Answer:
Explanation:
Given:
U = 590 V
I = 1.0 A
R₁ = 770 Ω
I* = 500 mA = 0.5 A
_________________
P - ?
A)
Power:
P = U·I = 590·1.0 = 590 W
B)
Сurrent:
I₁ = U / R ₁ = 590 / 770 = 0.770 A = 770 mA
C)
This current is life-threatening!!!
I₁ > I* (700 mA > 500 mA)
D)
Energy received:
E = I₁²·R₁ = 0.770²· 770 ≈ 460 W
Create a Graph : Include numbers, axes, and order pairs
a) Peak value = 20
b) Average value = 9.0
c) RMS value = 14.15
Explanation:The peak-to-peak value = 40
[tex]v_{p-p}=40[/tex][tex]\begin{gathered} v_{p-p}=2v_p \\ \\ 40=2v_p \\ \\ v_p=\frac{40}{2} \\ \\ v_p=20 \end{gathered}[/tex]c) The rms value
[tex]\begin{gathered} v_{rms}=\frac{v_p}{\sqrt{2}} \\ \\ v_{rms}=\frac{20}{\sqrt{2}} \\ \\ v_{rms}=14.14 \end{gathered}[/tex]b) Average value over alternation of the sine wave
[tex]\begin{gathered} v_{avg}=0.637v_p \\ \\ v_{avg}=0.637\times14.14 \\ \\ v_{avg}=9.0 \end{gathered}[/tex]Fernando, who has a mass of 43.0 kg, slides down
the banister at his grandparents' house. If the
banister makes an angle of 35.0° with the horizontal,
what is the normal force between Fernando and the
banister?
From the calculations, the normal force on the body is 345 N.
What is the normal force?Let us recall the theory of Newton that states that action and reaction are equal and opposite. We know that the force that is exerted on an object is equal to the reaction of that surface of the object on the source of the force.
We know that the normal force can be given by the formula;
N = mgcosθ
m = mass of the object
g = acceleration due to gravity
θ = angle involved
We would now have
N = 43.0 kg * 9.8 m/s^2 * cos 35
N = 345 N
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Where do you find valence electrons?
Answer:Valence electrons are the electrons in the outermost shell, or energy level, of an atom. For example, oxygen has six valence electrons, two in the 2s subshell and four in the 2p subshell.
Explanation:
...as shall be thought most meet and convenient for the general good of the colony, unto which we promise all due submission and obedience.
This quote from the Mayflower Compact may have influenced the colonists to believe that (1 point)
a
they needed to expand to created additional colonies
b
all citizens would have to abide by the new laws created
c
they needed to create a list of laws for all citizens to follow
d
each person in the colony must be obedient to the laws of the king
Mayflower Compact influenced the colonists to believe that all citizens would have to abide by the new laws created
The colonists who sailed for the New World created a set of self-government laws known as the Mayflower Compact. The Pilgrims and other immigrants hoped to settle in northern Virginia as soon as they arrived in America. The immigrants' ship was forced off track by perilous shoals and storms and instead came ashore in Massachusetts, close to Cape Cod, outside of Virginia's authority.
The colonial authorities knew that life without laws may be devastating, so they created the Mayflower Compact to assure that a workable social structure would prevail. It preserved the idea of a law enacted by the people. This is the essential tenet of democracy. The Declaration of Independence was influenced by the notion that all citizens, not just a small group of elites, should have their interests represented by the government.
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A ball with a mass of 1.50 kg traveling +2.00 m/s collides with a stationery ball with a mass of 1.00 kg. After the collision, the velocity of the 1.50-kg ball is +0.40 m/s. What is the velocity of the 1.00-kg ball after the collision?
The velocity of the 1.00-Kg ball after the collision with the 1.50 Kg ball is 2.4m/s.
The mass M₁ of first ball is 1.5 Kg. The initial velocity U₁ of the first ball is 2m/s.
The mass M₂ of the second ball is 1 Kg. The initial velocity U₂ if the second ball is 0m/s.
After collision, the final velocity V₁ of the first ball is +0.40m/s.
Let us say that the final velocity of the second ball after collision is V₂.
