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2 years ago

When the serving team makes a mistake giving the other team a point and the serve

Physics

1 answer:

taurus [48]2 years ago

4 0

Answer : whats the question

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What is the frequency of a wave that has a wavelength of 0.39 m and a speed

gogolik [260]

Answer:

<h3>The answer is option B</h3>

Explanation:

The frequency of a wave can be found by using the formula

$f = \frac{c}{ \lambda} \\$

where

c is the velocity

From the question

wavelength = 0.39 m

c = 86 m/s

We have

$f = \frac{86}{0.39} \\ = 220.512820...$

We have the final answer as

Hope this helps you

7 0

3 years ago

If potential energy at point A is 12 joules, what is the potential energy at D?

Irina-Kira [14]

Answer:

In D: 3J

Explanation:

Potential energy: Ep=mgh where m is the mass, h altitude.

In point A: h=20cm=0.2m

Epa=12=0.2×mg. Thus mg=12/0.2=60N

For point D: hd=5cm=0.05m

Epd=mg×0.05=60×0.05=3J

5 0

3 years ago

7–¹ plsss pa answer mga lods

Rasek [7]

Amigo no se lo que ablas

3 0

3 years ago

A soap bubble is essentially a thin film of water surrounded by air. The colors you see in soap bubbles are produced by interfer

drek231 [11]

Answer:

Violet

Explanation:

Wavelength spectrum of visible light for the range of 390nm to 460nm is seen as violet. So the violet colour will be seen and reflected as what the eyes will see

6 0

3 years ago

Two charges are located in the x x – y y plane. If q 1 = − 2.75 nC q1=−2.75 nC and is located at ( x = 0.00 m , y = 0.600 m ) (x

ludmilkaskok [199]

Answer:

$\vec{E}=(-15.78\hat{i}+63.81\hat{j})\frac{N}{C}$

Explanation:

In this case we have to work with vectors. Firs of all we have to compute the angles between x axis and the r vector (which points the charges):

$\theta_1=tan^{-1}(\frac{0.6m}{0m})=90\° \\\\\theta_2=tan^{-1}(\frac{0.4m}{1.3m})=17.10\°$

the electric field has two components Ex and Ey. By considering the sign of the charges we obtain that:

$\vec{E} = E_x \hat{i}+E_y\hat{j}\\\\\vec{E}=(E_1cos\theta_1-E_2cos\theta_2)\hat{i}+(E_1sin\theta_1-E_2sin\theta_2)\hat{j}\\\\E_1=k\frac{q}{r_1^2}=(8.99*10^{9}\frac{N}{m^2C^2})\frac{2.75*10^{-9}C}{(0.6m)^2}=68.67\frac{N}{C}\\\\E_2=k\frac{q}{r_2^2}=(8.99*10^{9}\frac{N}{m^2C^2})\frac{3.40*10^{-9}C}{((1.3m)^2+(0.4m)^2)}=16.52\frac{N}{C}$

Hence, by replacing E1 and E2 we obtain:

$\vec{E}=[(68.67N/C)cos(90\°)-(16.52N/C)cos(17.10\°)]\hat{i}+[(68.67N/C)sin(90\°)-(16.52N/C)sin(17.10\°)]\hat{j}\\\\\vec{E}=(-15.78\hat{i}+63.81\hat{j})\frac{N}{C}$

hope this helps!!

3 0

3 years ago