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3 years ago

12

what is the maximum force a hammer can apply to a steel nail 2.5 mm in diameter of the nails elastic limit is not to be exceeded

1 answer:

natita [175]3 years ago

7 0

Yield strength of mild steel is given as

$Y = 250 MPa$

here we know that

$Y = \frac{F}{A}$

$250 * 10^6 = \frac{F}{\pi r^2}$

now we can rearrange it as

$F = \pi r^2 * 250* 10^6$

here we know that

$r = 1.25 mm = 1.25 * 10^{-3} m$

$F = \pi*(1.25*10^{-3})^2* 250* 10^6$

$F = 1.23 * 10^3 N$

so maximum applied force will be F = 1.23 * 10^3 N

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A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10 m, and an

maksim [4K]

Answer:

The value is $\alpha = 0.002 Np/m$

Explanation:

From the question we are told that

The first amplitude of the wave is $E_{max}1 = 98.02 \ V/m$

The first depth is $D_1 = 10 \ m$

The second amplitude is $E_{max}2 = 81.87 \ (V/m)$

The second depth is $D_2 = 100 \ m$

Generally from the spatial wave equation we have

$v(x) = Ae^{-\alpha d}cos(\beta x + \phi_o)$

=> $\frac{v(x)}{v(x)} =\frac{ Ae^{-\alpha d}cos(\beta x + \phi_o)}{ Ae^{-\alpha d}cos(\beta x + \phi_o)}$

So considering the ratio of the equation for the two depth

$\frac{A}{A_S} = \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}$

=> $\frac{98.02}{81.87} = \frac{e^{-10 \alpha }}{e^{-100 \alpha }}$

=> $\alpha = \frac{0.18}{90}$

=> $\alpha = 0.002 Np/m$

4 0

3 years ago

Best Answer will receive BRAINLIEST One consequence of Newton's third law of motion is that __________. A. every object that has

In-s [12.5K]

One consequence of Newton's third law of motion is that all actions have equal and opposite reactions. <em>(C)</em>

In fact, that's pretty much what the law itself says in so many words.

7 0

3 years ago

Read 2 more answers

Your code returns a number of 99.123456789 +0.00455679 for your calculation. How should you report it in your lab write-up?

Aneli [31]

Answer: Your code returns a number of 99.123456789 +0.00455679

Ok, you must see where the error starts to affect your number.

In this case, is in the third decimal.

So you will write 99.123 +- 0.004 da da da.

But you must round your results. In the number you can see that after the 3 comes a 4, so the 3 stays as it is.

in the error, after the 4 comes a 5, so it rounds up.

So the final presentation will be 99.123 +- 0.005

you are discarding all the other decimals because the error "domains" them.

6 0

3 years ago

Any help would be great! Thank you x<br><br><br> Giving brainliest answer xoxo

garri49 [273]

Answer:

A vacuum would have been created. I hope this helps have a great day

3 0

3 years ago

Here, mA = 3.65 kg and mB = 7.05 kg. The string connecting the two objects is of negligible mass and the pulley is frictionless.

mr_godi [17]

Answer:

The magintude of the acceleration for both objects is $3.11m/s^2$

Explanation:

Drawing a free body diagram on the two boxes we can analyze the system more easily.

we can take the acceleration going up as positive for reference purposes.

for mA let's suppose that is ascending so:

$T_A-m_A*g=m_A*a$

and for mB (descending):

$T_B-m_B*g=-m_B*a$

$T_A=T_B$

because the two boxes has the same acceleration because they are attached together:

$m_B*g-m_B*a-m_a*g=m_a*a\\(m_B-m_A)*g=(m_a+m_B)*a\\a=\frac{(7.05kg-3.65kg)*9.8m/s^2}{3.65kg+7.05kg}\\\\a=3.11m/s^2$

So the magintude of the acceleration for both objects is $3.11m/s^2$

5 0

3 years ago