As there is no external force,
We can apply the concept of conservation of linear momentum,
According to which, in a system when no external force is in effect, the total linear momentum of the system is conserved.
We can assume both balls to be a single system,
Initial momentum = Final momentum
M₁U₁+M₂U₂ = M₁V₁+M₂V₂
Putting all the values,
(1.5x2)+(1x0) = (1.5x0.4)+(1V₂)
V₂ = 1.5(2-0.4)
V₂ = 2.4m/s.
So, the final velocity of the 1.00kg ball is 2.4m.s.
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A storage tank 20 m high is filled with pure water. (Assume the tank is open and exposed to the atmosphere at the top.)
(a) Find the gauge pressure at the bottom of the tank.
(b) Calculate the magnitude of the net force that acts on a square access hatch at the bottom of the tank that measures 0.6 m by 0.6 m.
PLEASE HELP!!!
The gauge pressure at the bottom of the tank is 297500 N/m².
The magnitude of the net force that acts on a square access hatch at the bottom of the tank is 107100 N.
What is the pressure at the bottom of the tank?The pressure at the bottom of the tank is the pressure due to the weight of the fluid above it and the atmospheric pressure.
The pressure due to the atmosphere = 101300 N/m²
The pressure due to the water above the tank is calculated with the formula below:
Pressure = height * density * g
height of tank = 20 m
the density of water = 1 * 10³ kg/m³
g = 9.81 m/s²
Pressure due to water = 20 * 1 * 10³ kg/m³ * 9.81
Pressure due to water = 192000 N/m²
Total pressure = 192000 + 101300 N/m²
Total pressure = 297500 N/m²
B. Force = pressure * area
area of the hatch = 0.6 * 0.6 = 0.36 m²
Force = 297500 N/m² * 0.36 m²
Force = 107100 N
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A 32.3 kg mass ( ) is suspended by the cable assembly as shown in the figure. The cables have no mass of their own. The cable to the left ( 1 ) of the mass makes an angle of 0.00∘ with the horizontal, and the cable to the right ( 2 ) makes an angle ( 2 ) of 38.5∘ . If the mass is at rest, what is the tension in each of the cables, 1 and 2 ? The acceleration due to gravity is =9.81 m/s2 .
The tension in each of the wires, number 1 and number 2, respectively
T_1=398.35
This is further explained below.
What is tension ?Generally, At equilibrium, in direction [tex]$\Sigma F_y=0$[/tex]
[tex]\Rightarrow T_2 \sin \theta_2-m g=0 \text {. }[/tex]
[tex]$\Rightarrow \quad T_2 \sin \theta_2=m g$[/tex]
tension, [tex]$T_2=\frac{m g}{\sin \theta_2}$[/tex]
[tex]T_2=\frac{32.3 \times 9.81}{\sin (38.5}^{\circ})} \\\\T_2= {509.00} \mathrm{~N}$.[/tex]
[tex]In x-dirn; $\quad \Sigma F_x=0 \Rightarrow T_2 \cos \theta_2-T_1=0$.[/tex]
[tex]\Rightarrow T_1 &=T_2 \cos \theta_2 \\[/tex]
[tex]T_1 &=509.00\times \cos (38.5^{\circ}\right) \\\\[/tex]
T_1=398.35
In conclusion, the tension in each of the cables, 1 and 2
T_1=398.35
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A fisherman in a stream 30 cm deep looks downward into the water and sees a rock on the stream bed. How deep does the stream appear to the fisherman? Let the index of refraction of the water equal 1.33O 26cmO 40cmO 34cmO 23cm
Given:
The depth of the stream, d=30 cm
The refractive index of water, n=1.33
To find:
The apparent depth of the stream.
Explanation:
Let the eye of the fisherman is at a large height from the surface of the water.
Thus,
[tex]\sin(i)=\frac{x}{d}[/tex]Where i is the angle of incidence and r is the opposite side of the angle of incidence.
And,
[tex]\sin(r)=\frac{x}{h}[/tex]Where r is the angle of refraction and h is the apparent depth of the stream.
The refractive index of the air is n_a=1.
