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ICE Princess25 [194]
2 years ago
12

A sky diver steps from a high-flying helicopter. if there were not air resistance, how fast would she be falling at the end of a

12 second jump?
Physics
1 answer:
Rama09 [41]2 years ago
8 0

If there is no air resistance and the body, in this case the sky diver simply falls from the helicopter then, the motion can be described as a free-falling body. The equation that would allow us to determine the speed of the body at any time, t is,

<span>                                    v(t)  = vo + gt</span>

<span>where v(t) is the unknown velocity, vo is the initial velocity which is equal to zero and g is the acceleration due to gravity which is equal to 9.8 m/s2.</span>

Substituting the known values,

<span>                                    v(t) = 0 + (9.8 m/s2)(12s)</span>

<span>                                                v(t) = 117.6 m/s</span>

<span>Hence, the speed of the sky diver at the end of 12 second is approximately 117.6 m/s.  </span>

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A coat rack weighs 65.0 lbs when it is filled with winter coats and 40.0 lbs when it is empty. The base of the coat rack has an
Whitepunk [10]

Answer:

0.056 psi more pressure is exerted by filled coat rack than an empty coat rack.

Explanation:

First we find the pressure exerted by the rack without coat. So, for that purpose, we use formula:

P₁ = F/A

where,

P₁ = Pressure exerted by empty rack = ?

F = Force exerted by empty rack = Weight of Empty Rack = 40 lb

A = Base Area = 452.4 in²

Therefore,

P₁ = 40 lb/452.4 in²

P₁ = 0.088 psi

Now, we calculate the pressure exerted by the rack along with the coat.

P₂ = F/A

where,

P₂ = Pressure exerted by rack filled with coats= ?

F = Force exerted by filled rack = Weight of Filled Rack = 65 lb

A = Base Area = 452.4 in²

Therefore,

P₂ = 65 lb/452.4 in²

P₂ = 0.144 psi

Now, the difference between both pressures is:

ΔP = P₂ - P₁

ΔP = 0.144 psi - 0.088 psi

<u>ΔP = 0.056 psi</u>

8 0
2 years ago
A 10.0 cm3 sample of copper has a mass of 89.6
Romashka-Z-Leto [24]
Density is mass divides by volume, so
89.6g / 10cm^3 =8.96g /cm^3

*cm^3 is a standard unit of volume*
4 0
2 years ago
refrigerant 134a enters a compressor operating at steady state as saturated vapor at 0.12 MPa and exits at 1.2 MPa and 70 C at a
Afina-wow [57]

Answer:

the power input to the compressor is 7.19Kw

Explanation:

Hello!

To solve this problem follow the steps below.

1. We will call 1 the refrigerant state at the compressor inlet and 2 at the outlet.

2. We use thermodynamic tables to determine enthalpies in states 1 and 2.

(note: Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  )

h1[quality=1, P=0.12Mpa)=237KJ/Kg

h2(P=1.2Mpa, t=70C)=300.6KJ/kg

3. uses the first law of thermodynamics in the compressor that states that the energy that enters a system is the same that must come out

Q=heat=0.32kJ/s

W=power input to the compressor

m=mass flow=0.108kg/S

m(h1)+W=Q+m(h2)

solving for W

W=Q+m(h2-h1)

W=0.32+0.108(300.6-237)=7.19Kw

the power input to the compressor is 7.19Kw

7 0
2 years ago
Most binary systems with an invisible companion contain a large, bright star and a small, dim star hidden by the light of its la
Vlada [557]

Answer:

Brain signals are converted

8 0
2 years ago
A swimming pool contains x (less than 0.02) grams of chlorine per cubic meter. the pool measures 5 meters by 50 meters and is 2
zubka84 [21]
The solution for this problem would be:(10 - 500x) / (5 - x) 
so start by doing: 
x(5*50*2) - xV + 5V = 0.02(5*50*2) 
500x - xV + 5V = 10 
V(5 - x) = 10 - 500x 
V = (10 - 500x) / (5 - x) 
(V stands for the volume, but leaves us with the expression for x)
3 0
2 years ago
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