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adelina 88 [10]
3 years ago
5

A rock is suspended by a light string. When the rock is in air, the tension in the string is 39.2 N. When the rock is totally im

mersed in water, the tension is 28.4 N. When the rock is totally immersed in an unknown liquid, the tension is 18.6 N. What is the Density of the unknown liquid. -When I looked at this problem, I though we needed to know the volume of the rock. Can someone show me how to do it without the volume of this rock?
Physics
1 answer:
adell [148]3 years ago
6 0

Answer:

1.1 g/cm^3

Explanation:

weight in air, Wa = 39.2 N

weight in water, Ww = 28.4 N

weight in liquid, Wl = 18.6 N

Let d be the density of unknown liquid.

By using the formula of relative density

relative density of unknown liquid

                            = (wt. in air - wt. in water) / (wt. in water - wt. in liquid)

d = (Wa - Ww) /(Ww - Wl)

d = (39.2 - 28.4) / (28.4 - 18.6)

d = 10.8 / 9.8

d = 1.1

Thus, the density of unknown liquid is 1.1 g/cm^3.

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Much energy as would Microraptor gui have to expend to fly with a speed of 10 m/s for 1.0 minutes is 486 J.

The first step is to find the energy that Microraptor must release to fly at 10 m/s for 1.0 minutes. The energy that Microraptor must expend to fly can be found using the relationship between Power and Energy.

P = E/t

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Now, a minimum of 8.1 W is required to fly at 10 m/s. So, the energy expended in 1 minute (60 seconds) is

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The ray diagram shows a vase that is placed beyond the center of curvature of a concave mirror.
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In an aqueous solution where the H+ concentration is 1 x 10-6 M, the OH concentration must be:
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A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur
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Check the explanation

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use, 1/u + 1/v = 1/f

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v = -0.446 cm <<<<<---------------Answer

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The image is virtual

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given

R = 1.5 cm

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