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4 years ago

Explain why the rust of an iron bar decreases the strength of the bar.

1 answer:

7 0

Rusting is a chemical change, which forms a new substance which is brittle. A solid that is formed by a chemical reaction.

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calcular la longitud de un péndulo que oscila a 10 Hz en santa fe de bogota, sabiendo que en esta ciudad la aceleracion de la gr

cricket20 [7]

Answer:

$L=2.48*10^{-3} m$

Explanation:

The period equation for a pendulum is given by:

$T=2\pi \sqrt{\frac{L}{g}}$

and we know that T = 1/f, where f is the frequency, so we will have:

$\frac{1}{f}=2\pi \sqrt{\frac{L}{g}}$

Now, we just need to solve this equation for L.

$\frac{1}{2\pi f}=\sqrt{\frac{L}{g}}$

$L=\frac{g}{(2\pi f)^{2}}$

g is the gravity in Bogota (g=9.78 m/s^{2})
f is 10 Hz
L is the lenght of the pendulum

$L=\frac{9.78}{(2\pi*10)^{2}}$

$L=2.48*10^{-3} m$

I hope it helps you!

5 0

3 years ago

Master of physics needed

Delicious77 [7]

Hey JayDilla, I get 1/3. Here's how:
Kinetic energy due to linear motion is:
$E_{linear}= \frac{1}{2}mv^2$
where
$v=r \omega$
giving
$E_{linear}= \frac{1}{2}mr^2 \omega ^2$

The rotational part requires the moment of inertia of a solid cylinder
$I_{cyl} = \frac{1}{2}mr^2$
Then the rotational kinetic energy is
$E_{rot}= \frac{1}{2}I \omega ^2= \frac{1}{4}mr^2 \omega ^2$
Adding the two types of energy and factoring out common terms gives
$\frac{1}{2}mr^2 \omega ^2(1+ \frac{1}{2})$
Here the "1" in the parenthesis is due to linear motion and the "1/2" is due to the rotational part. Since this gives a total of 3/2 altogether, and the rotational part is due to a third of this (1/2), I say it's 1/3.

8 0

4 years ago

Kindly Don't Spàm!<br>Thank uh !:)<br><img src="https://tex.z-dn.net/?f=%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20%20%5C%5C%20" id="Te

olga55 [171]

$\: \underline{ \boxed{ { \color{gray}\frak{\huge \: Answer : }}}}$

<h3> Collector Current at Saturation :</h3>

$\sf \large {I_{C(SAT)} = \frac{ V_{CC} }{R_C} }$

$\: \:$

$\sf \large {I_{C(SAT)} = \frac{12}{2.2 \times 10 ^{3} } }$

$\: \:$

$\bold{ \bf \large {I_{C(SAT)} =5.45mA }}$

$\: \:$

________________________________

<h3> Value Of Cut - off Voltage : </h3>

$\sf \large \: V_{CS(cut-off)} = V_{CC}$

$\: \:$

Therefore ,

$\: \:$

$\bf \large \: V_{CS(cut-off)} = \: 12v$

$\: \:$

________________________________

${ \bf\large \: Base \: Current , I_B = \frac{V_{CC}}{R_C} }$

$\sf \large \: I_B ={ \frac{12}{2.2 \times 10 ^{3} } }$

$\: \:$

$\bf \large \: I_B = 50μA$

$\: \:$

_______________________________

<h3>Collector Current ,</h3>

$\sf \large \: I_C = β * I_B$

$\: \:$

$\sf \large \: I_C = 50 * 50 * 10^{-6}$

$\: \:$

$\bf \large \: I_C = 2.5mA$

$\: \:$

________________________________

<h3> Collector to emitter Voltage : </h3>

$\sf \large \: V_{CE} = V_{CC} - ( I_C * R_C )$

$\: \: \:$

$\sf \large \: V_{CE} = 12 - ( 2.5 * 10-³ * 2.2 * 10³ )$

$\: \:$

$\bf \large \: V_{CE} = 6.5v$

________________________________

<h3>Q - point are :</h3>

$\sf \large \: I_{CEQ} = 2.5mA$

$\: \:$

$\sf \large \: V_{CEQ} = 6.5v$

________________________________

<h3>Q - point located on the DC load line as shown in fig ~</h3><h3 /><h3 /><h3 /><h3 />

________________________________

Hope Helps!:)

$\: \: \: \:$

5 0

2 years ago

The function of the eardrum in the middle ear is to

lozanna [386]

B. vibrate with the frequency of the received sound<span />

7 0

4 years ago

Read 2 more answers

In anticipation of a long 10o upgrade, a bus driver accelerates at a constant rate of 5 ft/s^2 while still on a level section of

Rashid [163]

Answer:

The distance (in miles) by the bus up the hill when its speed decreased to 50 mph is approximately 1.353 miles

Explanation:

The parameters of the motion of the driver are;

The upgrade of the road, θ = 10°

The rate of constant acceleration of the bus driver = 5 ft./s²

The speed of the bus as it begins to go up the hill, v₁ = 80 mph = 117.3228 ft./s

The speed of the driver at a point on the hill, v₂ = 50 mph ≈ 73.32677 ft./s

The acceleration due to gravity, g ≈ 32.1740 ft./s²

Therefore, we have;

The acceleration due to gravity down the incline plane, gₓ = g·sinθ

∴ gₓ = g·sin(θ) ≈ 32.1740 ft./s² × sin(10°) ≈ 5.587 ft/s²

The net acceleration of the bus, on the incline plane, $a_{Net}$ = gₓ - a = 5.587 ft./s² -5 ft./s² = 0.587 ft./s²

The vertical component of the velocity, $v_y$ = v × sin(θ)

∴ $v_y$ = 117.3228 ft./s × sin(10°) ≈ 20.37289 ft./s

vₓ = 117.3228 ft./s × cos(10°) ≈ 115.5404 ft./s

The velocity of the car, v₂, on the inclined plane is given as follows;

v₂ = v₁ - $a_{Net}$ × t

∴ t = (v₁ - v₂)/ $a_{Net}$ = (117.3228 ft./s - 73.32677 ft./s)/(0.587 ft./s²) ≈ 74.95 s

The distance covered, 's', is given as follows;

s = v₁·t - 1/2· $a_{Net}$ ·t²

∴ s = 117.3228 × 74.95 - 1/2 × 0.587 × 74.95² ≈ 7144.6069 ft.

The distance travelled up the hill, s ≈ 7144.6069 ft. ≈ 1.3531452 miles ≈ 1.353 miles

5 0

3 years ago