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3 years ago

18. Write conversion factors (as ratios) for the number of:

Physics

1 answer:

Romashka [77]3 years ago

6 0

Answer:

Conversion tables show:

1 m = 1.09361 yds

1 Lit = .26418 gal = 1.05672 qt or 1 qt = .944632 Lit

1 lb = .45359 kg = 2.2046 Lbs / Kg

So X yds = X m * 1.09361 yds / m = 1.09361 * X yds

Likewise X Lit = X qt / 1.05672 qt/ Lit = X / 1.05672 Lit = .94632 X Lit

So X Lbs = X kg * 2.2046 Lbs / Kg = 2.2046 Lbs

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konstantin123 [22]

Answer:

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(b) The maximum height of the projectile is 4,591.84 m

Explanation:

Given;

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angle of projection, θ = 30⁰

acceleration due to gravity, g = 9.8 m/s²

(a) The range of the projectile in meters;

$R = \frac{u^2sin \ 2\theta}{g} \\\\R = \frac{600^2 sin(2\times 30^0)}{9.8} \\\\R = \frac{600^2 sin (60^0)}{9.8} \\\\R = 31,813.18 \ m$

(b) The maximum height of the projectile in meters;

$H = \frac{u^2 sin^2\theta}{2g} \\\\H = \frac{600^2 (sin \ 30)^2}{2\times 9.8} \\\\H = \frac{600^2 (0.5)^2}{19.6} \\\\H = 4,591.84 \ m$

$v^2 = u^2 + 2gh\\\\v^2 = 600^2 + (2 \times 9.8)(4,591.84)\\\\v^2 = 360,000 + 90,000.064\\\\v = \sqrt{450,000.064} \\\\v = 670.82 \ m/s$

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Positive charge Q is distributed uniformly along the x-axis fromx=0 to x=a. A positive point charge q is located on the positive

dybincka [34]

Answer:

a. b- x= y

dx = -dy

b. F = $\frac{-kQqi}{r (a+r)}$

c. F = $\frac{-kQqi}{r^{2} }$

Explanation:

a. x components:

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= $\frac{kQdx}{(a(a+r-x)^2}$

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