All categories

3 years ago

The interstellar medium is composed of ________.

Physics

1 answer:

PolarNik [594]3 years ago

3 0

THE INTERSTELLAR MEDIUM COMPOSED GAS AND DUST

You might be interested in

The launching velocity of a missile is 20.0 m/s, and it is shot at 53º above the horizontal. What is the vertical component of t

Lyrx [107]

Given that

Velocity of missile (v) = 20 m/s ,

Angle of missile (Θ) = 53°

Determine , Vertical component = v sin Θ

= 20 sin 53°

= 15.97 m/s

6 0

3 years ago

An astronaut 100m from the spaceship observes a 200kg meteoroid that drifts toward the shop at 10m/sec. If the astronaut can gai

Fynjy0 [20]

Refer to the attachment for calculations

8 0

3 years ago

A car with a mass of 600 kg is traveling at a velocity of 10 m/s. How much kinetic energy does it have?

Viefleur [7K]

Answer:

KE = 30,000 J

Explanation:

KE = $\frac{1}{2}$ mv²

KE = $\frac{1}{2}$ (600)(10)²

KE = $\frac{1}{2}$ (600)(100)

KE = $\frac{1}{2}$ (60000)

KE = 30,000 J

5 0

3 years ago

Since moment of inertia is a scalar quantity, a compound object made up of several objects joined together has a moment of inert

Gwar [14]

What is the question?

8 0

3 years ago

A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she

lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

$f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L}$

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}} }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = 1.3225 \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz = 740.6 \ Hz$

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, $L_{new}$ = 2/3·L

The value of the fundamental frequency will then be given as follows;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \times \dfrac{2 \times L}{3} } =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz = 840 \ Hz$

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0

2 years ago