Jamison walks to a friends house. He walks 1000 meters north, then realizes he walked to far so he turns around and walks 200 meters south. The entire walk takes him 12 minutes. What is his average velocity in meters per second.

Answer:

C) The book's inertia carried it forward.

When the bus stops suddenly, the book tends to remain in motion due to its inertia. The book was at rest on the seat of the bus, and when the bus stopped suddenly, the book continued moving forward with the same speed and direction it had before the bus stopped. As a result, the book slid off the seat and onto the floor.

Answer:

r = 1.07 m

Explanation:

The maximum radius of the propeller that can be allowed is given by the following formula:

\(v = r\omega\\\\r = \frac{v}{\omega}\)

where,

r = maximum possible radius of the propeller = ?

v = maximum possible linear speed of the propeller = 270 m/s

ω = angular speed of the propeller = (2400 rpm)(2π rad/1 rev)(1 min/60 s)

ω = 251.33 rad/s

Therefore,

\(r = \frac{270\ m/s}{251.33\ rad/s}\)

r = 1.07 m

Here is a draft of the worksheet for the three main plate boundary types:

Plate Boundary (Movement)      Convergent (Colliding)

Diagram

        ||

        ||

        ||

Lithosphere (Created or Destroyed)  

       Created

Geologic Process      Mountain building

Real World Example

        Himalayas (Along India-Eurasia plate boundary in Asia)

References        

APA reference for research      

Plate Boundary (Movement)        Divergent (Separating)

Diagram

          |||||

Lithosphere (Created or Destroyed)      

        Destroyed  

Geologic Process        

         Volcanic eruptions and rift valleys

Real World Example  

        Mid-Atlantic Ridge (Between North America and Europe plates)

References        

         APA reference          

Plate Boundary (Movement)     Transform (Sliding)

Diagram

          |||||||||

Lithosphere (Created or Destroyed)  

       Neither          

Geologic Process  

       Earthquakes

Real World Example        

         San Andreas Fault (California, USA along Pacific-North America plates)

References

          APA reference

According to the statement, on the assumption of requesting help from a statistical software consultant, the given case that P(A) + P(B) = 1 is not true, since A and B are not mutually exclusive events.

It is possible for the next request to be related to both SPSS and SAS, in which case it would be included in both A and B.  Therefore, we cannot simply add their probabilities to get the total probability of either event.

B. P(A') represents the probability that the next request is NOT related to the SPSS package. We can find this by subtracting P(A) from 1, since the sum of the probabilities of an event and its complement is always equal to 1:

P(A') = 1 - P(A) = 1 - 0.30 = 0.70

Therefore, the probability that the next request is NOT related to SPSS is 0.70.

C. P(A ∪ B) represents the probability that the next request is related to either SPSS or SAS or both. We can find this by using the formula:

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

Where P(A ∩ B) represents the probability that the next request is related to both SPSS and SAS. Since we don't have information about the probability of this intersection, we cannot calculate P(A ∪ B) exactly. However, we know that the probability of the intersection must be between 0 and the minimum of P(A) and P(B), which in this case is 0.30. Therefore, we can say:

0 ≤ P(A ∩ B) ≤ 0.30

and use this range to get a lower and upper bound for P(A ∪ B):

P(A ∪ B) = P(A) + P(B) - P(A ∩ B) ≤ 0.30 + 0.50 - 0 = 0.80

P(A ∪ B) = P(A) + P(B) - P(A ∩ B) ≥ 0.30 + 0.50 - 0.30 = 0.50

Therefore, the probability that the next request is related to either SPSS or SAS (or both) is between 0.50 and 0.80.

D. P(A' ∩ B') represents the probability that the next request is NOT related to either SPSS or SAS. We can find this using the formula:

P(A' ∩ B') = P((A ∪ B)')

Where (A ∪ B)' represents the complement of the event that the next request is related to either SPSS or SAS (or both). We can find (A ∪ B)' using the formula:

(A ∪ B)' = A' ∩ B'

Which represents the event that the next request is NOT related to either SPSS or SAS. Therefore:

P(A' ∩ B') = P((A ∪ B)') = 1 - P(A ∪ B)

From part c, we know that 0.50 ≤ P(A ∪ B) ≤ 0.80, so:

P(A' ∩ B') = 1 - P(A ∪ B) ≤ 1 - 0.50 = 0.50

P(A' ∩ B') = 1 - P(A ∪ B) ≥ 1 - 0.80 = 0.20

Therefore, the probability that the next request is NOT related to either SPSS or SAS is between 0.20 and 0.50.

