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lorasvet [3.4K]
3 years ago
14

When the crests of 2 identical waves meet, what is the amplitude of the resulting wave?

Physics
1 answer:
djverab [1.8K]3 years ago
7 0
Twice the amplitude of each wave 
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I NEED HELP PLEASE, THANKS! :)
Zina [86]

Answer:

charge C = greatest net force

charge B = the smallest net force

ratio  = 9 : 1

Explanation:

we know that in Electrostatic Forces, when 2 charges are at same sign then they repel each other and if they are different signed charges then they attract each other

so as per Coulomb's formula of Electrostatic Forces

F = \frac{k\ q_1\ q_2}{r^2}     .....................1

and here k is 9 × 10^9 N.m²/c² and we consider each charge at distance d

so two charge force at A to B is

F1 = \frac{k\ q^2}{d^2}

and force between charges at A to C, at 2d distance

F1 = \frac{k\ q^2}{(2d)^2}  =  \frac{k\ q^2}{4d^2}

force between charges at A to D,  3d distance

F1 = \frac{k\ q^2}{(3d)^2}  = \frac{k\ q^2}{9d^2}  

so

Charge a It receives force to the left from b and c and to the right from d

so at a will be

F(a)  = -F1 - F2 + F3             ....................2

put here value

F(a) = -\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}

solve it

F(a) = \frac{k\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})  

F(a) = -\frac{41}{36}\ F1   = 1.13 F1  

and

Charge b It  receives force to the right from a and d and to the left from c

F(b) = F1 - F1 + F2            ....................3

F(b)  =  \frac{k\ q^2}{d^2}-\frac{k\ q^2}{d^2}+\frac{k\ q^2}{4d^2}    

F(b)  = \frac{1}{4} \ F1    =  0.25 F1

and

Charge c It receives forces to the right from all charges.

F(c) = F2 + F 1 + F 1      ....................4

F(c) = \frac{k\ q^2}{4d^2}+\frac{k\ q^2}{d^2}+\frac{k\ q^2}{d^2}      

F(c) =  \frac{9}{4} \ F1   = 2.25 F1

and

Charge d It receives forces to the left from all charges

F(d) = - F3 - F2 -F 1      ....................5

F(d) = -\frac{k\ q^2}{9d^2}-\frac{k\ q^2}{4d^2}-\frac{k\ q^2}{d^2}  

so

F(d) = -\frac{49}{36} \ F1    = 1.36 F1

and

now we get here ratio of the greatest to the smallest net force that is

ratio = \frac{2.25}{0.25}

 ratio  = 9 : 1

5 0
3 years ago
You drop a 20 Ton object and a
Alborosie

Explanation:

it dosent depend on the weights of the items. I'll reach the ground at same time taking as no air friction or restrictions.

i.e

v = u + gt

whte v is final velocity of the object

u is initial velocity of the object

g is acceleration due to gravity and

t is time. thanks

please if found helpful rate brainliest

7 0
1 year ago
The molecular weight of carbon dioxide (CO2) is 44.00 a.m.u., and the molecular weight of propane gas (C3H8) is 44.10 a.m.u. Thu
Nina [5.8K]

The molecular weight of carbon dioxide (CO2) is 44.00 a.m.u., and the molecular weight of propane gas (C3H8) is 44.10 a.m.u. Thus CO2 diffuses_______ C3H8.

Slower Than

Faster Than

The Same As

Slower than

6 0
3 years ago
Read 2 more answers
In a two-body collision, if the momentum of the system is conserved, then which of the
lakkis [162]

Answer:

c) may also be conserved

Explanation:

Momentum is conserved  in both elastic and inelastic type of collisions.

But the differences is that:

In an ELASTIC type of collisions, KINETIC ENERGY IS ALSO CONSERVED.

whereas, In an INELASTIC type of collision, KINETIC ENERGY IS NOT CONSERVED.

So unless until type of collision is specified, we can not say anything about the conservation of kinetic energy after collision.

Hence, may also be conserved is the appropriate option here.

5 0
3 years ago
A 4.40-kilogram hoop starts from rest at a height 1.70 m above the base of an inclined plane and rolls down under the influence
Anestetic [448]

Answer:

The linear velocity is  v=4.08m/s

Explanation:

According to the law of conservation of energy

   The potential energy possessed by the  hoop at the top of the inclined plane is converted to the kinetic energy at the foot of the inclined plane

        The kinetic energy can be mathematically represented as

                    KE = \frac{mv^2}{2} + \frac{Iw}{2}

Where I is the moment of inertia possessed by the hoop  which is mathematically represented as

                 I = mr^2

Here R is the radius of the hoop

         w is the angular velocity which the hoop has at the bottom of the lower part of the inclined plane which is mathematically represented as

                          w = \frac{v}{r}

Where v linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto a horizontal surface

      Now expressing the above statement mathematically

            potential \ energy = \frac{mv^2}{y} + \frac{Iw^2}{2}

               mgh = \frac{mv^2}{y} + \frac{Iw^2}{2}

=>            mgh =\frac{mv^2}{2} + \frac{(mr^2)(\frac{v}{r})^2 }{2}  

=>          mgh = \frac{mv^2}{2} + \frac{mv^2}{2}

=>           mgh = mv^2

=>              v = \sqrt{gh}

Substituting values

                v = \sqrt{9.81 * 1.7}

                  v=4.08m/s

4 0
3 years ago
Read 2 more answers
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