A closed cylinder air column must have a closed end and an open end.
Then, the answer is:
[tex]\text{False}[/tex]True or false. Iron fillings can indicate the presence of a magnetic field
Iron Filings are small shavings of ferromagnetic material.
Ferromagnetic material aligns with a magnetic field.
So, Iron fillings can indicate the presence of a magnetic field
True
A uniform rod of mass m=1.0 kg and length L=2 meters is free to rotate about its center as shown in the figure. A constant torque is applied by a constant force of magnitude F=5.0 N to one end of the rod, as shown. To be clear, the force is always perpendicular to the rod, which rotates about the axis indicated by the red line.a) If the rod is initially at rest, how long does it take to reach an angular velocity omega=30 radians/s. b) Assuming the torque stops when the angular velocity is 30 radians/s, what is the total rotational energy of the rod? c) The rod suddenly turns into a uniform sphere of radius R=1m and mass M=1.0 kg rotating about its center. Assuming angular momentum is conserved, what is the angular velocity of the sphere omegasphere?
a)
If the length is 2 meters and the rod rotates about its center, so 2 meters is the diameter, and the radius of rotation is 1 meter.
Then, if the force is 5 N, let's calculate the torque:
[tex]\begin{gathered} \tau=F\cdot r \\ \tau=5\cdot1=5\text{ Nm} \end{gathered}[/tex]Then, calculating the rotational inertia and the angular acceleration, we have:
[tex]\begin{gathered} I=\frac{1}{2}mr^2 \\ I=\frac{1}{2}\cdot1\cdot1^2=0.5 \\ \\ \alpha=\frac{\tau}{I}=\frac{5}{0.5}=10 \end{gathered}[/tex]The angular acceleration is 10 rad/s², so to reach an angular velocity of 30 rad/s, it takes 3 seconds.
b)
The rotational energy can be calculated with the formula below:
[tex]\begin{gathered} E_k=\frac{1}{2}I\cdot\omega^2 \\ E_k=\frac{1}{2}\cdot0.5\cdot30^2 \\ E_k=225\text{ J} \end{gathered}[/tex]c)
The angular momentum is given by:
[tex]\begin{gathered} L=I\cdot\omega \\ L=0.5\cdot30 \\ L=15 \end{gathered}[/tex]Then, since the rod turns into a sphere, the new rotational inertia is:
[tex]\begin{gathered} I=\frac{2}{5}mr^2 \\ I=\frac{2}{5}\cdot1\cdot1^2 \\ I=\frac{2}{5}=0.4 \end{gathered}[/tex]So the new angular velocity is:
[tex]\begin{gathered} L=I\cdot\omega \\ 15=0.4\cdot\omega \\ \omega=\frac{15}{0.4}=37.5\text{ rad/s} \end{gathered}[/tex]6) If a system has 48 J of work done on it and absorbs 22 J of heat, what is the value of ΔE for this change?
ANSWER
ΔE = 70 J
EXPLANATION
Given:
• The work done on the system, W = 48 J
,• The heat absorbed by the system, Q = 22 J
Find:
• The change of energy of the system, ΔE
We will use the convention where:
• Heat transferred to the system: positive (otherwise negative)
,• Work done on the system: positive (otherwise negative)
By the first law of thermodynamics,
[tex]\Delta E=Q+W[/tex]Replace with the known values and solve,
[tex]\Delta E=22J+48J=70J[/tex]Hence, the change in the system's internal energy is 70 J.
Hi I do know the answer to this it’s be answered for me twice already just looking for a bit of a deeper explanation physical and why and how?
4.1
The free body diagram is shown below
Newton's first law of motion states that a body at rest or in motion would continue to be at rest or in motion unles an external force acts on it causing it to start or stop moving.
Sum of forces in the horizontal direction, Fx = - Fr + F
Sum of forces in the vertical direction, Fy = N + FSinθ - mg
Since the motion is along the horizontal axis, There would not be an acceleration in the y axis. Thus, Fy = 0
Force = mgCosθ
4.3) From the information given,
mass, m = 50kg
Force =
Forcce = 50 x 9.81Cos20
4.4)
Fy = 0
N + FSinθ - mg = 0
N = mg - FSinθ
N = 50 * 9.81 - FSin20
N = 490.5 - FSin20
N = 490.5 - 0.342F
4.5)
Frictional force = coefficient of friction x normal force
From the information given,
coefficient of friction = 0.4
Frictional force = 0.4(490.5 - 0.342F)
I take my calculator from the top of a building and toss it straight upward at 9 m/s from a 23 m tall building.A. what is the speed of the Calculator right before it hits the ground? B. What is the acceleration of the calculator at peak height? C. How much time does it take for the calculator to reach peak height?
