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3 years ago

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Riding a bicycle on a flat, smooth surface is a lot easier than riding it along a bumpy surface or up a hill. A bumpy surface cr

eates more friction with your bike tires than a smooth surface does, and going up a hill means fighting gravity.
Write about a time you had to ride a bicycle on a difficult surface. What did you have to do to adjust your riding?

1 answer:

liberstina [14]3 years ago

6 0

Answer:

one time i was one the flat ground at my aunts house then we went on a hike so i brought my bike it had just rained that day so it was kinda muddy so there was sticks everywhere.i was riding up hill and noticed that it was very hard,then i rode down hill and it was much better

Explanation:

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Most energy obtained from water is converted from _____.

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Read 2 more answers

A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in

koban [17]

Answer:

a) The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) The electric force is $2.0\times 10^{-6}$ newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

$E = \frac{F_{e}}{q}$ (1)

Where:

$E$ - Electric field, measured in newtons per coulomb.

$F_{e}$ - Electric force, measured in newtons.

$q$ - Electric charge, measured in coulombs.

If we know that $F_{e} = 4.0\times 10^{-6}\,N$ and $q = 2.0\times 10^{-8}\,C$ , then the electric field at that point is:

$E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}$

$E = 2.0\times 10^{2}\,\frac{N}{C}$

The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) If we know that $E = 2.0\times 10^{2}\,\frac{N}{C}$ and $q = 1.0\times 10^{-8}\,C$ , then the electric force is:

$F_{e} = E\cdot q$

$F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)$

$F_{e} = 2.0\times 10^{-6}\,N$

The electric force is $2.0\times 10^{-6}$ newtons.

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2 years ago

The atomic radii of Mg2+ and F- ions are 0.072 and 0.133 nm, respectively.(a) Calculate the force of attraction between these tw

timurjin [86]

Answer:

$1.09527\times 10^{-8}\ N$

Explanation:

$q_1$ = Mg ion = $+2q$

$q_2$ = F ion = $-q$

q = Charge of electron = $1.6\times 10^{-19}\ C$

r = Distance between ions = $0.072+0.133\ nm$

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Electrical force is given by

$F=-\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=-\dfrac{8.99\times 10^9\times 2\times 1.6\times 10^{-19}\times -1\times 1.6\times 10^{-19}}{[(0.072+0.133)\times 10^{-9}]^2}\\\Rightarrow F=1.09527\times 10^{-8}\ N$

The attractive force is $1.09527\times 10^{-8}\ N$

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