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3 years ago

A meteorite strikes the Moon at the speed of 11 km/s. It has KE of 60,500,000 J. What will be its mass?

1 answer:

8 0

KE=1/2 mv²
60,500,000 = 1/2 m ( 11,000 m/s )²
60,500,000= 1/2 ( 11,000 )² m
60,500,000= 60,500,000 x m
m= 60,500,000/60,500,000
m= 1 kg

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What is the definition of physics

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3 years ago

A gold sphere of radius R=100 μm and density 19g/cm^3 falls through water. Given the viscosity of water is about 10^-3 Pa s and

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The terminal velocity of gold sphere is 39.2 cm/s

<h3>What is terminal velocity?</h3>

Terminal velocity is the maximum velocity attainable for an object as it falls through a fluid.

<h3>How to calculate the terminal velocity of the gold sphere?</h3>

The terminal velocity of the gold sphere is given by v = 2gr²(ρ - σ)/9η where

g = acceleration due to gravity = 9.8 m/s²,
r = radius of sphere = 100 μm = 100 × 10⁻⁶ m = 10⁻⁴ m = 10⁻² cm,
ρ = density of sphere = 19 g/cm³,
σ = density of water = 1.0 g/cm³ and
η = viscosity of water = 10⁻³ Pa-s

So, susbtituting the values of the variables into the equation, we have that

v = 2gr²(ρ - σ)/9η

v = 2 × 9.8m/s²× (10⁻² cm)²(19 g/cm³ - 1.0 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 9.8 m/s² × 10⁻⁴ cm² × (18 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 980 cm/s² × 10⁻⁴ cm² × 2 g/cm³/(1 × 10⁻³ Pa-s)

v = 3920 g/s² × 10⁻⁴/(1 × 10⁻³ Pa-s)

v = 392 cm/s × 10³ × 10⁻⁴

v = 392 × 10⁻¹ cm/s

v = 39.2 cm/s

So, the terminal velocity is 39.2 cm/s

Learn more about terminal velocity of sphere here:

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Question 15 of 25

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Ml(d^2θ/dt^2) =-mgθ

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The equation of motion of a pendulum is:

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$\sin\theta \simeq \theta.$

Additionally, let us define:

$\omega^2\equiv\dfrac{g}{\ell}.$

We can now write:

$\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\omega^2\theta.$

The solution to this differential equation is:

$\theta(t) = A\sin(\omega t + \phi),$

where $A$ and $\phi$ are constants to be determined using the initial conditions. Notice that they will not have any influence on the period, since it is given simply by:

$T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{g}{\ell}}.$

This justifies that the period depends only on the pendulum's length.

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