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lapo4ka [179]
3 years ago
8

If 50 mL of each of the liquids in the answer choices were poured into a 250 mL beaker, which layer would be directly above a sm

all rubber ball with a density of 0.960 g/mL? A. sea water – density of 1.024 g/mL B. mineral oil – density of 0.910 g/mL C. distilled water – density of 1.0 g/mL D. petroleum oil – density of 0.820 g/mL
Physics
1 answer:
shutvik [7]3 years ago
6 0

Answer:

B. mineral oil – density of 0.910 g/mL.

Explanation:

Hello,

In this case, since the density is known as the degree of compactness a body has (mass in the occupied volume), the higher the density, the higher the weight of the body, therefore, if submerged into a liquid it could float if less dense than the liquid or sink if more dense than the liquid.

In such a way, since the rubber is more dense than mineral (0.960 g/mL > 0.910 g/mL) oil but less dense than distilled water (0.960 g/mL < 1.0 g/mL) we can say that B. mineral oil – density of 0.910 g/mL is directly above it when submerged.

Best regards.

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calculate the work done in kilo joules in lifting a mass of 20kg at steady velocity through a vertical height of 20m
bulgar [2K]

Answer:

\huge\boxed{\sf Work\ done = 4 kJ}

Explanation:

Since work done is in the form of potential energy, we will use the formula of potential energy here.

We know that,

<h3>P.E. = mgh </h3>

Where,

m = mass = 20 kg

g = acceleration due to gravity = 10 m/s²

h = vertical height = 20 m

So,

<h3>Work done = mgh</h3>

Work done = (20)(10)(20)

Work done = 4000 joules

Work done = 4 kJ

\rule[225]{225}{2}

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Explanation:

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3 years ago
A satellite is spinning at 6.0 rev/s. The satellite consists of a main body in the shape of a sphere of radius 2.0 m and mass 10
bearhunter [10]

Answer:

605447.7066 kgm²/s

Explanation:

m_1 = Mass of sphere = 10000 kg

m_2 = Mass of rod = 10 kg

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\omega=6\times 2\pi\\\Rightarrow \omega=37.69911\ rad/s

Angular momentum is given by

L=I\omega

Moment of inertia of the satellite is

I_s=\frac{2}{5}m_1r^2

Moment of antenna of the satellite is

I_a=\frac{1}{3}m_2l^2

The angular momentum of the system is

L=I_s\omega+I_a\omega\\\Rightarrow L=\left(\frac{2}{5}m_1r^2+2\times \frac{1}{3}m_2l^2\right)\omega\\\Rightarrow L=\left(\frac{2}{5}10000\times 2^2+2\times \frac{1}{3}\times 10\times 3^2\right)\times 37.69911\\\Rightarrow L=605447.7066\ kgm^2/s

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3 years ago
A capacitor is charges with 9.6 nC and has a 120 V potential difference between its terminals. Compute
IRINA_888 [86]

Explanation:

Q = CV

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