If the 140 g ball is moving horizontally at 21 m/s , and the catch is made when the ballplayer is at the highest point of his leap, what is his speed immediately after stopping the ball?

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.

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

Answer:

we are on the same frame of reference moving with the earth with the same velocity.

Explanation:

Answer:

(a) I=0.01 kg.m²

(b) I=0.03 kg.m²

Explanation:

Given data

Mass of disk M=2.0 kg

Diameter of disk d=20 cm=0.20 m

To Find

(a) Moment of inertia through the center of disk

(b) Moment of inertia through the edge of disk

Solution

For (a) Moment of inertia through the center of disk

Using the equation of moment  of Inertia

[tex]I=\frac{1}{2}MR^{2}\\ I=\frac{1}{2}(2.0kg)(0.20m/2)^{2}\\ I=0.01 kg m^{2}[/tex]

For (b) Moment of inertia through the edge of disk

We can apply parallel axis theorem for calculating moment of inertia

[tex]I=(1/2)MR^{2}+MD\\ Here\\D=R\\I=(1/2)(2.0kg)(0.20m/2)^{2}+(2.0kg)(0.20m/2)^{2}\\ I=0.03kgm^{2}[/tex]

The moment of inertia of the disk for rotation about the center is 0.01 kg•m^2, and for rotation about the edge is 0.02 kg•m^2.

The moment of inertia of a solid disk about an axis through its center is given by the formula 1/2 * MR^2. In this case, the mass of the disk is 2.0 kg and the radius is 10 cm (half of the diameter). So, substituting the values into the formula, the moment of inertia about the center axis is:

1/2 * 2.0 kg * (0.1 m)^2 = 0.01 kg•m^2.

For rotation about an axis through the edge of the disk, the moment of inertia is:

MR^2 = 2.0 kg * (0.1 m)^2 = 0.02 kg•m^2.

The distance between the Sun and the nearest star, besides the Sun, is about 40,000 kilometers (25,000 miles), or 4 times the circumference of the Earth.

The distance between the Sun and the nearest star, besides the Sun, is incredibly vast. To put it in perspective, if the Sun is the size of a grapefruit and the Earth is 15 meters away, the nearest star would be about 40,000 kilometers away. This is equivalent to approximately 25,000 miles or about 4 times the circumference of the Earth.

It is important to note that the actual distance to the nearest star can vary as there are multiple stars that could be considered the nearest depending on how we define 'nearest.' The star Proxima Centauri, located in the Alpha Centauri system, is one of the closest stars to our Solar System, and it is approximately 4.24 light-years away.

Keep in mind that these distances are still minuscule compared to the vastness of the Universe, which contains billions of galaxies, each with billions of stars.

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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.

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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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.

Answer:

Both ball hit the ground about at the same time.

Explanation:

given information:

h = 2 m

the speed of red ball, v = 3 m/s

the time for red ball to reach the ground

h = [tex]\frac{1}{2}gt^{2}[/tex]

t = √2h/g

 = √2(2)/9.8

 = 0.64s

the time for yellow ball to reach the ground, it's considered as a vertical motion. thus

h = √2h/g

 = √2(2)/9.8

 = 0.64s

so, both ball hit the ground about at the same time.

The horizontal velocity does not influence how quickly an object falls to the ground. Thus, both the yellow and the red ball will hit the ground at the same time.

In the context of gravity and projectile motion, the horizontal velocity of an object does not influence the time it takes for that object to fall to the ground. The yellow ball and the red ball will both hit the ground at the same time. When you drop an object, it accelerates towards the ground due to gravity. The same process happens to an object moving horizontally; gravity also pulls it downwards. Therefore, the time each ball takes to fall to the ground is only dependent on their initial height and the acceleration due to gravity, not any horizontal velocity. Both balls start from the same height, therefore, they will hit the ground at the same time.

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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.

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.

Answer:

(A). The work done on the cart by the friction is -700 J.

(B). The work done on the cart by the gravitational force is 0 J.

(C). The work done on the cart by the shopper is 700 J.

(D). The force the shopper exerts 38.61 N.

(E). The total work done on the cart is zero.

