Answer:
Hi
Final temperature = 250.11 °C
Final volume = 0,1 m3.
Process work = 0
Explanation:
The specific volume in the initial state is: v = 0.1m3/2 kg = 0.05 m3/kg.
This volume is located between the volumes as saturated liquid and saturated steam at 20 °C. For this reason the water is initially in a liquid vapor mixture. As the piston was blocked the volume remains constant and the process is isometric, also known as isocoric process, so the final temperature will be the water temperature at a saturated steam of v=0.05m3/kg, which is obtained by using steam tables for water, by linear interpolation. As follows, using table A-4 of the Cengel book 7th Edition:
v=0.05 m3/kg
v1=0.057061 m3/kg
T1=242.56°C
v2=0.049779 m3/kg
T2=250.35°C
T=[tex]\frac{T2-T1}{v2-v1} x(v-v1)+T1=\frac{250.35°C-242.56°C}{0.049779m3/kg-0.057061m3/kg}x(0.05m3/kg-0.057061m3/kg)+242.56°C=250.11°C[/tex]
The process work is zero because there is no change in volume during heating:
W=PxΔv=Px0=0
where
W=process work
P=pressure
Δv=change of volume, is zero because the piston was blocked so the volume remains constant.
Using Charles's Law and considering the properties of water, in a locked piston, the volume remains constant as water vaporizes, despite temperature changes. The temperature is affected by specific heat and latent heat of vaporization. No work is done as the locked piston prevents expansion.
Explanation:In this scenario, we would need to use principles from physics to solve this problem, particularly the laws of thermodynamics and the properties of gases and liquids, specifically water. Despite the piston being locked, as the water heats up and turns into vapor, the volume would increase according to Charles's Law. However, since our piston is locked and cannot move, in this case, the volume does not change and stays constant at 0.1 m³.
Next, the temperature change can be calculated from the specific heat of water and the given mass of water. However, as the water turns into vapor, we also have to account for the latent heat of vaporisation which is energy needed to change the water to vapor without changing its temperature.
As for the process work, it is zero in this case because our system is not doing work on the surroundings because the piston is locked and no expansion occurred which normally forms the basis of work done.
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Joe and Max shake hands and say goodbye. Joe walks east 0.50 km to a coffee shop, and Max flags a cab and rides north 3.45 km to a bookstore. How far apart are their destinations?
Answer:
3.486 km
Explanation:
Suppose Joe and Max's directions are perfectly perpendicular (east vs north). We can calculate their distance at the destinations using Pythagorean theorem:
[tex]s = \sqrt{J^2 + M^2}[/tex]
where J = 0.5 km and M= 3.45 km are the distances between Joe and Max to their original parting point, respectively. s is the distance between them.
[tex]s = \sqrt{0.5^2 + 3.45^2} = \sqrt{12.1525} = 3.486 km [/tex]
One type of slingshot can be made from a length of rope and a leather pocket for holding the stone. The stone can be thrown by whirling it rapidly in a horizontal circle and releasing it at the right moment. Such a slingshot is used to throw a stone from the edge of a cliff, the point of release being 18.0 m above the base of the cliff. The stone lands on the ground below the cliff at a point X. The horizontal distance of point X from the base of the cliff (directly beneath the point of release) is thirty times the radius of the circle on which the stone is whirled. Determine the angular speed of the stone at the moment of release.
Answer:
15.66 rad/s
Explanation:
The vertical motion and horizontal motion are independent of each other.
t = √ ( 2 s/ g) where t = time for the ball to reach the ground and s is the height of the cliff = 18.0 m
t = √ ( 36 / 9.81 ) = 1.916 secs
horizontal distance travel = ut where u is the horizontal velocity of the stone = 30 × r (radius)
tangential velocity V = angular velocity ( ω) × radius
distance traveled = ω × r × t = 30 × r
radius cancelled on both side
ω = 30 / 1.9156 = 15.66 rad/s
A 38.0 kg child is in a swing that is attached to ropes 1.70 m long. The acceleration of gravity is 9.81 m/s 2 . Find the gravitational potential energy associated with the child relative to the child’s lowest position under the following conditions:
Answer:
at zero point : GPE = 0 J
at max height : GPE = 633.7 J
Explanation:
the gravitational potential energy at the lowest point is zero
maximum height relative to the lowest point = h =1.70 m
G potential energy at max height = mgh = (38kg)(9.81m/s^2)(1.7)
= 633.7 J
A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coefficient of kinetic friction between the stone and the surface?
Answer:
Coefficient of friction will be 0.296
Explanation:
We have given initial speed of the stone u = 8 m /sec
It comes to rest so final speed v = 0 m /sec
Distance traveled before coming to rest s = 11 m
According to third equation of motion
[tex]v^2=u^2+2as[/tex]
So [tex]0^2=8^2+2\times a\times 11[/tex]
[tex]a=\frac{-64}{22}=-2.90m/sec^2[/tex]
Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]
We know that acceleration is given by
[tex]a=\mu g[/tex]
So [tex]2.90=9.8\times \mu \\[/tex]
[tex]\mu =\frac{2.9}{9.8}=0.296[/tex]
So coefficient of friction will be 0.296
To calculate the coefficient of kinetic friction, we first use Newton's second law to calculate the force of friction. Then, we divide that force by the normal force, yielding our coefficient.
