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34kurt
2 years ago
5

The gap between electrodes in a spark plug is 0.060 cm. Producing an electric spark in a gasoline-air mixture requires an electr

ic field of 3.0 × 106 V/m. What minimum potential difference must be supplied by the ignition circuit to start a car?
Physics
1 answer:
VladimirAG [237]2 years ago
6 0

The minimum potential difference must be supplied by the ignition circuit to start a car is -1800 V

<u>Explanation:</u>

Given data,

E= 3 ×10 ⁶ Δx=0.06/100

We have to find the minimum potential difference

E= -ΔV/Δx

ΔV=- E × Δx

ΔV =-3 ×10 ⁶ . 0.06/100

ΔV=-1800 V

The minimum potential difference must be supplied by the ignition circuit to start a car is -1800 V

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The lighting needs of a storage room are being met by six fluorescent light fixtures, each fixture containing four lamps rated a
Amanda [17]

Answer:

amount of energy  = 4730.4 kWh/yr

amount of money = 520.34 per year

payback period = 0.188 year

Explanation:

given data

light fixtures = 6

lamp = 4

power = 60 W

average use = 3 h a day

price of electricity = $0.11/kWh

to find out

the amount of energy and money that will be saved and simple payback period if the purchase price of the sensor is $32 and it takes 1 h to install it at a cost of $66

solution

we find energy saving by difference in time the light were

ΔE = no of fixture × number of lamp × power of each lamp × Δt

ΔE is amount of energy save and Δt is time difference

so

ΔE = 6 × 4 × 365 ( 12 - 9 )

ΔE = 4730.4 kWh/yr

and

money saving find out by energy saving and unit cost that i s

ΔM = ΔE × Munit

ΔM = 4730.4 × 0.11

ΔM = 520.34 per year

and

payback period is calculate as

payback period = \frac{excess initial cost}{\Delta M}

payback period = \frac{32 + 66}{520.34}

payback period = 0.188 year

8 0
2 years ago
A rock is rolling down a hill. At position 1, it’s velocity is 2.0 m/s. Twelve seconds later, as it passes position 2, it’s velo
mr Goodwill [35]

Answer

Hi,

correct answer is {D} 3.5 m/s²

Explanation

Acceleration is the rate of change of velocity with time. Acceleration can occur when a moving body is speeding up, slowing down or changing direction.

Acceleration is calculated by the equation =change in velocity/change in time

a= {velocity final-velocity initial}/(change in time)

a=v-u/Δt

The units for acceleration is meters per second square m/s²

In this example, initial velocity =2.0m/s⇒u

Final velocity=44.0m/s⇒v

Time taken for change in velocity=12 s⇒Δt

a= (44-2)/12  = 42/12

3.5 m/s²

Best Wishes!

5 0
2 years ago
Find the kinetic of a 0.1 kilogram toy truck moving at a speed of 1.1 meters per second
omeli [17]
KE = kinetic energy
PE = potential energy
GPE = gravitational potential energy
energy is always measured in Joules (J)

KE = (0.5) times the mass times the velocity^2
square the velocity first

Mass = (KE x 2) / v^2
square the velocity first, then double the kinetic energy, then divide
mass is measured in kg

velocity = sqrt(KE x 2 / m)
velocity can be called speed, like in the the second problem
remember to find the square root after you double the KE and divide that by the mass.
for example: if after you doubled KE and divided it by the mass you got sqrt(20), the answer would be about 4.47

GPE = mass x gravitational pull (about 9.8 m/s^2 on earth) x height

height = (PE) / (g x m)
do g x m first

So for question 1:
KE = (0.5)0.1 x 1.1^2
always square the velocity first:
KE = (0.5)0.1 x 1.21
KE = 0.0605
so if you rounded it to the nearest hundreths you would get KE = 0.06 J
don't forget the unit of energy is in Joules
5 0
2 years ago
A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,00
Aleksandr-060686 [28]

Answer:

Explanation:

We have the following relation between power, P and intensity, I

I = P/(4*pi*r^2)

= 10^3/(4*pi*(35000*10^3))

= 6.5*10^-14 W/M^2

We also have the following relationship between electric field and electromagnetic radiation thus

I = (ceE^2)/2

Hence E = \sqrt{2I/ce}

substituting the values of I, c and e, we have

7*10^-6 V/m

3 0
2 years ago
Honey bees can acquire a small net charge on the order of 1 pC as they fly through the air and interact with plants. Estimate th
Inessa [10]

Answer:

F = 2.01*10^-16N -^k

Explanation:

In order to calculate the magnetic force perceived by the bee, you use the following formula:

F=qv\ X\ B            (1)

q: charge of the bee = 1pC = 1*10^-12 C

The average speed of a bee and the magnetic field of the earth are:

v = 6.70m/s

B = 30*10^-6 T

The bee is flying to the west (-^i). You consider that the magnetic field direction is to the north (^j). Then, the direction of the magnetic force is:

-^i X ^j = -^k

You replace the values of the parameters in the equation (1), in order to calculate the magnitude of the force:

|F|=qvB=(1*10^{-12}C)(6.70m/s)(30*10^{-6}T)=2.01*10^{-16}N

The magnetic force perceived by the bee is 2.01*10^-16N in the -^k direction, that is, toward the ground

5 0
2 years ago
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