, which statement describes what happens to the velocity between approximately 24 s and 25 s?
A) The lander's velocity increases away from the reference.
B) The lander's velocity decreases toward the reference.
C) The lander's velocity decreases away from the reference.
D) The lander's velocity increases toward the reference.
The velocity–time graph of the Lunar Landing Powered Descent that has a domain of 10 ≤ t ≤ 25, and a range of -2 ≤ v ≤ -40, between the 24th and the 25th seconds gives;
C) The Lander's velocity decreases away from the reference
What is a velocity–time graph?A velocity time graph is a graph that shows the variation of the velocity of an object with time.
The information with regards to the lunar landing powered descent graph are;
Coordinate points on the graph;
(10, -2); At time t = 10 seconds, the average velocity, v = 0
(10.5, -40); The velocity relative to the starting point changed and decreased from 0 m/s to -40 m/s
(15, -5); The velocity increased from -40 m/s to -5 m/s
(22, -3); The velocity increases from -5 m/s to -3 m/s
(24, -2); The velocity increases from -3 m/s to -2 m/s
(25, -15); The velocity decreases from -2 m/s to -15 m/s
Therefore, taking the reference point as the point where the velocity is zero, 0, we have that between approximately 24 seconds and 25 seconds, the Lander's velocity decreases away from the reference.
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A ball is released from rest and falls 125 meters in 5 seconds. What was the magnitude of its acceleration?
The magnitude of the acceleration of the ball is 10 m/s².
What is acceleration of an object?
The acceleration of an object is the change in the velocity of the object with time.
The acceleration of the object is calculated using the following kinematic equation as shown below,
h = vt + ¹/₂at²
where;
h is the height of the objecta is the acceleration of the objectt is the time of motionv is the initial velocity of the ball = 0Substitute the given parameters and solve for the acceleration of the ball.
125 = 0 + ¹/₂a(5)²
2(125) = 25a
250 = 25a
a = 250/25
a = 10 m/s²
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modulus of oil is 5 * 10 ^ 9 * Pa and its compressibility is 2 * 10 ^ - 5 * at * m ^ 1
Exercise 9: The brass plate of an outdoor sculpture experiences shear forces in an earthquake. The plate is 0.8m ^ 2 and 0.5 cm thick. What is the force exerted on each of its edges if the resulting displacement x is 0.16 mm? (Shear modulus = 3.5 * 10 ^ 10 * Pa ) .
Its edges are subjected to a force of 2.8 x 1010 N at the specified shear stress and area.
The force known as shear stress is the tendency for a material to deform by slippage along a plane or planes parallel to the applied stress. Shear stress or pressures, as their name suggests, tend to "shear" objects or produce material deformation.
The formula for shear stress is given as:
F = Aτ
where;
F is the force applied, A is the area of the surface, and τ is the shear stress
We have,
Shear modulus, τ = 3.5 × 10¹⁰ Pa
The area of the plate is A = 0.8 m²
Then,
F = ( 0.8 m² ) × (3.5 × 10¹⁰ Pa)
F = 2.8 x 10¹⁰ N
As a result, as the shear stress grows, so does the force applied to each of its edges that causes the specified displacement.
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Estamos bajando dos cajas de madera con frutos menores (una encima de la otra) por una rampa con una inclinación de 30 grados con la horizontal. La caja inferior tiene una masa de 80kg y su coeficiente de fricción con el suelo es de 0.3, y la caja superior tiene una masa de 30kg con un coeficiente de fricción con la caja inferior de 0.8. Calcula la fuerza que tenemos que ejercer para que ambas cajas bajen con velocidad constante de 0.5m/s.
La situación planteada se muestra a continuación:
(El diagrama mostrado es para la caja inferior). A continuación mostramos el diagrama para la caja superior:
Si consideramos el sistema de dos cajas como uno solo tenemos el siguiente sistemas de ecuaciones:
[tex]\begin{gathered} F+F_f-W_x=0 \\ N-W_y=0 \end{gathered}[/tex]De donde:
[tex]\begin{gathered} F=W_x-F_f \\ F=W_x-\mu N \\ F=W_x-\mu W_y \\ F=(80)(9.81)\sin 30-(0.3)(80)(9.81)\cos 30 \\ F=188.5 \end{gathered}[/tex]Ahora debemos determinar si la caja de arriba no se resbala al aplicar esta fuerza, para esto debemos recordar que la fricción es máxima con el coeficiente de fricción estático.
