What is one result of meiosis?

#4 HELP PLEASE! Find Gravitational Potential Energy! THANK YOU!

Scilla [17]

Answer:

$\Delta GPE=-3.92\ 10^9\ J$

Explanation:

Gravitational Potential Energy

It's the capacity of an object to do work due to its relative height from a fixed reference point.

It's computed as

GPE=m.g.h

Where m is the mass of the object, h is its height and g is the acceleration of gravity, $g=9.8 m/sec^2$

The mass of water is given as

$m=8,000,000\ kg$

The height above the rocks is h=50 m. Let's compute the GPE

$GPE_1=(8,000,000\ kg)(9.8\ m/s^2)(50\ m)$

$GPE_1=3,920,000,000\ Joule$

It should be expressed in scientific notation

$GPE_1=3.92\ 10^9\ J$

The GPE at the bottom, where h=0

$GPE_2=0$

The change of gravitational potential energy is:

$\Delta GPE=GPE_2-GPE_1$

$\boxed{\Delta GPE=-3.92\ 10^9\ J}$

8 0

3 years ago

A cylindrical block of mass M=50kg and height h=0.2m is hanging on a rope and is in equilibrium. Any difference in atmospheric p

agasfer [191]

Answer:

$\Delta P = 1961.4\,Pa$

Explanation:

The difference of pressure is given by gauge pressure:

$\Delta P = \rho_{w}\cdot g \cdot \Delta h$

$\Delta P = \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.2\,m)$

$\Delta P = 1961.4\,Pa$

8 0

4 years ago

Read 2 more answers

How to find input force

vfiekz [6]

Answer: Calculate the work input in a pulley by using the previous readings in the correct mathematical equation. : Work (W) equals force (f) multiplied by distance (d), or W=fd. The work done by the pulley is the equation of weight (w) multiplied by height traveled (h).

Explanation:

6 0

3 years ago

The Hi line of the Balmer series is emitted in the transition from n = 3 to n = 2. Compute the wavelength of this line for l H a

irina1246 [14]

Answer:

$\lambda=550\ nm$

Explanation:

The Hi line of the Balmer series is emitted in the transition from n = 3 to n = 2 i.e. $n_i=3$ and $n_f=2$

The wavelength of Hi line of the Balmer series is given by :

$\dfrac{1}{\lambda}=R(\dfrac{1}{n_f}-\dfrac{1}{n_i})$

$\dfrac{1}{\lambda}=1.09\times 10^7\times (\dfrac{1}{2}-\dfrac{1}{3})$

$\lambda=5.50\times 10^{-7}\ m$

$\lambda=550\ nm$

So, the wavelength for this line is 550 nm. Hence, this is the required solution.

6 0

3 years ago

Which of the following statements are true for electric field lines? Check all that apply. Check all that apply. Electric field

Scilla [17]

Answer:

Electric field lines point away from positive charges and toward negative charges. True

Electric field lines are continuous; they do not have a beginning or an ending. False

Electric field lines can never intersect. True

Electric field lines are close together in regions of space where the magnitude the electric field is weak and are father apart where it is strong. False

At every point in space, the electric field vector at that point is tangent to the electric field line through that point. True

Explanation:

Electric field lines point away from positive charges and toward negative charges. Always the field lines go to negative charges and leave from positive charges.

Electric field lines are continuous; they do not have a beginning or an ending. False

Because the field lines starts at positive charges and ends in negative charges.

Electric field lines can never intersect. True

It can not intercept the field lines because in that point the the field would have two directions. Besides, in that point the real value of the field should be found adding both field lines.

Electric field lines are close together in regions of space where the magnitude the electric field is weak and are father apart where it is strong. False

This fact is opposite to that so the regions of space where the magnitude the electric field is weak the lines are father apart and where the field is strong the lines are close together.

At every point in space, the electric field vector at that point is tangent to the electric field line through that point. True

This statement correspond to the definition of the field line.

5 0

3 years ago