Explanation:To find the coefficient of kinetic friction, you'll need to use the formula for kinetic friction (f = μkN), where 'f' is the force of friction, 'μk' is the coefficient of kinetic friction, and 'N' is the normal force. In this case, the force of friction can be found by using Newton's second law (f = ma), where 'm' is the mass of the stone and 'a' is its acceleration. The stone's acceleration can be found using the formula a = (vf - vi)/t, where 'vf' is the final velocity (0 m/s, since the stone comes to rest), 'vi' is the initial velocity (8.0 m/s), and 't' is the time it takes for the stone to stop. Once you've found the force of friction and the normal force (which is equal to the weight of the stone, or mg, where 'g' is the acceleration due to gravity), you can solve for the coefficient of kinetic friction.
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A projectile is launched at a diagonal angle and is in the air for 17.5 seconds. How long does it take the projectile to reach its apex?
Answer:
[tex]t_h=8.75\ s[/tex]
Explanation:
For a projectile the total time it stays in the air is called the flight time. The path traced during a projectile motion is parabolic with symmetry of time and distance.
Given that the projectile having a flight time of 17.5 seconds, will reach the top point in half of the flight time taken by it:
So, [tex]t_h=\frac{T}{2}[/tex]
[tex]t_h=\frac{17.5}{2}[/tex]
[tex]t_h=8.75\ s[/tex] is the time by the end of which the projectile will be at its apex having the vertical component of velocity as zero.
Do note that during a projectile analysis we neglect the air resistance due to which we have a constant horizontal component of the velocity and the vertical component of the velocity varies under the influence of gravity. The vertical velocity at the top becomes zero.
Desertification is a process whereby land degradation in arid, semi-arid, and dry sub humid areas results from climatic variations and human activites.
A. True
B. False
Answer:
TRUE
Explanation:
Desertification is usually defined as the process by which a productive land is transformed into a desert. This process occurs mainly in the arid, semi-arid as well as in the dry sub-humid areas, where there occurs less amount of rainfall and presence of less or no moisture content in the air.
It is caused by various processes, such as-
Cutting down of trees, in order to make urban areas. Lack of rainfall. Overgrazing of animals. Low level of groundwater. Forest-fire can also lead to desertification, as the trees are burned by this process. Soil erosion also is a major cause. It also occurs due to changes that occur in the climate of the area.Thus, the above-given statement is true.
Answer:
TRUE
Explanation:
Find the voltage change when (a) an electric field does 12 J of work on a 0.0001-C charge and (b) the same electric field does 24 J of work on a 0.0002-C charge.
Answer:
a. [tex]Change in Voltage=12*10^{4}V[/tex]
b. [tex]Change in Voltage=12*10^{4}V[/tex]
Explanation:
The work done in moving an electric charge round a circuit is express as
[tex]workdone=voltage*charge \\Wd=v*q[/tex]
The voltage is in-turn define as the electric potential energy per unit charge.
[tex]Voltage=\frac{potitntial energy }{charge}\\[/tex]
a. for a 12J work done on a charge of value 0.0001C, we can compute the voltage change as
[tex]Voltage=\frac{potitntial energy }{charge}\\Voltage=\frac{12J}{0.0001C}\\ Voltage=12,0000J/C\\ Change in Voltage=12*10^{4}V[/tex]
a. for a 24J work done on a charge of value 0.0002C, we can compute the voltage change as
[tex]Voltage=\frac{potitntial energy }{charge}\\Voltage=\frac{24J}{0.0002C}\\ Voltage=12,0000J/C\\Change in Voltage=12*10^{4}V[/tex]
The voltage change in (a) is 1.2 × 10⁵ J/C and in (b) is 1.2 × 10⁵ J/C.
a) Based on the given information,
• Work done (W) is 12 J and charge (q) is 0.0001 C.
The change in voltage is determined by using the formula,
= W/q
Now putting the values we get,
= [tex]\frac{12}{0.0001} J/C[/tex] or 1.2 × 10⁵J/C
b) Based on the given information,
• Work done (W) is 24J and charge (q) is 0.0002 C.
The change in voltage can be calculated as,
= W/q
Putting the values we get,
=[tex]\frac{24}{0.0002} J/C[/tex] or 1.2 × 10⁵J/C
Thus, the voltage change in both the conditions is 1.2 × 10⁵J/C.
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An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 78 cm long and has a mass of 3.8 kg, with the center of mass at 40% of the arm length from the shoulder. What is the magnitude of the torque about his shoulder due to the ball and the weight of his arm if he holds his arm.
a. Straight out to his side, parallel to the floor?
b. Straight, but 45° below horizontal?
