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The average annual snowfall values for four different cities are shown in the table below.

which of the following conclusions could be made from this data?

A. It snows more in Green Bay than in Buffalo on average.

B. It snows less in Green Bay than in Kalamazoo on average.

C. It snows more in Kalamazoo than in Buffalo on average.

D. It snows less in Buffalo than in Bristol on average. ​

To find the distance traveled, we just have to add all the side lengths of the square, in other words, the perimeter.

Answer:you think i know jesus im in y8 i barley know any g anyway someone asked me something that only albert eientsien knew .

Explanation:

I'm also cluless but i feel good for helpig other people out

The question involves calculating the resultant displacement of swimming across a river while also being carried north by the current. In this scenario, the eastward displacement is 19.0 m, and the northward displacement due to the current is 12.0 m. Using these vectors, the total displacement can be calculated as approximately 22.36 m.

The subject of this question is Physics, specifically the concept of calculating displacement, which is a vector quantity. To solve this, we start by setting up a vector diagram. One vector represents your direction of swimming (east), which is 19.0 m long. The other vector represents the direction of the river current (north), which is 12.0 m long. The resultant displacement is the hypotenuse of the right-angled triangle they form. To calculate this magnitude, express your displacement in components: The eastward displacement is (19.0 m, 0) and the northward displacement is (0, 12.0 m). Finally, we can determine the resultant displacement using the Pythagorean theorem (sqrt[(19.0m)^2 + (12.0m)^2]), which gives a magnitude of approximately 22.36 m.

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Answer:

Cool air sinks and warm air rises.

Explanation:

Answer:

\(R = \frac{V}{A}\)

Explanation:

See attachment for circuit

From the circuit, we have:

\(V = Voltage\)

\(A = Current\)

\(R = Resistance\)

Resistance is calculated as:

\(R = \frac{V}{A}\)

The steps are as follows:

Step 1: The student needs first to power on the circuit

Step 2: The student then take readings from both the ammeter (for current) and from the voltmeter (for voltage)

Step 3: Record new readings for current and voltage by adjusting the variable resistor

Step 4: Draw a voltage-current graph

Step 5: : Lastly, calculate the slope of the graph

The slope represents the resistance

Answer:

300

Explanation Hope I'm not wrong.

ANSWER:

10 eV

STEP-BY-STEP EXPLANATION:

Given:

Carga = 3.2 X 10^-19 C

Voltage = 5 V

Since energy is stored work, we can use the following formula:

The energy required is 10 electronvolts

Answer:

Winds are caused by differences in air pressure. Unequal heating of Earth’s surface creates areas of different pressure. Cooler areas have higher air pressure. Warmer areas have lower pressure. This causes air to move from high pressure to low pressure. This movement is the wind we feel.

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Given:

Initial speed of the motorcycle (u) = 35 m/s

Final speed of the motorcycle (v) = 0 m/s (Complete Stop)

Maximum deceleration of the motorcycle (a) = -1.2 m/s²

Required Equation:

\( \boxed{\bf{ v = u + at}}\)

Answer:

By substituting values in the equation, we get:

\( \rm \longrightarrow 0 = 35 + ( - 1.2)t \\ \\ \rm \longrightarrow 0 = 35 - 1.2t \\ \\ \rm \longrightarrow 35 - 1.2t = 0 \\ \\ \rm \longrightarrow 35- 35 - 1.2t = 0 - 35 \\ \\ \rm \longrightarrow - 1.2t = - 35 \\ \\ \rm \longrightarrow \dfrac{ - 1.2t}{ - 1.2} = \dfrac{ - 35}{ - 1.2} \\ \\ \rm \longrightarrow t = 29.167 \: s\)

\( \therefore \) Time taken by motorcycle to come to a complete stop (t) = 29.167 s

Hendry and Cathy will each throw an object, and the time and location at which they will collide can be determined using the laws of motion. Hendry's item had an initial velocity of 26.5 m/s, whereas Cathy's object had an initial velocity of 45 m/s. Hendry's object's equation of motion is given by: s = u*t + 0.5*a*t*2, where s is the displacement, u*t* is the starting velocity, t* is the time, and a*t* is the acceleration brought on by gravity.

