Pls help my mom is gonna kill me if i dont finish this today

Physics

2 answers:

iren [92.7K]3 years ago

7 0

Answer:

I believe it is a transfer of energy from the rubber band to the car

Explanation:

However it could be a transfer of potential energy from the stretched rubber band that is released into kinetic energy into the car. Depends as there is little context to the question.

Marizza181 [45]3 years ago

4 0

Answer:

I am pretty sure it’s a transformation

Explanation:

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I’ll give brainliest if it’s correct ;-;z

BlackZzzverrR [31]

Explanation:

what is the question? could you pls provide it

6 0

3 years ago

A cantilever beam with a width b=100 mm and depth h=150 mm has a length L=2 m and is subjected to a point load P =500 N at B. Ca

DerKrebs [107]

Answer:

Explanation:

Given that:

width b=100mm

depth h=150 mm

length L=2 m =200mm

point load P =500 N

Calculate moment of inertia

$I=\frac{bh^3}{12} \\\\=\frac{100 \times 150^3}{12} \\\\=28125000\ m m^4$

Point C is subjected to bending moment

Calculate the bending moment of point C

M = P x 1.5

= 500 x 1.5

= 750 N.m

M = 750 × 10³ N.mm

Calculate bending stress at point C

$\sigma=\frac{M.y}{I} \\\\=\frac{(750\times10^3)(25)}{28125000} \\\\=0.0667 \ MPa\\\\ \sigma =666.67\ kPa$

Calculate the first moment of area below point C

$Q=A \bar y\\\\=(50 \times 100)(25 +\frac{50}{2} )\\\\Q=250000\ mm$

Now calculate shear stress at point C

$=\frac{FQ}{It}$

$=\frac{500*250000}{28125000*100} \\\\=0.0444\ MPa\\\\=44.4\ KPa$

Calculate the principal stress at point C

$\sigma_{1,2}=\frac{\sigma_x+\sigma_y}{2} \pm\sqrt{(\frac{\sigma_x-\sigma_y}{2} ) + (\tau)^2} \\\\=\frac{666.67+0}{2} \pm\sqrt{(\frac{666.67-0}{2} )^2 \pm(44.44)^2} \ [ \sigma_y=0]\\\\=333.33\pm336.28\\\\ \sigma_1=333.33+336.28\\=669.61KPa\\\\\sigma_2=333.33-336.28\\=-2.95KPa$