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3 years ago

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Two Masses, a Pulley, and an Inclined Plane. . Block 1, of mass m1 = 0.700kg , is connected over an ideal (massless and friction

less) pulley to block 2, of mass m2, as shown. For an angle of θ = 30.0∘ and a coefficient of kinetic friction between block 2 and the plane of μ = 0.200, an acceleration of magnitude a = 0.200m/s2 is observed for block 2.. . find the mass of block 2, m2:

1 answer:

Umnica [9.8K]3 years ago

5 0

The answer to this is 1 kg.
You use this equation:
m2 = m1 (g - a) / [a + g u cos θ + g sin θ]

You plug in the given values:
m2 = 0.7 (9.81 - 0.2) / [0.2 + 9.81(0.20) cos 30 + 9.81 sin 30)

The result is
m2 = 1 kg

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A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca

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Answer:

$q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

Explanation:

Given that $L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V$ , we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

$E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000 \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}$

#To find the particular solution:

$Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

Hence the charge at any time, t is $q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C$

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3 years ago

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo

Serggg [28]

Before the engines fail $(0\le t\le3.00\,\rm s)$ , the rocket's horizontal and vertical position in the air are

$x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2$

$y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2$

and its velocity vector has components

$v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t$

$v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t$

After $t=3.00\,\rm s$ , its position is

$x=273\,\rm m$

$y=362\,\rm m$

and the rocket's velocity vector has horizontal and vertical components

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}$

After the engine failure $(t>3.00\,\rm s)$ , the rocket is in freefall and its position is given by

$x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t$

$y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

and its velocity vector's components are

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}-gt$

where we take $g=9.80\,\frac{\rm m}{\mathrm s^2}$ .

a. The maximum altitude occurs at the point during which $v_y=0$ :

$159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s$

At this point, the rocket has an altitude of

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m$

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve $y=0$ for $t$ , then add 3 seconds to this time:

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s$

So the rocket stays in the air for a total of $37.6\,\rm s$ .

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute $x$ for this time $t$ :

$273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m$

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3 years ago

A mass of 5 kg of saturated water vapor at 150 kPa is heated at constant pressure until the temperature reaches 200°C. Calculate

yulyashka [42]

Answer:

The work done by the steam is 213 kJ.

Explanation:

Given that,

Mass = 5 kg

Pressure = 150 kPa

Temperature = 200°C

We need to calculate the specific volume

Using formula of work done

$W=Pm\DeltaV$

$W=Pm(\dfrac{RT_{1}}{P_{atm}}-\dfrac{RT_{2}}{P_{atm}}$

$W=\dfrac{PmR}{P_{atm}}(T_{2}-T_{1})$

Where,R = gas constant

T = temperature

P = pressure

$P_{atm}$ =Atmosphere pressure

m = mass

Put the value into the formula

$W=\dfrac{150\times10^{3}\times5\times287.05}{1.01\times10^{5}}\times(473-373)$

$W=213\ kJ$

Hence, The work done by the steam is 213 kJ.

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3 years ago

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Archy [21]

Answer:

The acceleration of the earth is 7.05 * 10^-25 m/s²

Explanation:

<u>Step 1:</u> Data given

mass of the apple = 0.43 kg

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Following the third law of Newton F = m*a

F(apple) = F(earth) = m(apple)*a(apple)

F(apple) = 0.43 kg * 9.8 m/s² = 4.214 N

a(earth) = F(apple/earth)/m(earth)

a(earth) = 4.214N /5.98 * 10 ^24 kg

a(earth) = 7.05 * 10^-25 m/s²

The acceleration of the earth is 7.05 * 10^-25 m/s²

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Nadusha1986 [10]

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