From snell's law,
[tex]n_a\sin r=n\sin i[/tex]On substituting the known values,
[tex]\begin{gathered} 1\times\frac{x}{h}=1.33\times\frac{x}{d} \\ \implies h=\frac{d}{1.33} \\ =\frac{30}{1.33} \\ =23\text{ cn} \end{gathered}[/tex]Final answer:
The correct answer is option d.
A 1.2kg ball rolls forward with an acceleration of 1.11 m/s. What is the net force on the ball
Answer:
1.332 N
Explanation:
Net Force = Mass x Acceleration
1.2 x 1.11 = 1.332 N
I'm so sorry if I'm wrong.
Describe the relationship between the state of matter of a pure substance (gas, liquid, solid) and the motion of the particles.
I will brainliest you!
Answer:
Gas is one of the state of pure substance and it have no shape and also it is non comprsable
Liquid is state of pure substance and it occupies the shape of its body and its slightly comrisable
Solid is the other physical state and it have definite volume and shape and also it is highly comrisable
Question is in attachment!
Thanks!
Concept : Thermodynamics
A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?
Solution :
The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus, the process is adiabatic.
Initial pressure inside the cylinder =P1
Final pressure inside the cylinder =P2
Initial volume inside the cylinder =V1
Final volume inside the cylinder =V2
Ratio of specific heats, γ=CVCP=1.4
For an adiabatic process, we have:
P1V1γ=P2V2γ
The final volume is compressed to half of its initial volume.
∴V2=V1/2
P1V1γ=P2(V1/2)γ
P2/P1=V1γ/(V1/2)γ
=21.4=2.639
Hence, the pressure increases by a factor of 2.639.
Answer:
The cylinder is fully isolated from the rest of the environment.
There is no heat exchange between the system (cylinder) and its surroundings as a result of the design. As a result, the process is called adiabatic.
P1 represents the initial pressure inside the cylinder.
P2 is the final pressure within the cylinder.
V1 is the volume of the cylinder at its beginning.
The final volume of the cylinder is equal to V2.
The specific heat ratio, = Cp / Cv = 1.4,
We have the following for an adiabatic process:
P1V1γ = P2V2γ
After compression, the final volume is reduced to half of its original size.
Hence,
V2 = V1 / 2
P1V1γ = P2(V1 / 2)γ
P2 / P1 = V1γ / (V1 / 2)γ
= 21.4
We get,
= 2.639
Therefore, the pressure increases by a factor of 2.639
Explanation:
hope it helps you
In laser eye surgery, the average laser pulse power is 74 kW, and contains 1 mJ of light energy. Each laser pulse lasts
nanosecond.
If an average laser pulse power is 74 kW, and contains 1 mJ of light energy, each laser pulse lasts for a time period of 0.074 ns
E = P t
E = Light energy
P = Pulse power
t = Time
E = 74 KW = 74 * 10³ W
P = 1 * [tex]10^{-3}[/tex] J
t = E / P
t = 74 * 10³ / 1 * [tex]10^{-3}[/tex]
t = 74 * [tex]10^{6}[/tex] s
t = 0.074 ns
Power is the rate of work done. It can also be called as rate at which energy is used. This can be used to all forms of work and all types of energy.
Therefore, each laser pulse lasts 0.074 nanosecond.
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I need help with the answer
Answer:
Yes, of course
Explanation:
Ask the questionI'll answer itA 100-kg box is sliding down a frictional surface with an acceleration of -2.0m/s².
Determine the magnitude of the friction force on the object.
Answer: 1180 N
Explanation:
F - Fs = ma
⇒ (100 × 9.8) - Fs = 100 × (-2)
⇒ Fs = 1180 N
Using m1v1i + m2v2i = m1v1f + m2v2f how does one describe the concept of left or right mathematically?
The concept of left or right can be described mathematically by using ( m₁*v₁)initial + (m₂*v₂)initial = (m₁*v₁)final + (m₂*v₂)final because as per the conservation of the momentum the, momentum before and after the collision will remain conserved.
