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KiRa [710]
2 years ago
5

Rewrite the following sentence in the negative form >Each library contains 3000 brand new books

Physics
1 answer:
otez555 [7]2 years ago
3 0
None of the libraries have up to 3000 brand new books
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Two positive charges of magnitude 20 micro C and 100 micro C are placed at a distance of 150cm. Calculate the force of repulsion
Luba_88 [7]

first \: positive \: charge = q1 = 20

1micro \: =   {1}^{ - 6}

1q = 20 \times 10 {}^{ - 6}

second \: charge = q2 = 100 \times 10 {}^{ - 6}

formula = f =  \frac{k \times q1 × q2}{d {}^{2} }

remember

k = 9 \times  {10}^{9}

change distance 150cm to 1.5m

putting values

f =\frac{9 \times 10 {}^{9} \times 20  \times 10 {}^{ - 6} \times 100 \times  {10}^{ - 6}   }{1.5 {}^{2} }

<h2 /><h3>answer </h3><h3><u>8</u><u>N</u></h3>
4 0
3 years ago
The mass of a golf ball is 45.9 g . if it leaves the tee with a speed of 62.0 m/s , what is its corresponding wavelength?
inn [45]

The wavelength of the golf ball is <u>2.328×10⁻³⁴m.</u>

All moving particles with mass have a matter wave associated with it. These matter waves are called deBroglie waves.

The deBroglie wavelength λ of a particle is given by,

\lambda=\frac{h}{mv}

Here, h is the Planck's constant, m is the mass of the ball and v is its velocity.

Calculate the deBroglie wavelength of the moving golf ball by substituting 6.626×10⁻³⁴J s for h, 45.9×10⁻³kg for m and 62.0 m/s for v.

\lambda=\frac{h}{mv}\\ =\frac{6.626*10^-^3^4Js}{(45.9*10^-^3kg)(62.0m/s)} \\ =2.328*10^-^3^4m

The wavelength of the golf ball is  <u>2.328×10⁻³⁴m.</u>




4 0
4 years ago
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

<h3>200 Hz</h3>

Hope this helps you

7 0
4 years ago
A high fountain of water is located at the center of a circular pool. Not wishing to get his feet wet, a student walks around th
Alisiya [41]

Answer:

The fountain is 3.43 m high.

Explanation:

Circumference of the pool = 15 m.

C = 2\pir

where C is the circumference and r its radius.

r = \frac{C}{2\pi }

 = \frac{15}{2(\frac{22}{7}) }

r = 2.3864

radius of the pool = 2.40 m

So that the height of the fountain, h, can be determined by applying trigonometric function.

Tan θ = \frac{opposite}{adjacent}

Tan 55 = \frac{h}{2.4}

h = Tan 55 x 2.4

  = 1.4282 x 2.4

  = 3.4277

h = 3.43 m

The height of the fountain is 3.43 m.

8 0
3 years ago
How dose an exam question outed from text book​
Verizon [17]

Answer:

In which school you are???

Explanation:

4 0
3 years ago
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