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Oksana_A [137]
3 years ago
13

What is the primary force that holds things together on an astronomical scale is________?

Physics
1 answer:
Svetllana [295]3 years ago
5 0
The <span>primary force that holds things together on an astronomical scale is "Gravitational Force"

Hope this helps!</span>
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At the periphery of a hurricane the air is ____, and several kilometers above the surface, in the eye, the air is ____.
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An object accelerates in a direction that is always perpendicular to its motion.what is the effect if any of the acceleration on
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Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 23.5 m above water wit
Elodia [21]

As stated in the statement, we will apply energy conservation to solve this problem.

From this concept we know that the kinetic energy gained is equivalent to the potential energy lost and vice versa. Mathematically said equilibrium can be expressed as

\Delta KE = \Delta PE

\frac{1}{2}mv_f^2-\frac{1}{2} mv_0^2 = mgh_2-mgh_1

Where,

m = mass

v_{f,i} = initial and final velocity

g = Gravity

h = height

As the mass is tHe same and the final height is zero we have that the expression is now:

\frac{1}{2}v_f^2-\frac{1}{2} v_0^2 = gh_2

\frac{1}{2} (v_f^2-v_0^2) = gh_2

(v_f^2-v_0^2) = 2gh_2

v_f = \sqrt{2gh_2+v_0^2}

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v_f = 25.4m/s

7 0
3 years ago
Sapting Leng You have a string with a mass of 13.7 q. You stretch the string with a force of 8.39 N, giving it a length of 1.87
marshall27 [118]

Answer:

a)  \lambda=0.935\ \textup{m}

b) f=36.19\approx 36\ \textup{Hz}

Explanation:

Given:

String vibrates transversely fourth dynamic, thus n = 4

mass of the string, m = 13.7 g = 13.7 × 10⁻¹³ kg

Tension in the string, T = 8.39 N

Length of the string, L = 1.87 m

a) we know

L= n\frac{\lambda}{2}

where,

\lambda = wavelength

on substituting the values, we get

1.87= 4\times \frac{\lambda}{2}

or

\lambda=0.935\ \textup{m}

b) Speed of the wave (v) in the string is given as:

v =f\lambda

also,

v=\sqrt\frac{T}{(\frac{m}{L})}

equating both the formula for 'v' we get,

f\lambda=\sqrt\frac{T}{(\frac{m}{L})}

on substituting the values, we get

f\times 0.935=\sqrt\frac{8.39}{(\frac{13.7\times 10^{3}}{1.87})}

or

f=\frac{33.84}{0.935}

or

f=36.19\approx 36\ \textup{Hz}

5 0
2 years ago
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