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goblinko [34]
3 years ago
5

Which region of a longitudinal wave is this?

Physics
1 answer:
ipn [44]3 years ago
8 0
I think the correct answer is is D.
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Which scenario involves kinetic energy transforming into potential energy? A. a book lying on a shelf B. a train traveling at a
NeX [460]
The answer is C.

The Kinetic energy which was exerted and experience pulling the string of a bow is kept as a potential energy at the end of the arrow in contact with the string. Once release from aim at stationary position the potential energy is again transformed.
6 0
4 years ago
Find the time t2 that it would take the charge of the capacitor to reach 99.99% of its maximum value given that r=12.0ω and c=50
defon

Answer:

Explanation:

Given that, .

R = 12 ohms

C = 500μf.

Time t =? When the charge reaches 99.99% of maximum

The charge on a RC circuit is given as

A discharging circuit

Q = Qo•exp(-t/RC)

Where RC is the time constant

τ = RC = 12 × 500 ×10^-6

τ = 0.006 sec

The maximum charge is Qo,

Therefore Q = 99.99% of Qo

Then, Q = 99.99/100 × Qo

Q = 0.9999Qo

So, substituting this into the equation above

Q = Qo•exp(-t/RC)

0.9999Qo = Qo•exp(-t / 0.006)

Divide both side by Qo

0.9999 = exp(-t / 0.006)

Take In of both sodes

In(0.9999) = In(exp(-t / 0.006))

-1 × 10^-4 = -t / 0.006

t = -1 × 10^-4 × - 0.006

t = 6 × 10^-7 second

So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge

8 0
3 years ago
How many atoms are in each product silver sulfide ​
AlekseyPX

there are 3 atoms in each silver sulfide

5 0
3 years ago
A 50g ball is falling down. As the ball passes a certain distance its potential energy changes for 2J. Calculate this distance.
Paul [167]

potential energy = mass × gravity × height

so, change in potential energy = mass × gravity × change in height

2 = 50 × 10 × Δh

2 ÷ 500 = Δh

Δh = 0.004 m

This distance does depend on the initial velocity of the ball.

8 0
3 years ago
ASTM A229 oil-tempered carbon steel is used for a helical coil spring. The spring is wound with D = 50 mm d = 10.0 mm, and a pit
horrorfan [7]

Answer

given,

D = 50 mm = 0.05 m

d = 10 mm = 0.01 m

Force to compress the spring

F = \dfrac{d^4G\delta}{8D^3N}

\dfrac{\delta}{N} = p - d = 14 - 10 = 4 mm

F = \dfrac{d^4G}{8D^3}\times 0.004

F = \dfrac{0.1^4\times 79\times 10^9}{8\times 0.05^3}\times 0.004

     F = 3160 N

stress correction factor from stress correction curve is equal to 1.1

now, calculation of corrected stress

\tau = \dfrac{8FDk_s}{\pi d^3}

\tau = \dfrac{8\times 3160 \times 0.05 \times 1.1}{\pi \times 0.01^3}

              = 442.6 Mpa

The tensile strength of the steel material of  ASTM A229 is equal to 1300 Mpa

now,

\tau_s \leq 0.45 S_u

\tau_s \leq 0.45 \times 1300

\tau_s \leq 585\ Mpa

since corrected stress is less than the \tau_s

hence, spring will return to its original shape.

6 0
3 years ago
Read 2 more answers
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