Explanation:

Given that,

Distance =  20.0 m

Frictional force = 35.0 N

Angle = 25.0°

(A). We need to calculate the work done on the cart by the friction

Using formula of work done

[tex]W_{fr} = -F\cdot d[/tex]

Where, F = force

d = distance

Put the value into the formula

[tex]W_{fr}=-35.0\times20[/tex]

[tex]W_{fr}=−700\ J[/tex]

(B). The work done by the gravity is perpendicular to the direction of the motion

We need to calculate the work done on the cart by the gravitational force

Using formula of work done

[tex]W=fd\cos\theta[/tex]

Put the value into the formula

[tex]W=35.0\times20\cos90[/tex]

[tex]W=0\ J[/tex]

(C). We need to calculate the work done on the cart by the shopper

Using formula of work done

[tex]W_{sh}=W_{net}-W_{fr}[/tex]

Put the value into the formula

[tex]W_{sh}=0-(-700)[/tex]

[tex]W_{sh}=700\ J[/tex]

(D). We need to calculate the force the shopper exerts

Using formula of force

[tex]F_{sh}=\dfrac{W_{fr}}{d\cos\theta}[/tex]

Put the value into the formula

[tex]F_{sh}=\dfrac{700}{20\cos25}[/tex]

[tex]F_{sh}=38.61\ N[/tex]

(E). We need to calculate the total work done on the cart

Using formula of work done

[tex]W_{cart}=W_{fr}+W_{sh}[/tex]

Put the value into the formula

[tex]W_{cart}=700-(-700)[/tex]

[tex]W_{cart}=0\ J[/tex]

Hence, (A). The work done on the cart by the friction is -700 J.

(B). The work done on the cart by the gravitational force is 0 J.

(C). The work done on the cart by the shopper is 700 J.

(D). The force the shopper exerts 38.61 N.

(E). The total work done on the cart is zero.

The work done on a grocery cart by friction is -700 J, by gravitation is 0, by the shopper is designed to balance the friction, and the total net work considering only these factors tends to be 0, assuming the system is ideal.

(a) The work done on the cart by friction is given by the formula Work = force × distance × cos(θ). Since the force of friction is 35.0 N acting against the direction of the displacement (180° or π radians), and the distance is 20.0 m, the work done by friction = 35 × 20 × cos(180°) = -700 J.

(b) The work done on the cart by the gravitational force is zero because gravitational force acts vertically downwards and the displacement of the cart is horizontal, making the angle between the force and displacement 90° which leads to no work being done as cos(90°) = 0.

(c) The work done on the cart by the shopper can be found by calculating the component of the shopper's force in the direction of the displacement. Without the exact force the shopper applies, we directly cannot calculate the work done; however, it is designed to counteract friction and maintain constant speed.

(d) To find the force the shopper exerts using energy considerations, we equate the work done against friction to the work done by the shopper. Hence, 700 J of work is done by the shopper as well to overcome friction.

(e) The total work done on the cart is the sum of all works done by individual forces. However, without calculating the exact value of the shopper's force, we consider only the friction work, which is -700 J. If considering only friction, the external work (by the shopper) matches this in magnitude but is positive, suggesting a net work of 0 when only considering these two forces.

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.

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

Answer:

D. Violet

Explanation:

Answer:

violet

Explanation:

too samrt

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-

Thus, the above-given statement is true.

Answer:

TRUE

Explanation:

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.

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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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.

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.

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.

Explanation:

Power = Work ÷ Time

Work = Force x Displacement

Force = 22 N

Displacement = 3 m

Time = 6 seconds

Substituting

       Work = Force x Displacement

       Work = 22 x 3 = 66 J

       Power = Work ÷ Time

       Power = 66 ÷ 6

       Power = 11 W

Power is 11 W

Final answer:

To calculate power, divide the work done by the time taken. In this case, the power while carrying the laundry basket is 11 Watts.

Explanation:

To calculate power, we use the formula:

Power (P) = Work (W) / Time (t)

In this case, the work done is equal to the weight of the laundry basket multiplied by the distance it is lifted, which is

W = 22 N * 3.0 m = 66 J

Substituting the values into the formula:

P = W / t = 66 J / 6.0 s = 11 W

Therefore, your power while carrying the laundry basket is 11 Watts.

Completed Question:

Your neighbor Paul has rented a truck with a loading ramp. The ramp is tilted upward at 35 ∘,and Paul is pulling a large crate up the ramp with a rope that angles 10 ∘ above the ramp. Paul pulls with a force of 350 N. (Force is measured innewtons, abbreviated N.)

What is the magnitude of the horizontal component of his force?

What is the magnitude of the vertical component of his force?

Answer:

Both are 175√2 N.

Explanation:

The ramp does an angle of 35° if the soil and the rope that he's pulling does an angle of 10° with the ramp. So, the total angle that the rope does with the soil is 45° (10° + 35°).

The force is a vector, it means that it has a module, direction, and sense. So, if the force acts at an angle of 45° with the horizontal (the soil), the vector can be decomposed to vertical and horizontal vectors.

