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kvasek [131]
3 years ago
11

Calculate the percentage increase in length of a wire of diameter 2.2 mm stretched by a load of

Physics
1 answer:
vesna_86 [32]3 years ago
8 0

Answer:

0.21%

Explanation:

We are given;

Mass; m = 100 kg

Diameter; d = 2.2 mm = 2.2 × 10^(-3) m

Young's modulus; E = 12.5 x 10^(10) N/m².

Formula for area is;

A = πd²/4

A = (π/4) x (2.2 x 10^(-3))²

A = 3.8 x 10^(-6) m²

Force; F = mg

g is acceleration due to gravity and has a constant value of 9.8 m/s²

F = 100 × 9.8

F = 980 N

Formula for young's modulus is;

E = Stress/strain

Formula for stress = F/A

Formula for strain = ΔL/L

Thus;

E = (F/A)/(ΔL/L)

Making ΔL/L the subject, we have;

ΔL/L = (F/A)/E

Plugging in the relevant values;

ΔL/L = 980/(3.8 x 10^(-6) × 12.5 × 10^(10))

ΔL/L = 0.0021

Then percentage increase in length of a wire = 0.0021 × 100% = 0.21%

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What happens to an electromagnetic wave as it passes from space to matter?
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Two canoeists in identical canoes exert the same effort paddling and hence maintain the same speed relative to the water. One pa
Serga [27]

Answer:

Speed of river is 0.45 m/s

Speed of boat is 2.65 m/s

Explanation:

v_r = Speed of river

v_c = Speed of canoe

v_r+v_c=2.8\ m/s

v_r-v_c=-1.9\ m/s

Adding the equations we get

2v_r=0.9\\\Rightarrow v_r=\frac{0.9}{2}\\\Rightarrow v_r=0.45\ m/s

0.42+v_c=2.8\ m/s\\\Rightarrow v_c=2.8-0.45\\\Rightarrow v_c=2.65\ m/s

Speed of river is 0.45 m/s

Speed of boat is 2.65 m/s

6 0
3 years ago
You have an incident ray from a medium with a n1=1, through a medium with a n2 =3.325. If the incident angle is equal to 0.478 r
sesenic [268]

Answer:

0.139 rad

Explanation:

We use Snell's law n_1sin\theta_1=n_2sin\theta_2, where if n_1 is the <em>refractive index</em> of the medium containing the <em>incident ray</em>, \theta_1 would be the <em>incident angle</em>, and if n_2 is the <em>refractive index</em> of the medium containing the <em>refracted ray</em>, \theta_2 would be the <em>refraction angle</em>, which we want, so we do:

sin\theta_2=\frac{n_1}{n_2}sin\theta_1

And finally:

\theta_2=arcsin(\frac{n_1}{n_2}sin\theta_1)

We then insert our values:

\theta_2=arcsin(\frac{n_1}{n_2}sin\theta_1)=Arcsin(\frac{1}{3.325}sin(0.478rad))=arcsin(0.13834714686&#10;)=0.139 rad

6 0
3 years ago
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