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4 years ago

Some help please physics

Physics

2 answers:

salantis [7]4 years ago

3 0

My best guess is A There is missing a time unit in the question. I know that 4.4 *5 = 21 so that would be my best guess.

REY [17]4 years ago

3 0

The car's AVERAGE speed is (21/2)=10.5 m/s. (Zero at the beginning, 21 at the end.). In order to cover 50m at an average speed of 10.5 m/s, it takes (50/10.5)=4.76 seconds. It took 4.76 seconds to accelerate to 21 m/s^2, so the average acceleration is (21/4.76)=4.2 seconds. You probably had a formula in class that would take you to 4.4, but this is the way I think it through without a lot of formulas.

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stira [4]

Answer:

Explanation:

Given

acceleration is given by

$a=-g-c_Dv^2$

where $\ddot{y}=a$

$\dot{y}=v$

Also acceleration is given by

$a=v\frac{\mathrm{d} v}{\mathrm{d} s}$

$ds=\frac{v}{a}dv$

$\int ds=\int \frac{v}{-g-0.001v^2}dv$

$\Rightarrow Let -g-0.001v^2=t$

$-0.001\times 2vdv=dt$

$vdv=-\frac{dt}{0.002}$

$at\ v_0=50\ m/s,\ t=-g-0.001(50)^2$

$t=-g-2.5$

at $v=0,\ t=-g$

$\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}$

$\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}$

$s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}$

$s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})$

$s=113.608\ m$

when air drag is neglected maximum height reached is

$h=\frac{v_0^2}{2g}$

$h=\frac{50^2}{2\times 9.8}$

$h=127.55\ m$

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3 years ago

Identifying Advantages of Parallel Circuits

melisa1 [442]

Answer: If one bulb goes out the other bulbs stay lit.

If there is a break in one branch of the circuit, current can still flow through the other branches.

Explanation:

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professor190 [17]

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