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3 years ago

What will happen to a periodic wave acted upon an external damping force?

Physics

2 answers:

Svetllana [295]3 years ago

6 0

It'll decrease the frequency

OverLord2011 [107]3 years ago

4 0

The wave will decrease in its frequency due to a disturbing force acting upon it.

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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 angular momentum conservation, there should be no external torque before and after

As the asteroid is travelling directly towards the center of the Earth, after impact ,it does not impose any torque on earth's rotation, 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

2 years ago

Please answer will give brainliest

laila [671]

3b. No

4a. 50N

I hope this helps

8 0

2 years ago

Keisha finds instructions for a demonstration on gas laws. 1. Place a small marshmallow in a large plastic syringe. 2. Cap the s

lana [24]

The correct answer is option C. This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger.

Keisha follows the instructions for a demonstration on gas laws.
1. Place a small marshmallow in a large plastic syringe.
2. Cap the syringe tightly.
3. Pull the plunger back to double the volume of gas in the syringe.

Now, this activity is being done at the same temperature, because there is no mention of the temperature change. Thus, when the plunger is pulled back, the volume doubles, so pressure will decrease. Therefore, This is a demonstration of Boyle’s law. As the volume increases, the pressure decreases, and the marshmallow will grow larger.

7 0

3 years ago

Read 2 more answers

Which can make 4 covalent bonds? Carbon, Silicon, or both?

grin007 [14]

Answer:

I would say both

Explanation:

Each silicon atom has four valence electrons which are shared, forming covalent bonds with the four surrounding Si atoms.

Carbon contains four electrons in its outer shell. Therefore, it can form four covalent bonds with other atoms or molecules. The simplest organic carbon molecule is methane (CH4), in which four hydrogen atoms bind to a carbon atom (Figure 1).

5 0

3 years ago

Which element on the Periodic Table has fewer than 8 electrons but still has a full valence shell?

lora16 [44]

The answer to this is Helium :) it's in the farthest right columb and is a noble gas.

please mark as brainliest!

6 0

2 years ago

Read 2 more answers