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3 years ago

8

Please help! i'm horrible at this

1 answer:

frez [133]3 years ago

4 0

Answer:

a = Δv/t = (vf - vi)/t = (0 - 5)/4 = -1.25 m/s²

Explanation:

You may or may not need the negative sign, depending on how the question designer was thinking about the problem.

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In damped harmonic oscillation, the amplitude of oscillation becomes one third after 2 second. If A0 is initial amplitude of osc

Harman [31]

Answer:

$A=\frac{A_0}{\sqrt 3}$

Explanation:

Initial amplitude= $A_0$

We are given that

Amplitude after 2 s=A= $\frac{1}{3}A_0$

We have to find the amplitude after 1 s.

We know that amplitude at any time t

$A=A_0e^{-\alpha t}$

Using the formula

$\frac{A_0}{3}=A_0e^{-2\alpha}$

$\frac{1}{3}=e^{-2\alpha}$

$3=e^{2\alpha}$

$ln 3=2\alpha$

$\alpha =\frac{ln 3}{2}=ln\sqrt 3$

$e^{\alpha}=\sqrt 3}$

When t=1 s

$A=A_0e^{-\alpha}=\frac{A_0}{\sqrt 3}$

8 0

4 years ago

A reciprocating compressor is a device that compresses air by a back-and-forth straight-line motion, like a piston in a cylinder

QveST [7]

Answer:

$\Delta \theta = 47.57^{\circ} C$

Explanation:

given,

moles of air compressed, n = 1.70 mol

initial temperature, T₁ = 390 K

Power supply by the compressor, P = 7.5 kW

Heat removed = 1.3 kW

Angular frequency of the compressor, f = 110 rpm = 110/60 = 1.833 rps.

Time of compression = time of the hay revolution

= $\dfrac{1}{2}\ T$

= $\dfrac{1}{2}\times \dfrac{1}{f}$

= $\dfrac{1}{2}\times \dfrac{1}{1.833}$

=0.273 s

Using first law of thermodynamics

U = Q - W

now,

$\dfrac{\Delta U}{\Delta t} = \dfrac{\Delta Q}{\Delta t}- \dfrac{\Delta W}{\Delta t}$

Power supplied $\dfrac{\Delta W}{\Delta t}$ = 7.5 kW

heat removed $\dfrac{\Delta Q}{\Delta t}$ = 1.3 kW

now,

$\dfrac{\Delta U}{\Delta t} = 7.5 -1.3$

$\dfrac{\Delta U}{\Delta t} = 6.2 kW$

we know,

$\dfrac{\Delta U}{\Delta t}=\dfrac{nC_v\Delta \theta}{\Delta t}$

C_v for air = 5 cal/° mol

= 5 x 4.186 J/mol°C = 20.93 J/mol°C

now,

$\Delta \theta = \dfrac{\Delta U}{\Deta t}\times \dfrac{\Delta t}{n C_v}$

$\Delta \theta = 6.2\times 10^3 \times \dfrac{0.273}{1.7\times 20.93}$

$\Delta \theta = 47.57^{\circ} C$

the temperature change per compression stroke is equal to 47.57°C.

4 0

3 years ago

A charge of 240C flows through a wire in 3 minutes. Find the electric current flowing through it.

o-na [289]

<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ Use this equation:

Q = IT

Rearrange for I

I = Q/T

Substitute in the values:

I = 240/3

I = 80

The electric current is 80 amps.

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

7 0

3 years ago

Read 2 more answers

A voltmeter is a device that measures _____.

olga55 [171]

A voltmeter<span> its </span>instrument<span> used for </span>measuring<span> electrical potential difference between two points in an electric circuit. </span>An ammeter<span> is a </span>measuring device<span> used to</span>measure<span> the electric current in a circuit.

</span>

5 0

3 years ago

Read 2 more answers

calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end

sdas [7]

Answer: $0.8\ mm$

Explanation:

Given

length of wire $l=75\ cm$

change in length $\Delta l=1.85\ mm$

mass of wire $m=10\ kg$

Young's modulus for silver $E=7.9\times 10^{10}\ N/m^2$

load on wire $F=mg$

$F=10\times 9.8=98\ kg$

change in length is given by

$\Delta l=\dfrac{Pl}{AE}$

Where A=area of cross-section

$A=\dfrac{Pl}{\Delta lE}$

$A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}$

$A=\dfrac{73.5}{14.615\times 10^{7}}$

$A=5.029\times 10^{-7}\ m^2$

also wire is the shape of cylinder so cross-section is given by

$A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2$

$\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }$

$\Rightarrow d^2=64.02\times 10^{-8}$

$\Rightarrow d=8\times 10^{-4}\ m$

$\Rightarrow d=0.8\ mm$

4 0

3 years ago