Answer:
speed of eight ball speed after the collision is 3.27 m/s
Explanation:
given data
initially moving v1i = 3.4 m/s
final speed is v1f = 0.94 m/s
angle = θ w.r.t. original line of motion
solution
we assume elastic collision
so here using conservation of energy
initial kinetic energy = final kinetic energy .............1
before collision kinetic energy = 0.5 × m× (v1i)²
and
after collision kinetic energy = 0.5 × m× (v1f)² + 0.5 × m× (v2f)²
put in equation 1
0.5 × m× (v1i)² = 0.5 × m× (v1f)² + 0.5 × m× (v2f)²
(v2f)² = (v1i)² - (v1f)²
(v2f)² = 3.4² - 0.94²
(v2f)² = 10.68
taking the square root both
v2f = 3.27 m/s
speed of eight ball speed after the collision is 3.27 m/s
Final answer:
To find the eight ball's speed after the collision, we can use the principle of conservation of momentum and the fact that the collision is elastic. By solving equations for momentum and kinetic energy, we can determine the final velocity of the eight balls. Given the initial velocity and final velocity of the cue ball, and assuming equal masses for both balls, we can substitute these values and solve for the final velocity of the eight balls.
Explanation:
To find the eight ball's speed after the collision, we can use the principle of conservation of momentum. In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision. Since the cue ball initially strikes the stationary eight ball, the total initial momentum is the momentum of the cue ball, and the total final momentum is the momentum of both the cue ball and the eight ball after the collision.
Let's denote the mass of both balls as 'm'. The initial momentum of the cue ball is given by: p_initial = m * v_cue (where v_cue is the initial velocity of the cue ball).
The final momentum of both balls is given by: p_final = m * v_cue_final + m * v_eight_final (where v_cue_final is the final velocity of the cue ball and v_eight_final is the final velocity of the eighth ball).
Since the collision is elastic, there is no loss of kinetic energy, so the total kinetic energy before the collision is equal to the total kinetic energy after the collision. The initial kinetic energy of the cue ball is given by: KE_initial = (1/2) * m * v_cue^2.
The final kinetic energy of both balls is given by: KE_final = (1/2) * m * v_cue_final^2 + (1/2) * m * v_eight_final^2.
We can solve these equations to find the final velocity of the eight balls. Given that the cue ball's initial velocity (v_cue) is 3.4 m/s, its final velocity (v_cue_final) is 0.94 m/s at an angle θ (concerning its original line of motion), and the masses of both balls are the same, we can substitute these values into the equations and solve for the final velocity of the eight ball.
Friends Burt and Ernie stand at opposite ends of a uniform log that is floating in a lake. The log is 3.0 m long and has mass 20.0 kg. Burt has mass 30.0 kg; Ernie has mass 40.0 kg. Initially, the log and the two friends are at rest relative to the shore. Burt then offers Ernie a cookie, and Ernie walks to Burt's end of the log to get it. Relative to the shore, what distance has the log moved by the time Ernie reaches Burt? Ignore any horizontal force that the water exerts on the log, and assume that neither friend falls off the log.
Final answer:
Using conservation of momentum, the log moves 1.0 m relative to the shore as Ernie walks 3.0 m to reach Burt. No external horizontal forces are acting on the system, so Ernie's movement causes the log to move in the opposite direction to preserve the system's total momentum.
Explanation:
The student's question concerns the movement of a log with two people on it, one walking towards the other. We can use the conservation of momentum principle to answer this question, since there are no external horizontal forces acting on the system of Ernie, Burt, and the log.
In the initial state, the total momentum of the system is zero because they are at rest. As Ernie walks towards Burt, the log moves in the opposite direction to conserve momentum. Let's denote Ernie's displacement towards Burt as x and the log's displacement in the opposite direction as d. The momentum conservation equation for this system will be:
Ernie's momentum change = - Log's momentum change
(40.0 kg)(x) = - (20.0 kg)(d)
Because Ernie has moved the entire length of the log (3.0 m), we set x to 3.0 m, making our equation:
(40.0 kg)(3.0 m) = - (20.0 kg)(d)
120.0 kg·m = - (20.0 kg)(d)
d = - 6.0 m
However, this negative sign only indicates that the log's direction of movement is opposite to Ernie's. The log has moved 6.0 m relative to Ernie, but relative to the shore, the distances will be in the same ratio as their masses. Since the total length that can be covered by the log and Ernie combined is 3.0 m, the log will move a distance of:
d relative to shore = (20.0 kg / (40.0 kg + 20.0 kg)) × 3.0 m
d relative to shore = (1/3) × 3.0 m
d relative to shore = 1.0 m
The log moves 1.0 m relative to the shore while Ernie walks the 3.0 m length of the log to reach Burt.
Final answer:
The problem demonstrates conservation of momentum with Ernie walking across the log to reach Burt. The center of mass of the system remains stationary, and with the calculation, it's determined the log moves 1.2 m relative to the shore by the time Ernie reaches Burt.
Explanation:
The problem presented involves two friends, Burt and Ernie, on a floating log in a lake. This scenario illustrates a physics concept known as the conservation of momentum. Because there is no external horizontal force on the system composed of the log and the two friends, the center of mass of the system must remain stationary relative to the shore. When Ernie walks towards Burt, his movement will cause the log to slide in the opposite direction, ensuring the center of mass remains in the same place.