Answer:
a. 34.6 Nm
b.24.4 Nm
Explanation:
a.
78 cm = 0.78 m
W = F =mg
m1 = mass of steel ball = 3 kg
m2 = mass of long arm = 3.8 kg
moment due to steel ball = Fd =(m1*g)*(0.78)= (3*9.81)(0.78)=22.95 = 23 Nm
moment due to arm =Fd=(m2*g)*(0.78*0.4)= (3.8*9.81)(0.312)=11.63 = 11.6 Nm
net moment = 23 +11.6 = 34.6 Nm
b. now in this the angle will change the perpendicular moment arm
moment due to steel ball = (3*9.81)*(0.78cos45) = 16.23 =16.2 Nm
moment due to arm = (3.8*9.81)(0.4*0.78cos45) = 8.22 = 8.2 Nm
net moment = 16.2 +8.2 = 24.4Nm
(a) The magnitude of the torque about his shoulder due to the ball and the weight of his arm parallel to the floor is 34.55 Nm.
(b) The magnitude of the torque about his shoulder due to the ball and the weight of his arm straight, but 45° below horizontal is 24.43 Nm.
The given parameters;
mass of the ball, m = 3.0 kglength of the arm, L = 78 cm = 0.78 mmass of his arm, = 3.8 kgA sketch of the position of the ball and the arm;
---------------------------------------------------------------------------78cm
↓ 40% ↓
3 kg 3.8 kg
Take moment about the arm;
The moment due to arm (clockwise), is calculated as follows;
M₁ = Fd = (3.8 x 9.8) x (0.4 x 0.78) = 11.62 Nm
The moment due to ball (clockwise), is calculated as follows;
M₂ = Fd = (3 x 9.8) x (1 x 0.78) = 22.93 Nm
The magnitude of the torque about his shoulder due to the ball and the weight of his arm parallel to the floor is calculated as;
τ = M₁ + M₂
τ = 11.62 + 22.93 = 34.55 Nm
(b) The moment at angle 45⁰ below the horizontal is calculated as follows;
The moment due to arm (clockwise), is calculated as follows;
M₁ = Fd = (3.8 x 9.8) x (0.4 x 0.78) x (cos45) = 8.22 Nm
The moment due to ball (clockwise), is calculated as follows;
M₂ = Fd = (3 x 9.8) x (1 x 0.78) x cos(45) = 16.21 Nm
The magnitude of the torque about his shoulder due to the ball and the weight of his arm straight, but 45° below horizontal is calculated as;
τ = M₁ + M₂
τ = 8.22 + 16.21
τ = 24.43 Nm.
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You drive on Interstate 10 from San Antonio to Houston, half the time at 72 km/h and the other half at 98 km/h. On the way back you travel half the distance at 72 km/h and the other half at 98 km/h. What is your average speed (a) from San Antonio to Houston, (b) from Houston back to San Antonio, and (c) for the entire trip?
Answer: a. 85km/hr b.82.3km/hr
c. 84km/hr
Explanation: first let take the total time from San Antonio to Houston to be 2hr.
Half time 1hr was covered with speed of 72km/hr
Distance = speed*time=72km/hr *1hr
=72km
So too with the second half of 1hr covered with speed of 98km/hr
Distance = 98km
Total distance from Houston to San Antonio is 98+72 =170km
a. Average speed from San Antonio to Houston is
S1 =170/2
=85km/hr
b.half distance from Houston to San Antonio which is 170km/2
= 85km was covered with speed of 72km/hr first half, so time
t = dist/speed
t = 85/72 = 1hr 12 mins
Remaining 85 km covered with a speed of 98km/hr
Time = 85/98 = 0.88*60min
= 52 mins
Total time = 1hr +12mins +52mins
=2hr4mins= 124/60 hr
So average speed = distance/time
=170/124/60
Using reciprocal law
Average speed S2= 170*60/124
= 82.3km/hr
C. Average speed to and fro(entire tripe)
= (85+82.3)/2
=84km/hr
An object of height 2.2 cm is placed 5.1 cm in front of a diverging lens of focal length 19 cm and is observed through the lens from the other side 50% Part (a) How many centimeters beyond the lens does the observer locate the object's image? Grade Summa Deductions Potential ry 0% 100% cosO asin0a Sin Submissions Attempts remaining:J5 (5% per attempt) detailed view cotana acosO acotan0sinh0 atan cosh0 tanh cotanhO Degrees Radians END DELI CLEAR Submit Hint I give up! Hints: deduction per hint. Hints remaining: Feedback: 296 deduction per feedback. là 50% Part (b) What is the height of the image, in centimeters?
Answer:
a) i = -4.02 cm , b) h’= 1,576 cm
Explanation:
a) The constructor equation is
1 / f = 1 / i + 1 / o
Where f is the focal length, i and o are the distance to the image and the object
Let's clear the distance to the image
1 / i = 1 / f - 1 / o
1 / i = 1 / -19 - 1 / 5.1
1 / i = 0.2487
i = -4.02 cm
b) let's use the expression of magnification
m = h’/ h = - i / o
h’= - h i / o
h’= 2.2 4.02 /5.1
h’= 1,576 cm
The image of the object is located approximately -12.31 cm beyond the lens.
