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3 years ago

12

0. The volume of a cube is 27cm3. Find the length of edge of cube.

1 answer:

Dennis_Churaev [7]3 years ago

7 0

Answer:

27 cm3

Explanation:

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If a set of scores has a mean of 100.00 and

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Answer:

Variance is 25

Explanation:

Recall that the standard deviation is defined as the square root of the variance. therefore, if you know the standard deviation $(\sigma)$ , square it and you get the variance:

$Variance= \sigma^2\\Variance= 5^2= 25$

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3 years ago

A picture hangs evenly from two wires of equal length. The wires each make an angle of 45 degrees with the ceiling. If the mass

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41.578N

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3 years ago

The largest origin of nonpoint pollution in the united states is A) rain B) industry C) agriculture

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The answer would be : A

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3 years ago

The speed at which the package hits the ground is really fast! If a package hits the ground at such a speed, it can be crushed a

Fed [463]

Answer:

B. Decrease the plane’s speed and height

Explanation:

Decrease in speed of plane:

When the speed of the plane is decreased then initial velocity of the package will be lesser. As a result, final velocity will also be lesser because in this case final velocity depends upon the initial velocity.

Decrease in height:

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Loss in potential energy = Gain in Kinetic energy

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It means with the decrease in height less potential energy will be converted to kinetic energy. As package has constant mass so it will gain less velocity.

7 0

4 years ago

A block with mass m =6.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m.

Zanzabum

Answer

given,

mass of block (m)= 6.4 Kg

spring is stretched to distance, x = 0.28 m

initial velocity = 5.1 m/s

a) computing weight of spring

k x = m g

$k = \dfrac{mg}{x}$

$k = \dfrac{6.4 \times 9.8}{0.28}$

k = 224 N/m

b) $f = \dfrac{\omega}{2\pi}$

$\omega = \sqrt{\dfrac{k}{m}}= \sqrt{\dfrac{224}{6.4}} = 5.92 \ rad/s$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{224}{6.4}}$

$f =0.94\ Hz$

c) $v_b = -v cos \omega t$

$v_b = -5.1 \times cos (5.92 \times 0.42)$

$v_b = 4.04\ m/s$

d) $a_{max} = v \omega$

$a_{max} = 4.04 \times 5.92$

$a_{max} =23.94\ m/s^2$

e) $Y =- A sin (\omega t)$

$A = \dfrac{v}{\omega}$

$A = \dfrac{4.04}{5.92}$

A = 0.682 m

$Y =- 0.682 \times sin (5.92 \times 0.42)$

$Y =- 0.42$

Force = $m \omega^2 |Y|$

= $6.4 \times 5.92^2\times 0.42$

F = 94.20 N

4 0

4 years ago