5. The speed of a transverse wave on a string is 170 m/s when the string tension is 120 ????. To what value must the tension be

Answer:

Stars generate energy by the process of nuclear fusion.

They are large entities composed of gaseous elements.

The main constituents of stars are hydrogen and helium.

Explanation:

Stars are colossal objects with extensive gravitational forces causing them to contract, which allows fusion to take place: the atomic nuclei in the star's core are drawn very close together due to gravity and elevated temperatures, leading to the fusion reaction. This fusion serves as the energy output for a star.

Conversely, it is true that stars predominantly consist of hydrogen and helium (two hydrogen nuclei can fuse to become helium), which implies that a star is essentially an enormous ball of gas without a solid surface suitable for standing on.

As for the presence of water on a star, it is simply impossible. The extreme temperatures found in stars are far too high for water to exist in any liquid state on their surfaces.

Answer:

a)9.8

Explanation:

because any object falling from any height experiences 9.8

(sorry that's all I know)

Answer:3249.33 m/s

Explanation:

Provided:

Initial velocity = 3250 m/s

Acceleration

Displacement = 215 km

Applying the equation of motion:

where:

v = the final velocity

u = the initial velocity

a = acceleration

s = displacement

v = 3249.33 m/s

Answer:

The peak height can be expressed as y = b²/18g √(12L/b)³

Explanation:

We begin by assessing the scenario involving a projectile subject to time-dependent acceleration. Thus, we must define acceleration to determine the speed when at distance L, then apply the equations governing projectile launch.

Acceleration that varies with time:

a = dv / dt

dv = a dt

∫dv = ∫(bt²) dt

v = (b t³)/3

The starting speed is zero at t = 0.

Utilizing the definition of speed:

v = dy / dt

dy = v dt

∫dy = ∫(b t³/3) dt

y = (b/3)(t⁴/4)

y = (b/12)t⁴

From the initial instance where height is zero at time zero, we now compute the time needed to traverse the distance (y = L) across the canyon:

t = (12y/b)¼

t = (12L/b)¼

This time allows us to find the projectile's launch speed:

v = (b/3)

This speed acts as the initial speed for the projectile movement. To find its maximum height at zero final speed:

Vy² = v₀² - 2 g y

0 = v₀² - 2 g y

2 g y = v₀²

y = v₀²/2g

y = (1/2g)([b/3 ])²

y = (1/2g)(b²/9 )

y = b²/18g √(12L/b)³

Answer:

Explanation:

Given:

The pressure exerted by the cooker,

the area of the opening,

• It is understood that gauge pressure is assessed relative to atmospheric pressure, which acts on all parts of the observed system, requiring us to only counteract the gauge pressure.

Since the petcock is positioned at the opening's cross-section, it balances out the built-up pressure owing to its weight.

Thus:

where:

represents the mass of the petcock

denotes the acceleration due to gravity