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tiny-mole [99]
3 years ago
10

a car's engine is only 12 % efficient at converting chemical energy in gasoline to mechanical energy. If it takes 18000 N of for

ce to keep the car moving at a constant speed of 21 m/s, how much chemical energy would be needed to move the car a distance of 450m at this speed?
Physics
1 answer:
Georgia [21]3 years ago
6 0
I wish i could help but idk the answer

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How to find yield strength of a load vs deflection?
liraira [26]
Σ/ε
σ = F/A
ε = ΔL/L
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L = lenght
ΔL = |old lenght - new lenght|
6 0
3 years ago
What is this feature?
Viktor [21]

Answer: c Delta

Explanation:

A delta is a landform created by deposition of sediment that is carried by a river as the flow leaves its mouth and enters slower-moving or stagnant water. This occurs where a river enters an ocean, sea, estuary, lake, reservoir, or (more rarely) another river that cannot carry away the supplied sediment.

Hope this helps!!

3 0
3 years ago
Please solve for me am confused​
Lynna [10]

Answer:

what does it say

Explanation:

8 0
4 years ago
There are 5510 lines per centimeter in a grating that is used with light whose wavelegth is 467 nm. A flat observation screen is
Mademuasel [1]

Answer:

1.696 nm

Explanation:

For a diffraction grating, dsinθ = mλ where d = number of lines per metre of grating = 5510 lines per cm = 551000 lines per metre and λ = wavelength of light = 467 nm = 467 × 10⁻⁹ m. For a principal maximum, m = 1. So,

dsinθ = mλ = (1)λ = λ

dsinθ = λ

sinθ = λ/d.

Also tanθ = w/D where w = distance of center of screen to principal maximum and D = distance of grating to screen = 1.03 m

From trig ratios 1 + cot²θ = cosec²θ

1 + (1/tan²θ) = 1/(sin²θ)

substituting the values of sinθ and tanθ we have

1 + (D/w)² = (d/λ)²

(D/w)² = (d/λ)² - 1

(w/D)² = 1/[(d/λ)² - 1]

(w/D) = 1/√[(d/λ)² - 1]

w = D/√[(d/λ)² - 1] = 1.03 m/√[(551000/467 × 10⁻⁹ )² - 1] = 1.03 m/√[(1179.87 × 10⁹ )² - 1] = 1.03 m/1179.87 × 10⁹  = 0.000848 × 10⁻⁹ = 0.848 × 10⁻¹² m = 0.848 nm.

w is also the distance from the center to the other principal maximum on the other side.

So for both principal maxima to be on the screen, its minimum width must be 2w = 2 × 0.848 nm = 1.696 nm

So, the minimum width of the screen must be 1.696 nm

4 0
3 years ago
Hii! help asap. i’ll give brainliest thanks!
o-na [289]

I believe it’s the mass of the box but I don’t no if I’m right

Hope this helped
5 0
3 years ago
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