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3 years ago

The other name of eureka can

Physics

2 answers:

Softa [21]3 years ago

5 0

The other name is displacement vessels
(not sure tho)

34kurt3 years ago

3 0

Displacement vessels is correct

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Objects with masses of 130 kg and a 430 kg are separated by 0.300 m.(a) Find the net gravitational force exerted by these object

fredd [130]

Answer

given,

mass of object

m₁ = 130 Kg

m₂ = 430 Kg

distance between them = 0.3 m

a) net force when 35 kg is place in between them

$F = \dfrac{GMm}{R^2}$

now,

$F = - \dfrac{6.67 \times 10^{-11}\times 130 \times 30}{0.15^2} +\dfrac{6.67 \times 10^{-11}\times 430 \times 30}{0.15^2}$

$F = 3.12 \times 10^{-5}\ N$

Direction of force will be toward the mass of 430 Kg

b) position where force will be zero

$F =-\dfrac{6.67 \times 10^{-11}\times 130 \times 30}{(0.3-x)^2} +\dfrac{6.67 \times 10^{-11}\times 430 \times 30}{x^2} = 0$

$-\dfrac{130}{(0.3-x)^2} +\dfrac{430}{x^2} = 0$

$x^2- 0.6 x + 0.09=\dfrac{130}{430}x^2$

$0.7 x^2- 0.6 x + 0.09 =0$

solving the above equation

x = 0.1936 m

the distance of third mass will be at x = 0.1936 m from 430 Kg mass.

7 0

3 years ago

the density of ice is 917.what fraction of the volume of a piece of ice will be above the liquid when floating in fresh water

yulyashka [42]

Answer:

$8.3\,\%$ of that piece of ice would be above the freshwater. Assumptions:

the density of the ice is $\rho(\text{ice}) = 917\; \rm kg \cdot m^{-3}$ , and
the density of freshwater is $\rho(\text{water}) = 1.00 \times 10^3\; \rm kg \cdot m^{-3}$ .

Explanation:

The volume of that chunk of ice can be split into two halves: volume above water $V(\text{above})$ , and volume under water $V(\text{under})$ . The mass of the whole chunk of ice would be:

$m(\text{ice}) = \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under}))$ .

Let $g$ be the acceleration due to gravity. The gravity on the entire chunk of ice would be

$\begin{aligned}&W(\text{ice}) \\ &= m({\text{ice}}) \cdot g \\ &= \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) \cdot g\end{aligned}$ .

On the other hand, the size of buoyant force on an object is equal to the weight of the liquid that it displaces. That is: $F(\text{bouyancy}) = W(\text{water displaced})$ .

Recall that $V(\text{above})$ is the volume of the ice above the water, and $V(\text{under})$ is the volume of the ice under the water.

The mass of water displaced would be equal to:

$\begin{aligned}& m(\text{water displaced}) \\ &= \rho(\text{water}) \cdot V(\text{water displaced}) \\ &= \rho(\text{water}) \cdot V(\text{under})\end{aligned}$ .

The weight of that much water would be

$\begin{aligned} &W(\text{water displaced}) \\ &= m(\text{water displaced}) \cdot g \\ &= \rho(\text{water}) \cdot V(\text{under}) \cdot g \end{aligned}$ .

Apply the equation $F(\text{bouyancy}) = W(\text{water displaced})$ . The bouyant force on this chunk of ice would be equal to $\begin{aligned} &W(\text{water displaced}) = \rho(\text{water}) \cdot V(\text{under}) \cdot g \end{aligned}$ .

Since the ice is floating, the forces on it need to be balanced. In other words, $\begin{aligned}W(\text{ice}) &= F(\text{bouyancy}) \\ &= \rho(\text{water}) \cdot V(\text{under}) \cdot g\end{aligned}$ .

On the other hand, recall that

$\begin{aligned}&W(\text{ice}) = \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) \cdot g\end{aligned}$ .

Combine the two halves to obtain:

$\begin{aligned}& \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) \cdot g \\ &= W(\text{ice}) = \rho(\text{water}) \cdot V(\text{under}) \cdot g\end{aligned}$ .

$\begin{aligned}& \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) \cdot g = \rho(\text{water}) \cdot V(\text{under}) \cdot g\end{aligned}$ .

Divide both sides by $g$ (assume that $g \ne 0$ ) to obtain:

$\begin{aligned}& \rho(\text{ice}) \cdot (V(\text{above}) + V(\text{under})) = \rho(\text{water}) \cdot V(\text{under})\end{aligned}$ .

Rearrange to obtain:

$\begin{aligned}& \frac{V(\text{under})}{V(\text{above}) + V(\text{under})} = \frac{\rho(\text{water})}{\rho(\text{ice})}\end{aligned}$ .

However, the question is asking for $\displaystyle \frac{V(\text{above})}{V(\text{above}) + V(\text{under})}$ , the fraction of the volume above water. Note that

$\begin{aligned}& \frac{V(\text{under})}{V(\text{above}) + V(\text{under})} + \frac{V(\text{above})}{V(\text{above}) + V(\text{under})} = 1\end{aligned}$ .

Therefore,

$\begin{aligned} &\frac{V(\text{above})}{V(\text{above}) + V(\text{under})} \\ &= 1 - \frac{V(\text{under})}{V(\text{above}) + V(\text{under})} \\ &= 1 - \frac{\rho(\text{water})}{\rho(\text{ice})} = 1 - \frac{917}{10^3} = 0.083\end{aligned}$ .

That's equivalent to $8.3\,\%$ .

5 0

3 years ago

60 POINTS SO PLEASE HELP ME To add vectors in a straight line ______________ and ______________ are considered positive directio

tankabanditka [31]

Answer:

The answer to your question is

Explanation:

To add vectors in a straight line _right_________ and ___upwards_______ are considered positive directions. __left________ and __downwards____________ are considered negative directions.

I hope it helps you

7 0

3 years ago

a constant force acts on an object of mass 10 kg for a duration of 2 seconds it increases the object velocity from 4metre per se

inn [45]

Answer:

F=ma

a=v/t

a=vf-vi/t

a=8-4/2

a=4÷2

a=2msec*2

F=10×2

F=20N

3 0

3 years ago

Which of the following is not a function of PACs?

Katena32 [7]

Answer:

i think D

hope this helps

let me know if i'm wrong i will change the answer

Explanation:

5 0

3 years ago

Read 2 more answers