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2 years ago

How many nanoseconds does it take light to travel 1.00 ft in vacuum? (This result is a useful quantity to remember.)

1 answer:

4 0

To solve this problem, we will define speed as the amount of distance traveled per unit of time. We will clear the value of time and in parallel we will convert the Units to an international system to facilitate the calculation since we know the speed of light in a vacuum. The speed defined in terms of distance and time would be

$v = \frac{x}{t} \rightarrow t = \frac{d}{v}$

The speed of light in a vacuum is $3 * 10 ^ 8m / s$ and 1ft is equivalent to $0.3048m$ , so the estimated time would be

$t = \frac{d}{v}$

$t = \frac{0.3048m}{3*10^8m/s}$

$t = 1.02*10^{-9}s$

We know that 1 second is equivalent to $10 ^ 9ns$ , therefore

$t = 1.02*10^{-9}s (\frac{10^9ns}{1s})$

$t = 1.02ns$

Therefore the time is 1.02ns

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A rubber rod rubbed with fur acquires a charge of -4,8×10(-9)C. What is the charge on the fur? How much mass is transferred to r

mote1985 [20]

1) The charge left on the fur is equal and opposite to the charge transferred to the rod:
$Q=+4.8 \cdot 10^{9} C$
In fact, when the rod is rubbed with the fur, a net charge of $Q=-4.8 \cdot 10^{-9} C$ has been transferred to the rod, leaving it negatively charged. If we assume the fur was initially neutral, this means that we have now an excess of positive charges on the fur, and the amount of this charge must be equal (in magnitude, but with opposite sign) to the charge transferred to the rod.

2) The mass transferred to the rod is equal to the total mass of the electrons transferred to the rod.
The charge transferred to the rod is
$Q=-4.8 \cdot 10^{-9} C$
The charge of 1 electron is
$e=-1.6 \cdot 10^{-19} C$
So the number of electrons transferred is
$N= \frac{Q}{e}= \frac{-4.8 \cdot 10^{-9} C}{-1.6 \cdot 10^{-19} C}=3.0 \cdot 10^{10}$

The mass of 1 electron is $m=9.1 \cdot 10^{-31} kg$ , therefore the total mass transferred to the rod is
$M=Nm=(3 \cdot 10^{10})(9.1 \cdot 10^{-31} kg)=2.73 \cdot 10^{-20} kg$

7 0

3 years ago

A car went 60 km in 5/6 of an hour while a second car went 54 km in 2/3h. Which car was faster? How many times faster?

Nastasia [14]

Answer:

The car that went 54 km in 2/3h was faster, 5/4 times faster than the other car

Explanation:

Average speed of a car is the ratio between the displacement $\Delta x$ and the time (t) it takes to do that displacement:

$V_{avg}=\frac{\Delta x}{t}$

So, for the first car:

$V_{avg1}=\frac{60km}{\frac{5h}{6}} =72\frac{km}{h}$ (1)

for the second car we have:

$V_{avg2}=\frac{60km}{\frac{2h}{3}} =90\frac{km}{h}$ (2)

So, the second cart is faster than the first one. To find how many times divide speed 2 by speed 1:

$\frac{90}{72}=\frac{5}{4}$

4 0

3 years ago

Spheres of Charge: A metal sphere of radius 10 cm carries an excess charge of +2.0 μC. What is the magnitude of the electric fie

nadezda [96]

To solve this problem we will apply the concept related to the electric field. The magnitude of each electric force with which a pair of determined charges at rest interacts has a relationship directly proportional to the product of the magnitude of both, but inversely proportional to the square of the segment that exists between them. Mathematically can be expressed as,

$E = \frac{kV}{r^2}$