What is momentum?It can be defined as the product of the mass and the speed of the particle, it represents the combined effect of mass and the speed of any particle,
P = mv
As given in the problem we have described the concept of left or right mathematically by using( m₁*v₁)initial + (m₂*v₂)initial = (m₁*v₁)final + (m₂*v₂)final
The total momentum of the object before the collision would be equal to the total momentum of the objects after the collision.
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How is Newton's Third Law of Motion used in the sport Jokgu?
Newton's third law of motion states that every force has an equal and opposite reaction force. That is when two bodies interact with each other, they apply a force on each other which are equal in magnitude and opposite in direction.
In Jokgu, when you hit the ball it applies an equal force on the leg. And when a player presses his leg against the ground, the ground exerts an equal force on the player which helps him to move around.
If an object mass is 2 kg. And an applied force is acting on it at 15 N And the force of friction is 5 N and what is the objects acceleration
Given,
The mass of the object, m=2 kg
The applied force, F=15 N
The frictional force, f=5 N
The frictional force is a force that opposes the motion of an object. Hence it will be always be directed opposite to the direction of motion of the object.
The net force acting on the object is given by,
[tex]F_n=F-f[/tex]From Newton's second law of motion, the net force is given by,
[tex]F_n=ma[/tex]Where a is the acceleration of the object.
Therefore,
[tex]\begin{gathered} ma=F-f \\ \Rightarrow a=\frac{F-f}{m} \end{gathered}[/tex]On substituting the known values,
[tex]\begin{gathered} a=\frac{15-5}{2} \\ =5\text{ m/s}^2 \end{gathered}[/tex]Therefore the acceleration of the object is 5 m/s²
A car is driving away from you at a constant speed for a while, and then it
gradually slows to a stop. Imagine a speed-versus-time graph showing the
car’s motion. Describe the line on the graph.
The line on the speed-time graph will slant from up to down showing a state of coming to rest.
What is speed - versus - time graph?
A speed versus-time-graph is a type of graph that shows the relationship between the speed of an object and the time of motion of the object.
For upward acceleration of an object when the object is speeding up, the line on the graph for such motion increases upward in a direct relationship with time of motion.
For a downward motion in which the object in motion slows down, the line on the graph for such motion will slant downward showing a gradual decrease in the speed of the object.
Thus, for a car that is driving away from you at a constant speed for a while, and then it gradually slows to a stop, the line on the speed-time graph will slant from up to down showing a state of coming to rest.
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Calculate the force which will produce an extension of 0.30mm in a steel wire with a length of 4.0m and a cross section area of 2.0 x 10^(-6) m^2Youngs modulus of steel is 2.1 x 10 ^11 Pa
Given data:
* The extension of the steel wire is 0.3 mm.
* The length of the wire is 4 m.
* The area of cross section of wire is,
[tex]A=2\times10^{-6}m^2[/tex]* The young modulus of the steel is,
[tex]Y=2.1\times10^{11}\text{ Pa}[/tex]Solution:
The young modulus of the steel in terms of the force and extension is,
[tex]Y=\frac{F\times l}{A\times dl}[/tex]where F is the force acting on the steel wire,, l is the original length of the wire, dl is the extension of the wire, and A is the area,
Substituting the known values,
[tex]\begin{gathered} 2.1\times10^{11}=\frac{F\times4}{2\times10^{-6}\times0.3\times10^{-3}} \\ F=0.315\times10^2\text{ N} \\ F=31.5\text{ N} \end{gathered}[/tex]Thus, the force which produce the extension of 0.3 mm of the steel wire is 31.5 N.
How far will a car travel going at a speed of 18 m/s in 42 minutes?
O 756 meters.
O2,520 meters.
O 45,360 meters.
O 140 meters.
Answer:
d = rt
Explanation:
distance = ?
rate = 18 m/s
time = 42 x 60 = 2520 s
distance = 18 x 2520 = 45360 m
Situation: A .71-kg billiard ball moving at 2.5 m/s in the x-direction strikes a stationary ball of the same mass. After the collision, the first ball moves at 2.17 m/s, at an angle of 30.0° with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), answer the following questions to find the struck ball's momentum after the collision.1. Calculate the x-component of the first ball's final momentum after the collision in kg*m/s.2. Using the conservation of momentum in the x-direction, find the struck ball's x-component of momentum.3. Calculate the y-component of the first ball's final momentum after the collision in kg*m/s.4. Using the conservation of momentum in the y-direction, find the struck ball's y-component of momentum.