The acceleration caused by gravity is negative since the item is being flung upward. The item that Cathy threw has the following equation of motion: s = u * t - 0.5 * a * t2.where s is the distance travelled, u is the starting speed, t is the passage of time, and an is the acceleration brought on by gravity. The acceleration caused by gravity is negative since the item is being flung upward.

These equations allow us to determine the location and timing of the two items' collision. By figuring out the two equations for t, one may determine the moment when the objects will collide. By changing the value of t in either equation, one may determine the distance at which the objects will collide. Therefore, using the equations of motion, it is possible to determine the moment and distance at which the two objects will collide.

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Answer:

42.99°

Explanation:

To solve this problem, we can use vector addition. Let's assume that the boat is moving at an angle of θ with respect to the riverbank, as shown below:

             ^ current (1.80 m/s)

             |

             |

             |

             |

   boat      | Vboat (1.95 m/s)

---->---------+-------------> shore

The boat's velocity relative to the shore, Vtot, can be calculated as the vector sum of its velocity relative to the water, Vboat, and the velocity of the water relative to the shore, Vcurrent:

Vtot = Vboat + Vcurrent

To calculate the magnitude and direction of Vtot, we need to break down Vboat and Vcurrent into their x and y components. Let's assume that the angle between Vboat and the x-axis is α:

             ^

             |

       Vboat |

             |

   boat  ----+-------------> shore

         \   |

          \  |

           \ | α

            \|

The x and y components of Vboat can be calculated as follows:

Vboat,x = Vboat cos α

Vboat,y = Vboat sin α

The x component of Vcurrent is simply the speed of the current, since it is flowing horizontally to the right:

Vcurrent,x = 1.80 m/s

The y component of Vcurrent is zero, since the current is flowing parallel to the shore and does not affect the boat's vertical motion.

Now we can calculate the x and y components of Vtot:

Vtot,x = Vboat,x + Vcurrent,x = Vboat cos α + 1.80 m/s

Vtot,y = Vboat,y + Vcurrent,y = Vboat sin α

The magnitude of Vtot is given by:

|Vtot| = sqrt(Vtot,x^2 + Vtot,y^2)

Substituting the expressions for Vtot,x and Vtot,y, we get:

|Vtot| = sqrt((Vboat cos α + 1.80 m/s)^2 + (Vboat sin α)^2)

We can simplify this expression using the trigonometric identity sin^2 α + cos^2 α = 1:

|Vtot| = sqrt((Vboat^2 + 3.60 Vboat cos α + 3.24) + Vboat^2 sin^2 α)

Now we can substitute the values of Vboat (1.95 m/s) and Vcurrent (1.80 m/s) to get:

|Vtot| = sqrt((1.95 m/s)^2 + 3.60 (1.95 m/s) cos α + 3.24)

To find the direction of Vtot, we can use the tangent function:

tan β = Vtot,y / Vtot,x = Vboat sin α / (Vboat cos α + 1.80 m/s)

Solving for α, we get:

α = arctan((Vboat sin β) / (Vboat cos β + 1.80 m/s))

Substituting the values of Vboat and Vcurrent, we get:

α = arctan((1.95 m/s) sin β / (1.95 m/s cos β + 1.80 m/s))

Therefore, the boat's velocity relative to an observer on the shore, Vtot, is:

|Vtot| = sqrt((1.95 m/s)^2 + 3.60 (1.95 m/s) cos α + 3.24)

β = arctan((1.95 m/s) sin α / (1.95 m/s cos α + 1.80 m/s))

We can use the Pythagorean theorem to find the magnitude of the boat's velocity relative to an observer on the shore. The Pythagorean theorem states that the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides. In this case, the magnitude of the boat's velocity relative to an observer on the shore is the hypotenuse of a right triangle, with the velocity of the boat and the velocity of the current as the other two sides.

