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rewona [7]
2 years ago
10

The acronym used when referring to the Big Five Factors is OCEAN. True False

Physics
2 answers:
schepotkina [342]2 years ago
5 0

true

Explanation:

Because I did the test

TEA [102]2 years ago
4 0

Answer:

The right answer is True

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The rocket is fired vertically and tracked by the radar station shown. When θ reaches 66°, other corresponding measurements give
Flauer [41]

Answer:

velocity = 1527.52 ft/s

Acceleration = 80.13 ft/s²

Explanation:

We are given;

Radius of rotation; r = 32,700 ft

Radial acceleration; a_r = r¨ = 85 ft/s²

Angular velocity; ω = θ˙˙ = 0.019 rad/s

Also, angle θ reaches 66°

So, velocity of the rocket for the given position will be;

v = rθ˙˙/cos θ

so, v = 32700 × 0.019/ cos 66

v = 1527.52 ft/s

Acceleration is given by the formula ;

a = a_r/sinθ

For the given position,

a_r = r¨ - r(θ˙˙)²

Thus,

a = (r¨ - r(θ˙˙)²)/sinθ

Plugging in the relevant values, we obtain;

a = (85 - 32700(0.019)²)/sin66

a = (85 - 11.8047)/0.9135

a = 80.13 ft/s²

4 0
2 years ago
(a) Calculate the buoyant force on a 2.00-L helium balloon.
iragen [17]

Answer:

0.0239364 N

0.0057879 N

Explanation:

\rho = Density of the gas

g = Acceleration due to gravity = 9.81 m/s²

V = Volume

Mass of rubber = 1.5 g

Buoyant force is given by

F_b=\rho gV\\\Rightarrow F_b=1.22\times 9.81\times 2\times 10^{-3}\\\Rightarrow F_b=0.0239364\ N

The buoyant force is 0.0239364 N

Net vertical force is given by

F_n=F_b-W_{He}-W_{r}\\\Rightarrow F_n=0.0239364-0.175\times 2\times 10^{-3}\times 9.81-1.5\times 10^{-3}\times 9.81\\\Rightarrow F_n=0.0057879\ N

The net vertical force is 0.0057879 N

6 0
3 years ago
If a 1000-pound capsule weighs only 165 pounds on the moon, how much work is done in propelling this capsule out of the moon's g
Bas_tet [7]

Answer:

  W = 1,307 10⁶ J

Explanation:

Work is the product of force by distance, in this case it is the force of gravitational attraction between the moon (M) and the capsule (m₁)

              F = G m₁ M / r²

              W = ∫ F. dr

              W = G m₁ M ∫ dr / r²

we integrate

             W = G m₁ M (-1 / r)

                 

We evaluate between the limits, lower r = R_ Moon and r = ∞

           W = -G m₁ M (1 /∞ - 1 / R_moon)

            W = G m1 M / r_moon

Body weight is

             W = mg

             m = W / g

The mass is constant, so we can find it with the initial data

For the capsule

            m = 1000/32 = 165 / g_moon

            g_moom = 165 32/1000

            .g_moon = 5.28 ft / s²

I think it is easier to follow the exercise in SI system  

           W_capsule = 1000 pound (1 kg / 2.20 pounds)

           W_capsule = 454 N

           W = m_capsule g

           m_capsule = W / g

           m = 454 /9.8

           m_capsule = 46,327 kg

Let's calculate

          W = 6.67 10⁻¹¹ 46,327   7.36 10²² / 1.74 10⁶

          W = 1,307 10⁶ J

7 0
3 years ago
The density of a substance equals its mass divided by its volume. Talia listed the density of some common materials at 20 °C.
Zolol [24]

Answer:

mercury

Explanation:

6 0
3 years ago
Read 2 more answers
In a certain two slit diffraction experiment, two slits 0.02mm wide are spaced0.2mm between centers.(a) How many fringes appear
Kamila [148]

Answer:

a)   m = 10  and    b)  λ  = 3.119 10⁻⁷ m

Explanation:

In the diffraction experiments the maximums appear due to the interference phenomenon modulated by the envelope of the diffraction phenomenon, for which to find the number of lines within the maximum diffraction center we must relate the equations of the two phenomena.

Interference equation      d sin θ = m λ

Diffraction equation         a sin θ = n λ

Where d is the width between slits (d = 0.2 mm), a is the width of each slit (a = 0.02 mm). θ is the angle, λ the wavelength, m and n  are an integer.

Let's find the relationship of these two equations

    d sin θ / a sin θ = m Lam / n Lam

The first maximum diffraction (envelope) occurs for n = 1, let's simplify

    d / a = m

Let's calculate

    m = 0.2 / 0.02

    m = 10

This means that 10 interference lines appear within the first maximum diffraction.

b) let's use the interference equation, remember that the angles must be given in radians

    θ = 0.17 ° (π rad / 180 °) = 2.97 10⁻³ rad

    d sin  θ = m λ

    λ = d sin θ / m

    λ = 0.2 10⁻³ sin (2.97 10⁻³) / 2

    λ  = 3.119 10⁻⁷ m

8 0
2 years ago
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