We are given the following situation:
Ball 1 strikes ball 2 and after the collision forms an angle of 30 degrees. To determine the x-component of the final momentum of the first ball we use the following formula:
[tex]P_{1fx}=m_1v_{1f}\cos\theta_1[/tex]Where:
[tex]\begin{gathered} P_{1fx}=final\text{ x-component of the momentum of ball 1} \\ m_1=\text{ mass of ball 1} \\ v_{1f}=\text{ final velocity of ball 1} \\ \theta_1=\text{ angle of ball 1} \end{gathered}[/tex]Now, we substitute the values:
[tex]P_{1fx}=(0.71kg)(2.17\frac{m}{s})\cos(30)[/tex]Solving the operations we get:
[tex]P_{1fx}=1.33kg\frac{m}{s}[/tex]Therefore, the x-components of the momentum of the first ball is 1.33 kgm/s.
Part 2. To determine the x-component of the second ball we will do a balance of momentum in the x-direction:
[tex]P_{10x}+P_{20x}=P_{1fx}+P_{2fx}[/tex]Where:
[tex]\begin{gathered} P_{10x},P_{20x}=\text{ initial momentum in the x-direction of ball 1 and 2} \\ P_{2fx},P_{20x}=\text{ final momentum in the x-direction of ball 1 and 2} \end{gathered}[/tex]Since the second ball starts from rest we have that its initial momentum is zero, therefore:
[tex]P_{10x}=P_{1fx}+P_{2fx}[/tex]Now, we solve for the x-component of the momentum of the second ball:
[tex]P_{10x}-P_{1fx}=P_{2fx}[/tex]The initial momentum of the first ball is the product of its mass and velocity:
[tex]m_1v_{01}-P_{1fx}=P_{2fx}[/tex]Now, we plug in the values:
[tex](0.71kg)(2.5\frac{m}{s})-1.33kg\frac{m}{s}=P_{2fx}[/tex]Solving the operations:
[tex]0.44kg\frac{m}{s}=P_{2fx}[/tex]Part 3. To calculate the y-component of the first ball after the collision we will use the following formula:
[tex]P_{1fy}=m_1v_{1f}\sin\theta[/tex]Now, we plug in the values:
[tex]P_{1fy}=(0.71kg)(2.17\frac{m}{s})\sin(30)[/tex]Solving the operations we get:
[tex]P_{1fy}=0.77kg\frac{m}{s}[/tex]Therefore, the y-component of the momentum of the first ball is 0.77 kgm/s.
Part 4. To determine the y-component of the second ball we use a balance of momentum of the y-components of the balls:
[tex]P_{10y}+P_{20y}=P_{1fy}+P_{2fy}[/tex]Since the first ball is not moving in the y-direction this means that its y-component of the momentum is 0. Since ball 2 is not moving initially this means that its momentum in the y-direction is zero.
[tex]0=P_{1fy}+P_{2fy}[/tex]Now we solve for the y-component of the second ball, and we get:
[tex]-P_{1fy}=P_{2fy}[/tex]Therefore, the y-component of the second ball is:
[tex]-0.77kg\frac{m}{s}=P_{2fy}[/tex]Therefore, the y-component of the second ball is -0.77 kgm/s.
A penny, starting from rest at Position 1, rolls down and then up a curved track towards Position 5. When it reaches Position 5, it rolls back down the track. Which description below is most consistent with what you can expect from the penny's motion?
A. The penny will roll as high as Position 3 since that position has the least potential energy.
B. The penny will roll past Position 1 since energy cannot be created or destroyed.
C. The penny will roll as high as Position 2 since some of the energy will be transformed to heat.
D. The penny will roll as high as Position 1 since energy cannot be created or destroyed
The penny will roll past Position 1 since energy cannot be created or destroyed, therefore the correct answer is option B.
What is mechanical energy?The sum of all the energy in motion (total kinetic energy) and all the energy that is stored in the system (total potential energy) is known as mechanical energy.
