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marysya [2.9K]
3 years ago
5

When you float an ice cube in water, you notice that 90% of it is submerged beneath the surface. Now suppose you put the same ic

e cube in a glass of some liquid whose density is less than that of water. How much of the ice cube will be submerged below the surface of this liquid? a. More than 90% b. 90% c. Less than 90%
Physics
1 answer:
Gwar [14]3 years ago
6 0

Answer:

option (c)

Explanation:

90% of the body is submerged in water.

Now it is immersed in an unknown liquid whose density is less than the density of water.

Buoyant force acting on the body depends on the volume immersed, density of liquid and gravity.

As the density of liquid is less than the density of water, so the buoyant force acting on the body by the unknown liquid is less than water. So it is submerged less than 90% in this liquid.

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(7)Figure 4 shows three charges: Q₁, Q₂ and Q3 . Determine the net force (Fnet) acting on Q3. (Hint: Draw a free body diagram of
NISA [10]

Remember Coulomb's law: the magnitude of the electric force F between two stationary charges q₁ and q₂ over a distance r is

F = \dfrac{kq_1q_2}{r^2}

where k ≈ 8,98 × 10⁹ kg•m³/(s²•C²) is Coulomb's constant.

8.1. The diagram is simple, since only two forces are involved. The particle at Q₂ feels a force to the left due to the particle at Q₁ and a downward force due to the particle at Q₃.

8.2. First convert everything to base SI units:

0,02 µC = 0,02 × 10⁻⁶ C = 2 × 10⁻⁸ C

0,03 µC = 3 × 10⁻⁸ C

0,04 µC = 4 × 10⁻⁸ C

300 mm = 300 × 10⁻³ m = 0,3 m

600 mm = 0,6 m

Force due to Q₁ :

F_{Q_2/Q_1} = \dfrac{k (6 \times 10^{-16} \,\mathrm C)}{(0,3 \, \mathrm m)^2} \approx \boxed{6,0 \times 10^{-5} \,\mathrm N} = 0,06 \,\mathrm{mN}

Force due to Q₃ :

F_{Q_2/Q_3} = \dfrac{k (12 \times 10^{-16} \,\mathrm C)}{(0,6 \, \mathrm m)^2} \approx \boxed{3,0 \times 10^{-5} \,\mathrm N} = 0,03 \,\mathrm{mN}

8.3. The net force on the particle at Q₂ is the vector

\vec F = F_{Q_2/Q_1} \, \vec\imath + F_{Q_2/Q_3} \,\vec\jmath = \left(-0,06\,\vec\imath - 0,03\,\vec\jmath\right) \,\mathrm{mN}

Its magnitude is

\|\vec F\| = \sqrt{\left(-0,06\,\mathrm{mN}\right)^2 + \left(-0,03\,\mathrm{mN}\right)^2} \approx 0,07 \,\mathrm{mN} = \boxed{7,0 \times 10^{-5} \,\mathrm N}

and makes an angle θ with the positive horizontal axis (pointing to the right) such that

\tan(\theta) = \dfrac{-0,03}{-0,06} \implies \theta = \tan^{-1}\left(\dfrac12\right) - 180^\circ \approx \boxed{-153^\circ}

where we subtract 180° because \vec F terminates in the third quadrant, but the inverse tangent function can only return angles between -90° and 90°. We use the fact that tan(x) has a period of 180° to get the angle that ends in the right quadrant.

8 0
2 years ago
A child with a mass of 23 kg rides a bike with a mass of 5.5 kg at a velocity of 4.5 m/s to the south. Compare the momentum of t
serg [7]

Answer:

Explanation:

Given the following data;

Mass of child = 23 kg

Mass of bike = 5.5 kg

Velocity = 4.5 m/s

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

Momentum = mass * velocity

To find the momentum of each of them;

I. Momentum of the child

Momentum C = mass * velocity

Momentum C = 23 * 4.5

Momentum C = 103.5 Kgm/s

II. Momentum of the bike

Momentum B = mass * velocity

Momentum B = 5.5 * 4.5

Momentum B = 24.75 Kgm/s

Hence, we can deduce from the calculations that the momentum of the child is greater than that of the bike because of the higher mass possessed by the child.

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How has two dogs go awnser my other qestion first one will get brainly anser
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Yes, yea, yep, you don't make much sense.

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Frick and Frack are standing back-to-back, leaning on each other, but not moving. If Frick weighs
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Answer:

Frick is pushing harder

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Why might scientists measure the mass of object rather than the weight of an object?
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