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3 years ago

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the number of electrons in a copper penny is approximately 10 10^23. How large would the force be on an object if it carried thi

s charge and were repelled by an equal charge one meter away?

1 answer:

iVinArrow [24]3 years ago

8 0

Answer:

0.000230144 N

Explanation:

n = Number of electrons = $10\times 10^{23}$

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

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

r = Distance between obects = 1 m

The force would be

$F=\dfrac{nkq^2}{r^2}\\\Rightarrow F=\dfrac{10\times 10^{23}\times 8.99\times 10^9\times (1.6\times 10^{-19})^2}{1^2}\\\Rightarrow F=0.000230144$

The force would be 0.000230144 N

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A toy gun uses a spring to project a 4.5-g soft rubber sphere horizontally. The spring constant is 8.0 N/m, the barrel of the gu

marishachu [46]

Answer:

1.93 m/s

Explanation:

Parameters given:

Mass = 4.5g = 0.0045kg

Spring constant = 8.0 N/m

Length of barrel = 13 cm = 0.013m

Frictional force = 0.035N

Compression = 5.8 cm = 0.058m

First, we find the P. E. stored in the spring:

P. E. = ½*k*x²

P. E. = ½ * 8 * 0.058² = 0.013J

Then, we find the work done by the frictional force while the sphere is leaving the barrel of the gun:

Work = Force * distance

The distance here is the length of the barrel.

Work = 0.035 * 0.13 = 0.0046 J

The kinetic energy of the sphere can now be found:

K. E. = P. E. - Work done

K. E. = 0.013 - 0.0046 = 0.0084J

We can now find the speed using the formula for K. E.:

K. E. = ½*m*v²

0.0084 = ½ * 0.0045 * v²

v² = 0.0084/0.00255 = 3.733

=> v = 1.93 m/s

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3 years ago

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When a certain rubber band is stretched a distance x, it exerts a restoring force of magnitude f = ax, where a is a constant. th

Veseljchak [2.6K]

Given:
F = ax
where
x = distance by which the rubber band is stretched
a = constant

The work done in stretching the rubber band from x = 0 to x = L is
$W=\int_{0}^{L} Fdx = \int_{0}^{L}ax \, dx = \frac{a}{2} [x^{2} ]_{0}^{L} = \frac{aL^{2}}{2}$

Answer: $\frac{aL^{2}}{2}$

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4 years ago

A GPS signal travels at the speed of light 300,000,000 m/s. If it takes .05s to reach a phone on the surface of the Earth, how f

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3 years ago

A medical cyclotron used in the production of medical isotopes accelerates protons to 6.5 MeV. The magnetic field in the cyclotr

mart [117]

Answer:

diameter of largest orbit is 0.60 m

Explanation:

given data

isotopes accelerates KE = 6.5 MeV

magnetic field B = 1.2 T

to find out

diameter

solution

first we find velocity from kinetic energy equation

KE = 1/2 × m×v² ........1

6.5 × 1.6 × $10^{-19}$ = 1/2 × 1.672 × $10^{-27}$ ×v²

v = 3.5 × $10^{7}$ m/s

so

radius will be

radius = $\frac{m*v}{B*q}$ ........2

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so diameter = 2 × 0.30

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3 years ago

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Cheetahs can accelerate to a speed of 19.6 m/s in 2.45 s and can continue to accelerate to reach a top speed of 27.6 m/s . Assum

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Answer:

$v_{top} = 61.96 \frac{mi}{h}$

Explanation:

To express the cheetah's top speed in miles per hour, we just need to find the conversion factor.

We know that the top speed is

$v_{top} = 27.7 \frac{m}{s}$

So, we want to obtain miles from meters and hours from seconds.

<h3>miles from meters</h3>

First we write the equivalence:

$1609.34 \ m = 1 \ mi$

Now, we can divide by 1609.34 meters on both sides:

$\frac{1609.34 \ m}{ 1609.34 \ m} = \frac{1 \ mi}{ 1609.34 \ m}$

The left sides equals 1, so

$1 = \frac{1 \ mi}{ 1609.34 \ m}$

And this is our conversion factor from meters to miles. Now, we can multiply our top speed by this conversion factor, as the conversion factor equals one, and is dimensionless, the physical meaning will be the same.

$v_{top} = 27.7 \frac{m}{s} * \frac{1 \ mi}{ 1609.34 \ m}$

$v_{top} = 27.7 \frac{m}{s} * \frac{1 \ mi}{ 1609.34 \ m}$

$v_{top} = 0.0172120 \frac{mi}{s}$

This is the top speed in miles per second, now, for obtaining miles per hour:

<h3>hours from seconds</h3>

We can do pretty much the same, first, the equivalence:

$1 \ h = 3600 \ s$

as the seconds are dividing in the velocity, we know divide by 1 hour.

$\frac{1 \ h}{ 1 \ h} = \frac{3600 \ s}{ 1 \ h}$

$1 = \frac{3600 \ s}{ 1 \ h}$

and know we just multiply our top speed by this conversion factor

$v_{top} = 0.0172120 \frac{mi}{s} \frac{3600 \ s}{ 1 \ h}$

$v_{top} = 61.96 \frac{mi}{h}$

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3 years ago