After the pendulum is dropped, determine the height at which the kinetic energy is equal to the potential energy.

PIT_PIT [208]

' A ' is one crest of the wave. After every wavelength, there's another one.

' B ' . . . the vertical arrow under B shows the amplitude of the wave

' C ' is one trough of the wave. After every wavelength, there's another one.

' D ' . . . the horizontal arrow over D shows the wavelength.

3 0

3 years ago

Which of these devices does NOT contain a magnet or electromagnet?

Black_prince [1.1K]

I think it is D a light bulb

7 0

3 years ago

Read 2 more answers

If you place a balloon in the freezer it will______

matrenka [14]

B. Deflate because volume is inversely proportional to temperature

8 0

3 years ago

Consider a composite cube made of epoxy with fibers aligned along one axis of the cube (the fibers are parallel to four of the t

coldgirl [10]

The question is incomplete. The complete question is :

Consider a composite cube made of epoxy with fibers aligned along one axis of the cube (the fibers are parallel to four of the twelve cube edges). If the cube can only be loaded in axial tension such that the force is uniformly applied over - and is normal to - a cube face, what is the lowest possible positive length change the cube can experience under this tension? The applied tensile force is 102 KN. The unloaded cube edge length is 56 mm. The glass fibers have an elastic modulus of 200 GPa. The epoxy has an elastic modulus of 38 GPa. The cube is comprised of 18 vol% epoxy (the balancing vol % is glass fiber). Hint: The loading axis is intentionally unspecified. Answer Format: Lowest possible length increase (change of length) under tension.

Solution :

Given :

$E_{glass fibre}$ = 200 GPa

$V_{glass fibre} = 82\%$

$E_{epoxy}$ = 38 GPa

$V_{epoxy} = 82\%$

Edge length = 56 mm

Cube is loaded in axial tension such that the force is uniformly applied over a cube face.

$E_{\text{composite}}=\frac{E_{glass fibre} \times E_{epoxy}}{(E_{glass fibre .E_{epoxy}})+(E_{fibre}.V_{glass fibre})}$

$E_{composite} = \frac{200 \times 38}{(200 \times 0.18)+(38\times 0.82)}$

$=113.16$ GPa

Applied stress $=\frac{\text{applied load}}{\text{area}}$

$\sigma=\frac{102 \times 10^3 \ N}{56 \times 56 \times 10^{-6} \ m^2}$

= 32.5 MPa

By Hooke's law

$\sigma = E . \epsilon$

$\sigma = E. \frac{\Delta l}{l}$

$\Delta l = \frac{\sigma}{E}\times l$

Length change, $\Delta l =\frac{32.5 \times 10^6 \ Pa}{113.16 \times 10^9 \ Pa}\times 56 \times 10^{-2} \ m$

$\Delta l = \frac{32.5 \times 56}{113.16} \times 10^{-3} \ mm$

= 0.016 mm

7 0

2 years ago

An experiment in your science class lists the materials needed for the lab. It is your job, as a lab partner, to measure out 25

andreyandreev [35.5K]

Overview:
You would use a graduated cylinder to measure 25 ml of the liquid and use a scale to weigh the amount of Magnesium (it can be a balance but most labs now have small scales) and you would carefully put small amounts of magnesium until it states 2.5 grams.

Procedure:
Take the graduated cylinder and place it on the table. Carefully take the water and fill the graduated cylinder up to 25ml. If you took to much just simply pour some out and if you didnt take enough then put more. After you measured the water then you can pour the 25 ml of water into a beaker, but this is totally optional.

For the magnesium first you put on the scale and set it to grams. Afterwards just take small pieces of the magnesium and place it on the scale until you reach 2.5 grams. Afterwards you can carefully place the magnesium in another beaker or something like that.

5 0

3 years ago