Using the Pythagorean theorem, we get:

Vtot = √(Vb^2 + Vc^2)

Vtot = √((1.95 m/s)^2 + (1.80 m/s)^2)

Vtot = √(3.8025 m^2/s^2 + 3.24 m^2/s^2)

Vtot = √7.0425 m^2/s^2

Vtot = 2.655 m/s

So the magnitude of the boat's velocity relative to an observer on the shore is 2.655 m/s.

To find the direction of the boat's velocity relative to an observer on the shore, we can use trigonometry. Specifically, we can use the tangent function, which is defined as the opposite side divided by the adjacent side of a right triangle. In this case, the opposite side is the velocity of the current and the adjacent side is the velocity of the boat. Therefore, we have:

tan(θ) = Vc/Vb

tan(θ) = 1.80 m/s / 1.95 m/s

tan(θ) = 0.9231

θ = tan^-1(0.9231)

θ = 42.99°

So the direction of the boat's velocity relative to an observer on the shore is 42.99° to the right of straight ahead.

Answer:

the answer would be A. electricity don't specify the direction of any cardinal points the flow of charges moves.

Answer:

A

Explanation:

I did the test on ap3x

Answer:

about 20 pounds

Explanation:

If you weigh 120 pounds on Earth, you will weigh about 20 pounds on the moon.

Answer:

there are 1,000m in a km, so 200km is 200,000m

200,000m/10m/s = 20,000s

Explanation:

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Explanation:

the average speed = (4+2+4+2)m /24 s

= 20/24

= 0.83 m/s

the displacement = 0 (back to the origin position)

»average speed is

= (4m + 2m + 4m + 2m)/24s

= 12m/24s

= 0.5m/s✅

~Physics#4

Answer: I think it’s craters

Explanation:

Answer:

MOUNTAINS

CRATERS

PLATEAUS

Explanation:

The Moon has many familiar landforms like mountains, plains, and dark patches that look like seas. We now know that these areas are not seas filled with water. They are deep dry plains.

The ray does not bend when it enters the semi circular glass block - Light ray incident on semicircular block at 90 degrees, therefore there is no change in the direction of ray at P.

Electromagnetic radiation that falls within the region of the electromagnetic spectrum that the human eye can see is known as light or visible light.

Light is electromagnetic radiation that the human eye can perceive. From radio waves with wavelengths measured in meters to gamma rays with wavelengths shorter than around 1 1011 meters, electromagnetic radiation occurs throughout an incredibly broad range of wavelengths.

Light governs our sleep-wake cycle and is crucial to our health and wellbeing. In actuality, "light" that is visible is a type of radiation, which is just energy that moves in the form of electromagnetic waves. It can alternatively be explained as a flow of "wave-packets," or particles, known as photons.

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Answer:

In electromagnetic waves, energy is transferred through vibrations of electric and magnetic fields. In sound waves, energy is transferred through vibration of air particles or particles of a solid through which the sound travels. In water waves, energy is transferred through the vibration of the water particles.

Answer:

A wave can be thought of as a disturbance or oscillation that travels through space-time, accompanied by a transfer of energy. The direction a wave propagates is perpendicular to the direction it oscillates for transverse waves. A wave does not move mass in the direction of propagation; it transfers energy.

Thank you for reaching out to me with your question. From what I understand, you are curious about the importance of an assignment or exam that is worth 20% of your grade.

To put it simply, any assignment or exam that is worth a certain percentage of your grade is an indicator of how much weight that particular task carries in determining your overall grade for the course. In other words, if you were to score poorly on an assignment that is worth 20% of your grade, it could significantly impact your final grade.
It is important to note that each assignment or exam may be worth a different percentage, and it is up to the instructor to determine the weight of each task. Generally, assignments and exams that are worth a higher percentage of your grade carry more weight and have a greater impact on your final grade.
Therefore, it is crucial to take each assignment or exam seriously and give it your best effort, especially those that carry a higher percentage of your grade. It is also important to keep track of your grades throughout the semester and identify any areas that may need improvement, so you can work towards improving your overall grade.
I hope this explanation helps clarify the importance of an assignment or exam that is worth a certain percentage of your grade. Please let me know if you have any further questions or concerns.

