How to comment on brainly.com and I'm only 8-years-old so keep it simple please

8. Block A shown in the figure below weighs 2000 N. The chord attached to A passes over a

Kobotan [32]

Answer:

Read the passage. Then, answer the questions about the metaphor in boldface in the text.

Lately, I've been so overwhelmed with school and sports. There was a time when I enjoyed going to classes and going to practice every afternoon. Now, everything is piling up and wearing me down. Thankfully, I get to see you every day. You are truly the sunshine of my life. Thank you for making me laugh when I'm feeling down.

What is the context of the passage?

What is being compared in the metaphor?

What is the meaning of the metaphor?

Explanation:

4 0

2 years ago

A clean machine is a _______________ machine.

solniwko [45]

A clean machine is a clean machine :-)

4 0

2 years ago

A car is traveling at sea level at 78 mi/h on a 4% upgrade before the driver sees a fallen tree in the roadway 150 feet away. Th

Dmitrij [34]

Answer: V = 47.7 mi/hr

Explanation:

first we calculate elements of aero-dynamic resistance

Ka = p/2 * CD * A.f

p is the density of air(0.002378 slugs/ft^3) for zero altitude, CD is the drag coefficient(0.35) and A.f is the front region of the vehicle

so we substitute

Ka = 0.002378/2 * 0.35 * 18

Ka = 0.00749

Now we calculate the final speed of the vehicle (V2) using the relation;

S = (YbW/2gKa)In[ (UW + KaV1^2 + FriW ± Wsinθg) / (UW + KaV2^2 + FriW ± Wsinθg)

so

WE SUBSTITUTE

150 = (1.04 * 2700 / 2 * 32.2 * 0.0075) In [(0.8 * 2700 + 0.0075 *(78mil/hr * 5280ft/1min * 1hr/3600s)^2 + 0.017 * 2700 ± 2700 * 0.04) / (0.8 * 2700 + 0.0075 * V2^2 + 0.017 * 2700 ± 2700 * 0.04)]

150 = (2808/0.483) In [(2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

150 = 5813.66 In [ (2160 + 98.16 + 153.9) / ( 2160 + 0.0075V2^2 + 153.9)]

divide both sides by 5813.66

0.0258 = In [ (2412.06) / ( 0.0075V2^2 + 2313.9)]

take the e^ of both side

e^0.0258 = (2412.06) / ( 0.0075V2^2 + 2313.9)

1.0261 = (2412.06) / ( 0.0075V2^2 + 2313.9)]

(0.0075V2^2 + 2313.9) = 2412.06 / 1.0261

(0.0075V2^2 + 2313.9) = 2350.7

0.0075V2^2 = 2350.7 - 2313.9

0.0075V2^2 = 36.8

V2^2 = 36.8 / 0.0075

V2^2 = 4906.6666

V2 = √4906.6666

V2 = 70.0476 ft/s

converting to miles per hour

V2 = 70.0476 ft/s * 1 mil / 5280 ft * 3600s / 1hr

V = 47.7 mi/hr

8 0

3 years ago

It is said that Archimedes discovered his principle during a bath while thinking about how he could determine if KingHiero‘s cro

Rudiy27

Answer:

the crown is false densty= 12556kg/m^3[/tex]

Explanation:

Hello! The first step to solve this problem is to find the mass of the crown, this is found using the weight of the crown in the air by means of the equation for the weight.

W=mg

W=weight(N)=31.4N

M=Mass

g=gravity=9.81m/S^2

solving for M

m=W/g

$m=\frac{31.4N}{9.81m/S^2}=3.2kg$

The second step is find the volume of crown remembering that when an object is weighed in the water the result is the subtraction between the weight of the object and the buoyant force of the water which is the product of the volume of the crown by gravity by density of water

$F=mg-\alpha V g$

Where

F=weight in water=28.9N

m=mass of crown=3.2kg

g=gravity=9.81m/S^2

α=density of water=1000kg/m^3

V= crown´s volume

solving for V

$V=\frac{mg-F }{g \alpha } =\frac{(3.2)(9.81)-28.9}{9.81(1000)} =0.000254m^3$

finally, we remember that the density is equal to the index between mass and volume

$\alpha =\frac{m}{v} =\frac{3.2}{0.000254} =12556kg/m^3$

To determine the density of the crown without using the weight in the water and with a bucket we can use the following steps.

1.weigh the crown in the air and find the mass

2. put water in a cylindrical bucket and measure its height with a ruler.

3. Put the crown in the bucket and measure the new water level with a ruler.

4. Subtract the heights, and find the volume of a cylinder knowing the difference in heights and the diameter of the bucket, in order to determine the volume of the crown.

5. find density by dividing mass by volume

7 0

3 years ago

In DC electrode positive, how much power is at the work clamp?

Korolek [52]

Answer:

1/3 power

Explanation:

I'm just a smart guy

7 0

2 years ago