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sleet_krkn [62]
3 years ago
14

When sugar is poured from the box into the sugar bowl, the rubbing of sugar grains creates a static electric charge that repels

the grains, and causes sugar to go flying out in all directions. If each of two sugar grains acquires a charge of 1.79 x 10-11 C at a separation of 3.45 x 10-5 m, with what force will they repel each other?
Physics
1 answer:
Afina-wow [57]3 years ago
3 0

Answer:

2.6×10⁻³ N

Explanation:

From coulomb's law,

F = kq'q/r²................ Equation 1

Where F = Repulsive force, q' = charge on the first sugar grain, q = charge on the second sugar grain, r = distance of separation between the sugar grain, k = proportionality constant.

From the question,

since q' = q

Then,

F = kq²/r²..................... Equation 2

Given: q = 1.79×10⁻¹¹ C, r = 3.45×10⁻⁵ m,

Constant: k = 9×10⁹ Nm²/kg².

Substitute into equation 2

F = 9×10⁹(1.79×10⁻¹¹)²/(3.45×10⁻⁵ )²

F = 9×10⁹(3.2041×10⁻²²)/(11.9025×10⁻¹⁰)

F = (28.8369×10⁻¹³)/(11.9025×10⁻¹⁰)

F = 2.6×10⁻³ N.

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Thepotemich [5.8K]

Answer:

D. Burning paper

Explanation:

physical change is just changing the way it looks

chemical change is changing the properties of the object or thing

melting ice is just changing the way it looks it still has the same atoms and elements.

when you burn something it will change the properties of the atoms in the object

4 0
3 years ago
Which of the following describes the importance of cyanobacteria to early Earth atmosphere formation?
Dmitrij [34]

I do believe it is a
6 0
3 years ago
There is a 3-kg toy cart moving at 4 m/s. Calculate the kinetic<br> energy of the cart.
m_a_m_a [10]

The kinetic energy of the cart is 24 J.

<u>Explanation:</u>

The acceleration of a given mass from rest to the velocity is known as kinetic energy. It gains energy from acceleration and remains in this state until the speed of the object changes.  

The kinetic energy is the given by,

                           K.E = 1/2 mv^2

Given the mass m = 3 kg,        v = 4 m / s.

                            K.E = 1/2 \times 3 \times (4)^2          

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5 0
3 years ago
Consider an object with s=12cm that produces an image with s′=15cm. Note that whenever you are working with a physical object, t
Leni [432]

A. 6.67 cm

The focal length of the lens can be found by using the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}

where we have

f = focal length

s = 12 cm is the distance of the object from the lens

s' = 15 cm is the distance of the image from the lens

Solving the equation for f, we find

\frac{1}{f}=\frac{1}{12 cm}+\frac{1}{15 cm}=0.15 cm^{-1}\\f=\frac{1}{0.15 cm^{-1}}=6.67 cm

B. Converging

According to sign convention for lenses, we have:

- Converging (convex) lenses have focal length with positive sign

- Diverging (concave) lenses have focal length with negative sign

In this case, the focal length of the lens is positive, so the lens is a converging lens.

C. -1.25

The magnification of the lens is given by

M=-\frac{s'}{s}

where

s' = 15 cm is the distance of the image from the lens

s = 12 cm is the distance of the object from the lens

Substituting into the equation, we find

M=-\frac{15 cm}{12 cm}=-1.25

D. Real and inverted

The magnification equation can be also rewritten as

M=\frac{y'}{y}

where

y' is the size of the image

y is the size of the object

Re-arranging it, we have

y'=My

Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.

Also, the sign of s' tells us if the image is real of virtual. In fact:

- s' is positive: image is real

- s' is negative: image is virtual

In this case, s' is positive, so the image is real.

E. Virtual

In this case, the magnification is 5/9, so we have

M=\frac{5}{9}=-\frac{s'}{s}

which can be rewritten as

s'=-M s = -\frac{5}{9}s

which means that s' has opposite sign than s: therefore, the image is virtual.

F. 12.0 cm

From the magnification equation, we can write

s'=-Ms

and then we can substitute it into the lens equation:

\frac{1}{f}=\frac{1}{s}+\frac{1}{s'}\\\frac{1}{f}=\frac{1}{s}+\frac{1}{-Ms}

and we can solve for s:

\frac{1}{f}=\frac{M-1}{Ms}\\f=\frac{Ms}{M-1}\\s=\frac{f(M-1)}{M}=\frac{(-15 cm)(\frac{5}{9}-1}{\frac{5}{9}}=12.0 cm

G. -6.67 cm

Now the image distance can be directly found by using again the magnification equation:

s'=-Ms=-\frac{5}{9}(12.0 cm)=-6.67 cm

And the sign of s' (negative) also tells us that the image is virtual.

H. -24.0 cm

In this case, the image is twice as tall as the object, so the magnification is

M = 2

and the distance of the image from the lens is

s' = -24 cm

The problem is asking us for the image distance: however, this is already given by the problem,

s' = -24 cm

so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.

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by how much does the speed of a vehicle moving ina straight line change each second when it is accelerating at 2km/h/s? At 4 km/
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Answer:

At 2km/h*s it is changing speed every 2 seconds.

At 4km/h*s it is changing speed every 4 seconds.

At 10km/h*s it is changing speed every 10 seconds.

8 0
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