Entonces tenemos que la fricción es máxima para el bloque de arriba cuando:
[tex]F_f=(0.8)(30)(9.81)\cos 30=203.89[/tex]Dado que la fuerza que debemos aplicar no es mayor a la fricción máxima concluimos que ambas cajas bajaran al mismo tiempo. Por lo tanto la fuerza que debemos aplicar es de 188.5 N.
If you slide a box across the floor, it produces friction. The friction generates thermal energy. Which object/s would be warmer from the heat generated?
The temperature of the box and floor eventually rise as a result of the thermal energy flowing as heat inside these two things.
This is further explained below.
What is thermal energy?Generally, The energy present in a system that determines its temperature is referred to as thermal energy.
Thermal energy flows as heat. Thermodynamics is a whole field of physics that studies how heat is transmitted across various systems and how work is performed in the process (see the first law of thermodynamics).
In conclusion, This thermal energy is transferred in the form of heat throughout the box as well as the floor, which eventually results in an increase in temperature for both of these locations.
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A stone propelled from a catapult with a speed of 50m|s attains a height of 100m. Calculate:
a) the time taken
(b) the angle of projectile
(c) the range attained
For the projectile motion of the stone a) The time taken T = 9.04s b) the angle of the projectile = 62.31° c) the Range attained R = 209.923m
A stone propelled from a catapult is an example of projectile motion.
Given, the initial velocity of the stone u = 50 m/s; This initial velocity with which the stone is thrown has two components 1) vertical [tex]u_{y}[/tex] = u sinθ along the Y axis and 2) horizontal [tex]u_{x}[/tex]=u cosθ along the X axis
where θ is the angle of projection
maximum vertical displacement or height attained by stone = H
(Use sign convention +ve for upwards and -ve for downwards)
a) equation for the maximum vertical height reached by a body during projectile motion [tex]H=\frac{u^{2} sin^{2}\theta }{2g}[/tex] (1)
equation for the time of flight of projectile motion [tex]T=\frac{2usin\theta}{g}[/tex] (2)
Given H = displacement in the y direction = height attained = 100m
u = initial velocity = 50m/s ; acceleration due to gravity g =9.8 m/[tex]s^{2}[/tex]
using equation (1)
[tex]100=\frac{50^{2}sin^{2}\theta }{2*9.8}[/tex]
⇒[tex]sin^{2} \theta= \frac{100*2*9.8}{50^{2} }[/tex]
⇒sinθ[tex]=\sqrt{\frac{10*2*9.8}{50^{2} } }[/tex]
⇒sinθ = 0.885437
⇒θ = [tex]sin^{-1} (0.885437)[/tex]
⇒θ = 62.31°
Hence from equation (2) time of flight
[tex]T=\frac{2*50*sin\theta}{9.8}[/tex]
⇒[tex]T=\frac{2*50*0.885437}{9.8}[/tex]
⇒T = 9.04 s
Time is taken T =9.04s
b) Angle of projectile θ = 62.31°
c) The range attained [tex]R=\frac{u^{2}sin2\theta }{g}[/tex]
⇒[tex]R = \frac{50^{2}sin(2*62.31)}{9.8}[/tex]
⇒R = [tex]\frac{50^{2}*0.8229}{9.8}[/tex]
⇒R= 209.923m
horizontal range attained R= 209.923m
Components of a projectile motion:
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if potential difference across a lightbulb is 12 V, what amount of charge is required to transfer 60 J of energy to the lightbulb
Answer:
Explanation:
Given:
U =12 V
E = 60 J
________
q - ?
E = U·I·t
q = I·t
E = U·q
q = E / U = 60 / 12 = 5 С
In the district soccer championship finals, elizabeth kicks a 0.57 kg soccer ball with a force of 22.8 n. how much does she accelerate the soccer ball from rest in the process?
acceleration a = F/M therefore acceleration the soccer ball from rest in the process is 40 m/s².