Given,
The initial velocity with which the calculator was thrown, u=9 m/s
The height of the building, h=23 m
A.
When the calculator reaches the peak height, its velocity will become zero. That is v=0 m/s
And while it is going up the acceleration due to gravity will be acting downward. Thus the acceleration due to gravity will be a negative value.
From the equation of motion,
[tex]v^2-u^2=2gs_{}[/tex]Where s is the peak height reached by the calculator and g is the acceleration due to gravity.
On substituting the known values,
[tex]\begin{gathered} 0-9^2=2\times-9.8\times s \\ \Rightarrow s=\frac{-9^2}{2\times-9.8} \\ =4.13\text{ m} \end{gathered}[/tex]Thus the total height reached by the calculator from the ground is
[tex]\begin{gathered} H=h+s \\ =23+4.13 \\ =27.13 \end{gathered}[/tex]After reaching the peak height, the calculator starts descending. This descent starts with the initial velocity of v=0 m/s. And the acceleration due to gravity will be in the direction of motion of the calculator. Thus it will be a positive value.
From the equation of motion,
[tex]v^2_0-v^2=2gH[/tex]where v₀ is the velocity of the calculator right before it hits the ground.
On substituting the known values,
[tex]\begin{gathered} v^2_0-0=2\times9.8\times27.13 \\ \Rightarrow v_0=\sqrt[]{2\times9.8\times27.13}^{} \\ =23.06\text{ m/s} \end{gathered}[/tex]Thus the speed of the calculator right before it hits the ground is 23.06 m/s
B.
The acceleration of the calculator is a constant value. It is always equal to the acceleration due to gravity.
Thus the acceleration of the calculator at peak height is 9.8 m/s²
C.
From the equation of motion,
[tex]v=u+gt[/tex]Where t is the time it takes for the calculator to reach the peak height.
On substituting the known values,
[tex]\begin{gathered} 0=9+(-9.8)t \\ \Rightarrow t=\frac{-9}{-9.8} \\ =0.92\text{ s} \end{gathered}[/tex]Thus it takes 0.92 s for the calculator to reach the peak height.
Determine the image distance and image height for a 5.00 cm tall object placed 20.0 cm from a double convex lens with a focal length of 15.0 cm.
Given:
the height of the object is
[tex]h_0=5\text{ cm}[/tex]The distance of the object is
[tex]d_0=-20\text{ cm}[/tex]The focal length of the lens is
[tex]f=15\text{ cm}[/tex]Required: the distance of the image and height of the image.
Explanation:
the lens formula is given by
[tex]\frac{1}{f}=\frac{1}{d_i}-\frac{1}{d_0}[/tex]Plugging all the values in the above relation, we get:
[tex]\begin{gathered} \frac{1}{15\text{ cm}}=\frac{1}{d_i}-\frac{1}{-20\text{ cm}} \\ \frac{1}{d_i}=\frac{1}{15\text{ cm}}-\frac{1}{20\text{ cm}} \\ \frac{1}{d_i}=\frac{4-3}{60\text{ cm}} \\ d_i=60\text{ cm} \end{gathered}[/tex]Thus, the distance of the image is 60 cm.
now calculate the height of the image
we know that
[tex]\frac{h_i}{h_0}=\frac{d_i}{d_0}[/tex]substitute all the values in the above relation, we get:
[tex]\begin{gathered} h_i=5\text{ cm}\times\frac{60}{-20} \\ h_i=-15\text{ cm} \end{gathered}[/tex]Thus, the height of the image is 15 cm.
A pulse travels along a string that is fixed to a wall, as shown. The pulse reaches the wall and is reflected. Which of the following diagrams illustrates the orientation of the reflected pulse?
Using the concept of Simple Harmonic motion, we got that phase of the string wave changes by 180° and velocity gets reversed.
Simple Harmonic Motion or SHM is defined as motion in which the restoring force is directly proportional to the displacement of the body from its mean position. The direction of this restoring force is always towards a mean position. The acceleration of particle executing simple harmonic motion is given by a(t) = -ω2 x(t). Here, ω is the angular velocity of particle.