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Answer:

25.48

Explanation:

the earth's gravitational pull is 9.8 m/s. So to find the answer, multiply 2.6 x 9.8

The derivation of the  total energy Kinetic energy plus potential energy in rolling motion is shown in the image attached.

In rolling motion, the total energy is the sum of the kinetic energy and the potential energy. Rolling motion is a combination of rotational and translational motion, and the energy of rolling can be expressed as follows:

Total energy = Kinetic energy + Potential energy

The derivation has been done and shown in the image attached here.

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Answer:

Option (D).

Explanation:

For an ideal spring, the force is given in terms of displacement as follows :

F=-kx

\(x=\dfrac{-F}{k}\\\\\text{or}\\\\x=\dfrac{-1}{k}F\)...(1)

Where,

x is displacement in spring i.e. compression or stretching

k is spring constant of the spring

-ve sign shows that the force exerted by the spring is in opposite direction of the spring's displacement.

The equation of a line is : y=mx+c, m is slope

From equation (1)

Slope = 1/k

or

Slope = reciprocal of the spring's constant.

It would mean that the slope of the curve represents the reciprocal of the spring's constant.

Using the spring constant and displacement relation, the slope of the curve would be the reciprocal of the spring's constant.

Using the Force constant relation :

F = Force ; k = spring constant ; e = extension

Making the x the subject of the formula :

The slope, which is the spring constant can be expressed as ;

Hence, the slope is the reciprocal of the spring constant.

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Answer:

I don't think your appendix can explode because you ate too much honestly. It's not even possible to eat so much that your appendix explodes, and if you're feeling any pain it definitely isn't because your appendix is about to explode, believe me. Also you could just type it into the internet, that'd be a much faster solution.

Answer:

B

Explanation:

a) The value of t u = 140/t`b.

b) The y position of the cannonball at the time t is  55.5 mc.

c) The initial speed of the projectile is 52.4 m/s.

Given that a cannonball is fired horizontally from the top of a cliff. The cannon is at height H = 55.5 m above ground level, and the ball is fired with initial horizontal speed u. The projectile lands at a distance D = 140 m from the cliff. Assume that the cannon is fired at time t = 0 and that the cannonball hits the ground at time t.Now,We have to find the value of t, y position of the cannonball at the time t and the initial speed of the projectile.

a. To find the value of t:Here, we have to use the formula of distance

i.e.,S = ut + (1/2)gt², Where S = 140 m, u = u and g = 9.8 m/s².Hence,140 = u×t ………..(1)We know that, time taken by the cannonball to hit the ground can be calculated as,`(2H)/g`

Since the height of the cannon from the ground is 55.5m, the total height of the cannonball from the ground is

(2H) = 2 × 55.5

= 111 m`2H/g

= 111/9.8`

= 11.32653 s

From equation (1),u×t = 140u = 140/t

Therefore, `u = 140/t`b.

b)To find the y position of the cannonball at the time t:

Here, we have to use the formula of height i.e.,y = u×t – (1/2)gt²,

Where, y = height of the cannonball at time t, u = 140/t, t = time taken by the cannonball to hit the ground and g = 9.8 m/s².

We have already calculated the time taken by the cannonball to hit the ground in the previous step.`

y = 140 - (1/2) × 9.8 × t²`

On substituting the value of t as `t = 11.32653`,

we get,y = 140 - (1/2) × 9.8 × (11.32653)²= 55.5 mc.

c)  To find the initial speed of the projectile:

To calculate the initial speed of the projectile, we need to use the formula of range of projectile

.i.e.,R = u²sin2θ/g

Where R = 140 m, g = 9.8 m/s², θ = 0° (horizontal)

u² = R × g/sin2θ

   = 140 × 9.8/sin0°  

   = 2744m²/s²u  

   = \(\sqrt(2744m^2/s^2)\)

   = 52.4 m/s

Hence, the initial speed of the projectile is 52.4 m/s.

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Answer:

57 mm

Explanation:

57 mm is equivalent to 5.7 cm

Answer:

Choice 2: 6/17

Explanation:

4+7=11. 6+11= 17. so it's 6/17