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3 years ago

Ayuda por favor (archivo adjunto) con un ejercicio de expresión sobre periodo de oscilación de esta figura:

Physics

1 answer:

vodka [1.7K]3 years ago

8 0

Answer:

nolo se

Explanation:

no lo se

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I need help with this please

34kurt

The answer is Rubber

5 0

3 years ago

In your own words, explain what friction is

Andrew [12]

Answer:

Friction is a kind of force between two object when we rub or slide its surfaces together

for ex: when you rub the matchbox you get fire which is the result of friction

Explanation:

thats all i know

3 0

2 years ago

Read 2 more answers

E Which of the following particles that may be emitted in radioactive decay is not ionising?

mestny [16]

Explanation:

C. neutron.

it does not contain ionizing characteristics.

hope it helps. :)

5 0

3 years ago

A compressor receives air at 290 K, 100 kPa and a shaft work of 5.5 kW from a gasoline engine. It is to deliver a mass flow rate

Sladkaya [172]

Answer:

$P_2=4091\ KPa$

Explanation:

Given that

T₁ = 290 K

P₁ = 100 KPa

Power P =5.5 KW

mass flow rate

$\dot{m}= 0.01\ kg/s$

Lets take the exit temperature = T₂

We know that

$P=\dot{m}\ C_p (T_2-T_1)$

$5.5=0.01\times 1.005(T_2-290})\\T_2=\dfrac{5.5}{0.01\times 1.005}+290\ K\\\\T_2=837.26\ K$

If we assume that process inside the compressor is adiabatic then we can say that

$\dfrac{T_2}{T_1}=\left(\dfrac{P_2}{P_1}\right)^{0.285}$

$\dfrac{837.26}{290}=\left(\dfrac{P_2}{100}\right)^{0.285}\\2.88=\left(\dfrac{P_2}{100}\right)^{0.285}\\$

$2.88^{\frac{1}{0.285}}=\dfrac{P_2}{100}$

$P_2=40.91\times 100 \ KPa$

$P_2=4091\ KPa$

That is why the exit pressure will be 4091 KPa.

4 0

3 years ago

A particle executes simple harmonic motion with an amplitude of 3.00 cm. At what position does its speed equal half of is maximu

Mrac [35]

Answer:

x=±0.026m

Explanation:

In simple harmonic motion the maximum value of the magnitude of velocity

$v_{max}=wA=\sqrt{\frac{k}{m} }A$

The speed as a function of position for simple harmonic oscillator is given by

$v=w\sqrt{A^{2}-x^{2}$

where A is amplitude of motion

Given data

Amplitude A=3 cm =0.03 m

v=(1/2)Vmax

To find

We have asked to find position x does its speed equal half of is maximum speed

Solution

The speed of the particle the maximum speed as:

$v=\frac{V_{max} }{2}\\ w\sqrt{A^{2}-x^{2} }=\frac{wA}{2}\\ A^{2}-x^{2}=\frac{A^{2} }{4}\\ x^{2}=A^{2}- \frac{A^{2} }{4}\\ x^{2}=\frac{3A^{2}}{4} \\x=\sqrt{\frac{3A^{2}}{4}}$

x=±(√3(0.03)/2)

x=±0.026m

5 0

4 years ago