A pulse of string wave when travels along a stretched string and reaches the fixed end of the string, then it will be reflected back to the same medium and the reflected ray suffers a phase change of π with the incident wave but the wave's velocity after reflection reverses its direction.
To know more about Simple Harmonic motion, visit here:
https://brainly.com/question/17315536
#SPJ9
A fire is an example of ________________ reaction, because heat is _____________ the system.
A. exothermic; leaving
B. exothermic; entering
C. endothermic; leaving
D. endothermic; entering
Because heat is being introduced into the system, a fire is just an example of an exothermic reaction.
The right response is B.
What is exothermic example?Any process was considered to be exothermic if it produces heat while also undergoing a net decrease in standard enthalpy change. Examples include any type of combustion, iron rusting, and water freezing. Exothermic processes are those that discharge warmth and energy into the surrounding environment.
What process are exothermic?If heat is released from the system into the environment, the reaction or change is exothermic. The temperature of the environment rises because it is absorbing heat from the system.
To know more about exothermic visit:
brainly.com/question/4345448
#SPJ1
Calculate the power of a 2057 Ω night light plugged into a 120 V source.
Lets write the data down:
R = 2057 Ω
V = 120 V
The power of the night light is given by:
[tex]P=\frac{V^2}{R}[/tex]Plug in the data in the formula:
[tex]\begin{gathered} P=\frac{120^2}{2057} \\ P=\frac{14,400}{2057} \\ P\approx7.0W \end{gathered}[/tex]The power of the night light is 7.0 W.
A stone is thrown straight upward and reaches a maximum height of 32.1 m above itslaunch point. What was the initial speed with which the stone was thrown upwards?Answer:m/s
The initial speed = 25.08 m/s
Explanation:The maximum height, H = 32.1 m
The initial speed, u = ?
The acceleration due to gravity, g = 9.8 m/s²
Write out the maximum height formula and solve for u
[tex]H=\frac{u^2}{2g}[/tex]Substitute H = 32.1 and g = 9.8
[tex]\begin{gathered} 32.1=\frac{u^2}{2(9.8)} \\ \\ u^2=32.1(2)(9.81) \\ \\ u^2=629.16 \\ \\ u=\sqrt{629.16} \\ \\ u=25.08\text{ m/s} \end{gathered}[/tex]The initial speed = 25.08 m/s
24.A projectile is launched upward at an angle of 30° from thehorizontal with an initial velocity of 145 meters per second. Howfar does the projectile travel horizontally?(a) 1.86 x 10^3 m(b) 2.28 x 10^3 m(c) 4.55 x 10^3 m(d) 9.10 x 10^3 m
ANSWER:
(a) 1.86 x 10^3 m
STEP-BY-STEP EXPLANATION:
The movement that the projectile takes is a parabolic movement, therefore, the horizontal range is determined as follows:
[tex]x=\frac{v^2_0\cdot\sin (2\theta)}{g}[/tex]We know the initial velocity, the angle and the gravity, therefore, we replace and calculate the value of the horizontal distance:
[tex]\begin{gathered} x=\frac{145^2\cdot\sin (2\cdot30)}{9.8} \\ x=1857.97\cong1.86\cdot10^3 \end{gathered}[/tex]The horizontal distance is equal to approximately 1860 meters
Find the direction and magnitude of the vectors (a) A = (25m)x + (-12m)y(b) B = (2.0m)x + (15m)y and(c) A + B
ANSWER
[tex]\begin{gathered} a)\theta=-25.6\degree;|A|=27.73m \\ b)\theta=82.4\degree;|B|=15.13m \\ c)\theta=6.3\degree^{};|A+B|=27.17m \end{gathered}[/tex]EXPLANATION
To find the direction of a two-dimensional vector, we apply the formula:
[tex]\theta=\tan ^{-1}(\frac{a_y}{a_x})[/tex]To find the magnitude of a two-dimensional vector, we apply the formula:
