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3 years ago

a 55.0 kg person having a denity of 1010 kg/m^3 is underwater. What is the net force acting on the person?

Physics

1 answer:

bija089 [108]3 years ago

6 0

Answer:

The net force acting on the person is 5.34 N acting upwards

Explanation:

The information given are;

Mass of the person = 55.0 kg

Density of the person = 1010 kg/m³

Body volume of the person = Mass/Density = 55/1010 = 0.0545 m³

Weight of the person = Mass × Gravity = 55 × 9.81 = 539.55 N↓

Where the density of water = 1000 kg/m³, we have;

Upthrust = weight of the water displaced = Mass of the water displaced × Gravity

Mass of the water displaced = Density of water × Volume of the water

Mass of the water displaced = Density of water × Body volume of the person

= 1000 × 0.0545 = 54.5 kg

The Upthrust = 54.5 × 9.81 = 534.21 N↑

Given that the net force, $F_{Net}$ acting is the sum of the mass of forces acting on the person which are the weight acting downward and the upthrust acting upwards

We have;

F(Net) = 534.21↑ - 539.55↓ = 5.34 N↑

The net force acting on the person = 5.34 N upwards.

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Do noble gases lose or gain electrons to get a full valence shell?

alexdok [17]

They don't lose OR gain electrons as they've already achieved the octet rule and have 8 valence electronsn

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3 years ago

A 35-mm single lens reflex (SLR) digital camera is using a lens of focal length 35.0 mm to photograph a person who is 1.80 m tal

olganol [36]

Answer:

a) 35.44 mm

b) 17.67 mm

Explanation:

u = Object distance = 3.6 m

v = Image distance

f = Focal length = 35 mm

$h_u$ = Object height = 1.8 m

a) Lens Equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{35}-\frac{1}{3600}\\\Rightarrow \frac{1}{v}=\frac{713}{25200} \\\Rightarrow v=\frac{25200}{713}=35.34\ mm$

The CCD sensor is 35.34 mm from the lens

b) Magnification

$m=-\frac{v}{u}\\\Rightarrow m=-\frac{35.34}{3600}$

$m=\frac{h_v}{h_u}\\\Rightarrow -\frac{35.34}{3600}=\frac{h_v}{1800}\\\Rightarrow h_v=-\frac{35.34}{3600}\times 1800=-17.67\ mm$

The person appears 17.67 mm tall on the sensor

7 0

3 years ago

What is the mass of one mole of carbon dioxide in sl?

ra1l [238]

Answer:

Around 44.01g.

Explanation:

One mole of carbon dioxide molecules has a mass of 44.01g, while one mole of sodium sulfide formula units has a mass of 78.04g.

8 0

3 years ago

The y-component of the force F which a person exerts on the handle of the box wrench is known to be 70 lb. Determine the x-compo

Yuri [45]

Answer:

Fx = 121.24lb

F = 140lb

Explanation:

Since we are not given the angles subtended by the force, we can assume it to be 30 degrees.

The y component of the force expressed by the formula:

Fy = Fsintheta

Given the y-component of the force F to bee 70lb

70lb = Fsintheta

Get magnitude of the force

F = 70/sin theta

F = 70/sin 30

F = 70/0.5

F = 140lb

Get the x-component of the force

Fx = Fcos theta

Fx = 140cos 30

Fx = 140(0.8660)

Fx = 1,212.4lb

Hence the x-component of the force sis 121.24lb

Note that the angle used was assumed. Other values can as well be used

5 0

3 years ago

Q1. After three half-lives of an isotope, 1 billion of the original isotope's atoms still remain in a certain amount of this ele

77julia77 [94]

If 1 eighth equals 1 billion 7 eighth equals 7 billion.

The asker of the second question needs a tutorial in radiometric dating. There is little likelihood that the daughter isotope has the same atomic weight as the parent isotope. To measure the mass isotopes doesn't tell us how many atoms of each exist. To get around that let's pretend — which will likely serve the purpose ineptly intended — that the values give an the particle ratio, 125:875.

The original parent isotope count was 125 + 875 = 1000. The remaining parent isotope is 125/1000 or 1/8. 1/8 = (1/2)^h, where h is the number of half-lives.

h = log (1/8) ÷ log(1/2) = 3

And 3 half-lives • 150,000 years/half-life = 450,000 years.

5 0

3 years ago