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Karo-lina-s [1.5K]
2 years ago
12

Warm water is generally less dense than cold water. The mass of a ship increases as it is loaded with cargo. If a ship will be s

ailing through warm and cold water, people think about _____________as they load the ship with cargo.
Physics
2 answers:
KengaRu [80]2 years ago
6 0
If a ship will be sailing through warm and cold water, people think about making it less dense than the warmest water as they load the ship with cargo. I think you forgot to give the options along with the question. I hope that this is the answer that has actually come to your desired help.
7nadin3 [17]2 years ago
4 0

Answer: mass in relation to density

Explanation: the density of the water needs to be put into consideration because the mass of the ship is dependent on it

When the sheep reaches the warm water it will tend to sink if the mass of the ship is not considered with relation to the density

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devlian [24]

Answer:

d = 0.247 mm

Explanation:

given,

λ = 633 nm

distance from the hole to the screen = L = 4 m

width of the central maximum = 2.5 cm

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                                               y = 0.0125 m

For circular aperture

  sin \theta = 1.22\dfrac{\lambda}{d}

using small angle approximation

  \theta = \dfrac{y}{D}

now,

   \dfrac{y}{D} = 1.22\dfrac{\lambda}{d}

   y = 1.22\dfrac{\lambda\ D}{d}

   d = 1.22\dfrac{\lambda\ D}{y}

   d = 1.22\dfrac{633\times 10^{-9}\times 4}{0.0125}

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the diameter of the hole is equal to 0.247 mm

5 0
3 years ago
Which number below equals 129000? * *
solong [7]

Answer:

0.1

Explanation:

5 0
3 years ago
A 3.5 x 10-6 C charge is located 0.28 m from a 2.8 x 10-6 C charge. What is the magnitude of the force being exerted on the smal
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The electrostatic force between two charges is given by Coulomb's law:
F=k_e  \frac{q_1 q_2}{r^2}
where
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q1 is the first charge
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r is the separation between the two charges

By substituting the data of the problem into the equation, we can find the magnitude of the force between the two charges:
F=(8.99 \cdot 10^{9} Nm^2C^{-2} ) \frac{(3.5 \cdot 10^{-6} N/C)(2.8 \cdot 10^{-6}N/C)}{(0.28m)^2}=1.2 N
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Answer:

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