[tex]|a|=\sqrt[]{(a^{}_y)^2+(a_x)^2}[/tex]a) The direction of vector A is:
[tex]\begin{gathered} \theta=\tan ^{-1}(\frac{-12m}{25m}) \\ \theta=\tan ^{-1}(-0.48) \\ \theta=-25.6\degree \end{gathered}[/tex]The magnitude of vector A is:
[tex]\begin{gathered} |A|=\sqrt[]{(25m)^2+(-12m)^2} \\ |A|=\sqrt[]{625m^2+144m^2}=\sqrt[]{769m^2} \\ |A|=27.73m \end{gathered}[/tex]b) The direction of vector B is:
[tex]\begin{gathered} \theta=\tan ^{-1}(\frac{15m}{2m}) \\ \theta=\tan ^{-1}(7.5) \\ \theta=82.4\degree \end{gathered}[/tex]The magnitude of vector B is:
[tex]\begin{gathered} |B|=\sqrt[]{(2.0m)^2+(15m)^2} \\ |B|=\sqrt[]{4m^2+225m^2}=\sqrt[]{229m^2} \\ |B|=15.13m \end{gathered}[/tex]c) First, we have to find the sum of A and B:
[tex]\begin{gathered} A+B=(25m)x+(-12m)y+(2.0m)x+(15m)y \\ A+B=(25m+2.0m)x+(-12m+15m)y \\ A+B=(27m)x+(3m)y \end{gathered}[/tex]The direction of the vector (A + B) is:
[tex]\begin{gathered} \theta=\tan ^{-1}(\frac{3m}{27m}) \\ \theta=\tan ^{-1}(0.1111) \\ \theta=6.3\degree^{} \end{gathered}[/tex]The magnitude of the vector (A + B) is:
[tex]\begin{gathered} |A+B|=\sqrt[]{(27m)^2+(3m)^2} \\ |A+B|=\sqrt[]{729m^2+9m^2}=\sqrt[]{738m^2} \\ |A+B|=27.17m \end{gathered}[/tex]A 200-turn solenoid is 20.0 cm long and carries a current of 3.25 A.1. Find the magnetic field inside the solenoid in [mT].2. Find the force in [µN] exerted on a 15.0x10-6 C charged particle moving at 1050 m/s through the interior of the solenoid, at an angle of 11.5° relative to the solenoid’s axis.
Given,
Number of turns, N=200
Length of teh solenoid, l=20 cm=0.2 m
Current, I=3.25 A.
Charge is
[tex]q=15\times10^{-6}C[/tex]The velocity is v=1050 m/s
Angle is
[tex]\theta=11.5^o[/tex]To find
a. Magnetic field inside the solenoid
b. The force
Explanation
a. The magnetic field is
[tex]B=\mu_onI[/tex]n is the number of turns per unit length.
Thus,
[tex]\begin{gathered} n=\frac{N}{l} \\ \Rightarrow n=\frac{200}{0.2}=1000 \end{gathered}[/tex]So,
[tex]B=4\pi\times10^{-7}\times1000\times3.25=4.08\times10^{-3}=4.08\text{ mT}[/tex]b. The magnetic force is given by:
[tex]\begin{gathered} F=\text{qvBsin}\theta \\ \Rightarrow F=15\times10^{-6}\times1050\times4.08\times10^{-3}\sin 11.5=1.28\times10^{-5}N \end{gathered}[/tex]Conclusion
a.The magnetic field is 4.08 mT
b.The magnetic force is
[tex]1.28\times10^{-5}N[/tex]4. A 45.8 kg block is placed on an inclined plane that is 44.2 degrees from the horizontal. What isthe acceleration of the block? Ignore friction.Challenge Question: 5. The same block in problem #4 is placed on the same inclined plane from problem #4. However, now there is friction. If the coefficient of kinetic friction is 0.25, what is the Friction force?(I already answered question 4, i just need help with question 5)
Explanation
Step 1
free body diagram
Step 2
sum of the forces on x-axis
a)x
[tex]\begin{gathered} F=\text{mg}\cos \text{ }\theta=ma \\ \text{replace} \\ F=\text{ 45.8}\cdot9.81\cdot\cos \text{ 44.2=ma} \\ 322.10=45.8a \\ a=\frac{322.10}{45.8} \\ a=703\text{ }\frac{\text{m}}{s^2} \end{gathered}[/tex]
A 1000-watt kettle is connected to a 220-volt power source. Calculate the resistance of the kettle
The power of kettle is given as,
[tex]P=\frac{V^2}{R}[/tex]Plug in the known values,
[tex]\begin{gathered} 1000\text{ W=}\frac{(220V)^2}{R} \\ R=\frac{(220V)^2}{1000\text{ W}}(\frac{1\text{ ohm}}{1V^2W^{-1}})_{} \\ =48.4\text{ ohm} \end{gathered}[/tex]Thus, the resistance of the kettle is 48.4 ohm.
