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3 years ago

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A 0.1 kg magnet is floating above a second magnet. The magnetic force on the 0.1 kg magnet is

2 answers:

a_sh-v [17]3 years ago

8 0

It is C. I took the test and this was a question

Andru [333]3 years ago

7 0

C.) 1N Is the correct answer

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When an atom is involved in a nuclear reaction:

nataly862011 [7]

I believe the answer would be , it may change from one element to another. I’m saying this because during nuclear transmutation is when a subatomic particle fired at the nucleus of an atom changes into a heavier element , to break the nucleus apart into two nuclei and energy.

4 0

3 years ago

A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the a

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

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

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

$h=127.55\ m$

3 0

3 years ago

Gas hydrates are used _____.

Nonamiya [84]

Hope this helps mark brainest plz

4 0

3 years ago

Read 2 more answers

In a shuffleboard game, the puck slides a total of 12 m before coming to rest. If the coefficient ofkinetic friction between the

Artist 52 [7]

Answer:

puck decelerates due to the kinetic frictional force μk mg

Explanation:

given data

total distance = 12 m

coefficient of kinetic friction = 0.28

solution

we will apply equation of motion that is

v² - u² = 2 × a × s ................1

we know acceleration will be

a = $\frac{-u^2}{2\times S}$

Then we have

Force = mass × acceleration .................2

m × $\frac{-u^2}{2\times S}$ = -μk mg

The puck decelerates due to the kinetic frictional force μk mg

and frictional force is negative as it opposes the motion.

so we get initial velocity of the puck which is strike.

4 0

3 years ago

If you are reading this please help ASAP !!!!!

NISA [10]

Answer:

ok

Explanation:

5 0

3 years ago