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Misha Larkins [42]
3 years ago
15

A mountain climber is hanging from a vertical rope, far above the ground and far from the rock face. the rope is vertical. what

are the forces acting
Physics
1 answer:
notka56 [123]3 years ago
7 0
The weight of the climber and the tension of the rope
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The velocity of sound in air saturated with water vapour at 30°C
Luba_88 [7]

Explanation:

The velocity of sound depends on the density of the medium.  So we need to find the density of air at each set of conditions.  The density of air is:

ρ = (Pd / (Rd T)) + (Pv / (Rv T))

where Pd and Pv are the partial pressures of dry air and water vapor,

Rd and Rv are the specific gas constants of dry air and water vapor,

and T is the absolute temperature.

At the first condition:

Pv = 31.7 mmHg = 4226.3 Pa

Pd = 650 mmHg - 31.7 mmHg = 618.3 mmHg = 82433 Pa

Rv = 461.52 J/kg/K

Rd = 287.00 J/kg/K

T = 30°C = 303.15°C

ρ = (82433 / 287.00 / 303.15) + (4226.3 / 461.52 / 303.15)

ρ = 0.94746 + 0.03021

ρ = 0.97767 kg/m³

At the second condition:

Pv = 0 Pa

Pd = 650 mmHg = 86660 Pa

Rv = 461.52 J/kg/K

Rd = 287.00 J/kg/K

T = 0°C = 273.15°C

ρ = (86660 / 287.00 / 273.15) + (0 / 461.52 / 273.15)

ρ = 1.1054 + 0

ρ = 1.1054 kg/m³

The square of the velocity of sound is proportional to the ratio between pressure and density:

v² = k P / ρ

Since the atmospheric pressure is constant, we can say it's proportional to just the density:

v² = k / ρ

Using the first condition to find the coefficient:

(340)² = k / 0.97767

k = 113018.652

Now finding the velocity of sound at the second condition:

v² = 113018.652 / 1.1054

v = 319.75

6 0
3 years ago
Peer pressure is the influence that _________ may have on you
KonstantinChe [14]
Both c and a could be right but I am steering more towards a.
8 0
3 years ago
Read 2 more answers
The angular velocity of a process control motor is (13−12t2) rad/s, where t is in seconds. Part A At what time does the motor re
mihalych1998 [28]

Answer:

Explanation:

Given

\omega =13-\frac{1}{2}\cdot t^2

Motor reverse its direction when \omega =0

13-0.5t^2=0

26=t^2

t=\sqrt{26}=5.099\approx 5.1 s

(b)

\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega

\int d\theta =\int_{0}^{5.1}\omega dt

\int d\theta =\int_{0}^{t}(13-.05t^2)dt

\theta =(13t-0.1667\times t^3)_0^{5.1}

\theta =44.192^{\circ}

4 0
3 years ago
If you have lifted 10 pounds 2 feet, you have done ___ foot-pounds of work.
nasty-shy [4]
Work = Force x distance
          (10 pounds)(2 feet)
Work = 20 foot-pounds of work

hope this helps :)


8 0
3 years ago
A car travels 90 meters due north in 15 seconds. Then the car
tatyana61 [14]
I got you kid It’s A- 2.5m/sb
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