Given
mass of soccer ball, M = 0.57 kg
Force F = 22.8 n
acceleration a = F/M
a = 22.8 n/ 0.57 kg
a = 40m/s²
We refer to any process where the velocity changes as acceleration. You can only accelerate by changing your speed, direction, or both as velocity is a function of both speed and direction. When two things are moved against one another, instant friction results. The motion will be opposed by the friction, who will also be working against it. When an item is in motion and its velocity changes, the concept of acceleration takes place. Friction, which opposes motion, always lowers acceleration. An object and a surface come into contact and cause friction. The features of the surface and the item as well as whether or not the object is moving determine its magnitude.
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Bus A and Bus B leave school on a field trip when a stopwatch reads zero. Bus A travels at a constant 50 km/h, and Bus B travels at a constant 75 km/h. How far are the two buses from school when the stopwatch reads 1.5 h? a. Bus A is 76.5 km from school, Bus B is 51.5 km from school c. Bus A is 75 km from school, Bus B is 112.5 km from school b. Bus A is 112.5 km from school, Bus B is 75 km from school d. Bus A is 51.5 km from school, Bus B is 76.5 km from school
Answer:bus
Explanation:
A car traveling at a uniform speed of 100km per hour spends 15 minutes moving from point A to B along its route . Find the distance between A and B
distance between A and B25002 m
V = 100 km/h
V = (100 x 1000)/3600 = 27.78 m/s when converted to a unit of speed.
t = 15 minutes = 15 x 60 = 900 seconds
displacement =?
Displacement times time equals velocity
Velocity times time equals displacement
displacement = 27.78 x 900
Displacement = 25002 meters
The distance between A and B is 25002 meters since displacement and distance are measured in the same unit.
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a 20. g particle is moving to the left at 30. m/s. how much net work must be done on the particle to cause it to move to the right at 30. m/s?
The net work done will be zero on moving the ball to the right.
The net work done is equal to the change in kinetic energy. The initial velocity is - 30 m/s and the final velocity is + 30 m/s. Also, the formula for kinetic energy is: K.E. = 1/2m([tex] { v_{f} }^{2} [/tex]
- [tex] { v_{i} }^{2} [/tex]
Keep the values in formula to find the kinetic energy
K.E. = 1/2×20×(30² - (-30)²)
Taking square on Right Hand Side of the equation
K.E. = 10×(900 - 900)
Performing subtraction on Right Hand Side of the equation
K.E. = 0
Thus, the net work done will be zero. The work done on stopping the particle and then moving it to different direction will be equal. Thus, we get the value zero.
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You have someone in front of you that you are pulling hair samples from pre-emptively—not because you need them right now but just in case they might be relevant later. What is most likely true about the person in front of you?
The most likely true about the person in front of you is any of the above could be true and is denoted as option D.
What is a Sample?This is referred to as the matter or substance which is obtained so as to aid investigation processes and examples include hair, blood etc. It is useful as it helps to solve a crime and ensures that the culprit is punished.
Taking hair samples from pre-emptively could be because of any of the reasons listed such as being a suspect or a witness to a crime thereby making option D as the correct choice.
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The options are:
They are a suspect in a police interrogation.They are a witness of a crime.They are dead.Any of the above could be true.light travels through water with a speed of 2.25x10⁸ m/s. What is refractive index?
The refractive index of water is 1.3
Find the refractive index of water.
Lightspeed in water [tex]V_water}[/tex]= 2.25*10^8 m/s
We know that,
Speed of light in vaccum [tex]V_{vaccum}[/tex]=3*10^8 m/s
µ=[tex]\frac{C}{V_{m} }[/tex]
where,
µ= refractive index
C= speed of light in air/vaccum
[tex]V_{m}[/tex]=speed of light in any medium
µ=[tex]\frac{3*10^8 m/s}{2.25*10^8 m/s}[/tex]
µ=1.3
Hence, the refractive index of water is 1.3
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What is the independent variable of Measuring physical properties: the Tiny World
The independent variables of measuring physical properties are given below -
What is physical properties?
A physical properties in any property that is measurable, whose value describe a state of physical system.
The changes in the physical properties of a system can be used to describe its changes between the momentary that state of
physical properties are often referred to as observables.
They are not modal properties.
Any variable that can be attributed a value without attributes a value to any to any other variables is called independent variables.
