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kupik [55]
2 years ago
11

Suppose that, instead of the Coulomb force law, one finds experimentally that the force between any two charge q1 and q2 is Writ

e the appropriate electric field E surrounding a point charge q
Physics
1 answer:
denpristay [2]2 years ago
6 0

Answer: E= KQ/r^2

Explanation: An electric field is a region where an electric charge(positive or negative ) will experience a force.

The magnitude of an electric field E, at a point is given by Coulombs law as

E/ F/q

Where F= Coulombs force exertedon the charge and q= electric charge

E= F/q=(KQq)/r^2q

E=KQ/r^2

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B.) to determine that electric beams in cathode ray tubes were actually made of particles

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2 years ago
A rigid tank contains nitrogen gas at 227 °C and 100 kPa gage. The gas is heated until the gage pressure reads 250 kPa. If the a
aleksley [76]

Answer:

 T₂ =602  °C

Explanation:

Given that

T₁ = 227°C =227+273 K

T₁ =500 k

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The absolute pressure at condition 1 will be

P₁ = 100 + 100 KPa

P₁ =200 KPa

Gauge pressure at condition 2 given = 250 KPa

The absolute pressure at condition 2 will be

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P₂ =350 KPa

The temperature at condition 2 = T₂

We know that

\dfrac{T_2}{T_1}=\dfrac{P_2}{P_1}\\T_2=T_1\times \dfrac{P_2}{P_1}\\T_2=500\times \dfrac{350}{200}\ K\\

T₂ = 875 K

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Sachi wants to throw a water balloon to knock over a target and win a prize. The target will only fall over if it is hit with a
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3.1 m/s²

Explanation:

Given:

Mass of the balloon (m) = 11.4 g = 0.0114 kg ( 1 kg = 1000 g)

Force acting on the balloon (F) = 0.035 N

Acceleration with which the balloon must be hit (a) = ?

Now, we know that, from Newton's second law, net force acting on an object is equal to the product of its mass and acceleration.

Therefore, framing in equation form, we have:

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Rewriting in terms of acceleration 'a', we get:

a=\frac{F}{m}

Now, substitute the given values and solve for 'a'. This gives,

a=\frac{0.035\ N}{0.0114\ kg}\\\\a=3.07\approx 3.1\ m/s^2(Nearest\ tenth)

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