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3 years ago

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1)A rocket expels gas at high speed for a short period of time. We are going to treat the rocket as being far away from any grav

itational objects.a)Draw a momentum chart for the rocket expelling gas in space.Take the initial time before expelling gas and the final time after the rocket has finished expelling gas. The rocket has an initial constant speed.Put the rocket and the expelled gas on separate rows.b)Use your chart to explainhow the speed of the rocket changes. c)Does the rocket have to keep expelling gas to stay at a constant speed

1 answer:

lesya [120]3 years ago

5 0

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A 200.0 g block rests on a frictionless, horizontal surface. It is pressed against a horizontal spring with spring constant 4500

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Answer:

6 m/s

Explanation:

Given that :

mass of the block m = 200.0 g = 200 × 10⁻³ kg

the horizontal spring constant k = 4500.0 N/m

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To determine the speed the block will be traveling when it leaves the spring; we applying the work done on the spring as it is stretched (or compressed) with the kinetic energy.

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$kx^2 = mv^2$

$4500* 0.04^2 = 200*10^{-3} *v^2$

$7.2 =200*10^{-3}*v^{2}$

$v^{2} =\frac{7.2}{200*10^{-3}}$

$v =\sqrt{\frac{7.2}{200*10^{-3}}}$

v = 6 m/s

Hence,the speed the block will be traveling when it leaves the spring is 6 m/s

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4 years ago

When you rub two balloons on your sleeve, does it have positive or negative?

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When you rub the balloons, it will take electrons away, which is negative. So, the balloons are negative

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3 years ago

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A man stands still on a moving walkway that is going at a speed of 0.2 m/s to

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3 years ago

You throw a bowling ball of mass M and radius 1m has an initial speed v0 down the lane. It has an initial angular velocity of 3v

zhuklara [117]

Answer:

Explanation:

Given

Initial angular velocity $\omega_0=\frac{3v_0}{R}$

Initial speed $u=v_0$

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As No external torque is applied therefore angular momentum is conserved

$L_i=L_f$

$I_{cm}v_i+Mv_{cm}_iR=I_{cm}\omega _f+Mv_{cm}_f\cdot R$

For rolling without slipping $v_f=\omega _fR$

$\frac{2}{5}MR^2\cdot (\frac{3v_0}{R})+Mv_0\cdot R=\frac{2}{5}MR^2\cdot \frac{v_f}{R}+Mv_f\cdot R$

$\frac{6}{5}Mv_0R+Mv_0R=\frac{2}{5}Mv_fR+Mv_f\cdot R$

$v_f=\frac{11}{7}v_0$

acceleration provided by surface

$a=\mu _kg$

$a=0.08\times 9.8$

$a=0.784\ m/s^2$

Speed of ball after $t=2\ s$

using $v=u+at$

v=final velocity

u=initial velocity

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$\frac{11}{7}v_0-v_0=0.784\times 2$

$v_0=2.744\ m/s$

thus $v_f=\frac{11}{7}\times 2.74$

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3 years ago