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3 years ago

A highly volatile substance has an initial mass of 1200 g and its mass is reduced by 12% each second.

Physics

1 answer:

Softa [21]3 years ago

4 0

Answer:

Explanation:

a) 1.00 - 0.12 = 0.88

m = 1200(0.88)^t

b) t = ln(m/1200) / ln(0.88)

c) m = 1200(0.88)^10 = 334.20 g

d) t = ln(10/1200) / ln(0.88) = 37.451... = 37 s

e) t = ln(1/1200) / ln(0.88) = 55.463... = 55 s

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The age of the universe is thought to be about 14 billion years. Assuming two significant figures, write this in powers of ten i

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A ) 14 billion years = 14,000,000,000
$1.4 * 10^{10}$
b ) 1 year = 365 · 24 · 3,600 = 31,536 · 10³
$1.4 * 10 ^{10}*3.1536 * 10^{7} =4.4 * 10 ^{17}$

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4 years ago

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A uniform horizontal beam of weight 481 N and length 3.32 m has two weights hanging from it. One weight of 381 N is located 0.87

Andre45 [30]

Answer:

1143 N at 1.59 m from the left end

Explanation:

For the system to produce equilibrium, the total force and moment must be 0. Since the total weight downward is

481 + 381 + 281 = 1143 N

Therefore the magnitude of the force acting upward to balance this system must be the same of 1143 N

That alone is not enough, we also need the position of the force for the total moment to be 0.

Let x be the length from the this upward force to the left side. And let the left point be the point of reference for moment arm:

481 * 3.32/2 + 381 * 0.8798 + 281*(3.32 - 0.8798) - 1143*x = 0

x = (481*1.66 + 381 * 0.8798 + 281*2.4402)/1143 = 1.59m

5 0

4 years ago

A particle moves according to a law of motions=f(t),t≥0 Wheretismeasured in seconds andsin feet.a) Find the velocity at timet.b)

Ray Of Light [21]

Answer:

a) $v(t) = \frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2}$

b) 3s

c) t < 3s

d) 1.8ft

e) particle is speeding up when t > 5.385 s and vice versa, slowing down when t < 5.385 s

Explanation:

a)Suppose the equation for motion is:

$f(t) = \frac{9t}{t^2+9}$

Then the velocity is the derivative of the motion function

$v(t) = \frac{df(t)}{dt} = (9t(t^2+9)^{-1})^'$

From here we can apply product rule

$v(t) = (9t)^{'}(t^2+9)^{-1} + (9t)((t^2+9)^{-1})^'$

$v(t) = \frac{9}{t^2+9} - \frac{9t(2t)}{(t^2+9)^2}$

$v(t) = \frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2}$

b) The particle is at rest when v(t) = 0:

$\frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2} = 0$

$\frac{9}{t^2+9} = \frac{18t^2}{(t^2+9)^2}$

Multiply 2 sides by $(t^2+9)^2$ we have:

$t^2 + 9 = 2t^2$

$t^2 = 9$

$t = 3s$

$\frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2} > 0$

$\frac{9}{t^2+9} > \frac{18t^2}{(t^2+9)^2}$

Multiply 2 sides by $(t^2+9)^2$ we have:

$t^2 + 9 > 2t^2$

$t^2 < 9$

$t < 3s$

(d) As particle is moving in positive direction when t < 3s and negative direction when t > 3s, we can calculate the distance it's moving up to 3s and then after 3s

$f(3) = \frac{9*3}{3^2+9} = \frac{27}{18} = 1.5 ft$

At 3s, particle is changing direction to negative, so its position at 6s is

$f(6) = \frac{9*6}{6^2+9} = \frac{54}{45} = 1.2 ft$

Therefore from 3s to 6s it would have moved a distance of 1.5 - 1.2 = 0.3 ft

Then the total distance it has moved in the first 6 s is 1.5 + 0.3 = 1.8 ft

e) Acceleration is the derivative of velocity function:

$a(t) = \frac{dv(t)}{dt} = (\frac{9}{t^2+9} - \frac{18t^2}{(t^2+9)^2})^'$

$a(t) = -\frac{9(2(t^2+9))(2t)}{(t^2+9)^2} - \frac{36t}{(t^2+9)^2} + \frac{18t^2(2(t^2+9))(2t)}{(t^2+9)^3}$

$a(t) = -\frac{36t}{t^2+9} - \frac{36t}{(t^2+9)^2} + \frac{72t^3}{(t^2+9)^2}$

$a(t) = -\frac{54t}{t^2+9} + \frac{72t^3 - 36t}{(t^2+9)^2}$

Particle is speeding up when a(t) > 0:

$-\frac{54t}{t^2+9} + \frac{72t^3 - 36t}{(t^2+9)^2} > 0$

$\frac{72t^3 - 36t}{(t^2+9)^2} > \frac{54t}{t^2+9}$

as $t \& (t^2 + 9)^2 \geq 0$ we can multiply/divide both sides by it:

$8t^2 - 4 > 6(t^2+9)$

$8t^2 > 6t^2 + 58$

$t^2 > 29$

$t > \sqrt(29) \approx 5.385 s$

so particle is speeding up when t > 5.385 s and vice versa, slowing down when t < 5.385 s

8 0

3 years ago

Sound waves travel fastest through ______

ehidna [41]

Sounds waves travel fastest through solids

7 0

3 years ago

Read 2 more answers

Spiderman, whose mass is 74.0 kg, is dangling on the free end of a 11.0-m-long rope, the other end of which is fixed to a tree l

Anestetic [448]

Answer:

W = -1844.513 J

Explanation:

GIVEN DATA:

mass of spider man is m 74 kg

vertical displacement if spider is 11 m

final displacement = 11 cos 60.6 = - 6.753 m

change in displacement is = -6.753 - (-11) = 4.25 m

gravity force act on spiderman is f = mg = 74 × 9.8 = 725.2 N

work done by gravity is $W = F \delta r cos\theta$

$W = 725.2 \times 4.25 \times cos 180$

where 180 is the angle between spiderman weight and displacement

W = -1844.513 J

7 0

3 years ago