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Oliga [24]
3 years ago
11

If 20 beats are produce with one second which of the following frequency could possibly be held by two sound waves traveling thr

ough a medium a long a common path at the same time
Physics
1 answer:
musickatia [10]3 years ago
8 0

ANY two frequencies whose difference is 20 Hz.

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As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil
galina1969 [7]

Answer:

Second option 6.3 N at 162° counterclockwise from  

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

-F_3sin(b) + F_1 = 0

For the address and we have:

-F_3cos(b) + F_2 = 0

The forces F_1 and F_2 are known

F_1 = 5.7\ N\\\\F_2 = 1.9\ N

We have 2 unknowns (F_3 and b) and we have 2 equations.

Now we clear F_3 from the second equation and introduce it into the first equation.

F_3 = \frac{F_2}{cos (b)}

Then

-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°

Then we find the value of F_3

F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N

Finally the answer is 6.3 N at 162° counterclockwise from  

F1->

7 0
3 years ago
Define force and provide an example​
ollegr [7]

Answer:

force-strength,power or energy as an attribute of motion, movement or action. Example: Frictional force.

4 0
3 years ago
Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.300 m and carries a current
Marrrta [24]

Answer:

Explanation:

Since the wires attract each other , the direction of current will be same in both the wires .

Let I be current in wire which is along x - axis

force of attraction per unit length between the two current carrying wire is given by

\frac{\mu_0}{4\pi}  x \frac{2 I_1\times I_2}{d}

where I₁ and I₂ are currents in the wires and d is distance between the two

Putting the given values

285 x 10⁻⁶ = 10⁻⁷ x \frac{2\times25.5\times I_2}{.3}

I₂ = 16.76 A

Current in the wire along x axis is 16.76 A

To find point where magnetic field is zero due the these wires

The point will lie between the two wires  as current is in the same direction.

Let at y = y , the neutral point lies

k 2 x  \frac{16.76}{y} = k 2 x \frac{25.5}{.3-y}

25.5y = 16.76 x .3 - 16.76y

42.26 y = 5.028

y = .119

= .12 m

3 0
3 years ago
1. What did you observe about the magnitudes of the forces on the two charges? Were they the same or different? Does your answer
Aloiza [94]

Answer:

Following are the solution to the given question:

Explanation:

Its strength from both charges is equivalent or identical. The power is equal. And it is passed down

F=\frac{kq_1q_2}{r^2}

Therefore, the extent doesn't rely on the fact that charges are the same or different. Newton's third law complies with Electrostatic Charges due to a couple of charges. They are similar in magnitude, and they're in the other way.

|F_{12}| = |F_{21}|

7 0
3 years ago
What is the acceleration of a ball traveling horizontally with an initial velocity of 20 meters/second and, 2.0 seconds later, a
Nutka1998 [239]

The acceleration of the ball is 5 m/s^2. This can be calculated using a formula that relates the change in velocity, acceleration, and time. This formula is:

Vf = Vi + at

where:
Vf = final velocity
Vi = initial velocity
a = acceleration
t = time

Substituting the values gives:

30 = 20 + a(2)
<span>a = 5 m/s^2 --> Final Answer</span>
6 0
3 years ago
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