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Feliz [49]
2 years ago
5

A transverse wave has a frequency of 200 Hz with a wavelength of 1.0 m. Determine the speed

Physics
1 answer:
Lana71 [14]2 years ago
7 0

Answer:

200 m\ s Ans .....

Explanation:

Data:

f = 200 Hz

w = 1.0 m

v = ?

Formula:

v = f w

Solution:

v = ( 200)(1.0)

v = 200 m\s <em>A</em><em>n</em><em>s</em><em> </em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

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In economics what goods and services will be produced?​
poizon [28]

Answer:A market economy is a system where the laws of supply and those of demand direct the production of goods and services. 1  Supply includes natural resources, capital, and labor. Demand includes purchases by consumers, businesses, and the government. Businesses sell their wares at the highest price consumers will pay.

Explanation:

5 0
3 years ago
A ball is dropped from a height of 20 meters. At what height does the ball have a velocity of 10 meters/second?
borishaifa [10]

Answer:B

Explanation:

Initial velocity, u=0m/s

Distance,s=20m

a=+g=9.8m/s*s

Using v*v=u*u+2gs

v*v=0+2*9.8*20

v*v=392

v=19.8

When s=20m, v = 19.8m/s

Therefore when v = 10m/s, s= 10*20/19.8

s =10.1m

6 0
3 years ago
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
The dot or scalar product of two (3d) vec- tors ⃗a = ⟨a1,a2,a3⟩ and ⃗b = ⟨b1,b2,b3⟩ is defined as
Neko [114]

Yes, yes, we know all of that.  It certainly took you long enough to
get around to asking your question.

If
     a = (14, 10.5, 0)
and
     b = (4.62, 9.45, 0) ,

then, to begin with, neither vector has a z-component, and they
 both lie in the x-y plane.

Their dot-product  a · b = (14 x 4.62) + (10.5 x 9.45) =

                                             (64.68)   +   (99.225)  =  163.905 (scalar)          


I feel I earned your generous 5 points just reading your treatise and
finding your question (in the last line).  I shall cherish every one of them.                     

7 0
3 years ago
The same pipe is used to carry both air and water. For the same fluid velocity and friction factor for the air and water flows:
VladimirAG [237]

Answer:

the pressure drop for the water flow is greater than that for the air flow.

Explanation:

Detailed analysis of the problem is show below.

4 0
3 years ago
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