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2 years ago

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A 60-kg skate boarder is standing on his stationary board. A bigger, out of control 75-kg skate boarder crashes into him with a

velocity of 4 m/s. Their boards get stuck together in this inelastic collision. How fast are the moving after the collision?

1 answer:

andrezito [222]2 years ago

3 0

Answer:

2.22m/s

Explanation:

Given parameters:

M1 = 60kg

M2 = 75kg

V1 = 0m/s

V2 = 4m/s

Unknown;

Velocity after collision = ?

Solution:

To solve this problem, we must understand that the momentum before and after collision of the bodies must be the same;

M1 V1 + M2 V2 = v(M1 + M2)

So;

60 x 0 + 75 x 4 = v (60 +75)

300 = 135v

v = 2.22m/s

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A particle with charge 3.20×10−19 c is placed on the x axis in a region where the electric potential due to other charges increa

lys-0071 [83]

Answer:

-5 V

Explanation:

The charged particle (which is positively charged) moves from point A to B, and its kinetic energy increases: it means that the particle is following the direction of the field, so its potential energy is decreasing (because it's been converted into potential energy), therefore it is moving from a point at higher potential (A) to a point at lower potential (B). This means that the value

vb−va

is negative.

We can calculate the potential difference between the two points by using the law of conservation of energy:

$\Delta K+ \Delta U=0\\\Delta K + q\Delta V=0$

where:

$\Delta K=+1.6\cdot 10^{-18} J$ is the change in kinetic energy of the particle

$q=3.2\cdot 10^{-19} C$ is the charge of the particle

$\Delta V =V_b-V_a$ is the potential difference

Re-arranging the equation, we can find the value of the potential difference:

$\Delta V=V_b-V_a = -\frac{\Delta K}{q}=-\frac{1.6\cdot 10^{-18} J}{3.2\cdot 10^{-19} C}=-5 V$

8 0

3 years ago

Nerve cells transmit electric signals through their long tubular axons. These signals propagate due to a sudden rush of Na+ ions

iren [92.7K]

Answer:

$I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A$

Explanation:

For this case we have the following info given:

Number of Na+ ions $5.7 x10^{11} ions$

Each ion have a charge of +e and the crage of the electron is $1.6 x10^{-19}C$

The time is given $t = 7 ms$ if we convert this into seconds we got:

$t = 7ms * \frac{1s}{1000 ms}= 0.007s$

Now we can use the following formula given from the current passing thourhg a meter of nerve axon given by:

$Q = Ne$

Where N represent the number of ions, e the charge of the electron and Q the total charge

If we replace on this case we have this:

$Q= 5.7x10^{11} * (1.6 x10^{-19}C) = 9.12x10^{-8} C$

And from the general definition of current we know that:

$I =\frac{Q}{t}$

And since we know the total charge Q and the time we can replace:

$I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A$

The current during the inflow charge in the meter axon for this case is $13.02 \mu A$

3 0

2 years ago

The current that charges a capacitor transfers energy that is stored in the capacitor’s electric field. Consider a 2.0 μF capaci

lapo4ka [179]

Answer:

the capacitor voltage is V = 20 V

Explanation:

Given,

Capacitance of the capacitor = 2.0 μF

energy stored = 200 W

time (t) =2.0 μs

Capacitor voltage = ?

$\dfrac{dE}{dt}=200 W$

$E =200\times 2 \times 10^{-6}$

$E =400 \times 10^{-6}\ J$

we know,

$E = \dfrac{1}{2}CV^2$

$V =\sqrt{\dfrac{2E}{C}}$

$V =\sqrt{\dfrac{2\times 400 \times 10^{-6}}{2\times 10^{-6}}}$

$V =\sqrt{400}$

V = 20 V

so , the capacitor voltage is V = 20 V

7 0

2 years ago

Swimmers at the beach are tanning on towels. Which method of heat transfer is responsible for their tan? *

Free_Kalibri [48]

Answer:

ratiation

Explanation:

3 0

2 years ago

Read 2 more answers

xConsider the following reduction potentials: Cu2+ + 2e– Cu E° = 0.339 V Pb2+ + 2e– Pb E° = –0.130 V For a galvanic cell employi

slega [8]

Answer:

Approximately $\rm 90\; kJ$ .

Explanation:

Cathode is where reduction takes place and anode is where oxidation takes place. The potential of a electrochemical reaction ( $E^{\circ}(\text{cell})$ ) is equal to

$E^{\circ}(\text{cell}) = E^{\circ}(\text{cathode}) - E^{\circ}(\text{anode})$ .

There are two half-reactions in this question. $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ and $\rm Pb^{2+} + 2\,e^{-} \rightleftharpoons Pb$ . Either could be the cathode (while the other acts as the anode.) However, for the reaction to be spontaneous, the value of $E^{\circ}(\text{cell})$ should be positive.

In this case, $E^{\circ}(\text{cell})$ is positive only if $\rm Cu^{2+} + 2\,e^{-} \rightleftharpoons Cu$ is the reaction takes place at the cathode. The net reaction would be

$\rm Cu^{2+} + Pb \to Cu + Pb^{2+}$ .

Its cell potential would be equal to $0.339 - (-0.130) = \rm 0.469\; V$ .

The maximum amount of electrical energy possible (under standard conditions) is equal to the free energy of this reaction:

$\Delta G^{\circ} = n \cdot F \cdot E^{\circ} (\text{cell})$ ,

where

$n$ is the number moles of electrons transferred for each mole of the reaction. In this case the value of $n$ is $2$ as in the half-reactions.
$F$ is Faraday's Constant (approximately $96485.33212\; \rm C \cdot mol^{-1}$ .)

$\begin{aligned}\Delta G^{\circ} &= n \cdot F \cdot E^{\circ} (\text{cell})\cr &= 2\times 96485.33212 \times (0.339 - (-0.130)) \cr &\approx 9.0 \times 10^{4} \; \rm J \cr &= 90\; \rm kJ\end{aligned}$ .

5 0

2 years ago