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Tomtit [17]
3 years ago
14

A coin is placed at a depth of 15 cm in a beaker from the surface of water. Therefractive index of water is 4/3.Calculate height

through which the image of the coin is raised.​
Physics
1 answer:
Whitepunk [10]3 years ago
4 0

I found this answer on another website.

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There are six books in a stack, and each book weighs 5 N. The coefficient of static friction between the books is 0.2. With what
NikAS [45]

Answer:

b. 5 N

Explanation:

Each book weighs 5 N.  Therefore, five books weigh 25 N.  The friction force is:

Ff = Fn μ

Ff = (25 N) (0.2)

Ff = 5 N

3 0
3 years ago
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Which of the following statements are true? (mark all that apply)
Savatey [412]

Answer:

The answer(s) for this question are as followed: A, C, & D

Explanation:

I hope this helped, let me know if i missed any.

8 0
3 years ago
HELP!! ILL GIVE BRAINLIEST As we've learned so far, the popular model of the solar system shifted over
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3 years ago
A straightforward method of finding the density of an object is to measure its mass and then measure its volume by submerging it
Thepotemich [5.8K]

Answer:

<em>The density of rock = 3.37 g/cm³</em>

Explanation:

Density: Density can be defined as the ratio of the mass of a body to the volume. The S.I unit of density is kg/m³. It can be expressed mathematically as ,

D = M/V............................................... Equation 1

Where D = density of the body, M = mass of the body, V = volume of the body.

From Archimedes' principle, a body will displace a volume of water equal to  its volume.

Therefore, Volume of the object = volume of water displaced

<em>Given: M = 300 g, V = volume of water displace = 89.0 cm³.</em>

<em>Substituting these values into equation 1</em>

<em>D = 300/89</em>

<em>D = 3.37 g/cm³</em>

<em>The density of rock = 3.37 g/cm³</em>

4 0
4 years ago
The uncertainty in the position of an electron along an x axis is given as 68 pm. What is the least uncertainty in any simultane
umka21 [38]

Answer:

\Delta p_x=7.75\times 10^{-25}\ kg-m/s

Explanation:

Given that,

The uncertainty in the position of an electron along the x-axis is, \Delta x=68\ pm=68\times 10^{-12}\ m

We need to find the east uncertainty in any simultaneous measurement of the momentum component of this electron.

We know that the Heisenberg's uncertainty principle gives the relation between the uncertainty in position and the momentum of electron as :

\Delta p_x{\cdot}\Delta x\ge \dfrac{h}{4\pi }

Putting all the values, we get :

\Delta p_x{\cdot}\ge \dfrac{h}{4\pi \Delta x}\\\\\Delta p_x \ge \dfrac{6.63\times 10^{-34}}{4\pi \times 68\times 10^{-12}}\\\\\Delta p_x\ge 7.75\times 10^{-25}\ kg-m/s

So, the momentum component of this electrons is greater than 7.75\times 10^{-25}\ kg-m/s.

6 0
4 years ago
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