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zhuklara [117]
3 years ago
11

You wiggle a string, that is fixed to a wall at the other end, creating a sinusoidal wave with a frequency of 2.00 Hz and an amp

litude of 0.075 m. The speed of the wave is 12.0 m/s. At t=0 the string has a maximum displacement and is instantaneously at rest. Assume no waves bounce back from the far end of the wall. Find the angular frequency, period, wavelength, and wave number. Write a wave function describing the wave. Write equations for the displacement, as a function of time, of the end of the string that is being wiggled and at a point 3.00 m from that end. Determine the speed of the medium and draw history and snapshot graphs for the waves created.
Physics
1 answer:
allochka39001 [22]3 years ago
4 0

Answer:hhhkkzkxixuhhhdhhdhdhhhdhhhshsbbsbzhhdhhdhhhshhhdhhdusudhggdydyydhshdhgddhsjsudhdhhdh

Explanation:NzjzjhbxxhzhdghsjshshhHjajskajakakakakaiaiaiijhayayayagayahahah

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Now let’s apply Coulomb’s law and the superposition principle to calculate the force on a point charge due to the presence of ot
Mnenie [13.5K]

Answer:

F = - 1.68 10⁻⁴ N

, it is directed to the left of the x-axis

Explanation:

Coulomb's law is

     F = k q₁ q₂ / r²

Where K is the Coulomb constant that value 8.99 10⁹ N m²/ C², q are the electric charges and r is the distance between them. Let's apply to our problem for each pair of charges

Let's reduce the magnitudes to the SI system

    q₁ = 3.0 nc (1C / 10 9 nC) = 3.0 10⁻⁹ C

    x₁ = 2.0 cm (1m / 100cm) = 2.0 10⁻² m

    q₂ = -6.0 nC = -6.0 10⁻⁹ C

    x₂ = 4.0 cm = 4.0 10⁻² m

    q₃ = 5.0 nC = 5.0 10⁻⁹ C

    x3 = 0 m

Charges q1 and q3

    r = x₁ -x₃

    r = 2.0 10⁻² -0

    r = 2.0 10⁻² m

    F₁₃ = 8.99 10⁹ 3.0 10⁻⁹ 5.0 10⁻⁹ / (2.0 10⁻²)²

    F₁₃ = 33.7 10⁻⁵ N

As the charges are of the same sign, the force is repulsive, therefore it is directed to the left of the x-axis

Charges q2 and q3

    r = r₂ –r₃

    r = 4.0 10⁻² - 0 = 4.0 10⁻² m

    F₂₃ = 8.99 10⁹ 6.0 10⁻⁹ 5.0 10⁻⁹ / (4.0 10⁻²)²

    F₂₃ = 16.86 10⁻⁵ N

As the charges are of different sign, the force is attractive, therefore it is directed to the right of the x-axis

The force is a vector magnitude, so each component must be added independently, in this case all the forces are on the x-axis, let's take the right direction as positive

    F = F₂₃ - F₁₃

    F = 16.86 10⁻⁵ - 33.7 10⁻⁵

    F = - 16.84 10⁻⁵ N

    F = - 1.68 10⁻⁴ N

The negative sign means that it is directed to the left of the x-axis

5 0
3 years ago
A water balloon is thrown at 20 m/s from the top of a 20 m high building, what is its speed when it hits the ground? Does the an
Oksana_A [137]

Answer:

The final velocity is 28.14 m/s

Yes the angle of projection matters

Explanation:

Given;

initial velocity of the water balloon, u = 20 m/s

height of the building, h = 20 m

let the final speed of the ball when it hits the ground = v

The final speed is calculated as follows;

v² = u² + 2gh

v² = (20)²  +  2(9.8)(20)

v² = 400 + 392

v² = 792

v = √792

v = 28.14 m/s

Yes the angle matters, if the balloon had been dropped at a certain angle, the final velocity would have been estimated using the following formula;

v_y^2 = u_y^2 sin^2(\theta) + 2gh_y

where;

θ is the angle of projection, which accounts for the vertical component of the velocity.

6 0
2 years ago
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
Once a scientist has made a hypothesis, what would they typically do next? (2 points)
ipn [44]

Answer:

b-testing

Explanation:

First would be observation/research. Then the hypothesis. After that you would test your theory, conduct experiments. And finally, your conclusion- what you got from the whole process basically.

Hope this helps.

4 0
3 years ago
Write down the symbols of nitrogen and neon?​
statuscvo [17]

Answer:

Nitrogen is N, neon is Ne

5 0
2 years ago
Read 2 more answers
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