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Gemiola [76]
3 years ago
5

If the moon disappeared what effect would this have on the earths tides

Physics
2 answers:
iogann1982 [59]3 years ago
4 0
The moon has a huge impact on the tides. With the Moon gone, the oceans would become much calmer. The Sun still has its effect on them known as the solar tides. Surfers wouldn't be completely devoid of waves. 
Sliva [168]3 years ago
3 0

There would no longer be any tides.

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based on the creation of natural elements, what is most likely to be important to the modern iroquois culture?
Monica [59]

The Iroquois depended on the natural resources around them to meet all of their basic needs. Because they lived in the Eastern Woodlands of North America, their food, clothing, and shelter, as well as the materials for making their tools and weapons.Hunting tools were a bow and arrow, tomahawks and clubs. They ate corn, moose, beans and deer and many other foods. Beaver skin was used for capes.

They used reeds to make baskets and they used animal hides as clothing.

The Iroquois made weapons out of bones, wood, stones, and feathers. The Iroquois used wood, stone and feathers to make arrows.Hunting tools were a bow and arrow, tomahawks and clubs.

5 0
3 years ago
Two identical metal spheres A and B are in contact. Both are initially neutral. 1.0× 10 12 electrons are added to sphere A, then
Vilka [71]

Answer:

8 x 10^-8 C on both the spheres.

Explanation:

Number of electrons added to A = 1 x 10^12

As the sphere A and B are identical to each other so the charge is shared equally.

Charge of one electron = 1.6 x 10^-19 C

Charge on A after wards

= number of electrons after wards x charge of one electron

qA = 0.5 x 10^12 x 1.6 x 10^-19 = 8 x 10^-8 C

Similarly, the charge on sphere B afterwards

= number of electrons after wards x charge of one electron

qB = 0.5 x 10^12 x 1.6 x 10^-19 = 8 x 10^-8 C

8 0
4 years ago
Consider a turntable to be a circular disk of moment of inertia It rotating at a constant angular velocity ωi around an axis thr
Rom4ik [11]

Answer:

Note: Angular momentum is always conserved in a collision.

The initial angular momentum of the system is

L = ( It ) ( ωi )

where It = moment of inertia of the rotating circular disc,

ωi = angular velocity of the rotating circular disc

The final angular momentum is

L = ( It + Ir ) ( ωf )

where ωf is the final angular velocity of the system.

Since the two angular momenta are equal, we see that

( It ) ( ωi ) = ( It + Ir ) ( ωf )

so making ωf the subject of the formula

ωf = [ ( It ) / ( It + Ir ) ] ωi

Explanation:

7 0
3 years ago
The absolute pressure in water at a depth of 5m is read to be 145 kPa. Determine (a) the local atmospheric pressure, and (b) the
irga5000 [103]

Answer:

a) 95950 pascals

b) 137642.5 pascals

Explanation:

The absolute pressure (Pabs) on a fluid is:

P_{abs}=P_{gauge}+P_{atm} (1)

With Pgauge the pressure due depth on the fluid and Patm the atmospheric pressure. Pgauge is equal to:

P_{gauge}=\rho gh (2)

with ρ the fluid density, g the gravitational acceleration and h the depth on the fluid. Using (2) on (1) and solving for Patm:

P_{atm}=P_{abs}-P_{gauge}=P_{abs}-\rho_{water} gh

P_{atm}=(145000Pa)-(1000\frac{kg}{m^{3}})(9.81\frac{m}{s^{2}})(5m)

P_{atm}=95950Pa

b) Here we're going to use again (1) but now we have another value of density because it's other liquid, to know that value we should use the fact that specific gravity (S.G) for liquids is the ratio between fluid density and water density:

S.G=\frac{\rho_{fluid}}{\rho_{water}}

\rho_{liquid}=S.G*\rho_{water}

\rho_{liquid}=(0.85)*(1000\frac{kg}{m^{3}})=850\frac{kg}{m^{3}}

so:

P_{abs}=\rho_{liquid} gh+P_{atm}=(850\frac{kg}{m^{3}})(9.81\frac{m}{s})(5m)+95950Pa

P_{abs}=137642.5 Pa

3 0
3 years ago
Which of the following is not a good approximation of the relationship between independent and dependent variables? A. independe
KengaRu [80]

c would be the right answer for this problem

7 0
4 years ago
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