Complete question:

Let A denote the event that the next request for assistance from a statistical software consultant relates to the SPSS package, and let B be the event that the next request is for help with SAS. Suppose that P(A) =. 30 and P(B) = .50.

a. Why is it not the case that P(A) + (B) = 1?

b. Calculate P(A ′).

c. Calculate P(A ∪B).

d. Calculate P(A′ ∩B′).

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(a) Velocity with which the mug leaves the counter is approximately 5.02 m/s.

(b) The direction of velocity of mug just before it hits the floor is downwards, since mug is falling under influence of gravity.

Velocity is a quantity that designates how fast and also in what direction a certain point is moving.

(a) As we know, PE = m g h

\(\mathrm{PE = mgh }\)

\(\mathrm{= (m)(9.81 m/s^2)(1.36 m) }\)

= 13.4mJ

\(\mathrm{KE =\frac{1}{2} mv^2}\)

PE = KE

\(\mathrm{mgh = \frac{1}{2}mv^2}\)

\(\mathrm{v = \sqrt{2gh}}\)

\(\mathrm{v = \sqrt{2 \times 9.81 \times 1.36 m}}\)

v = 5.02 m/s

Therefore, the velocity with which the mug leaves the counter is approximately 5.02 m/s.

(b) The direction of the velocity of mug just before it hits the floor is downwards, since mug is falling under the influence of gravity. Velocity vector has a vertical component that points downwards and horizontal component that is parallel to counter.

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The object distance of the lens is 10 cm and the image distance of the lens is 30 cm.

The object and image distance formed by the lens is calculated by applying the following lens formula.

v + u = 40 ------- (1)

v/u = 3 ------------ (2)

v = 3u

Substitute v into equation (1);

3u + u = 40

4u = 40

u = 40/4

u = 10 cm

The image distance = 3u

= 3 x 10 cm

= 30 cm

Thus, the object distance is 10 cm and the image distance is 30 cm.

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Answer:

YESS well it is partly nessary but it depends on the situation

Explanation:

Answer:

Weathering, erosion, and deposistion are necessary processes in the rock cycle because:

Explanation:

First, start with igneous rocks. magma erupts (Extrusive igneous rocks) or solidifies in the sub-surface of the earth (Intrusive igneous rock). when they are exposed Weathering and erosion occur which is a slow breakdown of rock through the wind, water, or other processes. The weathered pieces (sediments) move to other places by wind or water and get deposited someplace else. When there are enough sediments and there is overburden pressure on these sediments, they become a sedimentary rock. Due to overburden pressure, they become metamorphic rocks. now the thing to understand here is that when metamorphic rocks are exposed, they too undergo weathering and erosion and their pieces also become sedimentary rocks.

Answer:

Explanation:

Let the required depth be d .

Sound will first travel trough 480 m deep lake . Then it will enter granite layer .   Sound travelling through granite will reach oil level , get reflected and come back to the surface of lake as echo . Total time taken by sound to travel total distance  is .93 s

Total distance = 2d + 2 x 480 m

= 2d + 960 m

speed of sound in granite is given as 6000 m / s and speed through water is 1480 m /s

total time taken

= 2d / 6000 + 960 / 1480 = .93

2d / 6000 + .6486 = .93

2d / 6000 = .2814

d= 844.2 m

(a) The tension in the string is determined as 19.6 N.

(b) The acceleration of each object is 5.3 m/s².

(c) The distance each object will move in the first second if it started from rest is 2.65 m.

(a) The tension in the string is the resultant weight of the masses and magnitude is calculated as follows;

T = ( 5.7 kg - 3.7 kg ) x 9.8 m/s²

T = 19.6 N

(b) The acceleration of each object is calculated as follows;

a = T / m

where;

a = 19.6 N / 3.7 kg

a = 5.3 m/s²

(c) The distance each object will move in the first second if it started from rest is calculated as;

s = ut + ¹/₂at²

where;

s = 0 + ¹/₂(5.3)(1²)

s = 2.65 m

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The diameter of the smaller end of the pipe is approximately 0.78 meters.

To determine the diameter of the smaller end of the pipe, we can use the principle of conservation of mass. According to this principle, the mass flow rate of water should remain constant throughout the pipe.