What is the power rating of a heating coil with a resis-tance of 12 12 that draws a current of 20 A?
According to the problem, we have the following
[tex]\begin{gathered} R=12 \\ I=20 \end{gathered}[/tex]We have to use Ohm's law
[tex]V=I\cdot R[/tex]Let's replace the given information
[tex]V=20\cdot12=240[/tex]Hence, the power is 20 Volts.A flashlight bulb uses 190.1W of power when the current in the bulb is 0.64 A. What is the voltage?
Given:
[tex]\begin{gathered} P=190.1\text{ W} \\ I=0.64\text{ A} \end{gathered}[/tex]The power is given as,
[tex]P=VI[/tex]Here, P is the power, V is voltage and I is the current.
Rearranging equation in order to get voltage,
[tex]V=\frac{P}{I}[/tex]Putting all known values,
[tex]\begin{gathered} V=\frac{190.1\text{ W}}{0.64\text{ A}} \\ =297.03\text{ V} \end{gathered}[/tex]Therefore, the voltage is 297.03 V.
A force acting on an object does no work if1. a machine is used to move the object.2. the force is greater than the force of friction.3. the force acts perpendicular to the direction of the object’s motion.4. the object accelerates.
The work done of any object is defined by
[tex]W=Fd\cos \theta[/tex]Here, F is the force acting on the object.
d is the displacement of the object due to the force applied.
[tex]\theta[/tex]is the angle between force and displacement.
For work done to become zero, the applied force can be zero, or displacement can be zero, or cosine angle is zero.
Cosine angle can be zero if force and displacement are perpendicular to each other.
Thus, if the force acts perpendicular to the direction of the object's motion, no work is done..
how does gravity affect the range of a projectile?
Gravity has an effect on a projectile because it reduces the height that it can achieve.
The object's upward movement is stopped by gravity, which pulls it back to earth, limiting the projectile's vertical component. As a projectile moves through the air it is slowed down by air resistance.
Which choice is a valid way to construct a motion diagram?1) Add together the average speeds of the various objects in motion.2) Take a series of photographs at equal time intervals of a moving object, perpendicular to the direction of motion; overlay the images to see how the position changes with time.3) Draw vectors to represent the speeds involved.4) Add vectors in a head-to-tail manner to determine the resultant vector.
A motion diagram displays the location of an object at various equally spaced time intervals in the same diagram. This allows us to visualize the motion of the object.
From the described procedures, the one which would be useful to make a motion diagram, is:
Option 2) Take a series of photographs at equal time intervals of a moving object, perpendicular to the direction of motion; overlay the images to see how the position changes with time.
Two charges are separated by 6.74 cm. Object A has a charge of 5.0 μ C , while object B has a charge of 7.0 μ C . What is the force on Object A?
ANSWER
69.34 N
EXPLANATION
Given:
• The charge of object A, qA = 5 x 10⁻⁶ C
,• The charge of object B, qB = 7 x 10⁻⁶ C
,• The distance between the objects, r = 6.74 cm = 0.0674 m
Find:
• The force on Object A, FA
The magnitude of the force between two charged particles, q₁ and q₂, separated by a distance r is given by Coulomb's Law,
[tex]F=k\frac{q_1q_2}{r^2}[/tex]Where k is Coulomb's constant and has a value of approximately 9x10⁹ Nm²/C²,
[tex]F=9\times10^9\frac{Nm^2}{C^2}\cdot\frac{5\times10^{-6}C\cdot7\times10^{-6}C}{0.0674^2m^2}\approx69.34N[/tex]Hence, the force on object A is 69.34 N.
A phonograph record has an initial angular speed of 33 rev/min . The record slows to 11 rev/min in 2.0 seconds. What is the records angular acceleration in rad/s2 during this time interval ?
In order to calculate the angular acceleration, we can use the following formula:
[tex]a=\frac{v_f-v_i}{t}[/tex]Where vf is the final angular speed, vi is the initial angular speed and t is the interval of time.
Since the speed is in rev/min, we need to convert to rad/s.
Knowing that 1 rev = 2π rad and 1 min = 60 s, we have:
[tex]\begin{gathered} 33\text{ rev/min}=33\cdot\frac{2\pi\text{ rad}}{60\text{ s}}=3.456\text{ rad/s} \\ 11\text{ rev/min}=11\cdot\frac{2\pi\text{ rad}}{60\text{ s}}=1.152\text{ rad/s} \end{gathered}[/tex]Now, using vf = 1.152, vi = 3.456 and t = 2, we have:
[tex]a=\frac{1.152-3.456}{2}=\frac{-2.304}{2}=-1.152\text{ rad/s2}[/tex]So the angular acceleration is -1.152 rad/s².