ME= PE + KE
As given in the problem if a penny, starting from rest at Position 1, rolls down and then up a curved track towards Position 5.
When it reaches Position 5, it rolls back down the track, then we have to find out Which description below is most consistent with what you can expect from the penny's motion,
Thus, the penny will roll past Position 1 since energy cannot be created or destroyed, therefore the correct answer is option B.
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please help!!
What is the name of the current theory that astronomers have developed that is used to describe the possible formation of the solar system?
Answer:
The answer of the question is nebular theory
An average froghopper insect has a mass of 12.7 mg and jumps to a maximum height of 278 mm when its takeoff angle is 57.0∘ above the horizontal. Find the takeoff speed of the froghopper.
Takeoff speed of the froghopper is 2.786 m/s
By analyzing the vertical motion of the froghopper,
The initial vertical velocity of the froghopper as it jumps from the ground is given by
[tex]U_{y}[/tex] = U₀Sinθ
here, θ = 57°
Therefore Sin57° = 0.838
[tex]U_{y}[/tex]= 0.838U₀
Vertical motion is the motion that occurs when the object is thrown all the way up, i.e. , the initial velocity or force acts only in the vertical axis, therefore this motion has only vertical motion.
Maximum height reached by the froghopper is h = 278 mm
h = 0.278 m
As we know that vertical velocity at the point of maximum height will be zero,
[tex]V_{y} = 0[/tex]
Since the vertical motion is an accelerated motion with constant (de)acceleration, so by using Equation of motion,
[tex]V^{2} _{y} - U^{2} _{y} = 2gh[/tex]
[tex]U_y = \sqrt{V^{2} _{y} - 2gh }[/tex]
[tex]U_y = \sqrt{0 - 2(-9.81) (0.278) }[/tex]
[tex]U_{y}[/tex] = 2.335 m/s
[tex]U_{y}[/tex] = 0.838U₀
U₀ = [tex]\frac{U_y}{0.838}[/tex]
U₀ = [tex]\frac{2.335}{0.838}[/tex]
U₀ = 2.786 m/s
The takeoff speed of the froghopper is 2.786 m/s.
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Question 2 of 25Which of the following is an effect of increasing the wavelength of anelectromagnetic wave?O A. Energy decreases.O B. Speed decreases.O C. Speed increases.O D. Energy increases.SUBMIT
Thesppeed of electromagnetic waves remains the same an is eqal to the speed of light. The wavelength is rearlated tothe energy bty the following ormula:
[tex]\lambda=\frac{c}{f}[/tex]Wher:
[tex]\begin{gathered} \lambda=\text{ wavelength} \\ c=\text{ speed of light} \\ f=\text{ frequency} \end{gathered}[/tex]Therefore, if the wavelength increases the frequency must decrease.-
Since the energy is proportiona to the frequency if the dfrequency decreases the energy decreases. Therefore, the right answer is A.
Answer:
energy decreases
Explanation:
hope this helps!
A 180 N force acts at 190 degrees and a 140 N force acts at 260 degrees. Determine the magnitude and direction (include angle) of the resultant. Scale is 1cm =10N
NEED HELPPP ASAPPPPPP
The magnitude of the force is found to be 274.19N.
Force (F1)acting on θ1(190°) is 180N
Force (F2)acting on θ2(260°) is 140N
Therefore the angle between F1 and F2 is
θ2-θ1 = 180°-140° =40°
Now we calculate magnitude of force,
Magnitude of force ,F= F1+F2
F. F = [tex]\sqrt{ ( F1+F2). (F1+F2) }[/tex]
|F|² = [tex]\sqrt{F1 ²+F2 ²+F1 F2 COS θ}[/tex]
F= [tex]\sqrt{ 180²+140²+2×180×140× cos40°}[/tex]
F= [tex]\sqrt{32400+19600+23184}[/tex]
F= [tex]\sqrt{75184}[/tex]
F= 274.19N.
Thus, the magnitude of the force is found to be 274.19N.
The magnitude of the force is the entire quantity of forces acting on an object. When all forces are pulling in the same direction, the force becomes stronger. The strength of a force decreases as it is applied to an object from various angles.
