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lutik1710 [3]
3 years ago
10

Humans evolved in Earth's atmosphere, therefore the pressures interior to the body are relatively close to atmospheric pressure.

Because the differences between internal body pressures and atmospheric pressure are so small, individuals have a difficult time gauging the magnitude of atmospheric pressure. Suppose an individual is lying on his stomach with sheets of paper stacked on his back. If each sheet of paper has a mass of 0.00320 kg and is the standard letter size of 8.5 in by 11 in ( 0.216 m by 0.279 m ), how many sheets must be stacked to produce a pressure on his back equal to atmospheric pressure (roughly 101325 Pa )?
Physics
2 answers:
Artist 52 [7]3 years ago
7 0

Answer:

193664 sheets of paper.

Explanation:

We know that the pressure is defined as

p=\frac{F}{A}

In words, the applied force divided by the area of application. The area of a single sheet of paper is

A=bh\\\\A=(0.216m)(0.279m)=0.060m^{2}

The force exerted by a single sheet of paper is its weight, given by teh expression:

F=mg

(Where m is the mass of a single sheet and g is the acceleration due to gravity on earth)

And, for n sheets of paper, the total weight is:

F=nmg

Now, substituting this in the definition of pressure and solving for n, we get:

p=\frac{nmg}{A}\\\\ \implies n=\frac{pA}{mg}

Finally, plugging in the known values, we can compute n:

n=\frac{(101325Pa)(0.060m^{2} )}{(0.00320kg)(9.81m/s^{2})}\\\\n=193664

This means that there must be 193664 sheets stacked to produce a pressure equal to atmospheric pressure.

harina [27]3 years ago
4 0

Answer:

194516 sheets

Explanation:

So the area of each sheet of paper is:

A = 0.216 * 0.279 = 0.060264 square meters

For the paper sheet to make the same effect as the atmospheric pressure P, then the gravity F from the paper sheet must be

F = AP = 0.060264 * 101325 = 6106 N

Let g = 9.81 m/s2, then the mass of paper needed to generate that gravity is

m = F/g = 6106 / 9.81 = 622.4 kg

If each sheet has a mass of 0.0032 kg, then the total number of sheets to have that much mass is

622.4 / 0.0032 = 194516 sheets

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804 n of force are applied to a 51.7 kg. What is the acceleration that the object experiences?
Andreyy89

We can use Newton II here  (where F=m*a), that F is the net (or resultant) force on the object, m is the mass of the object and a is the acceleration the object experiences.

This means, in this case there would be no friction and absolutely no other force which gives a component in the plane of motion, only then can you assume that F=804N.

Now using F= m*a

804 = 51.7*a

Therefore a = 804/51.7 = 15.55 m/s²


7 0
3 years ago
The mass of a lift is 600kg.the. The maximum tensile force that the cable supporting the lift can withstand is 7kN. Calculate th
Oksi-84 [34.3K]
Force is the product of mass and acceleration .
The question is ask to find acceleration.
But acceleration is the ratio of the force and the mass.
where 600kg is the mass and 7kN is the force
NB: kilo is 1000
now we have to multiply 7N by 1000
by doing so you will have 7000N
which is the force.
Now to find the acceleration: force/ mass
which is 7000/600
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5 0
3 years ago
A rope exerts a 280 N force while pulling an 80 Kg skier upward along a hill inclined at 12o. The rope pulls parallel to the hil
user100 [1]

Answer:

The speed of the skier after moving 100 m up the slope are of V= 25.23 m/s.

Explanation:

F= 280 N

m= 80 kg

α= 12º

μ= 0.15

d= 100m

g= 9,8 m/s²

N= m*g*sin(α)

N= 163 Newtons

Fr= μ * N

Fr= 24.45 Newtons

∑F= m*a

a= (280N - 24.5N) / 80kg

a= 3.19 m/s²

d= a * t² / 2

t=√(2*d/a)

t= 7.91 sec

V= a* t

V= 3.19 m/s² * 7.91 s

V= 25.23 m/s

4 0
3 years ago
what would happen to the size of the shadow if the distance between the light and the hand is increased.
Sergio039 [100]
I don’t think I will have any time to go
8 0
3 years ago
A large, metallic, spherical shell has no net charge. It is supported on an insulating stand and has a small hole at the top. A
tino4ka555 [31]

Answer:

(a) Negative Q

(b) Positive Q

Explanation:

Charge is the inherent property of matter due to the transference of electrons.

There are three methods of charging a body.

(i) Charging by friction: When two uncharged bodies rubbed together, then one body gets positive charged and the other is negatively charges it is due to the transference of electrons form one body to another.  

(ii) Conduction: when a charged body comes in contact with the another uncharged body, the uncharged body gets the same charge and the charge is distributed equally.

(iii) Induction: When a uncharged body keep near the charged body, the uncharged body gets the same amount of charge but opposite in sign.  

(a) When a small tack of charge Q is lowered into the hole, then due to the process of induction, the charge on the inner surface of the shell is - Q.

(b) Due to the process of conduction, the charge on the outer surface of the shell is Q.

7 0
3 years ago
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