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2 years ago

Please help question in photo

Physics

1 answer:

bonufazy [111]2 years ago

8 0

Answer:

A.) the car

Explanation:

hope this helped <3

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Suppose you found a mineral crystal that looked like topaz. what two minerals could you use in a scratch test to help determine

Vesna [10]

The two most important minerals you need to do the test are <span>Corundum and quartz. You use that to do scratch test</span>

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3 years ago

Objects in motion eventually stop moving due to a _____ .can someone help me with this question

Tom [10]

External force.

Air resistance, friction, grass rubbing on a rolling ball, etc.

If there is no external force on a moving object, it keeps going.

Forever !

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3 years ago

Which chemical reaction is most likely the slowest?

NemiM [27]

B. takes the longest!

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2 years ago

a ball rolls off the edge of a 2m high shelf at a speed of 5 m/s and hits the ground the taken to hit the ground is

RUDIKE [14]

Answer:

Explanation:

Use the one-dimensional equation

Δx = $v_0t+\frac{1}{2}at^2$ where delta x is the displacement of the object, v0 is the velocity of the object, a is the pull of gravity, and t is the time in seconds. That's our unknown.

Δx = -2 (negative because where it ends up is lower than the point at which it started),

$v_0=5$ , and

a = -9.8

Filling in:

$-2=5t+\frac{1}{2}(-9.8) t^2$ and simplified a bit:

$-2=5t-4.9 t^2$

this should look hauntingly familiar (a quadratic, which is parabolic motion...very important in physics!!). We begin by getting everything on one side of the equals sign and solving for t by factoring:

$-4.9 t^2+5t+2=0$ (the 0 is also indicative of the object landing on the ground! Isn't this a beautiful thing, how it all just works so perfectly together?)

When you factor this however your math/physics teacher has you factoring you will get that

t = 1.3 sec and t = -.31 sec

Since we all know that time can NEVER be negative, it takes the ball 1.3 sec to hit the ground from a height of 2 m if it is rolling off the shelf at 5 m/s.

3 0

2 years ago

An aluminum cylinder with a radius of 2.7 cm and a height of 67 cm is used as one leg of a workbench. The workbench pushes down

soldier1979 [14.2K]

Answer:

$1.9\times 10^{-4}$

$1.2\times 10^{-4}\ m$

Explanation:

r = Radius = 2.7 cm

F = Force = $3.2\times 10^4\ N$

A = Area = $\pi r^2$

$\sigma$ = Stress = $\frac{F}{A}$

E = Young's modulus = $7\times 10^{10}\ Pa$

$\epsilon$ = Strain

$L_0$ = Original length = 67 cm

$\Delta L$ = Change in length

Young's modulus is given by

$E=\frac{\sigma}{\epsilon}\\\Rightarrow \epsilon=\frac{\sigma}{E}\\\Rightarrow \epsilon=\frac{\frac{3.2\times 10^4}{\pi 0.027^2}}{7\times 10^{10}}\\\Rightarrow \epsilon=0.0001996=1.9\times 10^{-4}$

Strain is $1.9\times 10^{-4}$

Strain is given by

$\epsilon=\frac{\Delta L}{L_0}\\\Rightarrow \Delta L=\epsilon\times L_0\\\Rightarrow \Delta L=1.9\times 10^{-4}\times 0.67\\\Rightarrow \Delta L=0.0001273\\\Rightarrow \Delta L=1.2\times 10^{-4}\ m$

The cylinder height decreases by $1.2\times 10^{-4}\ m$

3 0

3 years ago