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nlexa [21]
3 years ago
9

Help plz i have until 4.20 plz

Physics
2 answers:
Setler [38]2 years ago
8 0

Answer:

Substance 1 because it became a liquid faster

Explanation:

Andrej [43]2 years ago
7 0
Substance 1 because it became a liquid faster
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)If a force of 5.00 N is needed to open a 90.0 cm wide door when applied at the edge opposite the hinges, what force must be app
masha68 [24]

Answer:

A force of 12.857 newtons must be applied to open the door.

Explanation:

In this case, a force is exerted on the door, a moment is performed and the door is opened. If moment remains constant, the force is inversely proportional to distance respect to axis of rotation passing through hinges. That is:

F \propto \frac{1}{r}

F = \frac{k}{r} (Eq. 1)

Where:

F - Force, measured in newtons.

k - Proportionality ratio, measured in newton-meters.

r - Distance respect to axis of rotation passing through hinges, measured in meters.

From (Eq. 1) we get the following relationship and clear the final force within:

F_{A}\cdot r_{A} = F_{B}\cdot r_{B}

F_{B}=\left(\frac{r_{A}}{r_{B}} \right)\cdot F_{A}(Eq. 2)

Where:

F_{A}, F_{B} - Initial and final forces, measured in newtons.

r_{A}, r_{B} - Initial and final distances, measured in meters.

If we know that F_{A} = 5\,N, r_{A} = 0.9\,m and r_{B} = 0.35\,m, then final force is:

F_{B}= \left(\frac{0.9\,m}{0.35\,m} \right)\cdot (5\,N)

F_{B} = 12.857\,N

A force of 12.857 newtons must be applied to open the door.

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How many electrons can a nucleas hold
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Give an example of a reputable website
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Read 2 more answers
a uniform rod is hung at onen end and is partially submerged in water. If the density of the rod is 5/9 than of wter, find the f
34kurt

Answer:

    \frac{h_{liquid} }{ h_{body} } = 5/9

Explanation:

This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.

         B = ρ_liquid  g V_liquid

let's write the translational equilibrium condition

         B - W = 0

let's use the definition of density

        ρ_body = m / V_body

        m = ρ_body  V_body

        W = ρ_body  V_body  g

we substitute

          ρ_liquid  g  V_liquid = ρ_body  g  V_body

          \frac{\rho_{body}   }{\rho_{liquid} } } =  \frac{V_{liquid}   }{V_{body} } }

In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar

          V = A h_bogy

Thus

          \frac{V_{liquid} }{V_{1body} } = \frac{ h_{liquid} }{h_{body} }

we substitute

           5/9 = \frac{h_{liquid} }{ h_{body} }

8 0
3 years ago
A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa
bearhunter [10]

Explanation:

Formula which holds true for a leans with radii R_{1} and R_{2} and index refraction n is given as follows.

          \frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

Since, the lens is immersed in liquid with index of refraction n_{1}. Therefore, focal length obeys the following.  

            \frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]  

             \frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]

and,       \frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}

or,          f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}

              f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}

                          = 32.4 cm

Using thin lens equation, we will find the focal length as follows.

             \frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}

Hence, image distance can be calculated as follows.

       \frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}

              s_{i} = \frac{fs_{o}}{s_{o} - f}

             s_{i} = \frac{32.4 \times 100}{100 - 32.4}

                       = 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

4 0
3 years ago
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