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a_sh-v [17]
2 years ago
15

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.
Physics
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

E=1100\times 45

E=49500\ J

Now the efficiency:

\eta=\frac{Q}{E}

\eta=0.5074\ or\ 50.74\%

8 0
2 years ago
In a certain clock, a pendulum of length L1 has a period T1 = 0.95s. The length of the pendulum
gulaghasi [49]

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

\frac{1}{0.95}=\sqrt{\frac{L_2}{L_1}}

Squaring both side

\frac{L_2}{L_1}=1.108

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2 years ago
You are learning about energy transformations in science class. Mel and Sam's built this set-up to see if light energy could be
Allisa [31]
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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Answer:

Explanation:

Given

Force P is acting upward

C is vertical contact Force

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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
elena-14-01-66 [18.8K]

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,

where

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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\%.

8 0
2 years ago
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