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3 years ago

Which is affected more by an earthquake, : deep sea or less deep sea???

Physics

2 answers:

lisabon 2012 [21]3 years ago

6 0

Answer:

deep sea will obviously be more affected because of the sea floor shaking

DaniilM [7]3 years ago

3 0

Answer:

deep see

because of the earth inside part highly pressure

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Nets used to catch falling boulders on the side of rocky hillside roadways

Natalija [7]

Answer:

more time to change the momentum of falling rocks

Explanation:

Momentum is related to "mass in motion." So, if an object is moving, then it has momentum as it has its mass in motion. The amount of momentum is dependent upon how much and how fast the object is moving.

If an object is moving slowly, it means that the object is losing momentum.

Nets used to catch falling boulders on the side of rocky hillside roadways are more effective than rigid fences because their breakage is reduced by more time to change the momentum of falling rocks.

3 0

4 years ago

Two plastic bowling balls, 1 and 2, are rubbed with cloth until they each carry a uniformly distributed charge of magnitude 0.50

galben [10]

Answer:

The right solution is " $4.5\times 10^{-10} \ Cm$ ".

Explanation:

Given that,

q = 0.50 nC

d = 900 mm

As we know,

⇒ $P=qd$

By putting the values, we get

⇒ $=0.50\times 900$

⇒ $=(0.50\times 10^{-9})\times 0.9$

⇒ $=4.5\times 10^{-10} \ Cm$

5 0

3 years ago

Read 2 more answers

The chart lists the masses of four planets.

notka56 [123]

Answer:

I didn't understand that all please tell me in Russia

4 0

3 years ago

If the ball leaves the projectile launcher at a speed of 2.2 m/s at an angle of 30ᴼ, and the projectile launcher is on a table a

IgorC [24]

Answer: 1.12 m

Explanation:

This situation is related to parabolic motion, hence we can use the following equations:

$y=y_{o}+V_{o}sin \theta t-\frac{g}{2}t^{2}$ (1)

$x=V_{o} cos \theta t$ (2)

Where:

$y=0 m$ is the ball final height (when it hits the ground)

$y_{o}=1.1 m$ is the ball initial height

$V_{o}=2.2 m/s$ is the initial velocity

$\theta=30\°$ is the angle at which the ball was launched

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the horizontal distance the ball travels

Rewriting (1) with the given values:

$0 m=1.1 m+(2.2 m/s)(cos 30\°)t-\frac{9.8 m/s^{2}}{2}t^{2}$ (3)

Multiplying all the eqquation by -1 and rearranging:

$4.9 m/s^{2} t^{2}-1.1 m/s t-1.1 m=0$ (4)

So, since we have a quadratic equation here (in the form of $0=at^{2}+bt+c$ , we will use the quadratic formula to find $t$ :

$t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ (5)

Where $a=4.9$ , $b=-1.1$ , $c=-1.1$

Substituting the known values and choosing the positive result of the equation, we have:

$t=\frac{-(-1.1)\pm\sqrt{(-1.1)^{2}-4(4.9)(-1.1)}}{2(4.9)}$

$t=0.59 s$ (6)

Now, substituting (6) in (2):

$x=(2.2 m/s)(cos 30\°)(0.59 s)$ (7)

$x=1.12 m$ (8) This is the horizontal distance at which the ball hits the ground.

3 0

4 years ago

Later research indicated that electric current is actually the flow of ____

iren [92.7K]

Answer:

Electrons.

Explanation:

Electricity was discovered before the discovery of electrons by J.J Thompson in 1896. Before the electron, it was thought that it is the positive ions that move through the wire and carry current—that's why today the conventional current represents the flow of positive charges.

After J.J Thompson's discovery of the electrons, it was realized that it is the electrons that actually carry the current through the conductor. But changing the direction of the conventional current didn't seem appropriate, and that's why the convention continues to be used to this day—reminding us that once it were the positive ions that were thought to carry the current.

6 0

3 years ago