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3 years ago

11

A steel rod has a length of 0.2 cm at 30 degrees C, what will be its length at 60 degrees C?

1 answer:

maks197457 [2]3 years ago

5 0

Answer:

2.00072 mm

Explanation:

So, we just have to calculate 2 mm x (1 + 0.36 x 10⁻³) = 2 mm + 0.00072 mm = 2.00072 mm

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3 years ago

A private aviation helicopter's main rotor blades rotate at approximately

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Answer: 7.5 rev/s

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$\omega=450 \frac{rev}{min} \frac{1 min}{60 s}$

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What is hydroelectric power ?<br><br> Answer quickly..!

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2 years ago

Read 2 more answers

Vector A has y-component Ay= +15.0 m . A makes an angle of 32.0 counterclockwise from the +y-axis. What is the x component of A?

Gemiola [76]

Answer:

x-component=-9.3 m

Magnitude of A=17.7m

Explanation:

We are given that

$A_y=+15 m$

$\theta=32^{\circ}$

We have to find the x-component of A and magnitude of A.

According to question

$A_y=\mid A\mid cos\theta$

Substitute the values then we get

$15=\mid A\mid cos32$

$\mid A\mid =\frac{15}{cos32}=\frac{15}{0.848}$

$\mid A\mid=17.7$ m

$tan\theta=\frac{perpendicular\;side}{Base}$

$tan32=\frac{A_x}{A_y}=\frac{A_x}{15}$

$0.62\times 15=A_x$

$A_x=9.3$

The value of x-component of A is negative because the vector A lie in second quadrant.

Hence, the x- component of A=-9.3 m

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3 years ago

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calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end

sdas [7]

Answer: $0.8\ mm$

Explanation:

Given

length of wire $l=75\ cm$

change in length $\Delta l=1.85\ mm$

mass of wire $m=10\ kg$

Young's modulus for silver $E=7.9\times 10^{10}\ N/m^2$

load on wire $F=mg$

$F=10\times 9.8=98\ kg$

change in length is given by

$\Delta l=\dfrac{Pl}{AE}$

Where A=area of cross-section

$A=\dfrac{Pl}{\Delta lE}$

$A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}$

$A=\dfrac{73.5}{14.615\times 10^{7}}$

$A=5.029\times 10^{-7}\ m^2$

also wire is the shape of cylinder so cross-section is given by

$A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2$

$\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }$

$\Rightarrow d^2=64.02\times 10^{-8}$

$\Rightarrow d=8\times 10^{-4}\ m$

$\Rightarrow d=0.8\ mm$

4 0

3 years ago