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ELEN [110]
3 years ago
9

What is the marble's range if it is fired horizontally from 1.6 m above the ground?

Physics
1 answer:
cupoosta [38]3 years ago
5 0

Answer:

The answer is 6.40 meters.

Explanation:

The speed v = √(2gh)

v = √( 2×9.8×6.4) = 11.2 m/s

After, finding the time it takes to hit the ground from a height of 1.6 meters.

time = √(2H÷g)

time = √(2×1.6÷9.8)

time = 0.5714 seconds.

Horizontal distance is speed × time = 11.2 × 0.5714 = 6.40 meters.

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i would say the heavier object

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Choose true or false for each statement regarding a parallel plate capacitor.. The voltage of a disconnected charged capacitor i
OlgaM077 [116]

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Explanation:

The voltage of a disconnected charged capacitor increases when the plate area is decreased.

When plate area decreases , capacitance C decreases , but charge Q remains constant .

Q = C V where C is capacitance and V is voltage .

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So the statement is true.

The electric field is dependent on the charge density on the plates.

This statement is true .

The voltage of a connected charged capacitor remains the same when the plate area is decreased .

For a connected capacitor , V or voltage is constant which is equal to voltage of charging battery .

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3 years ago
A light has wavelength of is used in the double slits experiment . Which of the following values represent the phase difference
Wewaii [24]

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Explanation:

8 0
3 years ago
Adhira loves to ride her bike around the neighborhood. She starts riding 1.2 miles at 30° S of E. Then, she rides another 2.0 mi
zepelin [54]

Answer:

D = 1.8677 miles , θ = 24.28º at South of West

Explanation:

This is an exercise in adding vectors, the easiest way to solve them is to decompose the vectors and add each component algebraically. Let's use trigonometry

first displacement. d = 1.2 miles to 30º south of East

     cos ( 360-30) = cos (-30) = x₁ / d

     sin (-30) = y₁ / d

     x₁ = d cos (-30)

     y₁ = d sin (-30)

     x₁ = 1.2 cos (-30) = 1,039 miles

     y₁ = 1.2 sin (-30) = -0.6 miles

second shift. d = 2.0 miles to 20º West of South

       cos (270-20) = x₂ / d

       cos (250) = y₂ / d

       x₂ = 2.0 cos 250 = -0.684 miles

       y₂ = 2.0 sin250 = -1.879 miles

Third displacement. d = 1.6 miles to 30º South of West

       cos (180 + 30) = x₃ / d

       sin (210) = y₃ / d

       x₃ = 1.6 cos 210 = -1.3856 miles

       y₃ = 1.6 sin 210 = -0.8 miles

Fourth displacement. d = 2.6 miles to 15º West of North

       cos (90 + 15) = x₄ / d

       sin (105) = y₄ / d

       x₄ = 2.6 cos 105 = -0.6729 miles

       y₄ = 2.6 sin 105 = 2,511 miles

having all the components we add

x-axis  (West-East direction)

       X = x₁ + x₂ + x₃ + x₄

       X = 1.039 -0.684 - 1.3846 - 0.6729

       X = -1.7025 miles

   

       Y = y₁ + y₂ + y₃ + y₄

       Y = -0.6 -1.879 -0.8 +2.511

       Y = -0.768

The modulus of this displacement is we use the Pythagorean theorem

      D = √ (X² + Y²)

      D = √ (1.7025² + 0.768²)

      D = 1.8677 miles

let's use trigonometry to find the direction

       tan θ = Y / X

       θ = tan⁻¹ Y / x

       θ = tan⁻¹ (0.768 / 1.7025)

       θ = 24.28º

as the two components are negative this angle is in the third quadrant

therefore in cardinal direction form is

         θ = 24.28º at South of West

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3 years ago
Question 11 of 11
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Answer:

D.

Explanation:

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