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svet-max [94.6K]

3 years ago

6

You leave your bicycle out in the rain and it rust is a physical or chemical change

2 answers:

Alexxx [7]3 years ago

3 0

It’s a chemical change

ivann1987 [24]3 years ago

3 0

Answer:

The process when the iron rusts in the rain is a chemical change.Since the iron is exposed to oxygen and water. Generally, in my opinion it's not a good idea to leave your bicycle in the rain,it'll start with chain,it can spread to other parts of a bicycle,so dont do that.

Good luck!

Intelligent Muslim,

From Uzbekistan.

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(NEED HELP BADLY)

spin [16.1K]

Answer:

The answer to your question is below

Explanation:

Data 1

mass 1 = 250

mass 2 = 250 kg

gravity constant = 6.67 x 10⁻¹¹ Nm²/kg²

distance = 8 m

Formula

$F = G\frac{m1m2}{r^{2} }$

Substitution

$F = 6.67 x 10^{-11} \frac{250 x 250}{8^{2} }$

Result

F = 0.000000065 N

Data 2

mass 1 = 1000 kg

mass 2 = 1000 kg

distance = 5 m

Substitution

$F = 6.67 x 10^{-11} \frac{1000 x 1000 }{5^{2} }$

Result

F = 0.000002667 N

8 0

4 years ago

You are on an airplane traveling 30° south of due west at 180 m/s with respect to the air. The air is moving with a speed 31 m/s

BlackZzzverrR [31]

1) 211m/s
2)240<span>°
3)759,600m or 759.6 km</span>

6 0

3 years ago

A catapult with a radial arm 3.81 m long accelerates a ball of mass 18.2 kg through a quarter circle. The ball leaves the appara

saw5 [17]

Answer:

(a) $\alpha = 53.73 m/s^2$

(b) I =428 $kgm^2$

(c) $\tau = 428 \times 53.73 = 22996 .44Nm$

Explanation:

GIVEN

mass = 18.2 kg

radial arm length = 3.81 m

velocity = 49.8 m/s

mass of arm = 22.6 kg

we know using relation between linear velocity and angular velocity

$\omega = \frac{v}{l}$

$\omega = \frac{49.8}{3.81} \\\omega = 12.99 rad/s$

for angular acceleration, use the following equation.

$\omega _{f}^2 = \omega_{i}^2+2\alpha\theta$

since $\omega _{i} = 0$

here for one circle is 2 π radians. therefore for one quarter of a circle is π/2 radians

so for one quarter $\theta = \frac{\pi }{2}$

$(12.99)^2 = 2\alpha(\frac{\pi }{2})$

on solving

$\alpha = \frac{168.74}{\pi }\\\alpha = 53.73 m/s^2$

(b)

For the catapult,

moment of inertia

$I = \frac{1}{2}MR^2$

$I = \frac{1}{2} \times 22.6\times 3.81 \times 3.81\I = 164kg m^2$

For the ball,

$I = MR^2$

$I = 18.2 \times 14.51$

$I = 264 kgm^2$

so total moment of inertia = 428 $kgm^2$

(c)

$\tau = I\alpha$

$\tau = 428 \times 53.73 = 22996 .44Nm$

3 0

3 years ago

Brandon buys a new Seadoo He goes 12 km north from the beach . He jumps wakes for 6 km to the east . What distance cover ? What

zmey [24]

Answer:

distance- I think it is 6 then the displacement is 13.41

Explanation:

4 0

3 years ago

Which of the following is the equivalent resistance of the circuit shown?

erica [24]

I believe the answer is going to B I hope this helps and good luck

6 0

3 years ago