A light-year is the distance light travels in one year, at a speed = 2.998 x 10^8 m/s.

The unit of measurement "mL" describes what property of a substance?

pychu [463]

D. density is your answer

8 0

3 years ago

From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. As the

Ivahew [28]

Answer:

The launching point is at a distance D = 962.2m and H = 39.2m

Explanation:

It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.

X axis

x = Vox t

t = x / vox

t = 7.1 / 340

t = 2.09 10-2 s

In this same time the height of the window fell

Y = Voy t - ½ g t²

Let's calculate the initial vertical speed, this speed is in the window

Voy = (Y + ½ g t²) / t

Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209

Voy = 27.7 m / s

We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s

Vy = Voy - gt₂

Vy = 0 -g t₂

t₂ = Vy / g

t₂ = 27.7 / 9.8

t₂ = 2.83 s

This is the time it also takes to travel the horizontal and vertical distance

X = Vox t₂

D = 340 2.83

D = 962.2 m

Y = Voy₂– ½ g t₂²

Y = 0 - ½ g t2

H = Y = - ½ 9.8 2.83 2

H = 39.2 m

The launching point is at a distance D = 962.2m and H = 39.2m

6 0

3 years ago

An open pipe, 0.29 m long, vibrates in the second overtone with a frequency of 1,227 Hz. In this situation, the fundamental freq

Andru [333]

Answer:

f = 409 Hz

Explanation:

We have,

Length of the open organ pipe, l = 0.29 m

Frequency of vibration of second overtone, $f_2 = 1227 Hz$

It is required to find the fundamental frequency of the pipe. For the open organ pipe, the frequency of second overtone is given by :

$f_2=\dfrac{3v}{2l}$

v is speed of sound

Let f is the fundamental frequency. It is given by :

$f=\dfrac{v}{2l}$

The relation between f and f₂ can be written as :

$f_2=3f\\\\f=\dfrac{f_2}{3}\\\\f=\dfrac{1227}{3}\\\\f=409\ Hz$

So, the fundamental frequency of the pipe is 409 Hz.

6 0

2 years ago

Hey there!

eimsori [14]

The time taken for the two balls to hit each other is 8 s.

The given parameters:

Acceleration of the rocket, a = 2 m/s²
Length of the chamber, s = 4 m
Speed of the first ball, = V1 = 0.3 m/s
Speed of the second ball, V2 = 0.2 m/s

The time taken for the two balls to hit each other is calculated by applying relative velocity formula as shown below;

$(V_1 - (-V_2) )t = s\\\\(V_1 + V_2) t = s\\\\(0.3 + 0.2) t = 4\\\\0.5t = 4\\\\t = \frac{4}{0.5} \\\\t = 8 \ s$

Thus, the time taken for the two balls to hit each other is 8 s.

Learn more about relative velocity here: brainly.com/question/17228388

8 0

2 years ago

Read 2 more answers

A 11.0 kg satellite has a circular orbit with a period of 1.80 h and a radius of 7.50 × 106 m around a planet of unknown mass. I

Anuta_ua [19.1K]

Answer:

Explanation:

Expression for times period of a satellite can be given as follows

Time period T = 1.8 x 60 x 60

= 6480

T² = $\frac{4\times \pi^2\times r^3}{GM}$ where T is time period , r is radius of orbit , G is gravitational constant and M is mass of the satellite.

6480² = 4 x 3.14² x 7.5³ x 10¹⁸ / GM

GM = 4 x 3.14² x 7.5³ x 10¹⁸ / 6480²

= 3.96 X 10¹⁴

Expression for acceleration due to gravity

g = GM / R² where R is radius of satellite

20 = 3.96 X 10¹⁴ / R²

R² = 3.96 X 10¹⁴ / 20

= 1.98 x 10¹³ m

R= 4.45 x 10⁶ m

8 0

3 years ago