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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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Delilah does 170 Joules of work in 30 seconds. Adam does 260 Joules of work in 20 seconds. Who was more powerful?​
sp2606 [1]

Power = \frac{Work}{Time}

Delilah: 170J/30s = 5.66 W

Adam: 260J/20s = 13 W

6 0
3 years ago
A spring with a spring constant of 120 J/m2 is fixed to a wall, free to oscillate. On the other end, a ball with a mass of 1500
Neporo4naja [7]

Answer:

A. 4.47 m/s

Explanation:

As the ball oscillates, it mechanical energy, aka the total kinetic and elastics energy stays the same. For the ball to be at maximum speed, its elastic energy i 0 and vice versa. When the ball is at rest, its kinetic energy is 0 and its elastic energy is at maximum at 50 cm, or 0.5 m

1500 g = 1.5 kg

E_e = E_k

kx^2/2 = mv^2/2

120*0.5^2/2 = 1.5*v^2/2

15 = 0.75v^2

v^2 = 15 / 0.75 = 20

v = \sqrt{20} = 4.47 m/s

5 0
3 years ago
How do muscles help the body move?
Finger [1]

Answer:

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A. The muscle contracts causing the tendon to pull the bone.

3 0
3 years ago
Read 2 more answers
A child is riding a merry-go-round, which has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745
Elena L [17]

w = instantaneous angular speed = 1.25 rad/s

r = radius = 4.65 m

α = angular acceleration = 0.745 rad/s²

centripetal acceleration is given as

a_{c} = rw²

a_{c} = (4.65 ) (1.25)² = 7.27 m/s²

tangential acceleration is given as

a_{t} = rα

a_{t} =4.65 x 0.745 = 3.46 m/s²

angle is given as

\theta = tan^{-1}(\frac{a_{c}}{a_{t}})

\theta = tan^{-1}(\frac{7.27}{3.46})

\theta = 64.5 deg

5 0
3 years ago
A uniform magnetic field of magnitude 0.72 T is directed perpendicular to the plane of a rectangular loop having dimensions 8.2
Liono4ka [1.6K]

Answer:

Explanation:

Magnetic flux is expressed as the product of magnetic field and cross sectional area.

Φ = BAsintheta

Given

B = 0.72T

A = 8.2cm×14cm

A = 0.082m × 0.14m

Area = 0.01148m²

Theta = 90°

Substitute into the formula

Φ = BAsintheta

Φ = 0.72(0.01148)sin90°

Φ = 0.72(0.01148)(1)

Φ = 0.0082656

Hence the magnetic flux through the loop is 8.2656 × 10^-3 Weber

8 0
3 years ago
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