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Dmitry_Shevchenko [17]

3 years ago

9

When a wool cloth is rubbed on the surface of a balloon, it tends to give away electrons causing the surface of the balloon to b

ecome :
positively charged
negativity charged
neutrally charged

2 answers:

Law Incorporation [45]3 years ago

4 0

Answer:

It is negativity charged

Shalnov [3]3 years ago

4 0

Answer:

positively charged

Explanation:

a wool has neutral charges hence when rubbed against a balloon it will generate positive charges

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The mechanic did 5406 Joules of work pushing the car.

That's the energy he put into the car. When he stops pushing, all the energy he put into the car is now the car's kinetic energy.

Kinetic energy = (1/2) (mass) (speed²)

And there we have it

The car's mass is 3,600 kg.
Its speed is 'v' m/s .
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1800 kg (v²) = 5406 joules

v² = (5406 joules) / (1800 kg)

v² = (5406/1800) (joules/kg)

= = = = = This section is just to work out the units of the answer:

v² = (5406/1800) (Newton-meter/kg)

v² = (5406/1800) (kg-m²/s² / kg)

v² = (5406/1800) (m²/s²)

= = = = =

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3 years ago

in a tall building, a certain window is 500 meters above the ground.A tennis ball is catapulted horizontally out of the window w

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The tennis ball lands at a point 40.4 m from the base of the building.

The tennis ball is projected with a horizontal velocity <em>u</em> from a window, which is at a height <em>y</em> from the ground. The ball lands at a distance <em>x</em> from the base of the building. Let the ball take a time <em>t</em> to reach the ground. In the time <em>t</em> ,the ball falls a vertical distance <em>y</em> and also travel a horizontal distance <em>x</em>.

The initial vertical velocity of the ball is zero, since the ball is projected in the horizontal direction. The ball falls down under the action of gravitational force.

Thus, use the equation of motion,

$y=\frac{1}{2} gt^2$

rewrite the expression for <em>t</em> and calculate the value of <em>t</em> using 9.81 m/s²for <em>g</em> and 500 m for <em>y</em>.

$t=\sqrt{\frac{2y}{g} } \\ =\sqrt\frac{(2)(500m)}{9.81m/s^2} \\ =10.096 s$

The horizontal distance <em>x</em> is traveled using the constant velocity <em>u </em>since no force acts on the ball in the horizontal direction.

Therefore,

$x=ut$

Substitute 4 m/s for <em>u</em> and 10.096 s for <em>t</em>

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3 years ago

A skydiver weighing 200 lbs with clothes that have a drag coefficient of .325 is falling in an area that has an atmospheric dens

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Answer:

The right solution is:

(a) 89.455 m/s

(b) 44.73 m/s

Explanation:

The given values are:

Mass,

m = 200 lbs

or,

= $\frac{200}{2.205} \ kg$

= $90.7 \ kg$

Air's density,

$\delta = 1.225 \ kg/m^3$

Drag coefficient,

$C_d=0.325$

When body is straight, area,

$A_1=6 \ ft^2$

As we know,

Terminal velocity,

⇒ $V_t=\sqrt{\frac{2W}{C_d \delta A} }$

or,

⇒ $=\sqrt{\frac{2mg}{C_d \delta A} }$

At straight orientation,

⇒ $V_t'=\sqrt{\frac{2\times 90.7\times 9.8}{0.325\times 1.225\times 0.558} }$

⇒ $=\sqrt{\frac{1777.72}{0.223}}$

⇒ $=89.455 \ m/s$

When belly flat,

⇒ $V_t''=\sqrt{\frac{2\times 90.7\times 9.8}{0.325\times 1.225\times 0.558\times 4} }$

⇒ $=\sqrt{\frac{1777.72}{0.889} }$

⇒ $=44.73 \ m/s$

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3 years ago