It is a variable that stands alone and isn't changed by other variables you are trying to measure.
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Hello, I am currently stuck on this question and I am very confused as to how to solve it, may I have some help?
Recall, Newton's gravitational law states that any particle of matter in the universe attracts any other particle with a force varying directly as the product of their masses and inversely as the square of the distance between them. It is expressed as
F = Gm1m2/r^2
where
F is the force in Newton
G is gravitational constant = 6.673 x 10^-11 Nm^2/kg^2
m1 and m2 are the masses in kg
r is the distance in meters
From the information given,
r = 1.5
acceleration = 2cm^2/s
Recall, 100 cm = 1m
2 cm = 2/100 = 0.02m
Thus,
acceleration = 0.02m/s^2
Since the masses are identical, then m1 = m2
Each of them is accelerating at 0.02m/s^2
Recall,
Force = mass x acceleration
Force = m1 x 0.02 = 0.02m1 N
By substituting the given values into the formula, we have
0.02m1 = (6.673 x 10^-11 x m1 x m1)/1.5^2
m1 on the left cancels out one m1 on the right. It becomes
0.02 = (6.673 x 10^-11 x m1)/1.5^2
By crossmultiplying,
0.02 x 1.5^2 = 6.673 x 10^-11 x m1
0.045 = 6.673 x 10^-11 x m1
m1 = 0.045/6.673 x 10^-11
m1 = 6.74 x 10^8 kg
The mass of each ball is 6.74 x 10^8 kg
a 30.0 kg child starting from rest slides down a water slide with a vertical height of 10.0m. what is the child's speed (a) halfway down the slides vertical distance and (b) three-fourths of the way down?
a)Child's speed halfway down the slide is 9.89m/s.
b)Child's speed three fourth of the way down is 12.12 m/s.
Potential energy that a massive object has in relation to another massive object due to gravity is called gravitational potential energy. Potential energy associated with the gravitational field is converted to kinetic energy when the objects falls towards each other.
As we know that, ΔPE=KE
mgΔh=mv²/2
Now, v=sqrt(2gΔh)
Halfway down the slide, Δh=5 m
gravity = 9.8 m/s²
v=sqrt(2gΔh)
v= √2* 9.8 * 5
= √98
= 9.89 m/s
For the three fourth of the way down, Δh=7.5 m.
v= √2* 9.8 * 7.5
= 12.12 m/s
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What is the resultant of two displacement vectors having the same direction?1) The resultant is the sum of the two displacements having the same direction as the original vectors.2) The resultant is the difference of the two displacements having the same direction as the original vectors.3) The resultant is the sum of the two displacements having the direction opposite to the direction of the original vectors.4) The resultant is the sum of the two displacements having the direction perpendicular to the direction of the original vectors.
When two vectors have the same direction, they can be added and the direction of the resultant is the same as the direction of the vectors. Additionally, the magnitude of the resultant is simply the sum of the corresponding magnitudes of the vectors.
Therefore, the answer is:
1) The resultant is the sum of the two displacements having the same direction as the original vectors.
a cureent of 0.22A flows through the secondary winding of the transformer. The terminal voltage of the winding is 2400V, and the primary winding is 120V . What is the current in the primary winding?
The current in the primary winding is 4.4 A.
What is primary winding?
A primary winding is the winding of a transformer that is connected to and recive energy from an external source of electron.
The primary winding is the coil that draws power from the source.
Sol- As per the given question -
Current Is=0.22A
Primary winding Vp=120v
Voltage of winding Vs=2400v
Current of primary winding Ip?
Np/Ns=Vp/Vs=Is/Ip
Vp/Vs= Is/Ip
120/2400 =0.22/Ip
Ip= 20×0.22
Ip= 2.2×2
Ip=4.4 A
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A ball is equipped with a speedometer and launched straight upward. The speedometer reading three seconds after launch is shown at the right; the ball is moving upward. At what approximate times would the ball be moving downward and display the following speedometer readings?
The time shown by the speedometer corresponding to speed of 10 m/s is 1 s.
The time shown by the speedometer corresponding to speed of 20 m/s is 1 s.
The time shown by the speedometer corresponding to speed of 30 m/s is 1 s.
1. As shown by the speedometer on the right.
The speed of 10 m/s indicated by the speedometer to the right may be used to calculate how long the ball has been falling from the maximum altitude it reached.