Explanation:In this case, we have a diverging lens with a focal length of 19 cm. The object is placed 5.1 cm in front of the lens. To determine the location of the image, we can use the lens formula: 1/f = 1/v - 1/u, where f is the focal length, v is the image distance, and u is the object distance.
Plugging in the values, we have: 1/19 = 1/v - 1/5.1. Solving this equation, we find v to be approximately -12.31 cm, which means the image is located 12.31 cm beyond the lens.
Why is thermal energy from the Sun transferred to Earth through electromagnetic waves instead of any other type of thermal energy transfer?
Answer:
Because electromagnetic waves can travel through empty space
Explanation:
The energy that is emitted from the sun is transferred to the earth in the form of radioactive waves. These waves are originated due to the vibration between the electric and magnetic fields. As this energy reaches the earth, it warms the earth's atmosphere, resulting in the transfer of heat energy in three possible ways namely the conduction, convection, and radiation.
This electromagnetic waves do not require any matter for the transmission of energy, and can easily travel in empty space from the core of the sun to the earth and other nearby planets. Whereas other types of waves cannot travel in space, so it is transferred in the form of electromagnetic waves only.
Answer:
The electromagnetic waves can pass through empty space.
Explanation:
A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 16.0 when the hand is 1.80 above the ground.How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)
There are mistakes in the question as the unit of speed and height is not mention here.The correct question is here
A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 16.0m/s when the hand is 1.80m above the ground.How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.)
Answer:
t=3.37s
Explanation:
Given Data
As we have taken hand at origin and positive upward
So given data are
[tex]y_{i}=0m\\y_{f}=-1.80m\\v_{i}=16.0m/s\\a=g=9.8m/s^{2}[/tex]
To find
time taken by the ball before it hits the ground
Solution
By using the common kinematic equation
[tex]y_{f}=y_{i}+v_{i}t+0.5at^{2}[/tex]
Put the given values and find for t
So
[tex]-1.80=0+16.0t+(0.5*(-9.8)t^{2} )\\-1.80=16.0t-4.9t^{2}\\ 4.9t^{2}-16.0t-1.80=0[/tex]
Apply quadratic formula to solve for t
[tex]t=\frac{-(-16.0)+\sqrt{(-16)^{2}+4(4.9)(-1.80)} }{2(4.9)}\\ t=3.37s[/tex]
a bucket filled with water has a weight of 150 N and is attached to a rope that is woud around a pully (solid disc) with a 0.5m radius and a mass of 15kg. I fthe bucket is allowed to fall, what will the angular acceleration of the pulley ?
Answer:
20 rad/s^{2}
Explanation:
weight / force (F) = 150 N
radius (r) = 0.5 m
mass (m) = 15 kg
angular acceleration = tangential acceleration / radius
where
tangential acceleration = force/ mass = 150/15 = 10 m/s^{2}substituting the above into the equation for angular acceleration we haveangular acceleration = 10 / 0.5 = [tex]20 rad/s^{2}[/tex]
Earth continually moves about 30 km/s through space, which means the wall you stand next to also is moving at 30 km/s. When you jump vertically the wall doesn't slam into you because ________
Answer:
we are on the same frame of reference moving with the earth with the same velocity.
Explanation:
Given that the earth is continuously moving at a speed of about 30 kilometers per second in the space. This means this is the observed speed from and external frame of reference in space being at rest.But when we jump from vertically on the earth we are already on the same moving frame of reference and bounded to it by the gravity and hence when we jump off its surface we jump with its velocity of motion and so does every other object present on the earth and hence we do not collide with the wall when taking a vertical jump beside it. To us the wall seems to be at rest because we both are on the same frame of the reference.A catapult launches a boulder with an upward velocity of 92 m/s. The height of the boulder, h, in meters after t seconds is given by the function h = –5t2+ 92t + 16. How long does it take to reach maximum height? What is the boulder’s maximum height? Round to the nearest hundredth, if necessary. A. Reaches a maximum height of 16.00 meters in 18.4 seconds. B. Reaches a maximum height of 18.57 meters in 9.2 seconds. C. Reaches a maximum height of 37.14 meters in 18.4 seconds. D. Reaches a maximum height of 439.20 meters in 9.2 seconds.
Answer:
D. Reaches a maximum height of 439.20 meters in 9.2 seconds.
Explanation:
Given
h = –5t²+ 92t + 16
then
h' = 0 when the boulder reaches its maximum height
(–5t²+ 92t + 16)' = - 10t + 92 = 0
⇒ t = 92/10
⇒ t = 9.2 s
the maximum height will be
h = –5(9.2)²+ 92(9.2) + 16
h = 439.20 m
Quaternary structure is achieved when a protein folds into a compact, three-dimensional shape stabilized by interactions between side-chain R groups of amino acids.True or False.
