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Nitella [24]
3 years ago
14

A steel guitar string with a diameter of 0.300 mm and a length of 70.0 cm is stretched by 0.500 mm while being tuned. How much f

orce is needed to stretch the string by this amount? Young's modulus for steel is 2.0 × 1011 N/m2
Physics
1 answer:
Ganezh [65]3 years ago
8 0

Answer:

10.1 N

Explanation:

Your answer is 10.1 N, I don't actually know how to do it but I hope it helps.

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PLEASE HELP:
mojhsa [17]
The marble's volume is equal to 1.2 cm^{3}
Density= mass/volume
Mass= 3.05 Grams
Volume= 1.2 cm^3
3.05/1.2= 2.92 <span>g/mL</span>

7 0
2 years ago
(8 points) Air is used as the working fluid in a simple ideal Brayton cycle that has a pressure ratio of 12, a compressor inlet
trapecia [35]

Answer:

a ) \dot m = 351.49 kg/s

b)  \dot m_{actual} = 1046.15 kg/s

Explanation:

given data:

pressure ration rp = 12

inlet temperature = 300 K

TURBINE inlet temperature  = 1000 K

AT the end of isentropic process (compression) temperature is

\frac{T_2'}{T_1} = rp ^{\frac{\gamma -1}{\gamma}}

\frac{T_2'}{300} = 12^{\frac{1.4 -1}{1.4}}

T_2' = 610.181 K

AT the end of isentropic process (expansion) temperature is

\frac{T_3}{T_4'} = rp ^{\frac{\gamma -1}{\gamma}}

\frac{1000'}{T_4'} = 12^{\frac{1.4 -1}{1.4}}

T_4' = 491.66 K

isentropic work is given as

w(compressor) = CP (T_2' -T_1)

w = 1.005(610.18 - 300)

w = 311.73 kJ/kg

w(turbine) = 1.005( 1000 - 491.66)

w(turbine) = 510.88 kJ/kg

a) mass flow rate for isentropic process is given as

\dot m = \frac{70000}{510.88 - 311.73}

\dot m = 351.49 kg/s

b) actual mass flow rate uis given as

\dot m_{actual} = \frac{70000}{51.088\times 0.85 - \frac{311.73}{0.85}}

\dot m_{actual} = 1046.15 kg/s

6 0
3 years ago
6th grade science pls help
oee [108]

Answer:

D

Explanation:

I hope you get a good grade!

6 0
2 years ago
1. My grass is dying, and I believe it's because it is not getting enough water. Sol
trasher [3.6K]

Answer: I actually need the same answer

Explanation:

5 0
3 years ago
A force of 50 newtons causes a sled to accelerate at a rate of 5 meters per second. What is the mass of the sled.
notka56 [123]
F=ma
50=m(5)
m=10kg
hence,ans is B
5 0
2 years ago
Read 2 more answers
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