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Free_Kalibri [48]
3 years ago
5

places with continental climates typically have _______ summers and ________ winters warm; cool hot; cool warm; cold hot; cold

Physics
1 answer:
ipn [44]3 years ago
8 0
Places with continental climates typically have hot summers and cold winters. As compared to places with mild climates, places with continental climates tend to experience the two extremes when it comes to the seasons. Summers can get very hot, and winters can get very cold. 
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A point charge is placed at each corner of square with side leanth a. The charges all have same magnitude q. My question, What i
nexus9112 [7]

Answer:

 E = k q / a²   (1.3535) (- i ^ + j ^)

  E = k q / a²  1.914  ,      θ’= 135

Explanation:

For this exercise we will use Newton's second law where we must add as vectors

        E_total = E₁₂ i ^ + E₁₄ j ^ + E₁₃

Let's look for the value of each term

On the x axis

       E₁₂ = k q / a²

On the y axis

       E₁₄ = k q / a²

For the charge in the opposite corner we look for the distance

        d = √ (a² + a²) = a √2

let's look for the field

      E₁₃ = k q / d²

      E₁₃ = k q / 2a²

let's use trigonometry to find the two components of this field

       cos 45 = E₁₃ₓ / E₁₃

       E₁₃ₓ = E₁₃ cos 45

       

       sin 45 = E_{13y} / E₁₃

       E_{13y} = E₁₃ sin 45

       E₁₃ₓ = k q / 2a²  cos 45

       E_{13y} = k q / 2a²  sin 45

let's find each component of the electric field

X axis

      Eₓ = -E₁₂ - E₁₃ₓ

      Eₓ = - k q / a² - k q / 2a² cos 45

      Eₓ = - k q / a² (1 + cos 45/2)

      cos 45 = sin 45 = 0.707

      Eₓ = - k q / a²   (1 + 0.707 / 2)

      Eₓ = - k q / a²    (1.3535)

Y axis  

      E_y = E₁₄ + E_{13y}

       E_y = k q / a² + k q / 2a²     sin 45

       E_y = k q / a² (1 + sin 45/2)

       E_y = k q / a²       (1.3535)

we can give the results in two ways

       E = k q / a²   (1.3535) (- i ^ + j ^)

In modulus and angle form, let's use Pythagoras' theorem for the angle

       E = √ (Eₓ² + E_y²)

        E = k q / a²    1.3535 √2

        E = k q / a²     1.914

we use trigonometry for the angle

        tan θ = E_y / Eₓ

         θ = tan⁻¹  (E_y / Eₓ)

         θ = tan⁻¹ (1 / -1)

         θ = 45

in the third quadrant, if we measure the angle of the positive side of the x-axis

           θ‘= 90 + 45

           θ’= 135

4 0
3 years ago
A sports car skids off of a wet mountain road at 48.5 m/s and lands in the river, 110
Oxana [17]

Answer:

  about 4.74 seconds

Explanation:

The time to fall distance d from height h is given by ...

  t = √(2d/g)

  t = √(2·110 m/(9.8 m/s^2)) ≈ 4.74 s

It will take the car about 4.74 seconds to fall 110 meters to the river.

__

We assume the car's speed is horizontal, so does not add or subtract anything to/from the time to fall from the height.

3 0
3 years ago
If the half-life of iodine-131 is 8.10 days, how long will it take a 50.00 g sample to decay to 6.25 g?
Sloan [31]

Answer:

your in mr langfords class

Explanation:

bruh moment

6 0
3 years ago
A vessel that contains a gas has two pressure gauges attached to it. One contains liquid mercury, and the other an oil such as d
castortr0y [4]

Answer:

Pressure of the gas = 12669 (Pa) and height of the oil is 1,24 meters

Explanation:

First, we can use the following sketch for an easy understanding, in the attached image we can see the two pressure gauges the one with mercury to the right and the other one with oil to left. We have all the information needed in the mercury pressure gauge, so we can determine the pressure inside the vessel because the fluid is a gas it will have the same pressure distributed inside the vessel (P1).

Since P1 = Pgas, we can use the same formula, but this time we need to determine the height of the column of oil in the pressure gauge.

The result is that the height of the oil column is higher than the height of the one that uses mercury, this is due to the higher density of mercury compared to oil.

Note: the information given in the units of the fluids is not correct because the density is always expressed in units of (mass /volume)

4 0
3 years ago
A 500 kg roller coaster moving at 15 m/s suddenly comes to a stop at the end of the ride. How much work energy was needed to sto
saveliy_v [14]

Answer:

energy required=-energy lost

energy lost=change in kinetic energy

EL=1/2 mv^2

4 0
3 years ago
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