Consider a Cassegrain-focus, reflecting telescope. Images recorded at Cassegrain-focus will be:

( Apply Concepts ) Nome , Alaska, lies at 64°N latitude. San Diego , California , lies at 32°N latitude. Which city receives mor

tatyana61 [14]

If your asking about the total amount of time the sun is visible, it will be equal over the course of a year. In Nome, Alaska, it will be daylight almost 24 hours a day in the summer and be dark almost 24 hours a day in the winter. In San Diego they are much closer to the equator and will receive sunlight roughly the same amount each day all year long. In the end these times will add up to the same amount.

However if you are asking about intensity of sun light received, San Diego will get considerably more since its near the equator and its getting almost direct sunlight. In Nome, the sunlight is hitting the earth at an angle so fewer rays of light are hitting the ground per square foot.

6 0

4 years ago

Two blocks are connected by a light string that passes over two frictionless pulleys. The block of mass m2 is attached to a spri

irina1246 [14]

(BELOW YOU CAN FIND ATTACHED THE IMAGE OF THE SITUATION)

Answer:

$d=\frac{2g(m1-m2)}{k}$

Explanation:

For this we're going to use conservation of mechanical energy because there are nor dissipative forces as friction. So, the change on mechanical energy (E) should be zero, that means:

$E_{i}=E_{f}$

$K_{i}+U_{i}=K_{f}+U_{f}$ (1)

With $K_{i}$ the initial kinetic energy, $U_{i}$ the initial potential energy, $K_{f}$ the final kinetic energy and $U_{f}$ the final potential energy. Note that initialy the masses are at rest so $K_{i} = 0$ , when they are released the block 2 moves downward because m2>m1 and finally when the mass 2 reaches its maximum displacement the blocks will be instantly at rest so $K_{f} =0$ . So, equation (1) becomes:

$U_{i}=U_{f}$ (2)

At initial moment all the potential energy is gravitational because the spring is not stretched so $U_{i}=U_{gi}$ and at final moment we have potential gravitational energy and potential elastic energy so $U_{f}=U_{gf}+U_{ef}$ , using this on (2)

$U_{gi}=U_{gf}+U_{ef}$ (3)

Additional if we define the cero of potential gravitational energy as sketched on the figure below (See image attached), $U_{gi}=0$ and we have by (3) :

$0= U_{gf}+U_{ef}$ (4)

Now when the block 1 moves a distance d upward the block 2 moves downward a distance d too (to maintain a constant length of the rope) and the spring stretches a distance d, so (4) is:

$0=-m1gd+m2gd+\frac{kd^{2}}{2}$