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2 years ago

An object weighing 2.7n while in air and 1.2n when completely immersed in water.find the relative density of the object

Physics

1 answer:

Elden [556K]2 years ago

8 0

Answer:

1.8 = relative density (there are no units for relative density)

Explanation:

It displaces water equal to it's volume and gets buoyancy equal to that amount of water

2.7 - 1.2 = 1.5 N of buoyancy

density of water = 1 gm /cc

1.5 N = m (9.81)

m of water displaced = .1529 kg

152.9 cc of water will produce this buoyancy....this is the volume of the object

find mass of object 2.7 = m (9.81) shows m = .2752 kg = 272.5 gm

density = mass/ volume = 272.5 / 152.9 = 1.8 gm / cc

Relative to water (which is 1 gm / cc) the relative density is 1.8

====> it is 1.8 times denser than water and will sink when in water....

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A snowball is thrown with an initial x velocity of 7.5 m/s and an initial y velocity of 8.4 m/s . Part A How much time is requir

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In order to calculate the time taken by the snowball to reach the highest point in its journey, we need to consider the variables along the y-direction.

Let us list out what we know from the question so that we can decide on the equation to be used.

We know that Initial Y Velocity $V_{iy}$ = 8.4 m/s

Acceleration in the Y direction $a_{y}$ = -9.8 m/ $s^{2}$ , since the acceleration due to gravity points in the downward direction.

Final Y Velocity $V_{fy}$ = 0 because at the highest point in its path, an object comes to rest momentarily before falling down.

Time taken t = ?

From the list above, it is easy to see that the equation that best suits our purpose here is $V_{fy} = V_{iy} + a_{y}t$

Plugging in the numbers, we get 0 = 8.4 - (9.8)t

Solving for t, we get t = 0.857 s

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4 years ago

A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10 m, and an

maksim [4K]

Answer:

The value is $\alpha = 0.002 Np/m$

Explanation:

From the question we are told that

The first amplitude of the wave is $E_{max}1 = 98.02 \ V/m$

The first depth is $D_1 = 10 \ m$

The second amplitude is $E_{max}2 = 81.87 \ (V/m)$

The second depth is $D_2 = 100 \ m$

Generally from the spatial wave equation we have

$v(x) = Ae^{-\alpha d}cos(\beta x + \phi_o)$

=> $\frac{v(x)}{v(x)} =\frac{ Ae^{-\alpha d}cos(\beta x + \phi_o)}{ Ae^{-\alpha d}cos(\beta x + \phi_o)}$

So considering the ratio of the equation for the two depth

$\frac{A}{A_S} = \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}$

=> $\frac{98.02}{81.87} = \frac{e^{-10 \alpha }}{e^{-100 \alpha }}$

=> $\alpha = \frac{0.18}{90}$

=> $\alpha = 0.002 Np/m$

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