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3 years ago

A 205 kg log is pulled up a ramp by means of a rope that

Physics

1 answer:

Volgvan3 years ago

7 0

Answer:2.737 kN

Explanation:

Given

mass of log(m)=205 kg

ramp inclination $=30^{\circ}$

coefficient of kinetic friction between log and ramp is ( $\mu _k$ )=0.9

log has an acceleration of 0.8 m/s^2

Let T be the tension in the rope

$T-mgsin\theta -f_r=ma$

Where $mgsin\theta$ =Sin component of weight

$f_r=friction\ Force=\mu _KN$ (Where N is Normal reaction)

$T-mgsin\theta -\mu _k\left ( mgcos\theta \right )=ma$

$T=m\left ( gsin\theta +\mu _kcos\theta +a\right )$

sin30=0.5

cos30=0.866

$T=205\times \left ( 9.81\times 0.5+0.9\times 9.81\times 0.866+0.8\right )$

$T=205\times 13.35=2.737 kN$

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A revolutionary war cannon, with a mass of 2260 kg, fires a 15.5 kg ball horizontally. The cannonball has a speed of 109 m/s aft

Anton [14]

Answer:

The gain in velocity is 0.37m/s

Explanation:

We need solve this problem though the conservation of momentum. That is,

$m_1 v_1 = m_2 v_2$

$m_1=2260Kg\\m_2=15.5Kg\\v_2= 109m/s$

Using the equation to find $v_1$ ,

$v_1=\frac{m_2 v_2}{m_1}\\v_1=\frac{15.5*109}{2260}\\v_1= 0.7475$

Using the conservation of energy equation, we have,

$KE= \frac{1}{2}m*v^2$

$KE_{cball}=\frac{1}{2}(15.5)(109)^2=92077.75J$

$KE_{cannon}=\frac{1}{2}(2260)(0.7475)^2=631.39J$

$Total KE= 92077.75+13425530=92708.9J$

Now this energy over the cannonball

$KE=\frac{1}{2}m*v_2^2$

$92708.9=\frac{1}{2}15.5v_2^2$

$V_2 = 109.37m/s$

The gain in velocity is 0.37m/s

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3 years ago

A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In

Daniel [21]

Answer: $81.57\mu V$

Explanation:

Given

radius of circular region r=1.50 mm

$A=\pi r^2=7.069\times 10^{-6}\ m^2$

Magnetic Field $B=1.50\ T$

time t=130 ms

Flux is given by

$\phi =B\cdot A$

change in Flux $d\phi =(B_f-B_i)A$

Emf induced is $e=\frac{\mathrm{d} \phi}{\mathrm{d} t}$

$e=\frac{(1.5)\cdot 7.069\times 10^{-6}}{130\times 10^{-3}}$

$e=81.57 \mu V$

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3 years ago

A 12 oz can of soda is left in a car on a hot day. In the morning the soda temperature was 60oF with a gauge pressure of 40 psi.

Neko [114]

In this case, volume of the can remains constant. The relationship between pressure and temperature at constant volume is given by:

P/T = Constant

Then
$\frac{ P_{1} }{ T_{1} } = \frac{ P_{2} }{ T_{2} }$

Where
P1 = 40 psi
P2 = ?
T1 = 60°F ≈ 289 K
T2 = 90°F ≈ 305 K (note, 363 K is not right)

Substituting;
$P_{2} = \frac{ P_{1} T_{2} }{ T_{1} } = \frac{40*305}{289} =42.21 psi$

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3 years ago