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4 years ago

13

A 2.00-gram air inflated balloon is given an excess negative charge, q1 = -3.25 × 10-8 C, by rubbing it with a blanket. It is fo

und that a charged rod can be held above the balloon at a distance of d = 5.00 cm to make the balloon float. In order for this to occur, what polarity of charge must the rod possess? How much charge, q2, does the rod have? Assume the balloon and rod to be point charges. The Coulomb force constant is 1/(4π ε0) = 8.99 × 109 N·m2/C2 and the acceleration due to gravity g = 9.81 m/s2.

1 answer:

Llana [10]4 years ago

3 0

Answer:

The rod`s charge must be positive, because the gravity force is pointing downwards and the electrostatic force must be pointing upwards (in order to balance the gravity force)

The charge is q_2 = 1.667 times 10^(-7) C

Explanation:

F_e = F_g

where F_g = m g and F_e= (1/4 pi e_0)*(q_1*q_2)/d^2,

please see the file attached for more details.

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If you have 240.0 mL of water at 25.00 °C and add 100.0 mL of water at 95.00 °C, what is the final temperature of the mixture? U

Rus_ich [418]

<u>Answer:</u> The final temperature of the mixture is 45.6°C

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of water = 1 g/mL

<u>When volume is 240.0 mL</u>

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{240.0mL}\\\\\text{Mass of water}=(1g/mL\times 240.0mL)=240g$

<u>When volume is 100.0 mL</u>

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{100.0mL}\\\\\text{Mass of water}=(1g/mL\times 100.0mL)=100g$

When two water solutions at different temperature are mixed, the amount of heat released by water present at higher temperature will be equal to the amount of heat absorbed by water present at lower temperature..

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]$ ......(1)

where,

q = heat absorbed or released

$m_1$ = mass of water solution 1 = 240 g

$m_2$ = mass of water solution 2 = 100 g

$T_{final}$ = final temperature = ?°C

$T_1$ = initial temperature of water solution 1 = 25°C

$T_2$ = initial temperature of water solution 2 = 95°C

c = specific heat of water= 4.186 J/g°C

Putting values in equation 1, we get:

$240\times 4.186\times (T_{final}-25)=-[100\times 4.186\times (T_{final}-95)]\\\\T_{final}=45.6^oC$

Hence, the final temperature of the mixture is 45.6°C

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3 years ago

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what kind of energy does a bow string have when it's stretched? a gravitational potential energy b.elastic potential energy c.ch

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Hi I believe it is b. sorry if this isnt found to be helpful.

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3 years ago

A student buys a plastic dart gun and tries to find the maximum horizontal range. The student shoots the gun straight up and it

ryzh [129]

Answer:

The value is $R_{max} = 33.54 \ m$

Explanation:

From the question we are told that

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So v is the velocity at maximum height and the value is v = 0 m/s

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Here u is the initial velocity of the dart as it leaves that gun

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$R = \frac{u ^2 sin 2\theta }{g}$

For maximum horizontal range the value of $\theta = 45^o$

So

$R_{max} = \frac{ 18.13 ^2 sin 2(45) }{9.8}$

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Answer:

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Explanation:

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The grainy appearance are produced by granules on the photosphere of the sun and granules are caused by convection currents of plasma within the sun's convection zone.

The interior of these granules are brighter (and thus hotter) than the exterior of the granules which are darker.

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