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Dmitry_Shevchenko [17]
2 years ago
10

What is adolescent chastity?​

Physics
1 answer:
navik [9.2K]2 years ago
6 0

Answer:Adolescent chastity can be defined as the absolute abstinence from sex and anything related to sex, such as, kissing, touching, fondling

Explanation:

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Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu
vlada-n [284]

Answer:

\frac{1}{10}M

Explanation:

To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after

As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

⇒I_{1} \times W_{1} =I_{2} \times W_{2}

  • I_{1} is the moment of interia of earth before impact
  • W_{1} is the angular velocity of earth about an axis passing through the center of earth before impact
  • I_{2} is moment of interia of earth and asteroid system
  • W_{2} is the angular velocity of earth and asteroid system about the same axis

let  W_{1}=W

since \text{Time period of rotation}∝\frac{1}{\text{Angular velocity}}

⇒ if time period is to increase by 25%, which is \frac{5}{4} times, the angular velocity decreases 25% which is \frac{4}{5}  times

therefore W_{1} = \frac{4}{5} \times W_{1}

I_{1}=\frac{2}{5} \times M\times R^{2}(moment of inertia of solid sphere)

where M is mass of earth

           R is radius of earth

I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where M_{1} is mass of asteroid

⇒ \frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}+ M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})

\frac{1}{2} \times M\times R^{2}= (\frac{2}{5} \times M\times R^{2}+ M_{1}\times R^{2})

M_{1}\times R^{2}= \frac{1}{10} \times M\times R^{2}

⇒M_{1}=}\frac{1}{10} \times M

3 0
3 years ago
Two students make the following claims:
antiseptic1488 [7]

Answer:

E. Student 1 is correct, because as θ is increased, h is the same.

Explanation:

Here we have the object of a certain mass falling under gravity so the force acting on the it will depend on mass of the object and the acceleration due to gravity.

Mathematically:

F=m.g

As we know that the work done is evaluated as the force applied on a body and the displacement of the body in the direction of the force.

And for work we have:

W=F.s\cos\theta

where:

s= displacement of the object

\theta= angle between the force and displacement vectors

Given that the height of the object is same in each trail of falling object under the gravity be it a free-fall or the incline plane.

  • In case of free-fall the angle between the force is and the displacement is zero.
  • In case when the body moves along the inclined plane the force applied by the gravity is same because it depends upon the mass of the object. And the net displacement in the direction of the gravitational force is the height of the object which is constant in both the cases.

So, the work done by the gravitational force is same in the two cases.

6 0
3 years ago
Please someone help me
beks73 [17]
New substances are formed by chemical reactions. When elements react together to form compounds their atoms join to other atoms using chemical bonds. For example, iron and sulfur react together to form a compound called iron sulfide. Hopefully this will help you decide...
5 0
3 years ago
Tina drove 3 miles north and 4 miles west to get to her grandmother’s house. Find the distance and displacement.​
TEA [102]

Answer:

distance = 3 + 4 = 7 mi

displacement is √(3² + 4²) = 5 mi

Explanation:

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2 years ago
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