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erastovalidia [21]
4 years ago
9

The two particles are both moving to the right. Particle 1 catches up with particle 2 and collides with it. The particles stick

together and continue on with velocity vf. Which of these statements is true?
A. vf=v1
B. vf=v2
C. vf is less than v2
D. vf is less than v2 but greater than v1
E. vf>v1

Physics
1 answer:
Irina18 [472]4 years ago
8 0

Answer:

See bolded below.

Explanation:

Consider the " Before " and " After. " " Before, " this particle 1 was trying to catch up with this particle 2, and " after " particle one had collided with particle two. Take a look at the attachment below for a more detailed examination.

Here is how this will play out. Particle 1, with great velocity, will hit particle 2, which would mean that Particle 2 has less velocity than Particle 1. Now after the collision, energy is transferred to Particle 2, and while Particle 1 has now stopped in it's tracks, Particle 2 - with more energy than before - will continue as long as it has to before friction eventually brings it to a stop.

_______________________________________________________

From this we can conclude that Vf, from the picture below, must have less energy than V1, but more energy than V2 - and vice versa.

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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
3 years ago
A proton travels with a speed of 5.02×10⁶ m/s in a direction that makes an angle of 60.0° with the direction of a magnetic ficld
Triss [41]

The magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.

Speed of the proton = 5.02 × 10 ⁶ m /a

Angel of between the velocity and the magnetic force = 60 °

The magnitude of magnetic field B = 0.180 T

The magnitude of the magnetic force on the proton is,

F = q(v \times B)

F = qvB \: sin \:  θ

F = 1.6 \times 10 ^{ - 19}  \times 5.02 \times 10 ^{6}  \times 0.180 \times  \: sin \: 60°

= 1.25 \times 10 ^{ - 13}  \: N

Therefore, the magnitude of the magnetic force on the proton is 1.25× 10—¹³ N.

To know more about magnetic force, refer to the below link:

brainly.com/question/23096032

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4 0
1 year ago
Can you explain to be the definition of Quantum Theory?
Vanyuwa [196]
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8 0
3 years ago
2. A ball is released from a vertical height of 20 cm. It rolls down a "perfectly
Paraphin [41]

Answer:

20 cm

Explanation:

Given that a ball is released from a vertical height of 20 cm. It rolls down a "perfectly frictionless" ramp and up a similar ramp. What vertical height on the second ramp will the ball reach before it starts to roll back down?

Since it is perfectly frictionless, the Kinetic energy in which the ball is rolling will be equal to the potential energy at the edge of the ramp.

Therefore, the ball will reach 20 cm before it starts to roll back down.

7 0
3 years ago
A bicycle wheel has a diameter of 63.0 cm and a mass of 1.75 kg. Assume that the wheel is a hoop with all of the mass concentrat
Masteriza [31]

Answer:

F2 = 834 N

Explanation:

We are given the following for the bicycle;

Diameter; d1 = 63 cm = 0.63 m

Mass; m = 1.75 kg

Resistive force; F1 = 121 N

For the sprocket, we are given;

Diameter; d2 = 8.96 cm = 0.0896 m

Radius; r2 = 0.0896/2 = 0.0448 m

Radial acceleration; α = 4.4 rad/s²

Now moment of inertia of the wheel which is assumed to be a hoop is given by; I = m(r1)²

Where r1 = (d1)/2 = 0.63/2

r1 = 0.315 m

Thus, I = 1.75 × 0.315²

I = 0.1736 Kg.m²

The torque is given by the relation;

I•α = F1•r1 - F2•r2

Where F2 is the force that must be applied by the chain to give the wheel an acceleration of 4.40 rad/s².

Thus;

0.1736 × 4.4 = (121 × 0.315) - (0.0448F2)

>> 0.76384 = 38.115 - (0.0448F2)

>> 0.0448F2 = 38.115 - 0.76384

>> F2 = (38.115 - 0.76384)/0.0448

>> F2 = 833.73 N

Approximately; F2 = 834 N

7 0
3 years ago
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