Answer:
False: Quaternary structure is achieved when multiple polypeptide chains in protein come together.
Explanation:
There are four levels of protein structure: primary, secondary, tertiary, and Quaternary structure.
Tertiary structure is a protein structure, which is achieved when a protein folds into a compact, three-dimensional shape stabilized by interactions between side-chain R groups of amino acids.
However, Quaternary structure is achieved when multiple polypeptide chains in protein come together.
The ____ method of assessing intake is time consuming, and the results may not be accurate if subjects modify their eating habits during the time of the study.
Answer:
Food record
Explanation:
A food record method is an assessment, study or act of collecting data related to food.
Although Food record method is time consuming and may not be accurate if subjects modify their eating habits during the time of the study the data collected here are of great importance. Some of the importance are;
(1). It helps in the registration of foods.
(2). The data can be used to describe a population's intake.
(3). The data can be used as a reference parameter in validation studies.
(4). The data can also be used with the Food Frequency Questionnaire.
It other disadvantage apart from its time consuming and error that might occur during the process of conducting the research is that it might be burdensome to the respondents.
The standard used by LEED in the Water Efficiency (WE) category to determine the baseline case for building water usage is the national standard for maximum flush and flow rates called:
Answer:
EPAct of 1992.
Explanation:
The Energy Policy Act (EPAct) is a United States Of America government act passed in the year 1992 and became effective in October 24, 1992.
The Energy Policy Act was established to address the energy needs in the United States of America by amending laws to increase and provide incentives for clean and renewable energy and also to decrease the dependence on imported energy.
A set of facts and relationships between facts that can explain and predict related phenomena is called a(n) ____.a. theoryb. experimentc. descriptive methodd. hypothesis
Answer:
a. theory
Explanation:
A scientific theory is a set of facts and rules, that is, scientific laws, which express relationships between observations of these facts. Therefore it is a set of principles to explain a certain type of natural phenomena. Thus, the strength of a scientific theory is related to the diversity of phenomena it can explain and its simplicity.
Final answer:
A set of facts and relationships that can explain and predict phenomena is known as a theory, which is a well-supported scientific explanation and the foundation of scientific knowledge.
Explanation:
A set of facts and relationships between facts that can explain and predict related phenomena is called a theory. A theory is a well-supported explanation of observations and is often used as the foundation of scientific knowledge. It goes beyond a hypothesis, which is a tentative explanation that can be tested, by being supported by a substantial amount of empirical evidence and experimentation. In contrast to a theory, a law summarizes the relationships between variables without explaining why they occur.
The process of discovery in science usually follows the scientific method, where hypotheses are made and then tested through experiments and observation to acquire new knowledge.
Two chargedparticles, with charges q1=q and q2=4q, are located at a distance d= 2.00cm apart on the x axis. A third charged particle,with charge q3=q, is placed on the x axis such that the magnitude of the force that charge 1 exerts on charge 3 is equal to the force that charge 2 exerts on charge 3. ~Find theposition of charge 3 when q = 2.00 nC . ~ Assuming charge 1 is located at the origin of the x axisand the positive x axis points to the right, find the two possible values x3,r and x3,l for the position of charge 3. I am stuck on this conversion x^2=2(2-x)^2 to find the value for x!
Answer:
Two possible points
x= 0.67 cm to the right of q1
x= 2 cm to the left of q1
Explanation:
Electrostatic Forces
If two point charges q1 and q2 are at a distance d, there is an electrostatic force between them with magnitude
[tex]\displaystyle f=k\frac{q_1\ q_2}{d^2}[/tex]
We need to place a charge q3 someplace between q1 and q2 so the net force on it is zero, thus the force from 1 to 3 (F13) equals to the force from 2 to 3 (F23). The charge q3 is assumed to be placed at a distance x to the right of q1, and (2 cm - x) to the left of q2. Let's compute both forces recalling that q1=1, q2=4q and q3=q.
[tex]\displaystyle F_{13}=k\frac{q_1\ q_3}{d_{13}^2}[/tex]
[tex]\displaystyle F_{13}=k\frac{(q)\ (q)}{x^2}[/tex]
[tex]\displaystyle F_{23}=k\frac{q_2\ q_3}{d_{23}^2}[/tex]
[tex]\displaystyle F_{23}=k\frac{(q)(4q)}{(0.02-x)^2}[/tex]
[tex]\displaystyle F_{23}=\frac{4k\ q^2}{(0.02-x)^2}[/tex]
Equating
[tex]\displaystyle F_{13}=F_{23}[/tex]
[tex]\displaystyle \frac{K\ q^2}{x^2}=\frac{4K\ q^2}{(0.02-x)^2}[/tex]
Operating and simplifying
[tex]\displaystyle (0.02-x)^2=4x^2[/tex]
To solve for x, we must take square roots in boths sides of the equation. It's very important to recall the square root has two possible signs, because it will lead us to 2 possible answer to the problem.
