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Dovator [93]
3 years ago
13

A rift valley forms at a

Chemistry
1 answer:
vovikov84 [41]3 years ago
4 0
D is the answer your looking for
You might be interested in
A steel beam that is 7.00 m long weighs 326 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending
GrogVix [38]

Answer:

The answer to the question is

Suki is 1.12 m close to the end of the steel beam before the beam begins to tip

Explanation:

To solve the question we list out the known variables first

Length of steel beam, L = 7.00 m

Weight  of steal beam W = 326 N

Distance between supports = 3.00 m

Weight of Suki = 555 N

By moments theory, the sum of moments about a point = 0

and assuming the steal beam is uniform, the weight of the beam will act at the center of the beam

As described the supports are 2.00 m from each end of the beam thus

taking moments about one of the support with Suki between  the end of the beam and the support we have

Distance from center of beam to support = 1.5 m

Distance of Suki to the nearest support = x m

therefore 555 × x + 326 × 1.5 = 0

or x = -0.881 m on the other side of the support

Therefore Suki is (2 - 0.881) m or 1.12 m close to the end before the beam begins to tip

7 0
2 years ago
How many ml of a 0.50m solution of hno3 solution are needed to make 500 ml of 0.15m hno3
Anna71 [15]

Answer:

150.0 mL.

Explanation:

  • It is known that the no. of millimoles of HNO₃ before dilution = the no. of millimoles of HNO₃ after dilution.

∵ (MV) before dilution = (MV) after dilution.

<em>∴ V before dilution = (MV) after dilution / M before dilution</em> = (0.15 M)(500.0 mL)/(0.50 M) = <em>150.0 mL.</em>

7 0
2 years ago
What was removed from the glucose molecules when they bonded to form maltose
steposvetlana [31]

Answer:

Two hydrogen atoms and one oxygen atom (water) was removed.

Explanation:

yw:))

5 0
3 years ago
A 50.0g g sample of 16n decays to 12.5g in 14.4 seconds. What is its half life
mixas84 [53]
T is amount after time t 
<span>Ao is initial amount </span>
<span>t is time </span>
<span>HL is half life </span>

<span>log (At) = log [ Ao x (1/2)^(t/HL) ] </span>
<span>log (At) = log Ao + log (1/2)^(t/HL) </span>
<span>log (At) = log Ao + (t/HL) x log (1/2) </span>

<span>( log At - log Ao) / log (1/2) = t / HL </span>
<span>log (At/Ao) / log (1/2) = t / HL </span>

<span>HL = t / [( log (At / Ao)) / log (1/2) ] </span>

<span>HL = 14.4 s / [ ( log (12.5 / 50) / log (1/2) ] </span>

<span>HL = 14.4 s / 2 = 7.2 seconds </span>
8 0
3 years ago
The half life of radon-222 is 3.8 days. How Much of a 100g sample is left after 15.2 days
professor190 [17]

Answer:  

6.2 g  

Explanation:  

In a first-order decay, the formula for the amount remaining after <em>n</em> half-lives is  

N = \frac{N_{0}}{2^{n}}  

where  

<em>N</em>₀ and <em>N</em> are the initial and final amounts of the substance  

1. Calculate the <em>number of half-lives</em>.  

If t_{\frac{1}{2}} = \text{3.8 da}  

n = \frac{t}{t_{\frac{1}{2}}} = \frac{\text{15.2 da}}{\text{3.8 da}}= \text{4.0}

2. Calculate the <em>final mass</em> of the substance.  

\text{N} = \frac{\text{100 g}}{2^{4.0}} = \frac{\text{100 g}}{16} = \text{6.2 g}

4 0
3 years ago
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