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nata0808 [166]
3 years ago
12

Waves do NOT carry. A)Weight B)Matter C)Energy D)Color (Choose one)

Physics
1 answer:
torisob [31]3 years ago
6 0
The answer is D color
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An arrow is launched with an initial velocity of 7.6 m/s at an angle of 42 degrees above the horizontal. What is the horizontal
AleksAgata [21]

Explanation:

At any point on the arrow's trajectory, the horizontal component of the velocity is the same. Therefore, the horizontal component of the velocity at the top of its trajectory is

v_x = v_{0x} = v_0\cos{42°} = (7.6\:\text{m/s})\cos{42°}

\:\:\:\:\:=5.6\:\text{m/s}

5 0
3 years ago
A hypothetical planet has a mass 2.81 times that of Earth, but the same radius.
patriot [66]

The acceleration due to gravity near the surface of the planet is 27.38 m/s².

<h3>Acceleration due to gravity near the surface of the planet</h3>

g = GM/R²

where;

  • G is universal gravitation constant
  • M is mass of the planet
  • R is radius of the planet
  • g is acceleration due to gravity = ?

g = (6.626 x 10⁻¹¹ x 2.81 x 5.97 x 10²⁴) / (6371 x 10³)²

g = 27.38 m/s²

Thus, the acceleration due to gravity near the surface of the planet is 27.38 m/s².

Learn more about acceleration due to gravity here: brainly.com/question/88039

#SPJ1

4 0
2 years ago
Now put the names of the planets in increasing order based on their distance from the sun
Anika [276]
The eight planets of the Solar System arranged in order from the sun:

Mercury: 46 million km / 29 million miles (.307 AU)
Venus: 107 million km / 66 million miles (.718 AU)
Earth: 147 million km / 91 million miles (.98 AU)
Mars: 205 million km / 127 million miles (1.38 AU)
Jupiter: 741 million km /460 million miles (4.95 AU)
Saturn: 1.35 billion km / 839 million miles (9.05 AU)
Uranus: 2.75 billion km / 1.71 billion miles (18.4 AU)
Neptune: 4.45 billion km / 2.77 billion miles (29.8 AU)

Astronomers often use a term called astronomical unit (AU) to represent the distance from the Earth to the Sun.

+ Pluto (Dwarf Planet): 4.44 billion km / 2.76 billion miles (29.7 AU)
5 0
3 years ago
Which of the following is NOT a example
solmaris [256]

Answer:

B. a piece of paper being torn

Explanation:

A chemical change is one that cannot be reversed. This means the original properties of the substance or object cannot be restored.

If you cook a raw egg, it would turn into a boiled egg (or a poached egg, however it is being cooked). The reaction is irreversible, so you cannot turn the cooked egg back into a raw egg - it is basically impossible to 'uncook' an already cooked egg.

When you toast a piece of bread, it turns into toast. You can't 'untoast' it back into bread. The chemical changes have already occurred and cannot be undone.

If you tear a piece of paper, it is still paper. You are only ripping it, not changing anything about it. You could simply tape the torn bit back to the original bit, or glue it - either way, it is still paper and nothing has occurred to drastically change the physical state of it.

Therefore, B is not a chemical change.

5 0
3 years ago
Scenario
Anvisha [2.4K]

Answer:

1) t = 23.26 s,  x = 8527 m, 2)   t = 97.145 s,  v₀ = 6.4 m / s

Explanation:

1) First Scenario.

After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.

Before starting, let's reduce the magnitudes to the SI system

         v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s

         y = 2650 m

Let's start by looking for the time it takes for the load to reach the ground.

         y = y₀ + v_{oy} t - ½ g t²

in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero

         0 = y₀ + 0 - ½ g t2

          t = \sqrt{ \frac{2y_o}{g} }

          t = √(2 2650/9.8)

          t = 23.26 s

this is the horizontal scrolling time

          x = v₀ t

          x = 366.67  23.26

          x = 8527 m

the speed at the point of arrival is

         v_y = v_{oy} - g t = 0 - gt

         v_y = - 9.8 23.26

         v_y = -227.95 m / s

Module and angle form

        v = \sqrt{v_x^2 + v_y^2}

         v = √(366.67² + 227.95²)

        v = 431.75 m / s

         θ = tan⁻¹ (v_y / vₓ)

         θ = tan⁻¹ (227.95 / 366.67)

         θ = - 31.97º

measured clockwise from x axis

We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.

2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º

we look for the components of speed

           cos θ = v₀ₓ / v₀

           sin θ = v_{oy} / v₀

            v₀ₓ = v₀ cos θ

            v_{oy} = v₀ sin θ

we look for the time for the arrival point that has coordinates x = 0, y = 0

            y = y₀ + v_{oy} t - ½ g t²

            0 = y₀ + vo sin θ t - ½ g t²

            0 = -50 + vo sin 50 t - ½ 9.8 t²

            x = x₀ + v₀ₓ t

            0 = x₀ + vo cos θ t

            0 = -400 + vo cos 50 t

podemos ver que tenemos un sistema de dos ecuación con dos incógnitas

          50 = 0,766 vo t – 4,9 t²

          400 =   0,643 vo t

resolved

          50 = 0,766 ( \frac{400}{0.643 \ t}) t – 4,9 t²

          50 = 476,52 t – 4,9 t²

          t² – 97,25 t + 10,2 = 0

we solve the quadratic equation

         t = [97.25 ± \sqrt{97.25^2 - 4 \ 10.2}] / 2

         t = 97.25 ±97.04] 2

         t₁ = 97.145 s

         t₂ = 0.1 s≈0

the correct time is t1 the other time is the time to the launch point,

         t = 97.145 s

let's find the initial velocity

         x = x₀ + v₀ cos 50 t

         0 = -400 + v₀ cos 50 97.145

         v₀ = 400 / 62.44

         v₀ = 6.4 m / s

5 0
3 years ago
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