The mass flow rate is given by the equation:
Mass flow rate = density of water * cross-sectional area * velocity

Since the density of the water remains constant, we can write:
Cross-sectional area1 * velocity1 = Cross-sectional area2 * velocity2

Given that the velocity1 is 13 m/s, the diameter1 is 1.2 m, and the velocity2 is 30 m/s, we can solve for the diameter2 using the equation:
(pi * (diameter1/2)^2) * velocity1 = (pi * (diameter2/2)^2) * velocity2

Simplifying the equation:
(1.2/2)^2 * 13 = (diameter2/2)^2 * 30

Calculating the equation:
(0.6)^2 * 13 = (diameter2/2)^2 * 30

0.36 * 13 = (diameter2/2)^2 * 30

4.68 = (diameter2/2)^2 * 30

Dividing both sides by 30:
0.156 = (diameter2/2)^2

Taking the square root of both sides:
0.39 = diameter2/2

Multiplying both sides by 2:
0.78 = diameter2

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A) There are 3.70 x 10¹⁶ Cs nuclei present initially and B) There are 2.27 x 10¹⁵ Cs nuclei present 2.6 min later.

A) To solve this problem, we can use the following equation,

\(N = N'2^{\frac{-t}{T_{1/2}}}\)), number of Cs nuclei present is N, initial number of Cs nuclei present is N', elapsed time is t, half-life of Cs-124 is T½. First, we need to convert the initial mass of Cs-124 to the number of nuclei present,

7.5 μg Cs-124(1g/10⁶μg)(6.022x10²³nuclei/1g)

= 4.52 x 10¹⁵ Cs-124 nuclei.

Using the equation above, we can find the number of Cs-124 nuclei present after 0 s,

\(N = 4.52 * 10^{15} * 2^{\frac{-0}{30.8}}}\)

= 4.52 x 10¹⁵ Cs-124 nuclei

Therefore, initially, there are 4.52 x 10¹⁵ Cs-124 nuclei present.

B) After 2.6 min (156 s), we can again use the same radioactivity equation to find the number of Cs-124 nuclei present,

\(N = 4.52 * 10^{15} * 2^{\frac{-156}{30.8}}}\)

N = 1.60 x 10¹⁴ Cs-124 nuclei

Therefore, 2.6 min later, there are 1.60 x 10¹⁴ Cs-124 nuclei present.

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Data mining can be used to organize large volumes of data into meaningful patterns and summaries.

Large data sets are sorted through in data mining in order to find patterns and relationships that may be used in data analysis to assist solve business challenges. Enterprises can forecast future trends and make more educated business decisions thanks to data mining techniques and technologies.

Data mining is a crucial component of data analytics as a whole and one of the fundamental fields in data science, which makes use of cutting-edge analytics methods to unearth valuable information in data sets. Data mining, at a more detailed level, is a step in the knowledge discovery in databases (KDD) procedure, a data science approach for obtaining, processing, and evaluating data. Although they are often used interchangeably, data mining and KDD are more frequently understood to be separate concepts.

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The total electric field produced by the charges at the midpoint is

8 x 10³ N/C.

Amount of charge on the first particle, q₁ = 40 x 10⁻⁹C

Amount of charge on the second particle, q₂ = 60 x 10⁻⁹C

Distance between the charges, r = 30 cm = 0.3 m

The expression for the electric field due to the charges is given by,

E = 1/4πε₀ x q/r²

At the midpoint between the two charges, d = 0.15 m

So, the electric field produced by the charge q₁ at the midpoint is,

E₁ = 1/4πε₀ x q₁/d²

E₁ = 9 x 10⁹ x 40 x 10⁻⁹/(0.15)²

E₁ = 1.6 x 10⁴ N/C

So, the electric field produced by the charge q₂ at the midpoint is,

E₂ = 1/4πε₀ x q₂/d²

E₂ = 9 x 10⁹ x 60 x 10⁻⁹C/(0.15)²

E₂ = 2.4 x 10⁴ N/C

Therefore, the total electric field produced by the charges at the midpoint is,

E = E₂ - E₁

E = 2.4 x 10⁴- 1.6 x 10⁴

E = 8 x 10³ N/C

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Explanation:

Fgravity = G*(mass1*mass2)/D²

G is the gravitational constant throughout the universe.

D is the distance between both objects.

D is now reduced by a factor of 5, meaning Dnew = D/5 we get

Fgravitynew = G*(mass1*mass2)/(D/5)² =

= G*(mass1*mass2)/(D²/25) =

= 25* G*(mass1*mass2)/D² = 25* Fgravity

the new force of gravity/attraction is 25×16 = 400 units.