Equation for free fall: Vf = Vo - gt
Vo=0, Vf=gt, and t=Vf/g
I suggest using an approximation of g for this problem: g = 10 m/s2 (I'll explain why in a moment).
1 second is equal to [10 m/s]/[10 m/s2].
The time is now dropping. From it has been four seconds since launch, the upward time was three seconds. You may use this to determine the launching speed.
2. The moment the speedometer registers 20 m/s
Determine the launching speed first:
Vo - gt = Vf
Given that the ball took 3 seconds to ascend and that its speed was zero at its highest point, you may calculate:
0 = Vo - gt
Vo=gt=10 m/s2 3 s = 30 m/s
Now, while the ball is moving upward and the speedometer reads 20 m/s, use the beginning velocity to compute.
t = [30 m/s - 20 m/s] / [10 m/s2] = 1 s 20 m/s = 30 m/s - 10 m/s2
Therefore, t = 1 s is the initial solution.
3. The moment the speedometer registers 30 m/s.
The projected speed for the launch is also 30 m/s.
This reading therefore reflects the precise moment the ball was released.
Time is therefore 0 and the launch moment remains the same.
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derive a formula for acceleration of the system in this lab (under the assumption that the pulley is massless and frictionless, so the two tension forces are equal). how could one use this formula (similar to the atwoods machine) to get a measurement of g?
What is acceleration ?
Acceleration is the rate at which the speed and direction of a moving object change over time. Something is said to be accelerating when it starts to move faster or slower
Let acceleration be a
For the hanging m, mg-T = ma
For the glider(mass M),T=Ma
Eliminating T,we get
mg-Ma=ma
a=mg/(m+M)
Using energy concept we will get the same result,
When m is lowered by h, M will move h distance in forward direction.
As there is no friction, both m and M will attain same speed (v).
So, mgh = 1/2(m+M)v2 [Word done by T will cancel out, as for m it is -hT while for M it is hT(see the direction of T and displacement in both cases)]
v2 =2mgh/(m+M)
Use formula v2 - u2 = 2as
2mgh/(m+M) - 0 =2ah
a = mg/(m+M)
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1. 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 variation of the components of the force with the angle of inclination of the direction of the applied force, as shown in the graph indicates that the Fₓ has a maximum, while [tex]F_y[/tex] has a minimum value given by Fₓ = 50 N × cos(0°) = 50 N, and [tex]F_y[/tex] = 50 N × sin(0°) at θ = 0°, while at 45°, Fₓ = [tex]F_y[/tex], with [tex]F_y[/tex] having a maximum value of 50 N, and Fₓ = 0 at 90°.
What is a component of a force?The component of a force are the mutually perpendicular forces to which a force acting at an angle, relative the coordinate axis can be resolved into.
The component of the force applied by the man to the force required to move the box in the horizontal direction, Fₓ, increases as the angle approaches zero, and decreases as the angle approaches, 90°, given that the horizontal component of the force is found from the equation, [tex]F_x = F \times cos(\theta)[/tex], where, [tex]\theta[/tex] is the angle the line of action of the force makes with the horizontal, and F, is the force the man applies' such that we have at θ = 0°, Fₓ = 50 N × cos(0°) = 50 N
When the angle is θ = 45°, Fₓ = 50 N × cos(45°) = 25·√2 N
When the angle θ = 90°, Fₓ = 50 N × cos(90°) = 0
The vertical component of the force applied [tex]F_y[/tex], however, increases as the angle made by the line of force with the horizontal, increases to 90°, as shown in the graph, and decreases to 0 as the inclination angle of the force decreases, such that the have:
When the angle made with the horizontal, θ = 0°, [tex]F_y[/tex] = 50 N × sin(0°) = 0
When the angle made with the horizontal, θ = 45°, [tex]F_y[/tex] = 50 N × sin(45°) = 25·√2 N
At θ = 90°, [tex]F_y[/tex] = 50 N × sin(90°) = 50 N
The graph of both components starts at 50 N and 0 N when θ = 0°, and both intersect when θ = 45°, while the values of the horizontal and vertical component changes, such that at 90°, the vertical force component is the largest.
Please see the attached graph.
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