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3 years ago

The following four waves are sent along strings with the same

Physics

1 answer:

likoan [24]3 years ago

3 0

Answer:

Explanation:

y_1 = (3 mm) sin(x - 3t)

comparing it with standard wave equation

y = A sin( ωt-kx )

we see

ω = -3 , k = -1

velocity = ω / k

= 3

y_2 = (6 mm) sin(2x - t)

we see

ω = -1 , k = -2

velocity = ω / k

= .5

y_3 = (1 mm) sin(4x - t)

we see

ω = -1 , k = -4

velocity = ω / k

= .25

y_4 = (2 mm) sin(x - 2t)

we see

ω = -2 , k = -1

velocity = ω / k

= 2

So greatest velocity to lowest velocity

y_1 = (3 mm) sin(x - 3t) , y_4 = (2 mm) sin(x - 2t) ,y_2 = (6 mm) sin(2x - t) , y_3 = (1 mm) sin(4x - t)

b )

Given the mass per unit length of wire the same , velocity is proportional to

√ T , where T is tension

so in respect of tension in the wire same order will exist for highest to lowest tension .

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PLEASE HELP ASAP Which one of the following is a step used for balancing chemical equations?

scZoUnD [109]

Answer:

C. Count the atoms in each substance in the reactants and products.

Explanation:

A chemical reaction can be defined as a chemical process which typically involves the transformation or rearrangement of the atomic, ionic or molecular structure of an element through the breakdown and formation of chemical bonds to produce a new compound or substance.

In order for a chemical equation to be balanced, the condition which must be met is that the number of atoms in the reactants equals the number of atoms in the products.

This ultimately implies that, the mass and charge of the chemical equation are both balanced properly.

In Chemistry, all chemical equation must follow or be in accordance with the Law of Conservation of Mass, which states that mass can neither be created nor destroyed by either a physical transformation or a chemical reaction but transformed from one form to another in an isolated (closed) system.

One of the step used for balancing chemical equations is to count the atoms in each substance in the reactants and products.

For example;

NH3 + O2 -----> NO + H2O

The number of atoms in each chemical element are;

For the reactant side:

Nitrogen, N = 1

Hydrogen, H = 3

Oxygen, O = 2

For the product side;

Nitrogen, N = 1

Hydrogen, H = 2

Oxygen, O = 2

When we balance the chemical equation, we would have;

NH3 + 3O2 -----> 4NO + 2H2O

3 0

3 years ago

A boat crossing a 153.0 m wide river is directed so that it will cross the river as quickly as possible. The boat has a speed of

Lynna [10]

We have the relation

$\vec v_{B \mid E} = \vec v_{B \mid R} + \vec v_{R \mid E}$

where $v_{A \mid B}$ denotes the velocity of a body A relative to another body B; here I use B for boat, E for Earth, and R for river.

We're given speeds

$v_{B \mid R} = 5.10 \dfrac{\rm m}{\rm s}$

$v_{R \mid E} = 3.70 \dfrac{\rm m}{\rm s}$

Let's assume the river flows South-to-North, so that

$\vec v_{R \mid E} = v_{R \mid E} \, \vec\jmath$

and let $-90^\circ < \theta < 90^\circ$ be the angle made by the boat relative to East (i.e. -90° corresponds to due South, 0° to due East, and +90° to due North), so that

$\vec v_{B \mid R} = v_{B \mid R} \left(\cos(\theta) \,\vec\imath + \sin(\theta) \, \vec\jmath\right)$

Then the velocity of the boat relative to the Earth is

$\vec v_{B\mid E} = v_{B \mid R} \cos(\theta) \, \vec\imath + \left(v_{B \mid R} \sin(\theta) + v_{R \mid E}\right) \,\vec\jmath$

The crossing is 153.0 m wide, so that for some time $t$ we have

$153.0\,\mathrm m = v_{B\mid R} \cos(\theta) t \implies t = \dfrac{153.0\,\rm m}{\left(5.10\frac{\rm m}{\rm s}\right) \cos(\theta)} = 30.0 \sec(\theta) \, \mathrm s$

which is minimized when $\theta=0^\circ$ so the crossing takes the minimum 30.0 s when the boat is pointing due East.

It follows that

$\vec v_{B \mid E} = v_{B \mid R} \,\vec\imath + \vec v_{R \mid E} \,\vec\jmath \\\\ \implies v_{B \mid E} = \sqrt{\left(5.10\dfrac{\rm m}{\rm s}\right)^2 + \left(3.70\dfrac{\rm m}{\rm s}\right)^2} \approx 6.30 \dfrac{\rm m}{\rm s}$