Answer:

Physics

Explanation:

The period of a pendulum is the time it takes for the pendulum to complete one full oscillation, or swing back and forth. The frequency of a pendulum, on the other hand, is the number of oscillations it completes in one second. The frequency is calculated by taking the reciprocal of the period, or 1/period.

If the period of a pendulum decreases from 8 seconds to 2 seconds, it means that the pendulum is swinging back and forth more quickly, as it is taking less time to complete one full oscillation. To find the new frequency of the pendulum, we can use the formula:

frequency = 1 / period

Initially, when the period was 8 seconds, the frequency was:

frequency = 1 / 8 = 0.125 Hz

After the period decreased to 2 seconds, the new frequency can be calculated as:

frequency = 1 / 2 = 0.5 Hz

So, the frequency of the pendulum increases from 0.125 Hz to 0.5 Hz when the period decreases from 8 seconds to 2 seconds. This means that the pendulum is oscillating at a faster rate, completing more oscillations in one second.

The researcher did not find strong evidence to support the idea that job applicants with popular names are viewed more favorably than equally qualified applicants with less popular names.

It sounds like the researcher conducted an experiment to investigate whether job applicants with popular names are viewed more favorably than equally qualified applicants with less popular names.

Based on the information provided, the researcher found that the differences in the evaluations of the applicants by the two groups were not significant at the .001 level.

The factors that play an important role in the hiring process for a job:

(1) Qualifications and experience: Employers typically look for candidates who possess the necessary qualifications and experience for the job. This includes education, training, certifications, and work experience.

(2) Skills and abilities: Employers also consider a candidate's skills and abilities related to the job. These may include technical, interpersonal, communication, and problem-solving skills.

(3) Personal characteristics: Personal characteristics, such as motivation, work ethic, and adaptability, can also play a role in the hiring process. Employers may look for candidates who demonstrate a positive attitude, a willingness to learn, and the ability to work well with others.

(4) Fit with company culture: Companies may also consider whether a candidate fits with their company culture, values, and mission. This can include factors such as teamwork, creativity, and innovation.

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Answer:

Atmosphere

Explanation:

Hydrogen,

H

2

, is one of the lightest gases that can be found on Earth. The other gas is helium, and these two gases are both lighter than air, because they have a high degree of buoyancy.

This means that the air below it is pushing up with a greater force than the air above it is forcing it down.

The new gravitational force between the two planets, when they are moved to two million miles apart, is 250,000 N

The gravitational force between two objects can be calculated using Newton's Law of Universal Gravitation, which states that the force is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

Given:

Initial distance between the planets = 1 million miles

Initial gravitational force = 1,000,000 N

Final distance between the planets = 2 million miles

To determine the new gravitational force, we need to compare the ratios of the distances and apply the inverse square law.

Let's denote the initial distance as d1, the initial gravitational force as F1, the final distance as d2, and the unknown final gravitational force as F2.

According to the inverse square law, the ratio of the gravitational forces is the square of the ratio of the distances:

(F2/F1) = (d1/d2)²

Substituting the given values:

(F2/1,000,000 N) = (1 million miles / 2 million miles)²

Simplifying:

(F2/1,000,000 N) = (1/2)²

(F2/1,000,000 N) = 1/4

F2 = (1/4) * 1,000,000 N

F2 = 250,000 N

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After considering the given data we come to the conclusion that the increase in the internal energy of the tie and spike is 16.322 J, under the condition that  a given weight of  0.76 kg spike had been jamed into a railroad tie .

In order to evaluate the increase in internal energy of the tie and spike we have to apply the formula of kinetic energy which is
1/2 × mv²
Here,
m = mass
v = velocity
Staging the values in the formula
Kinetic energy of the spike = (1/2)× (0.76)×(4.8)²= 87.2064 J
Energy absorbed by the tie and spike = 87.2064 J × 0.187 = 16.322 J

Hence, the evaluate rise in internal energy of the tie and spike is 16.322 J.
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Considering the Coulomb's Law, the magnitude of the charge on each particle is 4.2426 C.

Coulomb's law or law of electrostatics is the relationship between the interactions of electric charges, that is, it explains the force experienced by two electric charges at rest.

This law says that the electric force with which two point charges at rest attract or repel each other is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them, expressed mathematically as:

\(F=k\frac{Qq}{d^{2} }\)

where:

The force is attractive if the charges are of opposite sign and repulsive if they are of the same sign.