[tex]\displaystyle 0.02-x=\pm 2x[/tex]
Assuming the positive sign :
[tex]\displaystyle 0.02-x= 2x[/tex]
[tex]\displaystyle 3x=0.02[/tex]
[tex]\displaystyle x=0.00667\ m[/tex]
[tex]x=0.67\ cm[/tex]
Since x is positive, the charge q3 has zero net force between charges q1 and q2. Now, we set the square root as negative
[tex]\displaystyle 0.02-x=-2x[/tex]
[tex]\displaystyle x=-0.02\ m[/tex]
[tex]\displaystyle x=-2\ cm[/tex]
The negative sign of x means q3 is located to the left of q1 (assumed in the origin).
The two possible values for the position of charge 3 are [tex]\( x_{3,r} = \frac{2d}{3} \) and \( x_{3,l} = \frac{2d}{5} \)[/tex]. Substituting [tex]\( d = 2.00 \times 10^{-2} \) m and \( q = 2.00 \times 10^{-9} \) C[/tex], we get [tex]\( x_{3,r} = 2.67 \times 10^{-2} \) m[/tex] and [tex]\( x_{3,l} = 8.00 \times 10^{-3} \) m.[/tex]
To find the position of charge 3 (q3), we need to use Coulomb's law, which states that the force (F) between two point charges is directly proportional to the product of the magnitudes of the charges (q1 and q2) and inversely proportional to the square of the distance (r) between them. Mathematically, it is given by:
[tex]\[ F = k \frac{|q_1 q_2|}{r^2} \][/tex]
Given that the magnitude of the force that charge 1 (q1) exerts on charge 3 (q3) is equal to the force that charge 2 (q2) exerts on charge 3 (q3), we can set up the following equation:
[tex]\[ k \frac{|q_1 q_3|}{r_{13}^2} = k \frac{|q_2 q_3|}{r_{23}^2} \][/tex]
Since [tex]\( q_1 = q \), \( q_2 = 4q \), and \( q_3 = q \)[/tex], we can simplify the equation to:
[tex]\[ \frac{1}{r_{13}^2} = \frac{4}{r_{23}^2} \][/tex]
We know that [tex]\( r_{23} = d - r_{13} \)[/tex] because charge 2 is located at a distance [tex]\( d \)[/tex] from the origin where charge 1 is situated. Substituting [tex]\( r_{23} \) with \( d - r_{13} \)[/tex], we get:
[tex]\[ \frac{1}{r_{13}^2} = \frac{4}{(d - r_{13})^2} \][/tex]
Taking the square root of both sides, we have:
[tex]\[ \frac{1}{r_{13}} = \frac{2}{d - r_{13}} \][/tex]
Cross-multiplying gives us:
[tex]\[ d - r_{13} = 2r_{13} \] \[ d = 3r_{13} \] \[ r_{13} = \frac{d}{3} \][/tex]
[tex]\[ x_{3,r} = \frac{d}{3} \][/tex]
Substituting [tex]\( d = 2.00 \times 10^{-2} \)[/tex] m, we get:
[tex]\[ x_{3,r} = \frac{2.00 \times 10^{-2} \text{ m}}{3} \] \[ x_{3,r} = 6.67 \times 10^{-3} \text{ m} \][/tex]
For the left side, we have:
[tex]\[ r_{23} = \frac{d}{5} \][/tex]
Therefore, we have:
[tex]\[ x_{3,l} = d - \frac{d}{5} \] \[ x_{3,l} = \frac{4d}{5} \][/tex]
Substituting [tex]\( d = 2.00 \times 10^{-2} \) m[/tex], we get:
[tex]\[ x_{3,l} = \frac{4 \times 2.00 \times 10^{-2} \text{ m}}{5} \] \[ x_{3,l} = 1.60 \times 10^{-2} \text{ m} \][/tex]
However, the correct expressions for [tex]\( x_{3,r} \) and \( x_{3,l} \)[/tex] are:
[tex]\[ x_{3,r} = \frac{2d}{3} \] \[ x_{3,l} = \frac{2d}{5} \][/tex]
Substituting [tex]\( d = 2.00 \times 10^{-2} \)[/tex]m, we get:
[tex]\[ x_{3,r} = \frac{2 \times 2.00 \times 10^{-2} \text{ m}}{3} \] \[ x_{3,r} = 2.67 \times 10^{-2} \text{ m} \] \[ x_{3,l} = \frac{2 \times 2.00 \times 10^{-2} \text{ m}}{5} \] \[ x_{3,l} = 8.00 \times 10^{-3} \text{ m} \][/tex]
These are the two possible positions for charge 3 where the forces exerted by charges 1 and 2 are equal in magnitude.
A 0.145-kg baseball pitched horizontally at 27.0 m/s strikes a bat and pops straight up to a height of 31.5 m. If the contact time between bat and ball is 2.5 ms, calculate the average force between the ball and bat during contact.
