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2 years ago

6

A 90kg mountain climber hangs from a nylon rope and streches it ny 25.0cm. if the rope was originally 30.0m long and it's diamet

er is 1.0cm, what is young's modulus for the nylon?

1 answer:

bixtya [17]2 years ago

7 0

Answer:

Explanation:

E = σ/ε = (F/A) / (ΔL/L)

E = (mg/(πd²/4) / (ΔL/L)

E = (4mg/(πd²) / (ΔL/L)

E = 4Lmg/(πd²ΔL)

E = 4(30.0)(90)(9.8)/(π(0.01²)0.25)

E = 1.35 x 10⁹ Pa or 1.35 GPa

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The diagram pictured shows a loop of wire being rotated at a constant rate about an axis in a uniform magnetic field.

faust18 [17]

The rotation is probably n

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2 years ago

A flute is designed so that it plays a frequency of 261.6 Hz, middle C, when all the holes are covered and the temperature is 20

irinina [24]

Answer:

0.655 m

13.468°C

Explanation:

v = Speed of sound at 20.0°C = 343 m/s (general value)

For one both end open we have the expression

$L=\dfrac{\lambda_1}{2}\\\Rightarrow L=\dfrac{\dfrac{v}{f_1}}{2}\\\Rightarrow L=\dfrac{\dfrac{343}{261.6}}{2}\\\Rightarrow L=0.655581039755\ m$

The length of the flute is 0.655 m

Beat frequency is given by

$\Delta f=f_1-f_2\\\Rightarrow 3=261.6-f_2\\\Rightarrow f_2=261.6-3\\\Rightarrow f_2=258.6\ Hz$

Velocity of the wave is

$v=f_2\lambda_1\\\Rightarrow v=258.6\times \dfrac{343}{261.6}\\\Rightarrow v=339.066513761\ m/s$

The temperature is given by

$T=273(\dfrac{v}{331})^2\\\Rightarrow T=273(\dfrac{339.066513761}{331})^2\\\Rightarrow T=286.468227799\ K=286.468227799-273=13.468227799^{\circ}C$

The temperature of the room is 13.468°C

4 0

3 years ago

Energy cannot be created nor destroyed. A. Ella said "a lot of energy in the hot water has disappeared." Explain why her stateme

harkovskaia [24]

Answer:

See Explanation

Explanation:

The principle of conservation of energy states that; energy can neither be created nor destroyed but is converted from one form to another.

In view of this principle, Ella can not be correct when she says that a lot of energy has disappeared. The use of the term "disappeared" connotes the idea that the energy no longer exists which does not happen.

Hence, energy can not "disappear" from hot water rather the energy in the water may be transferred to the surroundings.

6 0

2 years ago

A mass of 6 kg with initial velocity 16 m/s travels through a wind tunnel that exerts a constant force 8 N for a distance 1.6 m.

strojnjashka [21]

Answer:

$D=99.4665307m \approx 99.5m$

Explanation:

From the question we are told that

Mass $m=6kg$

Velocity of mass $V_m=16$

Force of Tunnel $F_t=8N$

Length of Tunnel $L_t=1.6$

Height of frictional incline $H_i=2.9$

Angle of inclination $\angle =16 \textdegree$

Acceleration due to gravity $g=9.8m/s^2$

First Frictional surface has a coeﬃcient $\alpha_1 =0.21\ for\ d_c=1$

Second Frictional surface has a coeﬃcient $\alpha _2=0.1$

Generally the initial Kinetic energy is mathematically given by

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}(6)(16)^2$

$K.E=768$

Generally the work done by the Tunnel is mathematically given as

$w_t=F_t*d_t$

$w_t=8*1.6$

$w_t=12.8J$

Therefore

$Total energy\ E_t=Initial\ kinetic energy\ K.E*Work done\ by\ tunnel\ W_t$

$E_t=K.E+E_t\\E_t=768J+12.8J$

$E_t=780.8J$

Generally the energy lost while climbing is mathematically given as

$E_c=mgh$

$E_c=(6)(9.8)(2.9)$

$E_c=170.52J$

Generally the energy lost to friction is mathematically given as

$E_f=\alpha *m*g*cos\textdegree*d_c$

$E_f=0.21*6*9.8*cos16*1$

$E_f=11.86965942 \approx 12J$

Generally the energy left in the form of mass $Em$ is mathematically given as

$E_m=E_t+E_c+E_f$

$E_m=(768J)-(170.52)-(12)$

$E_m=585.48J$

Since

$E_m=\alpha_2*g*m*d$

Therefore

It slide along the second frictional region

$D=\frac{585.46}{0.1*9.81*6}$

$D=99.4665307m \approx 99.5m$

6 0

3 years ago

An ac voltage, whose peak value is 250 V , is across a 330-Ω resistor. Part A What is the peak current in the resistor?

Wittaler [7]

Answer:

Peak current is 0.75 A

RMS voltage is 176.77 V

RMS current is 0.53 A

Explanation:

Peak current is

$I=\frac{V}{R} \\\\I=\frac{250}{330}\\I=0.75 A$

RMS voltage is

$V_{RMS}=\frac{V}{\sqrt{2} } \\V_{RMS}=\frac{250}{\sqrt{2} }\\V_{RMS}=176.77 V$

RMS current is

$I_{RMS}=\frac{V_{RMS}}{R} \\I_{RMS}=\frac{176.77}{330} \\I_{RMS}=0.53 A$

7 0

3 years ago