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3 years ago

13

suppose an electrically charged ruler transfers some of its charge by contact to a tiny plastic sphere. will the ruler and the s

phere attract or repel afterwards? explain your reasoning

1 answer:

poizon [28]3 years ago

5 0

Answer:

Here ball and rod will repel each other as they are of similar charges

Explanation:

As we know that the two charges attract or repel each other by electrostatic force

This force is given as

$F = \frac{kq_1q_2}{r^2}$

so we know if two charges are similar in nature then they will repel each other and if the two charges are opposite in nature then they will attract each other

So here when rod touch the ball then it transfer its charge to the ball and due to similar charges in ball and rod they both repel each other

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A student measures that 81,500 J of thermal energy were added to 0.5 kg of water. If the specific heat of water is 4,184 J/kg 0C

Effectus [21]

Answer:

$\Delta T=38.95^{\circ} C$

Explanation:

Given that,

Heat measured, Q = 81500 J

Mass of water, m = 0.5 kg

The specific heat of water is 4,184 J/kg °C

We need to find the change in temperature. The heat measured is given by :

$Q=mc\Delta T$

Where

$\Delta T$ is the change in temperature

$\Delat T=\dfrac{Q}{mc}\\\\\Delat T=\dfrac{81500}{0.5\times 4184 }\\\\\Delta T=38.95^{\circ} C$

So, the change in temperature is $38.95^{\circ} C$ .

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3 years ago

zlopas [31]

Bronchial Tubes
hope this helps :)

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4 years ago

Read 2 more answers

Water is traveling through a horizontal pipe with a speed of 1.7 m/s and at a pressure of 205 kPa. This pipe is reduced to a new

Arturiano [62]

Answer:

The velocity is $v_2 = 6.8 \ m/s$

The pressure is $P_2 = 204978 Pa$

Explanation:

From the question we are told that

The speed at which water is travelling through is $v = 1.7 \ m/s$

The pressure is $P_1 = 205 k Pa = 205 *10^{3} \ Pa$

The diameter of the new pipe is $d = \frac{D}{2}$

Where D is the diameter of first pipe

According to the principal of continuity we have that

$A_1 v_1 = A_2 v_2$

Now $A_1$ is the area of the first pipe which is mathematically represented as

$A_1 = \pi \frac{D^2}{4}$

and $A_2$ is the area of the second pipe which is mathematically represented as

$A_2 = \pi \frac{d^2}{4}$

Recall $d = \frac{D}{2}$

$A_2 = \pi \frac{[ D^2]}{4 *4}$

$A_2 = \frac{A_1}{4}$

So $A_1 v_1 = \frac{A_1}{4} v_2$

substituting value

$1.7 = \frac{1}{4} * v_2$

$v_2 = 4 * 1.7$

$v_2 = 6.8 \ m/s$

According to Bernoulli's equation we have that

$P_1 + \rho \frac{v_1 ^2}{2} = P_2 + \rho \frac{v_2 ^2}{2}$

substituting values

$205 *10^{3 }+ \frac{1.7 ^2}{2} = P_2 + \frac{6.8 ^2}{2}$

$P_2 = 204978 Pa$

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