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3 years ago

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A spark is generated in an automobile spark plug when there is an electric potential of 3000 V across the electrode gap. If 60 W

of power is generated in a single spark that delivers a total charge of 3 nC, how long does it take for the spark to travel across the gap?
A. 50 ns
B. 75 ns
C. 125 ns
D. 150 ns
E. 225 ns 5

1 answer:

Paha777 [63]3 years ago

7 0

Answer:

The correct option is d

Explanation:

From the question we are told that

The electric potential is $V = 3000 \ V$

The power is $P = 60 \ W$

The charge delivered is $q = 3nC = 3.0 *10^{-9} \ C$

Generally the power generated is mathematically represented as

$P = I V$

=> $I = \frac{P}{V }$

=> $I = \frac{60 }{3000 }$

=> $I = 0.02 \ A$

This current flow is mathematically represented as

$I = \frac{q -q_o}{\Delta t }$

Where $q_o$ is the charge delivered at t=0 s which is 0s

So

$0.02 = \frac{ (3.0 *10^{-9}) -0 }{t - 0 }$

$t = 1.50 *10^{-7 } \ s$

$t = 150 *10^{-9 } \ s$

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A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

rjkz [21]

Answer:

a) $F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

b) $\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

Explanation:

In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).

After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:

$\Sigma F_{y}=0$

We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:

$-F_{y}-W+N=0$

When solving for the normal force we get:

$N=F_{y}+W$

we know that

W=mg

and

$F_{y}=Fsin \theta$

so after substituting we get that

N=F sin θ +mg

We also know that the kinetic friction is defined to be:

$f_{k}=\mu_{k}N$

so we can find the kinetic friction by substituting for N, so we get:

$f_{k}=\mu_{k}(F sin \theta +mg)$

Now we can find the sum of forces in x:

$\Sigma F_{x}=0$

so after analyzing the diagram we can build our sum of forces to be:

$-f+F_{x}=0$

we know that:

$F_{x}=Fcos \theta$

so we can substitute the equations we already have in the sum of forces on x so we get:

$-\mu_{k}(F sin \theta +mg)+Fcos \theta=0$

so now we can solve for the force, we start by distributing $\mu_{k}$ so we get:

$-\mu_{k}F sin \theta -\mu_{k}mg)+Fcos \theta=0$

we add $\mu_{k}mg$ to both sides so we get:

$-\mu_{k}F sin \theta +Fcos \theta=\mu_{k}mg$

Nos we factor F so we get:

$F(cos \theta-\mu_{k} sin \theta)=\mu_{k}mg$

and now we divide both sides of the equation into $(cos \theta-\mu_{k} sin \theta)$ so we get:

$F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

which is our answer to part a.

Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:

$f_{s}=\mu_{s}(F sin \theta +mg)$

and

F cos θ = f

when substituting one into the other we get:

$F cos \theta=\mu_{s}(F sin \theta +mg)$

which can be solved for the static friction coefficient so we get:

$\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

which is the answer to part b.

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3 years ago

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Calcium chloride is used to make pickles. The calcium atom transfers electrons to chlorine atoms, leading to the formation of ch

GarryVolchara [31]

Calcium chloride contains ionic bonds.

Pennies contain metallic bonds.

Hydrochloric acid contains covalent bonds.

You're welcome.

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3 years ago

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An 18.5-cm-diameter loop of wire is initially oriented perpendicular to a 1.5-T magnetic field. The loop is rotated so that its

nikdorinn [45]

Answer:

The average induced emf in the loop is 0.20 V

Explanation:

Given:

Radius of loop $r = \frac{d}{2} = 9.25 \times 10^{-2}$ m

Magnetic field $B = 1.5$ T

Change in time $\Delta t = 0.20$ sec

According to the faraday's law,

Induced emf is given by

$\epsilon = -\frac{\Delta \phi}{\Delta t}$

Where $\phi =$ magnetic flux

$\phi = BA\cos0$ ( here $\theta = 0$ )

Where $A = \pi r^{2}$

We neglect minus sign because it's shows lenz law

$\epsilon = \frac{B \pi r^{2} }{\Delta t}$

$\epsilon = \frac{1.5 \times 3.14 \times (9.25 \times 10^{-2} )^{2} }{0.20}$

$\epsilon = 0.20$ V

Therefore, the average induced emf in the loop is 0.20 V

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Whats the kinetic energy of an object that has a mass of 12 kilograms and moves with a velocity of 10 m/s

Vesnalui [34]

Answer:

600J

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1/2mv^2

1/2(12)(10^2)

6(100)

600

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An aluminum cylinder weighing 30 N, 6 cm in diameter and 40 cm long, is falling concentrically through a long vertical sleeve of

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Answer: Velocity terminal = 0.093m/s

Explanation:

1. We start by evaluating the gap distance between the two cylinders as h = R(sleeve) - R(cylinder)

= (0.0604/2 - 0.06/2)m

= 2×10^-4

Surface are of the cylinder in the drop, which is required in order to evaluate the shearing stress can be expressed as A(cylinder) = π.d.L

= (π×0.06×0.4)m²

= 0.075m²

Since the force of the cylinder's weight is going to balance the shearing force on the walls, we can express the next equation and derive terminal velocity from it.

Shearing stress = u×V.terminal/h = 0.86×V/0.0002

= 4300Vterminal

Therefore, Fw = shearing stress × A

30N = 4300Vterminal × 0.075

V. terminal = 30/4300 m.s

V. terminal = 0.093m/s

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4 years ago