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3 years ago

14

Hey! I found this question quite interesting. Check it out - https://www.meritnation.com/ask-answer/question/to-raise-a-200kg-st

eel-bean-to-a-height-of-10m-on-a-bridge-b/numbers/16795077

1 answer:

Yanka [14]3 years ago

6 0

Answer:

yes yes so do I

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A student decides to give his bicycle a tune up. He flips it upside down (so there's no friction with the ground) and applies a

aleksklad [387]

Answer:

$V=9.2565m/s$

Explanation:

From the question we are told that:

Force $F = 34 N$

Time $t = 0.6 s$

Length of pedal $l_p=16.5cm \approx0.165m$

Radius of wheel $r = 33 cm = 0.33 m$

Moment of inertia, $I = 1200 kgcm2 = 0.12 kg.m2$

Generally the equation for Torque on pedal $\mu$ is mathematically given by

$\mu=F*L\\\mu=34*0.165$

$\mu=5.61N.m$

Generally the equation for angular acceleration $\alpha$ is mathematically given by

$\alpha=\frac{\mu}{l}$

$\alpha=\frac{5.61}{0.12}$

$\alpha=46.75$

Therefore Angular speed is \omega

$\omega=\alpha*t$

$\omega=(46.75)*(0.6)$

$\omega=28.05rad/s$

Generally the equation for Tangential velocity V is mathematically given by

$V=r\omega$

$V=(0.33)(28.05)$

$V=9.2565m/s$

5 0

3 years ago

What equation will I need for this problem?

Yuri [45]

The formulas you will need are:

   Kinetic energy = (1/2) (mass) (speed)² .
and
   Potential energy = (mass) (gravity) (height) .

And of course,

   Total energy = constant .

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4 years ago

What part of an atom determines its size

velikii [3]

The answer is it's middle the nueculs

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3 years ago

EMERGENCY!! Doing my Physics homework now and I need help on this one please - all information provided, I'd be very VERY gratef

IrinaK [193]

(a) The vertical motion is accelerated by gravity. The horizontal component is constant (neglecting air resistance, if this is not a projectile motion, the horizontal component would also be accelerated)

(b) Vertical:

$v_y = 30\sin 40^\circ = 19.3\frac{m}{s}$

Horizontal:

$v_x = 30\frac{m}{s}\cos 40^\circ = 23.0 \frac{m}{s}$

(c) Use the kinematic equation for distance. Calculate only the vertical component (horizontal is irrelevant):

$s_y = -\frac{1}{2}gt^2 +vt\,,\,\,\,s_y=0\\0 = -\frac{1}{2}gt^2 +vt\\\frac{1}{2}gt =v\\t = \frac{2v}{g}= \frac{2\cdot30\sin 40^\circ\frac{m}{s}}{9.8\frac{m}{s^2}}=3.9s$

The ball will be in the air for about 3.9 s.

(d) The range is the horizontal distance traveled. We know the ball is in the air for 3.9s and it moves with a horizontal velocity of 23 m/s. So:

$s_x = 23\frac{m}{s}\cdot 3.9 s = 89.7m$

The range is 89.7 meters.

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3 years ago

HELP QUICK! Please can anyone do the other half for me

IrinaVladis [17]

Answer:

9) c

10) c

11) d

12) d

Explanation:

I think it's helps you

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3 years ago