Why are metals the best conductors of electricity? Why are metals the best magnetisers ? What is the key characteristic of metals that allows us to answer these two apparently different questions?
Question: What is the key characteristic of metals that allows us to answer these two apparently different questions?
Answer:
Explanation:
The atoms of a metal contains at least one free electron. Conduction of electricity is due to the these free electrons.
Magnetism is also affected by the availability of free electrons.
Thus, metals are good conductors of electricity and also the best magnetisers because of the availability of free electrons in metals
Use g = 10 m/s/s for this problem. A force is applied to a block through tension in a rope, at an angle to the horizontal as shown. The block is being pulled across a rough surface. The mass of the block is 71 kg. The tension in the rope is 1197 N. The angle from horizontal is 15 degrees. The coefficient of friction μ is 0.3.
Answer:
14.6 m/s²
Explanation:
First, we will make the free body diagram
Since the net vertical force is equal to 0 because the block is not moving up, we can write the following equation
[tex]\begin{gathered} F_{nety}=F_n+F_{Ty}-mg=0 \\ F_n+F_T\sin15-mg=0 \end{gathered}[/tex]Then, we can solve for the normal force and replace Ft = 1197 N, m = 71 kg and g = 10 m/s²
[tex]\begin{gathered} F_n=mg-F_T\sin15 \\ F_n=(71\text{ kg\rparen\lparen10 m/s}^2)-(1197N)(0.26) \\ F_n=400.2\text{ N} \end{gathered}[/tex]Now, we can write the following equation for the net horizontal force
[tex]\begin{gathered} F_{net}=F_{Tx}-F_f=ma \\ F_T\cos15-\mu F_n=ma \end{gathered}[/tex]Where μ is the coefficient of friction and a is the acceleration of the block. Solving for a and replacing Ft = 1197 N, m = 71 kg, Fn = 400.2 N and μ = 0.3, we get
[tex]\begin{gathered} a=\frac{F_T\cos15-\mu F_n}{m} \\ \\ a=\frac{(1197\text{ N\rparen}\cos15-0.3(400.2N)}{71\text{ kg}} \\ \\ a=14.6\text{ m/s}^2 \end{gathered}[/tex]Therefore, the acceleration of the block is 14.6 m/s²
which property justifies the statement below? x(y-3)=xy-3xis it associative, transitive, distributive, or communitive
Given the equation:
x(y - 3) = xy - 3x
Let's identify the property which justifies the statement.
Here, the property which was applied is the distributive property.
The law of distributive property states that multiplying the sum of two or more addends by a given value will amount to the same result as multiplying the addends individually by the number and adding the products.
In this equation, we can see x is distributed into the values in the parentheses.
Therefore, the property that justifies this statement is the distributive property.
ANSWER:
Distributive property
Let f(x) = 5* Let g(x) = 54 - 7 = - g Which statement describes the graph of g(x)with respect to the graph of f(x)? O g(x)is translated 7 units right from f(x). O g(x)is translated 7 units down fromf (x). O g(x)is translated 7 units left fromf (x). O g(x)is translated 7 units up fromf (x).
Answer:
[tex]\text{Second Option}[/tex]Explanation: The two functions are:
[tex]\begin{gathered} f(x)=5^x \\ g(x)=5^x-7 \end{gathered}[/tex]The graphs of these two functions are as follows:
From this we can infer that the answer is:
Study the following statements regarding energy and the Law of Energy Conservation: 1. The Law of Energy Conservation means it is saved for another time often for environmental reasons. 2. The Law of Energy Conservation means that there is the same amount of energy before the transfer took place as after 3. When energy is dissipated it means that it disappears. 4. When energy is dissipated it means some of the energy is converted into less useful forms. Which two of the above statements are true?
Answer:
2. The Law of Energy Conservation means that there is the same amount of energy before the transfer took place as after
4. When energy is dissipated it means some of the energy is converted into less useful forms.
Explanation:
Conservation means that energy is conserved through the transfer. It is equal at the start and at the end.
