What did both newton and einstein believe directly affects gravity? mass of objects distance between objects curvature of object
1 answer:
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Answer:
v = 127.66 m / s θ’= 337.59
Explanation:
For this exercise we must use the speed composition of the drone.
The first speed is 63 m / s in direction 29, three minutes later the speed reaches 22 m / s and direction 233, finally it returns to the launch point with 98 m / s in direction 321, in the attachment you can see a diagram of these speeds.
To find the resulting average velocity, the easiest thing is to decompose each velocity into the x and y coordinate system, then add each velocity
let's break down the speeds
cos 29 = v₁ₓ / v₁
sin 29 = / v₁
v₁ₓ = v₁ cos 29
v_{1y} = v₁ sin 29
v₁ₓ = 63 cos 29 = 55.10 m / s
v_{1y} = 63 sin 29 = 30.54 m / s
speed 2
cos 22 = v₂ₓ / v₂
sin 22 = v_{2y} / v₂
v₂ₓ = v₂ cos 233
v_{2y} = v₂ sin 233
v₂ₓ = 22 cos 233 = -13.24 m / s
v_{2y} = 22 sin 233 = -17.57 m / s
speed 3
cos 321 = v₃ₓ / v₃
sin 321 = v_{3y} / v₃
v₃ₓ = v₃ cos 321
v_{1y} = vₐ sin 321
v₃ₓ = 98 cos 321 = 76.16 m / s
v_{3y} = 98 sin 321 = -61.67 m / s
We already have all the component of the speeds, the resulting speed is
vₓ = v₁ₓ + v₂ₓ + v₃ₓ
vₓ = 55.10 -13.24 +76.16
vₓ = 118.02 m / s
v_{y} = v_{1y} + v_{2y} + v_{3y}
v_{y} = 30.54 -17.54 - 61.67
v_{y} = -48.67 m / s
there are two ways to give the result
v = (118.02 i -48.67 j) m / s
or in the form of magnitud and angle.
We use the Pythagorean theorem for the module
v = √ (vₓ² + v_{y}²)
v = RA (118.02² + 48.67²)
v = 127.66 m / s
let's use trigonometry for the angle
tan θ = v_{y} / vₓ
θ = tan⁻¹ (v_{1} / vₓ)
θ = tan⁻¹ (-48.67 / 118.02)
θ = -22.41
if we want to measure the angles with respect to the positive side of the x axis
θ’= 360 - 22.41
θ’= 337.59
This problem provides information about the pressure and temperature ideal gases are studied at. The answer to the questions are that all molecules have the same density, 2.43x10²⁵ mol/m³ and 2.43x10¹⁹ mol/cm³.
<h3>Idela gases</h3>In science, we can start studying gases with the concept of ideal gas, as they do not collide one to another and are assumed to be perfect spheres with no relevant interactions.
In such a way, one can conclude that the <u>number density of all ideal gasses at SATP is the same</u>, as they are assumed to be perfect spheres with equal volumes per molecule.
Moreover, when calculating the number of molecules per cubic meter, one must use the ideal gas equation as:
And plug in the numbers we are given:
Lastly, we can calculate the molecules per cubic centimeter by performing the following conversion:
Learn more about ideal gases: brainly.com/question/26450101