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oee [108]
3 years ago
11

I Need help with this problem i don’t know what to do

Physics
1 answer:
Mazyrski [523]3 years ago
8 0

Answer:

The density of the sample is 36 g/cm³

Explanation:

m= 972g

l=3cm

V = l³ = 3³ = 27 cm³

density = mass/volume

= 972/27

= 36 g/cm³

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An object with a non-zero speed must be _________.
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It would be either A or C if its still moving and not stopping
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The tension in the rope securing the upper pulley and the force that must be applied to keep the system in equilibrium ​
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Answer:

Explanation:

In a frictionless system with no acceleration, the tension in the rope must be F along its entire length

FBD analysis of the lower pulley has two upward acting tension vectors F and one downward acting weight vector W

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FBD analysis of the upper pulley has one upward acting support vector T and three downward acting tension vectors F

T = 3F

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T = 1.5W

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3 years ago
15) In which of the following cases would sound reach each ear out of phase? A. You are standing directly in front of the sound
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In one of the greatest coincidences to arise on Brainly in quite some time, Choice-C is the correct choice for BOTH #15 and #16 .

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If the sun is scaled down to 9.5 inches in diameter, the earth would be about _____ inches in diameter
vovangra [49]

Answer: 0.087 inches

Explanation:

This problem can be solved applying the <u>Rule of three</u>, which is a  mathematical rule to find out an amount that is with another quantity given in the same relation as other two also known.

In this sense, the actual diameter of the Sun is 1,391,020 km and the actual diameter of the Earth is 12,742 km.

Now, applying the <u>Rule of three:</u>

<u />

Sun 1,391,020 km ------ 9.5 inches

Earth 12,742 km --------- ? inches

? inches=\frac{(12,742 km)(9.5 inches)}{1,391,020 km}

? inches=0.087 inches

Therefore:

If the sun is scaled down to 9.5 inches in diameter, the earth would be about <u>0.087 inches </u>in diameter.

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3 years ago
A 50.0 mg sample of an unknown radioactive substance was placed in storage and its mass measured periodically. After 19.7 days,
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<h2>Answer: 4.928 days </h2><h2 />

Explanation:

This problem can be solved using the <u>Radioactive Half Life Formula: </u>

<u></u>

A=A_{o}.2^{\frac{-t}{h}} (1)

Where:

A=3.13mg is the final amount of the material

A_{o}=50mg is the initial amount of the material

t=19.7days is the time elapsed

h is the half life of the material (the quantity we are asked to find)

Knowing this, let's substitute the values and find h from (1):

3.13mg=(50mg)2^{\frac{-19.7days}{h}} (2)

\frac{3.13mg}{50mg}=2^{\frac{-19.7days}{h}} (3)

Applying natural logarithm in both sides:

ln(\frac{3.13mg}{50mg})=ln(2^{\frac{-19.7days}{h}}) (4)

-2.77=-\frac{19.7days}{h}ln(2) (5)

Clearing h:

h=\frac{-19.7days}{-2.77}(0.693) (6)

Finally:

h=4.928days

6 0
3 years ago
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