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Vitek1552 [10]
3 years ago
7

A vertical wire carries a current vertically upward in a region where the magnetic field vector points toward the north. What is

the direction of the magnetic force on this current due to the field? A) downward B) toward the north C) toward the east D) toward the south E) toward the west Two long parallel wires placed side-by-side on a horizontal table carry identical size currents in opposite directions. The wire on your right carries current toward you, and the wire on your left carries current away from you. From your point of view, the magnetic field at the point exactly midway between the two wires A) is zero. B) points upward. C) points downward. D) points away from you. E) points toward you.
Physics
1 answer:
vesna_86 [32]3 years ago
7 0

Answer:

a. C or E.. Cyclic

b. A

Explanation:

A. Using the right hand rule... I the thumb is the current direction, and the other fingers coil to rep the magnetic field direction.

B. Two parallel wires carrying the same current but in opposing direction repels..

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What is the effect of sunlight on the earth? A. Sunlight warms the earth evenly. B. Sunlight warms the poles more than the equat
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C) <span>Sunlight warms the earth unevenly.</span>
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3 years ago
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A galvanic cell based on these half-reactions is set up under standard conditions where each solutions is 1.00 L and each electr
Marat540 [252]

Complete Question

    Fe^{2+}  + 2e^-----> Fe \ \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \   E^0_{red}  = - 0.441 V

   

    Cd^{2+} + 2e^-  -----> Cd  \ \ \ \ \ \ \ \ \ \ \ \ \ \  E^0_{red} = -0.403V

A galvanic cell based on these half-reactions is set up under standard conditions where each solutions is 1.00 L and each electrode weighs exactly 100.0 g. How much will the Cd electrode weigh when the non-standard potential of the cell is 0.03305 V?

Answer:

The mass is M= 117.37g

Explanation:

The overall reaction  is as follows

             Cd^{2+} + Fe  Fe^{2+} + Cd

The reaction is this way because the potential  of Cd^{2+} \ reducing \  to \ Cd is higher than the potential  of Fe^{2+} \ reducing \  to \ Fe so the the Fe would be oxidized and Cd^{2+} would be reduced

  At equilibrium the rate constant of the reaction is

                Q = \frac{concentration \ of \ product  }{concentration \ of  reactant }

                      = \frac{[Fe^{2+}[Cd]]}{[Cd^{2+}][Fe]}

The Voltage of the cell E_{cell} = E_{Cd^{2+}/Cd } + E_{Fe^{2+} /Fe}

     Substituting the given values into the equation

                         E_{cell} = -0.403 -(-0.441)

                                 = 0.038V

The voltage of the cell at any point can be calculated using the equation

               E = E_{cell} - \frac{0.059}{n_e}  Q

Where n_e \ is \ the \ number\  of\  electron

Substituting for Q

           E = E_{cell} - \frac{0.059}{n_e} \frac{[Fe^{2+}[Cd]]}{[Cd^{2+}][Fe]}

 When E = 0.03305 V

            E = E_{cell} - \frac{0.59}{n_e} \frac{Fe^{2+}}{Cd^{2+}}

Since we are considering the Cd electrode the equation becomes

            E= E_{cell} - \frac{0.059}{n_e}  [\frac{1}{Cd^{2+}} ]

Substituting values and making [Cd^{2+}]  the subject

          [Cd^{2+}] =\frac{1}{e^{[\frac{0.03305- 0.038}{\frac{0.059 }{2} }] }}

                      = 0.8455M

Given from the question that the volume is 1 Liter

   The number of mole = concentration * volume

                                       = 0.8455 * 1

                                        = 0.8455 moles

At the standard state the concentration of Cd^{2+} is  =1 mole /L

  Hence the amount deposited on the Cd electrode would be

              =  Original amount - The calculated amount

              =   1 - 0.8455

              = 0.1545 moles

The mass deposited is mathematically represented as

             mass = mole * molar \ mass

The Molar mass of Cd = 112.41 g/mol

          Mass  = 0.1545 *112.41

                    = 17.37g

Hence the total mass of the electrode is = standard mass + calculated mass

            M= 100+ 17.37

            M= 117.37g

                               

6 0
3 years ago
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In an 8.00km race, one runner runs at a steady 11.8 km/hr and another runs at 15.0 km/hr. How far from the finish line is the sl
Aleks [24]

Answer:

The slower runner is 1.71 km from the finish line when the fastest runner finishes the race.

Explanation:

Given;

the speed of the slower runner, u₁ = 11.8 km/hr

the speed of the fastest runner, u₂ = 15 km/hr

distance, d = 8 km

The time when the fastest runner finishes the race is given by;

Time = \frac{Distance }{speed}\\\\Time = \frac{8}{15} \\\\Time = 0.533 \ hr

The distance covered by the slower runner at this time is given by;

d₁ = u₁ x 0.533 hr

d₁ = 11.8 km/hr x 0.533 hr

d₁ = 6.29 km

Additional distance (x) the slower runner need to finish is given by;

6.29 km + x = 8km

x = 8 k m - 6.29 km

x = 1.71 km

Therefore, the slower runner is 1.71 km from the finish line when the fastest runner finishes the race.

5 0
4 years ago
A 12 kg object speeds up from an initial velocity of 10 m:s-1
amid [387]

Momentum = m • v

Original momentum = m • 10 m/s north

Final momentum = m • 15 m/s north

Change = m • (15 - 10) m/s north

Change = m • +5 m/s north

Change = +60 kg-m/s north

5 0
3 years ago
PLEASE HELP!! ITS URGENT!!!​
Natasha2012 [34]

Answer:

F = 800 [N]

Explanation:

To be able to calculate this problem we must use the principle of momentum before and after the impact of the hammer.

We must summarize that after the impact the hammer does not move, therefore its speed is zero. In this way, we can propose the following equation.

ΣPbefore = ΣPafter

(m_{1}*v_{1}) - F*t = (m_{1}*v_{2})

where:

m₁ = mass of the hammer = 0.15 [m/s]

v₁ = velocity of the hammer = 8 [m/s]

F = force [N] (units of Newtons)

t = time = 0.0015 [s]

v₂ = velocity of the hammer after the impact = 0

(0.15*8)-(F*0.0015) = (0.15*0)\\F*0.0015 = 0.15*8\\F = 1.2/(0.0015)\\F = 800 [N]

Note: The force is taken as negative since it is exerted by the nail on the hammer and this force is directed in the opposite direction to the movement of the hammer.

6 0
3 years ago
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