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2 years ago

Anybody knows ? mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm .

Physics

1 answer:

notsponge [240]2 years ago

6 0

Answer:

I think it's because it's an insulator and like charges attract

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In an open system, the vapor pressure is equal to the outside air pressure. true false

Anvisha [2.4K]

False if the pressure is in an open system it could not be equal to the outside

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3 years ago

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If you double the voltage in a circuit and reduce the resistance by a factor of four, what will happen to the current?

Kamila [148]

doubling or tripling the voltage will cause the current to be doubled or tripled. On the other hand, any alteration in the resistance will result in the opposite or inverse alteration of the current. So doubling or tripling the resistance will cause the current to be one-half or one-third the original value.

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2 years ago

A tennis player player receives a shot with the ball (.06kg) travelling horizontally at 50.0 m/s and returns the shot with the b

ahrayia [7]

Let:

m- 0.0600Kg 
v1- +50.0 m/s 
v2- -42 m/s 

a) 

momentum1 - momentum2 = impulse 
p1 - p2 = Ft 
m(v1 + v2) = Ft 
Ft = 0.54 N-s 

b) 

Work is traditionally force times distance but an easy way to get through this one is to think logically and keep your units straight: 

The racket creates an overall velocity of 90 m/s for the ball: 

W=(Ft)*(V) 
W=0.54 N-s * 90 m/s 
W=48.6 N-m = 48.6 J

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3 years ago

"A circuit consists of a battery and two resistors connected in parallel. If the resistance in both resistors is doubled, and th

sveticcg [70]

Let's call R the value of the resistance of the two resistors. In the first situation, the resistors are connected in parallel, so their equivalent resistance is given by:
$\frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{R}=\frac{2}{R}$
which means:
$R_{eq}=\frac{R}{2}$
And calling I the current in the circuit, the voltage is given by Ohm's law:
$V=IR_{eq}=\frac{IR}{2}$

In the second situation, the resistance of each resistor is doubled: R'=2R. So, the equivalent resistance in this case is given by
$\frac{1}{R_{eq}}=\frac{1}{2R}+\frac{1}{2R}=\frac{2}{2R}=\frac{1}{R}$
which means
$R_{eq}=R$
And the new voltage is given by:
$V'=IR_{eq}=IR=2V$
which is twice the original voltage, so the voltage has doubled.

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2 years ago

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A positively charged particle Q1 = +15nC is held fixed at the origin. A second charge Q2 of mass m = 9.5g is floating a distance

Paladinen [302]

Answer:

$Q_2=4.4293\times 10^{22}\ nC$

Explanation:

Given

charge on the fixed particle, $Q_1=15\times 10^{-9}\ C$

mass of the second charge, $m_2=9.5\times 10^{-3}\ kg$

distance between the fixed charge and the floating charge on the top, $d=0.25\ m$

According to the question the second charge is floating just above the fixed positive charge despite of gravity this means that the floating charge is also positive in nature and hence feels the repulsion from the fixed charge which is equal in magnitude to the gravitational force on the charge.

$m_2.g=\frac{1}{4\pi\epsilon_0} \times \frac{Q_1.Q_2}{d^2}$

where:

$\epsilon_0=$ permittivity of free space

g = acceleration due to gravity

$9.5\times 10^{-3}\times 9.8=8.85\times 10^{-9}\times \frac{15\times 10^{-9}\times Q_2}{0.25^2}$

$Q_2=4.4293\times 10^{22}\ nC$

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3 years ago