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3 years ago

What change would occur in the kinetic energy of vapor if a container of vapor is placed at 0 Kelvin?

Physics

1 answer:

n200080 [17]3 years ago

7 0

Answer:

The kinetic energy of gas particles will increase. The kinetic energy of gas particles will decrease. The kinetic energy of gas particles will remain unchanged

Explanation:

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When two or more identical capacitors are connected in series across a potential difference?

OlgaM077 [116]

<span>When two or more identical capacitors (or resistors) are connected
in series across a potential difference, the potential difference divides
equally among them.

For example, if you have nine identical capacitors (or resistors) all
connected end-to-end like elephants in a circus parade, and you
connect the string to a source of 117 volts (either AC or DC), then
you will measure

(117v / 9) = 13 volts

across each unit in the string.</span>

6 0

3 years ago

This exercise uses the radioactive decay model. After 3 days a sample of radon-222 has decayed to 58% of its original amount. (a

Vadim26 [7]

Answer: a) 3.85 days

b) 10.54 days

Explanation:-

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = ?

t = time taken for decomposition = 3 days

a = let initial amount of the reactant = 100 g

a - x = amount left after decay process = $\frac{58}{100}\times 100=58g$

First we have to calculate the rate constant, we use the formula :

Now put all the given values in above equation, we get

$k=\frac{2.303}{3}\log\frac{100}{58}$

$k=0.18days^{-1}$

a) Half-life of radon-222:

$t_{\frac{1}{2}}=\frac{0.693}{k}$

$t_{\frac{1}{2}}=\frac{0.693}{0.18}=3.85days$

Thus half-life of radon-222 is 3.85 days.

b) Time taken for the sample to decay to 15% of its original amount:

where,

k = rate constant = $0.18days^{-1}$

t = time taken for decomposition = ?

a = let initial amount of the reactant = 100 g

a - x = amount left after decay process = $\frac{15}{100}\times 100=15g$

$t=\frac{2.303}{0.18}\log\frac{100}{15}$

$t=10.54days$

Thus it will take 10.54 days for the sample to decay to 15% of its original amount.

3 0

3 years ago

The angular speed of the rotor in a centrifuge increases from 420 to 1420 rad/s in a time of 5.00 s. (a) Obtain the angle throug

9966 [12]

Answer:

a) θ = 2500 radians

b) α = 200 rad/s²

Explanation:

Using equations of motion,

θ = (w - w₀)t/2

θ = angle turned through = ?

w = final angular velocity = 1420 rad/s

w₀ = initial angular velocity = 420

t = time taken = 5s

θ = (1420 - 420) × 5/2 = 2500 rads

Again,

w = w₀ + αt

α = angular accelaration = ?

1420 = 420 + 5α

α = 1000/5 = 200 rad/s²

6 0

3 years ago

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PLSSS HELP MEEEE

vekshin1

Answer:

Cu2+

Explanation:

7 0

3 years ago

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You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$

hence, the change in Er is about 1.52J times the initial rotational energy

5 0

3 years ago

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