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Scorpion4ik [409]
3 years ago
5

. How far apart must two point charges of 75.0 nC (typical of static electricity) be to have a force of 1.00 N between them?

Physics
1 answer:
o-na [289]3 years ago
6 0

Answer:

0.00712 m

Explanation:

Given:

Charge on first particle (q₁) = 75 nC = 75\times 10^{-9}\ C

Charge on second particle (q₂) = 75 nC = 75\times 10^{-9}\ C

Force (F) = 1.00 N

Separation (d) = ?

The magnitude of force is given by Coulomb's law which states that, the magnitude of force acting between two charged particles separated by a distance is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them.

Therefore, the magnitude of force is given as:

F=\dfrac{k|q_1||q_2|}{d^2}

Where, k=9\times 10^9\ N\cdot m^2/ C^2 is the coulomb's constant.

Plug in the given values and solve for 'd'. This gives,

1.00=\frac{9\times 10^9\times 75.0\times 10^{-9}\times 75.0\times 10^{-9} }{d^2}\\\\d^2=\frac{9\times 10^9\times 75.0\times 10^{-9}\times 75.0\times 10^{-9} }{1.00}\\\\d=\sqrt{\frac{9\times 10^9\times 75.0\times 10^{-9}\times 75.0\times 10^{-9} }{1.00}}\\\\d=0.00712\ m

Therefore, the distance between the charges is 0.00712 m.

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Korvikt [17]

Answer:

Straight line parallel to time axis.

Explanation:

The slope of the position time graph gives the velocity.

As the man is still, that means the velocity is zero. So, the slope of the graph is zero. It is a straight line parallel to time axis.

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3 years ago
A stone is thrown vertically into the air at an initial velocity of 96 ft/s. On Mars, the height s (in feet) of the stone above
vladimir1956 [14]

Answer:

240 ft

Explanation:

t = Time taken

u = Initial velocity = 96 ft/s

v = Final velocity

s = Displacement

a = Acceleration = 12 m/s² on Mars 32 ft/s² on Earth negative due to upward direction

Mars

s=ut+\frac{1}{2}at^2\\\Rightarrow s=96\times t+\frac{1}{2}\times -12\times t^2\\\Rightarrow s=96t-6t^2\ ft

Earth

s=ut+\frac{1}{2}at^2\\\Rightarrow s=96\times t+\frac{1}{2}\times -32\times t^2\\\Rightarrow s=96t-16t^2\ ft

Differentiating the first equation with respect to time we get

\frac{ds}{dt}=96-12t

Equating with zero

0=96-12t\\\Rightarrow t=\frac{96}{12}=8\ s

Differentiating the second equation with respect to time we get

\frac{ds}{dt}=96-32t

Equating with zero

0=96-32t\\\Rightarrow t=\frac{96}{32}=3\ s

Applying the time taken to the above equations, we get

s=96t-6t^2\ ft\\\Rightarrow s=96\times 8-6\times 8^2\\\Rightarrow s=384

s=96t-16t^2\\\Rightarrow s=96\times 3-16\times 3^2\\\Rightarrow s=144

Difference in height = 384-144 = 240 ft

The stone will travel 240 ft higher on Mars

6 0
3 years ago
A woman is walking at 4 m/s. She is accelerating at a rate of 1 m/s2. To find out what her velocity is after 3 seconds, what els
liq [111]

Answer:

distance

Explanation:

8 0
2 years ago
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If a small sports car collides head-on with a massive truck, which vehicle experiences the greater impact force? Which vehicle e
garri49 [273]

Answer:

Small sports car.

Explanation:

Lets take

mass of the small car = m

mass of the truck = M

As we know that when car collide with the massive truck then due to change in the moment of the car both car as well as truck will feel force.We also know that from Third law of Newton's ,it states that every action have it reaction with same magnitude but in the opposite direction.

Therefore

F = m a

a=Acceleration of the car

a=\dfrac{F}{m}

F= M a'

a'=Acceleration of the massive truck

a'=\dfrac{F}{M}

Here given that M > m that is why a > a'

Therefore car will experiences more acceleration.

5 0
3 years ago
Imagine that an electron in an excited state in a nitrogen molecule decays to its ground state, emitting a photon with a frequen
mash [69]
Since energy cannot be created nor destroyed, the change in energy of the electron must be equal to the energy of the emitted photon.

The energy of the emitted photon is given by:
E=hf
where
h is the Planck constant
f is the photon frequency
Substituting f=8.88 \cdot 10^{14}Hz, we find
E=hf=(6.6 \cdot 10^{-34} Js)(8.88 \cdot 10^{14} Hz)=5.86 \cdot 10^{-19} J

This is the energy given to the emitted photon; it means this is also equal to the energy lost by the electron in the transition, so the variation of energy of the electron will have a negative sign (because the electron is losing energy by decaying from an excited state, with higher energy, to the ground state, with lower energy)
\Delta E= -5.86 \cdot 10^{-19} J
6 0
3 years ago
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