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3 years ago

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Group 1A elements have similar properties because they ______. Select all that apply

1 answer:

Thepotemich [5.8K]3 years ago

8 0

Answer:

D............................

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Suppose that it takes 0.6 seconds for a mass on a spring to move from its highest position to its lowest position. What is the p

o-na [289]

The mass on the spring is bouncing.

We would call it a wave-like motion, except that it all stays in the same place. But, just like a wave, moving from the highest position to the lowest position
is one-half of a full wiggle.

(The other half consists of moving from the lowest position back up to the
highest position, where it started from.

So, half of the wave-like motion takes 0.6 seconds.

A full cycle of the wave motion ... the actual period of the bounce,
is double that much time . . .
1.2 seconds.

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4 years ago

contractor will use a package emulsion having a specific gravity of 1.25 and relative bulk strength of 130 to open an excavation

tester [92]

Answer:

The burden distance is 7 ft

Solution:

As per the question:

Specific gravity of package emulsion, $SG_{E} = 1.25$

Specific gravity of diabase rock, $SG_{R} = 2.76$

Diameter of the packaged sticks, d = 3 in

Now,

To calculate the first trail shot burden distance, B:

$B = [\frac{2SG_{E}}{SG_{R}} + 1.5]\times d$

$B = [\frac{2\times 1.25}{2.76} + 1.5]\times 3 = 7.22$

B = 7 ft

5 0

3 years ago

Please help! this this timed!!!

VMariaS [17]

Answer:

Thermal heat

Explanation:

The water heats up which it means it is thermal (heat/hot)

4 0

3 years ago

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An object moving due to gravity can be described by the motion equation y=y0+v0t−12gt2, where t is time, y is the height at tha

LenaWriter [7]

Answer:

$t=6.4534 s$

Explanation:

This is an exercise where you need to use the concepts of <em>free fall objects</em>

Our <u>knowable variables</u> are initial high, initial velocity and the acceleration due to gravity:

$y_{0}=75m$

$v_{oy} =20m/s$

$g=9.8 m/s^{2}$

At the end of the motion, the <u><em>rock hits the ground</em></u> making the final high y=0m

$y=y_{o}+v_{oy}*t-\frac{1}{2}gt^{2}$

If we <em>evaluate the equation</em>:

$0=75m+(20m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}$

This is a classic form of <u><em>Quadratic Formula</em></u>, we can solve it using:

$t=\frac{-b ± \sqrt{b^{2}-4ac } }{2a}$

$a=-4.9\\b=20\\c=75$

$t=\frac{-(20) + \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=-2.37s$

$t=\frac{-(20) - \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=6.4534s$

Since the <u><em>time can not be negative</em></u>, the <em>reasonable answer</em> is

$t=6.4534s$

8 0

3 years ago

Juan whose weight is 500 N is standing on the ground. The force the ground exerts on

Anna71 [15]

Answer:

more than 500 n i think the answer will

8 0

3 years ago

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