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34kurt
3 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]3 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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What is the correct answer?
pentagon [3]

Answer:

B) x^2+6x+8

Explanation:

x-4 | x^3+2x^2-16x-32

    -  x^3-4x^2             <-- (x-4)(x^2)

_________________

              6x^2-16x-32

           -  6x^2-24x     <-- (x-4)(6x)

_________________

                          8x-32

                       -  8x-32 <- (x-4)(8)

___________________________

                                 0 | x^2+6x+8

This means the answer is B) x^2+6x+8

             

3 0
3 years ago
A baseball pitcher throws a ball across home plate. The ball travels 13.40m in 0.357s Determine the
Tanya [424]

Answer: I'm not sure what it needs to be rounded to, but I got 37.53501401 m/s

Explanation: The formula for speed is speed = distance/time. You plug in the distance (13.40) and the time (0.357), then divide 13.40 by 0.357

I hope this helps! :)

4 0
2 years ago
Select answer about motion and newtons law and explain why it's correct 23POINTS NEED ASAP
zlopas [31]

The train would need the greatest amount of force due to weight! If you think of it, a baseball won't need much force to stop it, but if you have a heavy train, it will need excessive force to stop the train. The answer would be #3


I hope this answer helps!

Sorry if it doesn't make sense, as I don't know that much about physics! I am just thinking of what makes sense.

3 0
3 years ago
Let v1, , vk be vectors, and suppose that a point mass of m1, , mk is located at the tip of each vector. The center of mass for
g100num [7]

Answer:

Explanation:

Center of mass is give as

Xcm = (Σmi•xi) / M

Where i= 1,2,3,4.....

M = m1+m2+m3 +....

x is the position of the mass (x, y)

Now,

Given that,

u1 = (−1, 0, 2) (mass 3 kg),

m1 = 3kg and it position x1 = (-1,0,2)

u2 = (2, 1, −3) (mass 1 kg),

m2 = 1kg and it position x2 = (2,1,-3)

u3 = (0, 4, 3) (mass 2 kg),

m3 = 2kg and it position x3 = (0,4,3)

u4 = (5, 2, 0) (mass 5 kg)

m4 = 5kg and it position x4 = (5,2,0)

Now, applying center of mass formula

Xcm = (Σmi•xi) / M

Xcm = (m1•x1+m2•x2+m3•x3+m4•x4) / (m1+m2+m3+m4)

Xcm = [3(-1, 0, 2) +1(2, 1, -3)+2(0, 4, 3)+ 5(5, 2, 0)]/(3 + 1 + 2 + 5)

Xcm = [(-3, 0, 6)+(2, 1, -3)+(0, 8, 6)+(25, 10, 0)] / 11

Xcm = (-3+2+0+25, 0+1+8+10, 6-3+6+0) / 11

Xcm = (24, 19, 9) / 11

Xcm = (2.2, 1.7, 0.8) m

This is the required center of mass

6 0
3 years ago
A woman rides a carnival Ferris wheel at radius 20 m, completing 6.0 turns about its horizontal axis every minute. What are (a)
natta225 [31]

Answer:

Explanation:

Given

radius of ferris wheel is 20 m

It completes 6 turns in 1 min

i.e. 1 turn in 10 sec

Therefore its angular velocity is \omega =\frac{2\pi }{10} rad/s

(a)Period of motion is 10 s

Magnitude of centripetal acceleration is \omega ^2r

=\left ( \frac{\pi }{5}\right )^2\times 20=7.89 m/s^2

(b)Highest point will be 40 m

(c) lowest point 0 m i.e. at ground

7 0
3 years ago
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