To calculate the displacement of the log, we have to consider the initial center of mass of the system. Before any movement, the center of mass is located based on the masses of Burt, Ernie, and the log itself. With Burt (30.0 kg) located at one end, Ernie (40.0 kg) at the opposite end, and the log (20.0 kg) between them, the center of mass can be calculated using a weighted average:
Total mass of system = Mass of Burt + Mass of Ernie + Mass of log = 30.0 kg + 40.0 kg + 20.0 kg = 90.0 kg
Distance from Burt to the center of mass of the system, before movement, can be found by (30.0 kg * 0 m + 40.0 kg * 3.0 m) / 90.0 kg = 1.33 m from Burt's end.
After Ernie walks to Burt's end, they are at the same point, and the log has moved beneath them. To keep the center of mass stationary, the log moves a distance such that requires solving the collective mass times its original location equal to the combined mass of Burt and Ernie times the distance the log has moved. Through this calculation, we find that the log moves 1.2 m relative to the shore as Ernie reaches Burt.
Three identical capacitors are connected in series across a potential source (battery). If a charge of Q flows into this combination of capacitors, how much charge does each capacitor carry? Explain.
A) 3Q
B) Q
C) Q/3
D) Q/9
Answer:
C.Q/3
Explanation:
The total capacitances in series
1/C=1/C1+1/C2+1/C3
=1 /C+1/C+1/C
3/C
Ctotal=C/3
Charge in each capacitances
1/3*Q
Q/3
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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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 shortest wavelength within the visible spectrum is _______ light. A. orange B. blue C. red D. violet
Answer:
D. Violet
Explanation:
Answer:
violet
Explanation:
too samrt
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:
Two stars M1 and M2 of equal mass make up a binary star system. They move in a circular orbit that has its center at the midpoint of the line that separates them. If M1-M2-6.95 sm (solar mass), and the orbital period of each star is 2.20 days, find their orbital speed. (The mass of the sun is 1.99x 1030 kg.) km/s M2
The orbital speed of stars in a binary system can be calculated using Kepler's laws and principles of circular motion, given the combined mass and orbital period. In a binary system, both stars orbit around their common center of mass, and their speed indicates how quickly they complete their orbit.
Explanation:The problem involves calculating the orbital speed of two stars in a binary system. For this, we will apply the principles of Kepler's laws of planetary motion and principles of circular motion. We are given that the two stars M1 and M2 are of equal mass and their total mass is 6.95 solar masses.
The semi-major axis of the orbit, 'D', can be found using the formula D³ = (M₁ + M₂)P². Plugging in the given mass and orbital period, we can find D. Then, we can calculate the speed, 'v', based on the principle of circular motion: v = 2πD / P.
An important point in a binary star system is that both stars move around their common center of mass. In case of equal masses, it is at the exact midpoint of the line that separates them. Remember that the orbital speed of the stars is a measure of how fast they travel in their orbit, completing it in the given period of 2.20 days.
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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.
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 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 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]
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]
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.
Why, based on what you learned in the module, might an institution like a university choose to use cast, steel-reinforced concrete as a structural material for a new, important building, rather than wood or stone?
Institutions like universities may choose cast, steel-reinforced concrete for important buildings due to its strength, fire resistance, and cost-effectiveness.
Explanation:An institution like a university might choose to use cast, steel-reinforced concrete as a structural material for a new, important building rather than wood or stone for several reasons:
Strength and Durability: Concrete is a strong and durable material that can withstand heavy loads and extreme weather conditions. Steel reinforcement provides additional strength to the structure, making it suitable for important buildings that require long-term stability.Fire Resistance: Concrete is a non-combustible material that offers excellent fire resistance. This makes it a safer choice for buildings compared to wood, which is highly flammable.Cost-effectiveness: Concrete is a widely available and relatively affordable material, especially when compared to stone. Its ease of construction and lower maintenance costs make it an attractive option for institutions with limited budgets.Learn more about Benefits of using cast, steel-reinforced concrete for important buildings here:https://brainly.com/question/31607449
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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.
If we monitor a point on a wire where there is a current for a certain time interval, which gives the charge that moves through the point in that interval?
a. the product of the current and the time interval
b. the ratio of the current to the time interval
c. the ratio of the time interval to the current
Answer:
a. the product of the current and the time interval
Explanation:
the basic formula is: [tex]Q = It[/tex]
current means the time rate of flow of charge: [tex]I =\frac{Q}{t}[/tex]
where Q - charge
I - current
t - time
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 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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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
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.
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.
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
Find the direction of the net electric force exerted on a charge Q, located at the center of the square, for the following arrangement of charge: the two positive charges are on the top corners, and the two negative charges are on the bottom corners.
Answer:
Downwards towards the square bottom
Explanation:
The direction of the net electric force exerted on the charge +Q is downwards, towards the bottom of the square as the two positive charges at the top are repulsive on the charge Q pushing it down while the negative charges on the bottom corners tend to attract the positive charge Q downwards pulling it towards them
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 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.
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
The electric field 1.5 cm from a very small charged object points toward the object with a magnitude of 180,000 N/C. What is the charge on the object?
The charge on the object is 0.45 nanocoulombs (nC).
Explanation:The electric field 1.5 cm from a very small charged object can be calculated using the equation: E = kQ/r², where E is the electric field, k is the electrostatic constant (8.99 x 10⁹ Nm²/C²), Q is the charge on the object, and r is the distance from the object. In this case, we know that E = 180,000 N/C and r = 1.5 cm = 0.015 m.
Substituting the given values into the equation, we can solve for Q:
E = kQ/r²
180,000 N/C = (8.99 x 10⁹ Nm²/C²)(Q)/(0.015 m)²
Q = (180,000 N/C)(0.015 m)^2 / (8.99 x 10⁹ Nm²/C²)
Q = 0.45 x 10⁻⁹C
Therefore, the charge on the object is 0.45 nanocoulombs (nC).
Learn more about the Electric field here:https://brainly.com/question/8971780
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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]
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.