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STALIN [3.7K]
3 years ago
14

An object of mass 11kg is falling in air and experiences a force due to air resistance of 35N. Determine the magnitude of net fo

rce acting on the object.
Physics
1 answer:
ivanzaharov [21]3 years ago
4 0

Answer:

<em>The net force acting on the object is 72.8 Nw</em>

Explanation:

<u>Net Force in Free Falling Objects</u>

When an object is falling in the air, the default force applied to it is the gravitational force or its weight. If air resistance is considered, then the net force acting on the object must take into consideration two opposed forces

F_n=W-F_a

Where F_n is the net force, F_a is the air resistance, and W is the weigh of the object

W=mg

F_n=mg-F_a

F_n=(11)(9.8)-35=72.8\ N

The net force acting on the object is \boxed{72.8 Nw}

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An object with a mass of 2.0 kg accelerates 2.0 m/s 2when an unknown force is applied to it. What is the amount of force?
Musya8 [376]

Answer:

4 N

Explanation:

mass = 2 kg

acceleration = 2 m/s^2

Force = mass * acceleration

         = 2 *2

         = 4 N

5 0
3 years ago
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rjkz [21]

Answer:

The one you have selected is correct.  :)

Explanation:

7 0
2 years ago
Which would you expect to freeze faster pure water or salt solution why?
krek1111 [17]
Pure water.

A salt solution contains impurities whereas pure water will not contain any impurities.

Impurities increase the boiling point (freezing point) of a substance.

Thus, I would expect the pure water solution to freeze faster than the salt solution.
6 0
3 years ago
Give the relationship(s) for any pair of protons with the proper term(s). Label – your choice. A.Heterotopic B.Heterotopic, dias
Afina-wow [57]

Answer and Explanation

• Heterotopic protons are those that when substituted by the same substituent, are structurally different. They are not similar, diastereotopic or enantiotopic.

• Diastreotopic protons refers to two protons in a molecule which, if replaced by the same substituent, would generate compounds that are diastereomers. Diastereotopic groups are often, but not always, identical groups attached to the same atom in a molecule containing at least one chiral center.

For example, the two hydrogen atoms of the C3 carbon in (S)-2-bromobutane are diastereotopic (shown in the attached image). Replacement of one hydrogen atom with a bromine atom will produce (2S,3R)-2,3-dibromobutane. Replacement of the other hydrogen atom with a bromine atom will produce the diastereomer (2S,3S)-2,3-dibromobutane.

• Homotopic protons in a compound are equivalent protons. Two protons A and B are homotopic if the molecule remains the same (including stereochemically) when the protons are interchanged with some other atom (substituent) while the remaining parts of the molecule stay fixed. Homotopic atoms are always identical, in any environment.

For example, ethane, the two H atoms on C1 and C2 carbons on the same side (as shown in the attached image) are homotopic as they exhibit the phenomenon described above.

• Enantiotopic protons are two protons in a molecule which, if one or the other were replaced (by the same substituent), would generate a chiral compound. The two possible compounds resulting from that replacement would be enantiomers.

For example, in the attached image to this answer, the two hydrogen atoms attached to the second carbon in butane are enantiotopic. Replacement of one hydrogen atom with a bromine atom will produce (R)-2-bromobutane. Replacement of the other hydrogen atom with a bromine atom will produce the enantiomer (S)-2-bromobutane.

Hope this helps!!!

7 0
3 years ago
A square plate of side 9 m is submerged in water at an incline of 60∘ with the horizontal. Its top edge is located at the surfac
Tpy6a [65]

Answer:

The force on one side of  the plate is 3093529.3 N.

Explanation:

Given that,

Side of square plate = 9 m

Angle = 60°

Water weight density = 9800 N/m³

Length of small strip is

y=\dfrac{\Delta y}{\sin60}

y=\dfrac{2\Delta y}{\sqrt{3}}

The area of strip is

dA=\dfrac{9\times2\Delta y}{\sqrt{3}}

We need to calculate the force on  one side of  the plate

Using formula of pressure

P=\dfrac{dF}{dA}

dF=P\times dA

On integrating

\int{dF}=\int_{0}^{9\sin60}{\rho g\times y\times 6\sqrt{3}dy}

F=9800\times6\sqrt{3}(\dfrac{y^2}{2})_{0}^{9\sin60}

F=9800\times6\sqrt{3}\times(\dfrac{(9\sin60)^2}{2})

F=3093529.3\ N

Hence, The force on one side of  the plate is 3093529.3 N.

7 0
3 years ago
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