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3 years ago

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If a net force of 25 N is exerted over a distance of 4 m to the right on a 2 kg mass initially at rest and moves it, what is the

final velocity of the object? v = _____ m/s

1 answer:

pashok25 [27]3 years ago

5 0

Answer: 10 m/s

Given:

Force, F = 25 N

Initial velocity $u=0 m/s^{2}$

Distance moved, s= 4 m

mass, m = 2 kg

Formula Used:

$F = ma$

where a is the acceleration.

equation of motion:

$2as=v^{2}-u^{2}$

First, we would calculate the acceleration from the formula for force.

$a= \frac{F}{m}=\frac{25}{2}m/s^{2}$

Using the equation of motion, we can find the final velocity now:

$2$ × $\frac{25}{2}$ × $4= v^{2}-0^{2}$

taking the square root on both sides and ignoring the negative sign:

v= 10 m/s

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Answer:

Surely Achilles will catch the Tortoise, in 400 seconds

Explanation:

The problem itself reduces the interval of time many times, almost reaching zero. However, if we assume the interval constant, then it is clear that in two hours Achilles already has surpassed the Tortoise (20 miles while the Tortoise only 3).

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2 years ago

A certain quantity of steam has a temperature of 100.0 oC. To convert this steam into ice at 0.0 oC, energy in the form of heat

KonstantinChe [14]

Answer:

2452.79432 m/s

Explanation:

m = Mass of ice

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$s_w$ = Specific heat of water

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v = Velocity of ice

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$Q_1=mL_s\\\Rightarrow Q_1=m(2.256\times 10^6)$

Heat released from water at 100 °C

$Q_2=ms_w\Delta T\\\Rightarrow Q_2=m4186\times (100-0)\\\Rightarrow Q_2=m0.4186\times 10^6$

Heat released from water at 0 °C

$Q_3=mL_i\\\Rightarrow Q_3=m(333.5\times 10^3)\\\Rightarrow Q_3=m(0.3335\times 10^6)$

Total heat released is

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The kinetic energy of the bullet will balance the heat

$K=Q\\\Rightarrow \frac{1}{2}mv^2=3008100m\\\Rightarrow v=\sqrt{2\times 3008100}\\\Rightarrow v=2452.79432\ m/s$

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3 years ago

Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du

Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time $t$ is

$\theta=\dfrac\alpha2t^2$

It rotates 44.5 rad in this time, so we have

$44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}$

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$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2$

where $\omega_f$ is the angular velocity achieved after 6.00 s. The velocity of the disk at time $t$ is

$\omega=\alpha t$

so we have

$\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}$

making the average velocity

$\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}$

Another way to find the average velocity is to compute it directly via

$\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}$

c. We already found this using the first method in part (b),

$\omega=14.8\dfrac{\rm rad}{\rm s}$

d. We already know

$\theta=\dfrac\alpha2t^2$

so this is just a matter of plugging in $t=12.0\,\mathrm s$ . We get

$\theta=179\,\mathrm{rad}$

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

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kirza4 [7]

Answer:

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