The average force between the ball and the bat during the contact is 1,566 N.
The given parameters;
mass of the ball, m = 0.145 kghorizontal velocity of the ball, u = 27 m/sheight of the ball's displacement, h = 31.5 mthe contact time of the ball, t = 2.5 msThe average force between the ball and the bat during the contact is determined by applying Newton's second law of motion.
F = ma
where;
a is the acceleration of the ballm is the mass of the ballThe average force is calculated as follows;
[tex]F = ma = m\frac{v}{t} \\\\F = \frac{mv}{t} \\\\F = \frac{0.145 \times 27}{2.5 \times 10^{-3}} \\\\F = 1,566 \ N[/tex]
Thus, the average force between the ball and the bat during the contact is 1,566 N.
Learn more here:https://brainly.com/question/16200276
To calculate the average force between the ball and bat during contact, you first calculate the speed the ball leaves the bat using the gravitational potential energy formula. Find the change in velocity, then use the formula for force using this change in velocity and the given time.
Explanation:The subject of this question is physics, specifically involving the concepts of force, mass, speed, and time. This problem first requires using the equation for gravitational potential energy to find the speed at which the ball leaves the bat. That is Potential energy = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. This will give you the final speed of the baseball after the contact with the bat.
Next, find the change in velocity (or 'delta v') by subtracting the initial speed of the baseball from its final speed. The average force can then be calculated using Newton's second law transformed in impulse form: Force = delta p / delta t = m * delta v / delta t, where delta p is the change in momentum, delta v is the change in speed, and delta t is the change in time.
This calculation should provide the answer to your question, and requires understanding of physics concepts such as gravitational potential energy, impulse, and force.
Learn more about Physics of force here:https://brainly.com/question/14215609
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If the distance between the levers input force and the fulcrum is 8 cm, and the distance between the fulcrum and the output force is 24 cm, what is the ideal mechanical advantage of the lever?
Answer:
Ideal mechanical advantage of the lever is 3.
Explanation:
Given that,
The distance between the levers input force and the fulcrum is 8 cm, [tex]d_i=8\ cm[/tex]
The distance between the fulcrum and the output force is 24 cm, [tex]d_o=24\ cm[/tex]
To find,
The ideal mechanical advantage of the lever.
Solution,
The ratio of the distance between the fulcrum and the output force to the distance between the levers input force and the fulcrum is called the ideal mechanical advantage of the lever. It is given by :
[tex]m=\dfrac{d_o}{d_i}[/tex]
[tex]m=\dfrac{24}{8}[/tex]
m = 3
So, the ideal mechanical advantage of the lever is 3.
Gina walks 3 mph for 4 hrs. How far did she walk? 1.3 mi 7 mi 0.75 mi 12 mi
Answer: [tex]12 mi[/tex]
Explanation:
Velocity [tex]V[/tex] is mathematically defined as:
[tex]V=\frac{d}{t}[/tex] (1)
Where:
[tex]V=3 \frac{mi}{h}[/tex] is Gina's velocity
[tex]t=4 h[/tex] is the time Gina spends walking
[tex]d[/tex] is the distance Gina has walked
Isolating [tex]d[/tex] from (1):
[tex]d=Vt[/tex] (2)
[tex]d=(3 \frac{mi}{h})(4 h)[/tex] (3)
Finally:
[tex]d=12 mi[/tex] This is the distance Gina has walked
Match the characteristic or descriptive phrase to the type of application it describes.
Part A
a. sound waves
b. soft-tissue imaging
c. electromagnetic wave
d. fetal imaging
Part B
1. ultra sound
2. mri
Answer:
Sound waves and fetal imaging is related to the ultrasound.
Electromagnetic wave and soft tissue imaging is related to the MRI.
Explanation:
Ultrasound scans is basically a process in which high frequency sound waves are used for the fetal imaging or examining the internal organs like liver, kidneys etc. It's the safe procedure as no radiations are involved in this case.
However the MRI stands for Magnetic resonance imaging that use the electromagnetic waves and is best for the soft tissue imaging etc.
Sound waves and electromagnetic waves are key to ultrasound and MRI imaging. Ultrasound uses sound waves for applications like fetal imaging, while MRI uses electromagnetic waves for soft-tissue imaging.
Explanation:To match the characteristic or descriptive phrase to the type of application it describes:
a. sound waves - 1. ultrasoundb. soft-tissue imaging - 2. MRIc. electromagnetic wave - 2. MRId. fetal imaging - 1. ultrasoundUltrasound is a medical imaging technique that uses high-frequency sound waves to create images of the inside of the body. It is particularly useful for fetal imaging during pregnancy as it is non-invasive and safe for both the mother and fetus. On the other hand, Magnetic Resonance Imaging (MRI) uses electromagnetic waves in the radio-frequency range to create detailed images of the body's soft tissues, making it invaluable for detecting conditions within the body's interior.
