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Nostrana [21]
2 years ago
13

a wave in a rope is traveling at 6 m/s and at a frequency of 2 Hz. what is the wavelength of the wave producced

Physics
1 answer:
Tems11 [23]2 years ago
4 0

Answer:

3

Explanation:

You might be interested in
which force field can accelerate an electron, but never change its speed? a) electric field b) magnetic field c) both of these d
Drupady [299]

Option a; Electric field can accelerate an electron, but never change its speed

An electric field (also known as an E-field) is a physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them. It can also refer to the physical field of a charged particle system. Electric fields are created by electric charges and time-varying electric currents. Electric and magnetic fields are both aspects of the electromagnetic field, one of nature's four fundamental interactions (also known as forces). Electric fields are significant in many areas of physics and are used in electrical technology. In atomic physics and chemistry, for example, the electric field is the attractive force that holds the atomic nucleus and electrons together in atoms. It is also the driving force behind chemical bonds between atoms.

Learn more about Electric field here:

brainly.com/question/15800304

#SPJ4

4 0
1 year ago
You pull a solid nickel ball with a density of 8.91 g/cm3 and a radius of 1.40 cm upward through a fluid at a constant speed of
Sunny_sXe [5.5K]

Answer:

P = 1.090\,N

Explanation:

The constant speed means that ball is not experimenting acceleration. This elements is modelled by using the following equation of equilibrium:

\Sigma F = P - W + F_{D}

\Sigma F = P - \rho \cdot V \cdot g + c\cdot v = 0

Now, the exerted force is:

P = \rho \cdot V \cdot g - c\cdot v

The volume of a sphere is:

V = \frac{4\cdot \pi}{3}\cdot R^{3}

V = \frac{4\cdot \pi}{3}\cdot (0.014\,m)^{3}

V = 1.149\times 10^{-5}\,m^{3}

Lastly, the force is calculated:

P = (8910\,\frac{kg}{m^{3}} )\cdot (1.149\cdot 10^{-5}\,m^{3})\cdot (9.81\,\frac{m}{s^{2}} )+(0.950\,\frac{kg}{s})\cdot (0.09\,\frac{m}{s} )

P = 1.090\,N

5 0
3 years ago
If τ=r×F then F.τ is equal
Mademuasel [1]
Yes that is correct.
4 0
3 years ago
Find the value of currents through each branch
Irina-Kira [14]

Answer:

the branch currents are as follows:

  top left: I2 = 0.625 A

  middle left: I1 = 2.500 A

  bottom left: I1-I2 = 1.875 A

  top center: I2+I3 = 2.500 A

  bottom center: I2+I3-I1 = 0 A

  right: I3 = 1.875 A

Explanation:

You can write the KVL equations:

Top left loop:

  I2(4) +(I2 +I3)(2) +I1(1) = 10

Bottom left loop:

  (I1-I2)(4) +(I1-I2-I3)(2) +I1(1) = 10

Right loop:

  (I2+I3)(2) +(I2+I3-I1)(2) = 5

In matrix form, the equations are ...

  \left[\begin{array}{ccc}1&6&2\\7&-6&-2\\-2&4&4\end{array}\right]\cdot\left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}10\\10\\5\end{array}\right]

These equations have the solution ...

  \left[\begin{array}{c}I_1\\I_2\\I_3\end{array}\right] =\left[\begin{array}{c}2.500\\0.625\\1.875\end{array}\right]

This means the branch currents are as follows:

  top left: I2 = 0.625 A

  middle left: I1 = 2.500 A

  bottom left: I1-I2 = 1.875 A

  top center: I2+I3 = 2.500 A

  bottom center: I2+I3-I1 = 0 A

  right: I3 = 1.875 A

_____

This can be worked almost in your head by using the superposition theorem. When the 5V source is shorted, the 10V source is supplying (I1) to a circuit that is the 4 Ω and 2 Ω resistors in parallel with their counterparts, and that 2+1 Ω combination in series with 1 Ω for a total of a 4Ω load on the 10 V source. That is, I1 due to the 10V source is 2.5 A, and it is nominally split in half through the upper and lower branches of the circuit. There is no current flowing through the (shorted) 5 V source branch.

When the 10V source is shorted, the 5V source is supplying a 4 +4 Ω branch in parallel with a 2 +2 Ω branch, a total load of 8/3 Ω. This makes the current from that source (I3) be 5/(8/3) = 15/8 = 1.875 A. There is zero current from this source through the 1 Ω resistor.

Nominally, the current from the 5V source splits 2/3 through the 2 Ω branch and 1/3 through the 4 Ω branch.

Using superposition, I2 = I1/2 -I3/3 = (2.5 A/2) -(1/3)(15/8 A) = 0.625 A. This is the same answer as above, without any matrix math.

  (I1, I2, I3) = (2.5 A, 0.625 A, 1.875 A)

__

It helps to be familiar with the formulas for resistors in series and parallel.

8 0
3 years ago
A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand
enyata [817]

Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

v=gt

And the distance traveled downwards is:

\displaystyle y=\frac{gt^2}{2}

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

\displaystyle t=\sqrt{\frac{2y}{g}}

Replacing into the first equation:

\displaystyle v=g\sqrt{\frac{2y}{g}}

Rationalizing:

\displaystyle v=\sqrt{2gy}

Let's call v1 the final speed of the package dropped from a height H. Thus:

\displaystyle v_1=\sqrt{2gH}

Let v2 be the final speed of the package dropped from a height 4H. Thus:

\displaystyle v_2=\sqrt{2g(4H)}

Taking out the square root of 4:

\displaystyle v_2=2\sqrt{2gH}

Dividing v2/v1 we can compare the final speeds:

\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}

Simplifying:

\displaystyle v_2/v_1=2

The final speed of the second package is twice as much as the final speed of the first package.

5 0
3 years ago
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