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3 years ago

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A twig from a tree drops from a 200m high cliff on to a beach below.during its fall 40% twig's energy is converted into thermal

energy.what is the speed with which the twig hits the beach?

1 answer:

kolezko [41]3 years ago

7 0

<span>Calculating for potential energy P.E=mgh
m=mass
g=acceleration due to gravity
h=height
</span>Let the twig's mass be M<span>
</span><span>
therefore at the start, the twig has 200*9.8*M potential energy.

</span>During its fall 40% twig's energy is converted into thermal energy and the remaining <span>60% of this will turn into kinetic energy during the fall, giving us 1176M
Now the equation for kinetic energy, K.E is 0.5*m*v^2
m is mass
v is velocity

Therefore 1176M=0.5*M*v^2 (mass will divide out)
velocity is </span><span>48.49 m/s</span>

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The poet is expressing her belief that the second, more natural option is far more desirable than the first option.

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Some hold the Sabbath going to Church is set faith and meditation.

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1 year ago

If the speed of light in a substance is 2.26 x 10^8 m/s, what is the index of refraction of that substance?

vladimir1956 [14]

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1.33

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2 years ago

A ball is hit so that it accelerates at 10 m/s2. The ball hits a glove with a force of 5 N. What is the mass of the ball?

jenyasd209 [6]

Answer:

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» <u>Concepts</u>

Newton's second law, the Law of Acceleration, states that F = ma, where F = Force in Newtons, m = mass in kg, and a = acceleration in m/s^2.

» <u>Application</u>

We are asked to find the mass of the ball using the equation F = ma. We're also given the force and acceleration, so the equation looks like 5 = 10(m).

» <u>Solution</u>

Step 1: Divide both sides by 10.

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Thus, the mass of the ball is 0.5 kg.

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2 years ago

Is the magnitude of the acceleration of the center of mass of spool A greater than, less than, or equal to the magnitude of the

Anon25 [30]

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2 years ago

A cylinder with a piston contains 0.200 mol of oxygen at 2.00×105 Pa and 360 K . The oxygen may be treated as an ideal gas. The

ikadub [295]

Answer:

A. $W=600\ J$

B. $Q=2112\ J$

C. $\Delta U=1512\ J$

D. $W=0\ J$

Explanation:

Given:

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initial pressure in the cylinder, $P_i=2\times 10^5\ Pa$

initial temperature of the gas in the cylinder, $T_i=360\ K$

<em>According to the question the final volume becomes twice of the initial volume.</em>

<u>Using ideal gas law:</u>

$P.V=n.R.T$

$2\times 10^5\times V_i=0.2\times 8.314\times 360$

$V_i=0.003\ m^3$

A.

<u>Work done by the gas during the initial isobaric expansion:</u>

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C.

<u>we have the specific heat capacity of oxygen at constant pressure as:</u>

$c_v=21\ J.mol^{-1}.K^{-1}$

Now we apply Charles Law:

$\frac{V_i}{T_i} =\frac{V_f}{T_f}$

$\frac{0.003}{360} =\frac{0.006}{T_f}$

$T_f=720\ K$

<u>Now change in internal energy:</u>

$\Delta U=n.c_p.(T_f-T_i)$

$\Delta U=0.2\times 21\times (720-360)$

$\Delta U=1512\ J$

B.

<u>Now heat added to the system:</u>

$Q=W+\Delta U$

$Q=600+1512$

$Q=2112\ J$

D.

Since during final cooling the process is isochoric (i.e. the volume does not changes). So,

$W=0\ J$

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2 years ago