The decomposition helps to study the movement because an inclined force acts both horizontally and vertically. By the sin and cos of a triangle, the horizontal (x) and vertical (y) forces are:

Fx = F*cosα

Fy = F*sinα

Where α is the angle with the horizontal, and sin45° = cos45° = √2/2

Fx = 350*cos45° =350*√2/2

Fx = 175√2 N

Fy = 350*sin45° = 350*√2/2

Fy = 175√2 N

Answer:

The correct option was not given that is External organizational environment.

Explanation:

It's the external organizational environment which contains the entities that exist outside of it but still have a significant impact upon it's development and growth etc.

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.

The problem involves calculating the height of a cliff by considering both the rock's fall time and the sound's travel time back up. The actual height is found by equating the distance sound travels to the free fall distance, leading to an equation that, when solved, gives the cliff's height. Ignoring the sound's travel time results in overestimating the cliff's height.

To solve this problem, we must first calculate the time it took for the rock to reach the ground and then use this to find the height of the cliff. The total time of 7.00 seconds includes both the time the rock is falling and the time the sound of the impact takes to travel back up to the top of the cliff.

Part A: Calculating the height of the cliff

Let's denote the time it takes for the rock to fall as t, and the remaining time for the sound to travel up as (7 - t) seconds. Knowing the speed of sound is 330 m/s, the distance the sound travels (which equals the height of the cliff) can be given by Distance = Speed × Time, hence 330 × (7 - t).

To find the height using the falling time, we use the formula for an object in free fall: Height = 0.5 × g × t^2 where g is the acceleration due to gravity (9.8 m/s^2). As both expressions describe the height of the cliff, they can be set equal to each other giving us 0.5 × 9.8 × t^2 = 330 × (7 - t). Solving for t and then substituting back to find the height will give us the answer.

Part B: Effect of Ignoring the Sound Travel Time

If you ignored the time for the sound to reach you, you would be overestimating the time it took for the rock to hit the ground. Since a part of the 7.00 seconds is actually the sound traveling back up, the actual falling time is less than 7.00 seconds. Consequently, the calculated height of the cliff would be *overestimated* because you would be applying the entire 7.00 seconds to the falling distance calculation.

Essentially, ignoring the sound's travel time makes it seem like the rock was in free fall for longer, leading to a larger computed height, which is inaccurate.

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

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.

Answer:

Yes, it's true. Computers do work that way. It's experienced by one of the authors of the book how computers work.

Explanation:

A. True, Flash memory is a type of non-volatile computer memory that can be electrically erased and reprogrammed. If a software update fails due to issues with the flash memory, it's possible that the memory itself is corrupted or malfunctioning. However, it's also possible that the software update process itself encountered errors.

Given the scenario, if the flash memory is the only issue and the hardware component is otherwise functional, manually downloading the software directly onto the flash memory could potentially bypass any issues encountered during the automated update process.

Therefore, the user's suggestion of manually downloading the software could indeed solve the problem, making the statement true.

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

The charge on the object is 0.45 nanocoulombs (nC).

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).

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Institutions like universities may choose cast, steel-reinforced concrete for important buildings due to its strength, fire resistance, and cost-effectiveness.

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:

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Answer:

54 n C

Explanation:

given,

Electric field strength = 60,000 N/m

diameter of the sphere = 10 cm

distance of the electric field = 4 cm

distance of the point form the center of ball

 r = 10/2 + 4

 r = 9 cm = 0.09 m

the electric field strength

 [tex]E = \dfrac{kq}{r^2}[/tex]

 [tex]60000 = \dfrac{9\times 10^9\times q}{0.09^2}[/tex]

 [tex]q = \dfrac{60000\times 0.09^2}{9\times 10^9}[/tex]

        q = 54 x 10⁻⁹ C

        q = 54 n C

the charge on the ball is equal to 54 n C

The magnitude of charge in the ball is 54 nC.

Given data:

The strength of Electric field is, E = 60,000 N/C.

The diameter of ball is, d = 10 cm = 0.1 m.

The distance is, r = 4.0 cm = 0.04 cm.

The region where the electric force has its significance is known as electric field strength. And the expression for electric field strength is given as,

[tex]E = \dfrac{kq}{(r+d/2)^{2}}[/tex]

Here, k is the Coulomb's constant and q is the magnitude of charge.

Solving as,

[tex]60,000 = \dfrac{9 \times 10^{9} \times q}{(0.04+0.1/2)^{2}}\\\\q =54 \times 10^{-9} \;\rm C\\\\q = 54 \;\rm nC[/tex]

Thus, we can conclude that the magnitude of charge in the ball is 54 nC.

Learn more about the electric field strength here:

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