In this case, you know that:

Replacing in the Coulomb's Law:

\(1.8x10^{-8} N=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{Qq}{(0.3 m)^{2} }\)

Being Q=q:

\(1.8x10^{-8} N=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{q^{2} }{(0.3 m)^{2} }\)

Solving:

1.8×10⁻⁸ N÷ 9×10⁹ \(\frac{Nm^{2} }{C^{2} }\)= q²÷ (0.3 m)²

2×10⁻¹⁸ C²/m²= q²÷ (0.3 m)²

2×10⁻¹⁸ C²/m²= q²÷ 0.09 m²

2×10⁻¹⁸ C²/m² ×0.09 m²= q²

1.8×10⁻¹⁹ C²= q²

√1.8×10⁻¹⁹ C²= q

4.2426 C= q= Q

Finally, each charge has a value of 4.2426 C.

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Answer:

Since they all have the same mass and radius they all experience the same amount of torque.

Torque = moment of inertia * angular acceleration

The one with the smallest moment of inertia will experience the greatest angular acceleration and thus reach the bottom first.

I = M r^2     ring

I = 1/2 M r^2  cylinder

I = 2/5 M r^2   sphere

The sphere will reach the bottom first.

The thrust force that is acting on the rocket engine is obtained as  5782 kN.

We know that the thrust is the force that is acting on an object that enables the object to be able to move higher. The thrust force can be applied to a rocket such that the rocket can be able to move higher and then pass through the air.

This implies that the thrust force is about the most important force when we want to be able to launch a rocket or when we are trying to move an aeronautical engine. In this case, we are asked to be able to obtain the thrust force as we can see from the parameters that have been given.

We now know that the thrust force can be obtained as; 118kg * 9.8 m/s^2 * 5000m/s = 5782 kN

There would be a thrust of  5782 kN working on the engine.

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Answer:

most likely be included in the analysis section of a lab report

Explanation:

The answers are 4. The distance from where the ball was kicked is 38.06 meters, 5. The speed of the ball 2.5 seconds after it was kicked is 13.82 m/s, and  6. The ball is 21.88 meters above the ground 2.5 seconds after it is kicked.

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Answer:

meeee

Explanation:

Answer:

Me

Explanation:

:)

Answer:

Explanation:

The speed in the medium is always slower than the speed in a vacuum.

The statement" The direction of the force on the charge placed in the electric field is upward" is false because the direction of the force on a negative charge (-6 C) placed in an upward-directed uniform electric field of magnitude 24 N/C would be downward.

The direction of the force on a charged particle placed in an electric field is determined by the charge of the particle and the direction of the electric field. In this case, a charge of -6 C is placed in an electric field directed upward with a magnitude of 24 N/C.

The force on a charged particle in an electric field can be calculated using the formula:

F = q * E

Where F is the force, q is the charge of the particle, and E is the electric field.

Since the charge q in this case is negative (-6 C) and the electric field E is directed upward, we can substitute the values into the formula:

F = (-6 C) * (24 N/C)

F = -144 N

The negative sign in the force value indicates that the force is in the opposite direction to the electric field. Therefore, the force on the charge placed in the electric field is downward, not upward.

The force on a negative charge is always opposite to the direction of the electric field. This is because negative charges experience an attractive force towards positive charges, and electric fields are directed from positive charges to negative charges.

Therefore, the statement "The direction of the force on the charge placed in the electric field is upward." is false.

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Power P is the amount of energy E transferred or converted per unit time t, and is expression is:

P = E/t

The necessary energy to lift the box its the work of against the Gravity Force, and is given by:

E = mgh

Then the power is:

P = E/t

P = mgh/t

P = (45 kg)(9.8 m/s²)(1.7 m )/ 30

P = 24.99 W

If the fundamental frequency of the harmonic is 7 Hz, the frequency at which it is created depends on the harmonic being referred to.

Frequency of harmonics

If the fundamental frequency of a harmonic is 7 Hz, then the frequency at which the harmonic is created depends on the harmonic series being used.

The frequency of the nth harmonic in a harmonic series is equal to n times the fundamental frequency. Therefore, the first five harmonics of a harmonic series with a fundamental frequency of 7 Hz would be:

First harmonic (fundamental): 7 Hz

Second harmonic: 2 x 7 Hz = 14 Hz

Third harmonic: 3 x 7 Hz = 21 Hz

Fourth harmonic: 4 x 7 Hz = 28 Hz

Fifth harmonic: 5 x 7 Hz = 35 Hz

So, the frequency at which the harmonic with a fundamental frequency of 7 Hz is created depends on which harmonic is being referred to.

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