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3 years ago

14

Which of the following would be the least reliable source of information? A science textbook B government internet site C weekly

news magazine D reference journal at the library

2 answers:

ddd [48]3 years ago

7 0

In my opinion D ....because science textbook is unbeatable..... government internet site can give you a lot of info..and weekly magazine is all about news happening around you and it also contains a lot of information on science...history

so ans is D

Neporo4naja [7]3 years ago

3 0

The correct answer should be C

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A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates as functions of time are given by x = 18.0t and y =

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First derivative: x' = 18 y' = 4 - 9.8t

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Velocity vector: V = (18) i + (4 - 9.8t) j

Acceleration vector A = (- 9.8) j

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You have a rod with a length of 146.4 cm. You prop up one end on a brick which is 3.8 cm thick. Your uncertainty in measuring th

AfilCa [17]

Answer:

$\partial \theta = 0.003$

Explanation:

we know that

$sin\theta = \frac{3.8}{146.4}$

$\theta = sin^{-1} \frac{3.8}{146.4}$

$\theta = 1.484°$

$\theta = 1.484° *\frac{\pi}{180} = 0.0259 radians$

as we see that $sin\theta = \theta$

relative error $\frac{\partial \theta}{\theta} = \frac{\partial X}{X_1} +\frac{\partial X}{X_2}$

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$X_2$ IS THE HEIGHT OF ROAD

$\partial X$ = uncertainity in measuring distance

$\partial X = 0.05$

Putting all value to get uncertainity in angle

$\frac{\partial \theta}{0.0259} = \frac{0.05}{3.8} +\frac{0.05}{146.4}$

solving for $\partial \theta$ we get

$\partial \theta = 0.003$

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3 years ago

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Elanso [62]

Answer:

The value is $v = 2.3359 *10^{4} \ m/s$

Explanation:

From the question we are told that

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$T__{E}} = KE__{i}} + KE__{e}}$

Here $KE_i$ is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

$KE_i = \frac{1}{2} * m * u^2$

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And $KE_e$ is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

$KE_e = \frac{1}{2} * m * v_e^2$

Here $v_e$ is the escape velocity from earth which has a value $v_e = 11.2 *10^{3} \ m/s$

=> $KE_e = \frac{1}{2} * m * (11.3 *10^{3})^2$

=> $KE_e = 6.272 *10^{7} \ \ m \ \ J$

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

$KE_p = \frac{1}{2} * m * v^2$

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$T__{E}} = KE_p$

So

$2.101 *10^{8} m + 6.272 *10^{7} m = \frac{1}{2} * m * v^2$

=> $5.4564 *10^{8} = v^2$

=> $v = \sqrt{5.4564 *10^{8}}$

=> $v = 2.3359 *10^{4} \ m/s$

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3 years ago