All categories

3 years ago

What would happen to ocean tides on earth if earth stopped rotating and why?

Physics

1 answer:

saul85 [17]3 years ago

7 0

We would lose tides. We'd still have some from the moon but it'd be a few week thing instead of everyday.

You might be interested in

There are two parallel conductive plates separated by a distance d and zero potential. Calculate the potential and electric fiel

taurus [48]

Answer:

The total electric potential at mid way due to 'q' is $\frac{q}{4\pi\epsilon_{o}d}$

The net Electric field at midway due to 'q' is 0.

Solution:

According to the question, the separation between two parallel plates, plate A and plate B (say) = d

The electric potential at a distance d due to 'Q' is:

$V = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d}$

Now, for the Electric potential for the two plates A and B at midway between the plates due to 'q':

For plate A,

$V_{A} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Similar is the case with plate B:

$V_{B} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{\frac{d}{2}}$

Since the electric potential is a scalar quantity, the net or total potential is given as the sum of the potential for the two plates:

$V_{total} = V_{A} + V_{B} = \frac{1}{4\pi\epsilon_{o}}.q(\frac{1}{\frac{d}{2}} + \frac{1}{\frac{d}{2}}$

$V_{total} = \frac{q}{4\pi\epsilon_{o}d}$

Now,

The Electric field due to charge Q at a distance is given by:

$\vec{E} = \frac{1}{4\pi\epsilon_{o}}.\frac{Q}{d^{2}}$

Now, if the charge q is mid way between the field, then distance is $\frac{d}{2}$ .

Electric Field at plate A, $\vec{E_{A}}$ at midway due to charge q:

$\vec{E_{A}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Similarly, for plate B:

$\vec{E_{B}} = \frac{1}{4\pi\epsilon_{o}}.\frac{q}{(\frac{d}{2})^{2}}$

Both the fields for plate A and B are due to charge 'q' and as such will be equal in magnitude with direction of fields opposite to each other and hence cancels out making net Electric field zero.

3 0

3 years ago

A spherical Christmas tree ornament is 8.00 cm in diameter. What is the magnification of an object placed 12.0 cm away from the

LiRa [457]

The magnification of the ornament is 0.25

To calculate the magnification of the ornament, first, we need to find the image distance.

Formula:

1/f = u⁻¹+v⁻¹.................... Equation 1

Where:

f = Focal length of the ornament
u = image distance
v = object distance.

make u the subject of the equation

u = fv/(f+v)................ Equation 2

From the question,

Given:

f = 8/2 = 4 cm
v = 12 cm

Substitute these values into equation 2

u = (12×4)/(12+4)
u = 48/16
u = 3 cm.

Finally, to get the magnification of the ornament, we use the formula below.

M = u/v.................. Equation 3

Where

M = magnification of the ornament.

Substitute these values above into equation 3

M = 3/12
M = 0.25.

Hence, The magnification of the ornament is 0.25

8 0

3 years ago

Which characteristic do all minerals have in common?

Korolek [52]

Answer:

Five Characteristics of a Mineral

Minerals Are Natural. You must find minerals in nature; substances concocted in laboratories don't qualify. ...

Minerals Are Inorganic. ...

Minerals Are Solids. ...

Definite Chemical Composition. ...

Crystalline Structure.

These are just some examples.

Explanation:

7 0

3 years ago

Read 2 more answers

How many sig figs are in 3252.6

jekas [21]

The number 3252.6 has 5 significant figures

7 0

3 years ago

Which of the following is a heterogeneous mixture?

Reil [10]

That would be Italian dressing because it does not mix completely therefore it is the definition of a heterogeneous mixture

8 0

3 years ago