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RideAnS [48]
3 years ago
15

Objects 1 and 2 attract each other with a electrostatic force

Physics
2 answers:
daser333 [38]3 years ago
6 0

Electrostatic force changes like the inverse square of the distance (just like gravity).

If you double the distance, you change the force to 1/4 of what it used to be.

After the move, Objects 1 and 2 attract each other with a force of (18/16) = 1.125 units .

horrorfan [7]3 years ago
5 0

Answer:

1.125

Explanation:

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The moon completes one (circular) orbit of the earth in 27.3 days. The distance from the earth to the moon is 3.84×108 m. What i
I am Lyosha [343]

Answer:

2.72\cdot 10^{-3} m/s^2

Explanation:

Let's start by calculating the angular velocity of the Moon. We know that the period is:

T=27.3 d \cdot 24 \cdot 60 \cdot 60 =2.36\cdot 10^6 s

So now we can calculate its angular velocity:

\omega=\frac{2\pi}{T}=\frac{2\pi}{(2.36\cdot 10^6)}=2.66 \cdot 10^{-6} rad/s

The centripetal acceleration is given by

a=\omega^2 r

where

\omega=2.66\cdot 10^{-6}rad/s

r=3.84\cdot 10^8 m is the radius of the orbit

Substituting,

a=(2.66\cdot 10^{-6})^2(3.84\cdot 10^8)=2.72\cdot 10^{-3} m/s^2

4 0
3 years ago
Suppose a ray of light traveling in a material with an index of refraction n a reaches an interface with a material having an in
KATRIN_1 [288]

Answer: C and D

Explanation: One of the first rule for total internal reflection to occur is that the ray must move from a dense to a less dense medium, hence refractive index of medium a must be greater than that of b.

When a ray moves from a dense to a less dense medium, the refracted ray moves away from the normal thus increasing the size of the angle of refraction (total internal refraction occurs when the angle of refraction is 90° and the angle of incidence at this point is known as the critical angle), hence the angle of incidence must be greater than the critical angle.

These points verifies option C and D

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3 years ago
PLS HELP FOR THIS PHYSICS TEST <br><br> (pls only answer if u rlly know, this is an important test)
lesya692 [45]

Answer:

C. 8.01 m/s²

Explanation:

vf²= vi² + 2 • a • d

2ad = vf² - vi²

a = (vf²- vi²)/2d

d=25.00 -5.00=20.00 m

vi =0

vf=17.90 m/s

a =(17.90² -0²)/(2*20) = 8.01 m/s²

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3 years ago
- When an object is dropped what happens to potential and kinetic energy as it falls?
aev [14]
The answer is D: the GPE goes down and the KE goes up
7 0
3 years ago
Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton
solmaris [256]

Answer:

a

\lambda = 3.68 *10^{-36} \  m

b

\lambda_p = 1.28*10^{-14} \ m

Explanation:

From the question we are told that

   The mass of the person is  m =  180 \  kg

    The speed of the person is  v  =  1 \  m/s

    The energy of the proton is  E_ p =  5 MeV = 5 *10^{6} eV  = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \  J

Generally the de Broglie wavelength is mathematically represented as

      \lambda = \frac{h}{m * v }

Here  h is the Planck constant with the value

      h = 6.62607015 * 10^{-34} J \cdot s

So  

     \lambda = \frac{6.62607015 * 10^{-34}}{ 180  * 1  }

=> \lambda = 3.68 *10^{-36} \  m

Generally the energy of the proton is mathematically represented as

         E_p =  \frac{1}{2}  *   m_p  *  v^2_p

Here m_p  is the mass of proton with value  m_p  =  1.67 *10^{-27} \  kg

=>     8.0*10^{-13} =  \frac{1}{2}  *   1.67 *10^{-27}  *  v^2

=>   v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }

=>   v = 3.09529 *10^{7} \  m/s

So

        \lambda_p = \frac{h}{m_p * v_p }

so    \lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }

=>     \lambda_p = 1.28*10^{-14} \ m

     

5 0
3 years ago
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