Let at time t1 means?

Sound waves can be modeled by the equation of the form y=20sin(3t+theta). Determine what type of interference results when sound

Mariulka [41]

Answer:

go to the link quizzlet it will give you tha answer

Explanation:

4 0

2 years ago

If a substance is found to be reactive flammable soluble and explosive what observation is also a physical property

Ainat [17]

Answer:

soluble

Explanation:

Chemical properties only manifest when a chemical reaction occurs. Being reactive, flammable and explosive are chemical properties, because they involve chemical reactions: the substances are changed; the chemical bonds of some substances, called reactants, are broken, and the chemical bonds are created, forming other substances, called products.

Solubility is a physical property because during dissolution no new substances are formed. You can prove it when the solvent evaporates leaving behind the same original substance.

The the observation that the substance is soluble is describing a physical property.

3 0

3 years ago

Infrared observations of the orbits of stars close to the galactic center indicate a small object at the center with a mass of a

MariettaO [177]

Answer:

(4.31±0.38) million Solar masses.

Explanation:

The galactic center is the center of the milky way around which the galaxy rotates. It is most likely the location of a supermassive black hole which has a mass of (4.31±0.38) million Solar masses. The location is called Sagittarius A*.

As there is interstellar dust in our line of sight from the Earth infrared observations need to be taken.

5 0

2 years ago

Two identical conducting spheres, having charges of opposite sign, attract each otherwith a force of 0.108 n when separated by 5

lisov135 [29]

Let's start from the final situation. After the two spheres are connected with the conducting wire, the total charge distributes equally between the two spheres (because they are identical). We can call the charge on each sphere $Q/2$ , with Q being the total charge.
The electrostatic force in this situation is 0.0360 N, so we can write
$F=k \frac{( \frac{Q}{2} )^2}{r^2}$
where k is the Coulomb's constant and $r=50.0 cm=0.50 m$ is the separation between the two spheres. Using F=0.0360 N, we can find the value of Q, the total charge shared between the two spheres:
$Q= \sqrt{ \frac{4Fr^2}{k} } = \sqrt{ \frac{4(0.0360 N)(0.50 m)^2}{8.99 \cdot 10^9 Nm^2C^{-2}} }=2.0 \cdot 10^{-6}C$

Now let's go back to the initial situation, before the conducting wire was attached; in this situation, the two spheres have a charge of $q_1$ and $q_2$ , whose sum is Q:
$Q=q_1 + q_2$
The electrostatic force between the two spheres in the initial situation is:
$F=k \frac{q_1 q_2}{r^2}$
And since we know F=0.108 N, we find
$q_1 q_2 = \frac{Fr^2}{k} = \frac{(0.108 N)(0.50 m)^2}{8.99 \cdot 10^9 N m^2 C^{-2}}=3.0 \cdot 10^{-12} C$
But the problem tells us that the two spheres have charges of opposite sign, so we must put a negative sign:
$-3.0 \cdot 10^{-12} C = q_1 q_2$

So now we have basically a system of 2 equations:
$2.0 \cdot 10^{-6} C = Q = q_1 + q_2$
$-3.0 \cdot 10^{-12} C = q_1 q_2$
If we solve it, we find the initial charge on the two spheres:
$q_1 = -1 \cdot 10^{-6}C$
$q_2 = +3 \cdot 10^{-6 } C$

6 0

2 years ago

A light-rail train going from one station to the next on a straight section of track accelerates from rest at 1.1 m/s^2 for 20s.

svet-max [94.6K]

Answer:

A) The distance between the stations is 1430m

B) The time it takes the train to go between the stations is 80s

Explanation:

First we will calculate the distance covered for the first 20s.

From one the equations of kinematics for linear motion

$S = ut + \frac{1}{2}at^{2} \\$

Where $S$ is distance traveled

$u$ is the initial velocity

$t$ is time

and $a$ is acceleration

Since the train starts from rest, $u$ = 0 m/s

Hence, for the first 20s

$a =$ 1.1 m/s²; $t =$ 20s, $u$ = 0 m/s

∴ $S = ut + \frac{1}{2}at^{2} \\$ gives

$S = (0)(20) + \frac{1}{2}(1.1)(20)^{2}$

$S = \frac{1}{2}(1.1)(20)^{2}$

$S =$ 220m

This is the distance covered in the first 20s.

The train then proceeds at constant speed for 1100m.

Now, we will calculate the speed attained here

From

$v = u +at$

Where $v$ is the final velocity

Hence,

$v = 0 + 1.1(20)$

$v = 1.1(20)$

$v =$ 22 m/s

This is the constant speed attained when it proceeds for 1100m

The train then slows down at a rate of 2.2 m/s² until it stops

We can calculate the distance covered while slowing down from

$v^{2} = u^{2} + 2as$

The initial velocity, $u$ here will be the final velocity before it started slowing down

∴ $u$ = 22 m/s

The final velocity will be 0, since it came to a stop.

∴ $v$ = 0 m/s

$a = -$ 2.2 m/s² ( - indicates deceleration)

Hence,

$v^{2} = u^{2} + 2as$ gives

$0^{2} =22^{2} +2(-2.2)s$

$0=22^{2} - (4.4)s\\4.4s = 484\\s = \frac{484}{4.4} \\s = 110m$

This is the distance traveled while slowing down.

A) The distance between the stations is

220m + 1100m + 110m

= 1430m

Hence, the distance between the stations is 1430m

B) The time it takes the train to go between the stations

The time spent while accelerating at 1.1 m/s² is 20s

We will calculate the time spent when it proceeds at a constant speed of 22 m/s for 1100m,

From,

$Speed =\frac{Distance}{Time}\\$

Then,

$Time = \frac{Distance}{Speed}$

$Time = \frac{1100}{22}$

Time = 50 s

And then, the time spent while decelerating (that is, while slowing down)

From,

$v = u + at\\0 = 22 +(-2.2)t\\2.2t = 22\\t = \frac{22}{2.2} \\t= 10 s$

This is the time spent while slowing down until it stops at the station.

Hence, The time it takes the train to go between the stations is

20s + 50s + 10s = 80s

The time it takes the train to go between the stations is 80s

6 0

2 years ago