how much gravitational potential energy does a 5 kg rock have if is sitting on the edge of a cliff that is 10m high

Prisms separate <br> light, such as that from the Sun, by wavelength

Sorry I'm so late, but I just took this test and the answer is white (for people who didn't study well ;) )

6 0

2 years ago

Calculate the change in length of a Pyrex glass dish (Coefficient of linear expansion for Pyrex is 3 * 10-6 / oC) that is 0.25 m

DIA [1.3K]

9* $10^{-5}$ m

Explanation:

Step 1:

We are given the initial length of the Pyrex glass dish at a particular temperature and need to calculate the change in the length when the temperature changes by 120° C. The coefficient of linear expansion of Pyrex is provided.

Step 2:

Change in length = Coefficient of linear expansion * Change in temperature * Initial length

Step 3:

Coefficient of linear expansion = 3* $10^{-6}$ /°C

Change in temperature = 120°C = 120 K

Initial length = 0.25 m

Step 4:

Change in length = 3* $10^{-6}$ * 120 * 0.25 = 9* $10^{-5}$ m

8 0

3 years ago

Environment A is warm and gets lots of rain. Environment B is warmer and gets more rain year round. Identify Environment A.(2 po

zlopas [31]

The Swamp
There are many websites that say the rainforest but the rainforest is warmer and gets rain year round and swamps are warm and gets lots of rain but not year round.

5 0

2 years ago

In this graph, what is the displacement of the particle in the last two seconds?

nikklg [1K]

In this graph, what is the displacement of the particle in the last two seconds?of the particle in the last two seconds?

<span>0.2 meters
2 meters
4 meters
6 meters</span>
In this graph, the displacement of the particle in the last two seconds is 2 meters.

7 0

3 years ago

A circular cross section, d = 25 mm, experiences a torque load, T = 25 N·m, and a shear force, V = 85 kN. Calculate the shear st

Maru [420]

Answer:

The correct answer is 231 Mpa i.e option a.

Explanation:

using the equation of torsion we Have

$\frac{T}{I_{p}}=\frac{\tau }{r}\\\\\therefore \tau =\frac{T}{I_{p}}\times r$

where,

$\tau$ = shear stress at a distance 'r' from the center

T = is the applied torque

$I_{p}$ = polar moment of inertia of the section

r = radial distance from the center

Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.

We know that in case of a circular section the maximum shearing stresses due to a shear force occurs at the center and equals

$\tau _{max}=\frac{4}{3}\times \frac{V}{A}$

Applying values we get

$\tau _{max}=\frac{4}{3}\times \frac{85\times 10^{3}}{0.25\times \pi \times (25\times 10^{-3})^{2}}\\\\\therefore \tau _{max}=230.88Mpa\approx 231Mpa$

3 0

2 years ago