Chemical vapor deposition can be defined as the interaction between a mixture of gases and the surface of a heated substrate, causing chemical decomposition of some of the gas constituents and formation of a solid film on the substrate.
a) true
b) false
Answer: The given statement is true.
Explanation:
When a substrate is exposed to one more more number of volatile substances that react together on the surface of substrate to produce a suitable deposit of a thin non-volatile film is known as chemical vapor deposition.
This type of reaction generally occurs in heat flux.
Therefore, we can conclude that the statement chemical vapor deposition can be defined as the interaction between a mixture of gases and the surface of a heated substrate, causing chemical decomposition of some of the gas constituents and formation of a solid film on the substrate, is true.
Car drag racing takes place over a distance of a 1 4 mile (402 m) from a standing start. If a car (mass 1500 kg) would be propelled forward with a pulling force equal to that of gravity, what would be the change in kinetic energy and the terminal speed of the car (in mph) at the end of the race be?
Answer:
The change in kinetic energy is 5,909,400J and the terminal speed of the car is 160,800mph
Explanation:
Change in kinetic energy = mg(h2 - h1)
Mass (m) = 1500kg, g = 9.8m/s^2, final distance (h2) = 402m, initial distance (h1) = 0m
Change in kinetic energy = 1500×9.8×(402 - 0) = 1500×9.8×402 = 5,909,400J
From equations of motion
h = ut + 1/2gt^2 [u is initial speed and is equal to zero because the car drag racing began from a standing start (rest)]
h = 1/2gt^2
402 = 1/2 × 9.8t^2
402 = 4.9t^2
t^2 = 402/4.9 = 82.04
t = √82.04 = 9.06s × 1h/3600s = 0.0025h
Terminal speed = distance/time = 402m/0.0025h = 160,800mph
An observer sees a flower pot sail up and then back down past a window 2.45 m high. If the total time the pot is in the sight is 1 s, what is the height above the window up to which the pot rises?
Answer: 2.55meter
Explanation: Using the second equation of motion.
S{hieght} = U*t + {g*t²}/2
Where U is initial velocity =0m/s
g is acceleration due to gravity 10m/s²
t is time 1secs
So we have,
hieght = 0 + {g*t²}/2
hieght = {10*(1)²}/2
Total hieght travelled is 10/2
Which is 5 meter.
But we are asked to find the hieght above the window which as a hieght of 2.45meter.
So,
hieght above window would be
{5 - 2.45}meter
Which is 2.55 meter.
A constant eastward horizontal force of 70 newtons is applied to a 20 -kilogram crate moving toward the east on a level floor. If the frictional force on the crate has a magnitude of 10 newtons, what is the magnitude of the crate's acceleration?
Answer:
a = 3 m/s²
Explanation:
given,
mass of crate = 20 Kg
horizontal force on crate = 70 N
frictional force on the crate = 10 N
acceleration of crate = ?
now, calculating net force acting on the crate.
F = horizontal force - frictional force
F = 70 - 10
F = 60 N
net force on the crate is equal to 60 N.
We also know that
F = m a
60 = 20 x a
a = 3 m/s²
Hence, the acceleration of the crate is equal to 3 m/s²
Final answer:
The magnitude of the crate's acceleration is 3 m/s² toward the east, calculated using Newton's second law with the net force of 60 N (70 N applied force minus 10 N frictional force) divided by the mass of 20 kg.
Explanation:
To calculate the magnitude of the crate's acceleration, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object times its acceleration (F = ma). The net force is the difference between the applied force and the frictional force. In this case, the applied eastward horizontal force is 70 newtons and the frictional force is 10 newtons, acting in the opposite direction.
Net force = applied force - frictional force = 70 N - 10 N = 60 N
Now, using the equation F = ma, we can solve for the acceleration (a) as follows:
a = F / m = 60 N / 20 kg = 3 m/s²
Therefore, the magnitude of the crate's acceleration is 3 m/s² toward the east.
Lord Kelvin's calculations for dating the earth were correct, but his assumption that earth was undergoing radiant cooling from combustion of the sun was incorrect, thus he arrived at an incorrect age of 15-20 million years old.
a) true
b) false
Answer:GiGi
Explanation:
A department store sells an ""astronomical telescope"" with an objective lens of 30 cm focal length and an eyepiece lens of focal length 5 mm. What is the magnifying power of this telescope?
Answer:
The magnifying power of this telescope is (-60).
Explanation:
Given that,
The focal length of the objective lens of an astronomical telescope, [tex]f_o=30\ cm[/tex]
The focal length of the eyepiece lens of an astronomical telescope, [tex]f_e=5\ mm=0.5\ cm[/tex]
To find,
The magnifying power of this telescope.
Solution,
The ratio of focal length of the objective lens to the focal length of the eyepiece lens is called magnifying of the lens. It is given by :
[tex]m=\dfrac{-f_o}{f_e}[/tex]
[tex]m=\dfrac{-30}{0.5}[/tex]
m = -60
So, the magnifying power of this telescope is 60. Therefore, this is the required solution.