Force has a magnitude and a direction, thus a vector quantity. The outcomes of two equal-sized forces acting in opposite directions one in the east and the other in the west are not the same.
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ASAPPPPPPPPPPPPPPPPPPPPPPPP!!!!!!
Answer:
its A
Explanation:
Answer:
it A bc it makes the most sense
Explanation:
7. Which wave property increases as the energy of a wave increases? *O periodfrequencywavelengthamplitude
Energy is directly proportional to the amplitude.
Thus, the amplitude of the wave increases as the energy of the wave increases.
3. A uniform light beam is pivoted halfway along its length. At one end it supports a load of 5 kN while the other end is tethered to a fixed point by a rope inclined at 45° to the horizontal. If the beam is in equilibrium, what is the tension in the rope?
Answer:
Approximately [tex]7.1\; {\rm kN}[/tex] ([tex]5 \sqrt{2}\; {\rm kN}[/tex].)
Explanation:
Let [tex]F_{1}[/tex] and [tex]F_{2}[/tex] denote the two forces that act on this beam. Let [tex]s_{1}[/tex], [tex]s_{2}[/tex], [tex]\theta_{1}[/tex], and [tex]\theta_{2}[/tex] denote the distance from pivot and angle relative to the beam of the two forces, respectively. The magnitude of the torques that the two forces exert on this beam will be [tex]F_{1}\, s_{1}\, \sin(\theta_{1})[/tex] and [tex]F_{2}\, s_{2}\, \sin(\theta_{2})[/tex], respectively.
The two forces in this question act on the beam from opposite sides of the pivot. Hence, for the beam to be in equilibrium, the torque from the two forces need to be equal in magnitude. In other words:
[tex]F_{1}\, s_{1}\, \sin(\theta_{1}) = F_{2}\, s_{2}\, \sin(\theta_{2})[/tex].
Let [tex]F_{1}[/tex] denote the [tex]5\; {\rm kN}[/tex] force that the load exerts on this beam; [tex]\theta = 90^{\circ}[/tex] since this load is placed directly on the beam. The normal force from the load will be perpendicular to the beam.
Let [tex]F_{2}[/tex] denote the force that the rope exerts on this beam; [tex]\theta = 45^{\circ}[/tex].
Note that [tex]s_{1} = s_{2}[/tex] since the pivot is exactly halfway between the two forces.
Rearrange the equation [tex]F_{1}\, s_{1}\, \sin(\theta_{1}) = F_{2}\, s_{2}\, \sin(\theta_{2})[/tex] to find the unknown [tex]F_{2}[/tex]:
[tex]\begin{aligned}F_{2} &= \frac{F_{1}\, s_{1}\, \sin(\theta_{1})}{s_{2}\, \sin(\theta_{2})} \\ &= \frac{F_{1}\, \sin(\theta_{1})}{\sin(\theta_{2})} && (\text{since $s_{1} = s_{2}$}) \\ &= \frac{5\; {\rm kN}\, \sin(90^{\circ})}{\sin(45^{\circ})} \\ &= \frac{5\; {\rm kN}}{(1 / \sqrt{2})} \\ &= 5 \sqrt{2}\; {\rm kN} \\ &\approx 7.1\; {\rm kN} \end{aligned}[/tex].
The tension in the rope will be equal in magnitude to the force exerted on the beam: approximately [tex]7.1\; {\rm kN}[/tex] ([tex]5 \sqrt{2}\; {\rm kN}[/tex].)
How long must a 400 W electrical engine work in order to produce 300 kJ of work?
In order to calculate the time needed, let's use the formula below:
[tex]P=\frac{E}{t}[/tex]Where P is the power in Watts, E is the energy/work in Joules and t is the time in seconds.
First, let's convert the energy from kJ to J (1 kJ = 1000 J):
[tex]300\text{ kJ}=300000\text{ J}[/tex]So, for P = 400, we have:
[tex]\begin{gathered} 400=\frac{300000}{t} \\ t=\frac{300000}{400} \\ t=750\text{ s} \end{gathered}[/tex]Therefore the time needed is 750 seconds.