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2 years ago

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Arm abcd is pinned at b and undergoes reciprocating motion such that θ=(0.3 sin 4t) rad, where t is measured in seconds and the

argument for the sine is in radiaus. determine the largest speed of point a during the motion and the magnitude of the acceleration of point d at this instant.

1 answer:

storchak [24]2 years ago

8 0

<span>θ=0.3sin(4t)
w=0.3cost(4t)(4)=1.2cost(4t)
a=-4.8sin(4t)

cos4t max will always be 1 (refer to cos graph), for same reason, sin4t will always be 0

therefore, wmax=1.2rad/s

vAmax=r*w=250*1.2=300mm/s
(may be different if your picture/radius is from a different picture)

adt=a*r=200*-4.8sin(4t)=0 (sin(4t)=0)

adn=r*w^2=200*1.2^2=288

ad= square root of adt^2+adn^2 = 288mm/s^2</span>

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Suppose that you have left a 200-mL cup of coffee sitting until it has cooled to 30∘C , which you find totally unacceptable. You

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Answer:

$\eta=0.5074\ or\ 50.74\%$

Explanation:

<em><u>Considering the density & specific heat capacity of coffee to be equal to that of water.</u></em>

<em><u>GIVEN:</u></em>

density $\rho=1\ g.mL^{-1}$

specific heat $c=4.186\ J.g^{-1}.K^{-1}$

mass of coffee, $m=200\times 1=200\ g$

initial temperature of coffee, $T_i=30^{\circ}C$

final temperature of coffee, $T_f=60^{\circ}C$

power rating of oven, $P=1100\ W$

time taken to reach the final temperature, $t=35\ s$

<u>Heat released by the coffee to come to 60°C:</u>

$Q=m.c.\Delta T$

$Q=200\times 4.186\times 30$

$Q=[tex]\eta=\frac{25116}{49500}$ \ J[/tex]

<u>Now the energy used by the oven in the given time:</u>

$E=P.t$

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$E=49500\ J$

Now the efficiency:

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$\eta=0.5074\ or\ 50.74\%$

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2 years ago

In a certain clock, a pendulum of length L1 has a period T1 = 0.95s. The length of the pendulum

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Answer:

Ratio of length will be $\frac{L_2}{L_1}=1.108$

Explanation:

We have given time period of the pendulum when length is $L_1$ is $T_1=0.95sec$

And when length is $L_2$ time period $T_2=1sec$

We know that time period is given by

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

So $0.95=2\pi \sqrt{\frac{L_1}{g}}$ ----eqn 1

And $1=2\pi \sqrt{\frac{L_2}{g}}$ -------eqn 2

Dividing eqn 2 by eqn 1

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I think it might be heat energy. light transforms into heat energy

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A woman is straining to lift a large crate, without success because it is too heavy. We denote the forces on the crate as follow

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An archer draws her bow and stores 34.8 J of elastic potential energy in the bow. She releases the 63 g arrow, giving it an init

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Answer:

Approximately $71\%$ .

Explanation:

The formula for the kinetic energy $\rm KE$ of an object is:

$\displaystyle \mathrm{KE} = \frac{1}{2}\, m \cdot v^2$ ,

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$m$ is the mass of that object, and
$v$ is the speed of that object.

Important: Joule ( $\rm J$ ) is the standard unit for energy. The formula for $\rm KE$ requires two inputs: mass and speed. The standard unit of mass is $\rm kg$ while the standard unit for speed is $\rm m \cdot s^{-1}$ . If both inputs are in standard units, then the output (kinetic energy) will also be in the standard unit (that is: joules,

Convert the unit of the arrow's mass to standard unit:

$m = 63\; \rm g = 0.063\; \rm kg$ .

Initial $\rm KE$ of this arrow:

$\begin{aligned}\mathrm{KE} &= \frac{1}{2} \, m \cdot v^2 \\ &= \frac{1}{2}\times 0.063\; \rm kg \times \left(\rm 28 \; m \cdot s^{-1}\right)^2 \\ &\approx 24.696\; \rm J\end{aligned}$ .

That's the same as the energy output of this bow. Hence, the efficiency of energy transfer will be:

$\displaystyle \frac{24.696\; \rm J}{34.8\; \rm J} \times 100\% \approx 71\%$ .

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2 years ago