Energy cannot be created or destroyed, so it doesn't disappear, it changes (converted)
Point A is on the ground, and points B and C are h = 20 meters above the ground. Point B is directly above point A, and point C is L = 40 meters away from point A as shown.(a) How much work must be done by an external agent to move a 2 kg object from rest at point A to rest at point B?(b) How much work must be done to move the same object from rest at point A to rest at point C?
Given the mass of the object, m = 2 kg
Distance between point A and B, h= 20 m
(a) To find work done in moving the object from A to B
Work done is
[tex]\begin{gathered} W1=Force\times displacement\text{ } \\ =mgh \end{gathered}[/tex]Here, g is the acceleration due to gravity whose value is 9.8 m/s^2.
Substituting the values, we get
[tex]\begin{gathered} W1=2\times9.8\times20 \\ =\text{ 392 J} \end{gathered}[/tex](b) To find work done in moving the object from A to C
Gravitational force is a conservative force and work done depends only on the initial and final position and not on the path.
So, the work done will be
[tex]\begin{gathered} W2=\text{mgh} \\ =2\times9.8\times20 \\ =\text{ 392 J} \end{gathered}[/tex]The vertices of two rectangles are A(−5,−1),B(−1,−1),C(−1,−4),D(−5,−4) and W(1,6),X(7,6),Y(7,−2),Z(1,−2). Compare the perimeters and the areas of the rectangles. Are the rectangles similar? Explain.Perimeter of ABCD: , Area of ABCD: Perimeter of WXYZ: , Area of WXYZ:
In order to compare the perimeters and areas, let's first find two adjacent sides of each rectangle.
From ABCD, let's calculate AB and BC:
A and B have the same y-coordinate, so the length is the difference in x-coordinate:
AB = -1 - (-5) = -1 + 5 = 4
B and C have the same x-coordinate, so the length is the difference in y-coordinate:
AB = -1 - (-4) = -1 + 4 = 3
Therefore the perimeter and area are:
[tex]\begin{gathered} P=4+3+4+3=14 \\ A=4\cdot3=12 \end{gathered}[/tex]Now, for rectangle WXYZ, let's use WX and XY:
W and X have the same y-coordinate, so the length is the difference in x-coordinate:
WX = 7 - 1 = 6
X and Y have the same x-coordinate, so the length is the difference in y-coordinate:
XY = 6 - (-2) = 6 + 2 = 8
So the perimeter and area are:
[tex]\begin{gathered} P=6+8+6+8=28 \\ A=6\cdot8=48 \end{gathered}[/tex]In order to check if the rectangles are similar, let's check the following relation:
[tex](\frac{P_1}{P_2})^2=\frac{A_1}{A_2}[/tex]So we have:
[tex]\begin{gathered} (\frac{14}{28})^2=\frac{12}{48} \\ (\frac{1}{2})^2=\frac{1}{4} \\ \frac{1}{4}=\frac{1}{4}\text{ (true)} \end{gathered}[/tex]Since the relation is true, so the rectangles are similar.
Finding the coefficient of friction,I need how I would find the friction coefficient
the coefficient is 0.267
ExplanationFree-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object
Step 1
so, to graph the forces we use the Cartessian plane
let the x-axis paralle to the inclided plane
therefore,the weigth is a force acting directly to the center of the earth , ( bottomof page), so we have to find the x and y components of the weitgh
hence
the Free body diagram would be
Step 2
how to find the friction coefficient
the force of friction (ff) is given by
[tex]\begin{gathered} ff=\mu *N \\ where\text{ }\mu\text{ is the coefficient } \\ Nis\text{ the normal force} \end{gathered}[/tex]so, to find the coefficient we need the Normal force
if we do the sum of forces in y -axis equal zero
[tex]\begin{gathered} N-wcos(15)=0 \\ N=wcos(15) \end{gathered}[/tex]now, x -axis
[tex]\begin{gathered} ff-wsin(15)=0 \\ ff=wsin(15) \end{gathered}[/tex]finally, replace and solve for the coefficient
[tex]\begin{gathered} ff=\mu N \\ w*sin(15)=\mu *wcos(15) \\ divide\text{ both sides by w cos\lparen15\rparen} \\ \frac{wsin(15)}{wcos(15)}=\frac{\mu *w*cos(15)}{wcos(15)} \\ \frac{sin(15)}{cos(15)}=\mu \\ tan(15)=\mu \\ 0.267=\mu \end{gathered}[/tex]therefore, the coefficient is 0.267
I hope this helps you