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deff fn [24]
3 years ago
15

Whenever you work a physics problem you should get into the habit of thinking about whether the answer is physically realistic.

Think about how far off the ground a typical small truck is. Is the answer to Part B physically realistic?
Physics
2 answers:
Marta_Voda [28]3 years ago
4 0

Answer:

Yes, the answer on <em>how far off the ground a typical truck is happens to be</em> physically realistic.

Explanation:

This concept could be explained with when someone is viewing a picture from a particular distance. The farther the person is from the picture, the smaller the details of the picture and the higher they picture would be unappriciated and vice versa.

Likewise the small truck on ground. <em>The distance off ground the truck is being viewed, the smaller the truck would become.</em>

Alisiya [41]3 years ago
4 0

Answer:

the is how far off the ground a typical small truck is it is happen to be physically realistic

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An isolated conducting sphere has a 17 cm radius. One wire carries a current of 1.0000020 A into it. Another wire carries a curr
notsponge [240]

14 ms is required to reach the potential of 1500 V.

<u>Explanation:</u>

The current is measured as the amount of charge traveling per unit time. So the charge of electrons required for each current is determined as the product of current with time.

       Charge = Current \times Time

As two different current is passing at two different times, the net charge will be the different in current.  So,

        \text { Charge }=(1.0000020-1.0000000) \times t=2 \times 10^{-6} \times t

The electric voltage on the surface of cylinder can be obtained as the ratio of charge to the radius of the cylinder.

        V=\frac{k q}{R}

Here k = 9 * 10^9, q is the charge and R is the radius. As q=2 \times 10^{-6} \times t and R =17 cm = 0.17 m, then the voltage will be

        V=\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}

The time is required to find to reach the voltage of 1500 V, so

1500 =\frac{9 \times 10^{9} \times 2 \times 10^{-6} \times t}{0.17}

\begin{aligned}&t=\frac{1500 \times 0.17}{\left(9 \times 10^{9} \times 2 \times 10^{-6}\right)}\\&t=14.1666 \times 10^{-3} s=14\ \mathrm{ms}\end{aligned}

So, 14 ms is required to reach the potential of 1500 V.

3 0
3 years ago
Which is a characteristic of the atom marked A
erastovalidia [21]
Its very dense. Hey, are you homeschooled?
6 0
3 years ago
A book falls of the table and hits the floor. Is work done? Why?​
morpeh [17]

<u>Yes, work is done when a book falls of the table.</u>

This is because:

When the book falls, it's potential energy is converted into kinetic energy.   As it reaches the floor down, this kinetic energy is converted to heat energy and sound energy due to the impact.

When a force is imposed on an object to cause displacement of that object, work is done on that object. For a force to do work on an object, there should be a displacement and this force should cause the displacement. So here, since the book falls from the table and causes the displacement of the book from the table to the floor. It is said that work is done.

Work can be given by the formula:

W = F • d

where F is the force and d is the displacement.

6 0
2 years ago
Read 2 more answers
Two motorcyclists are riding side-by-side at night and the distance between their center-mounted headlights is 1.40 m. (a) If th
dezoksy [38]

Answer:

θ = 1.591 10⁻² rad

Explanation:

For this exercise we must suppose a criterion when two light sources are considered separated, we use the most common criterion the Rayleigh criterion that establishes that two light sources are separated census the central maximum of one of them coincides with the first minimum of the other source

         

Let's write the diffraction equation for a slit

       a sin θ = m λ

The first minimum occurs for m = 1, also field in these we experience the angles are very small, we can approximate the sin θ = θ

             θ = λ / a

In our case, the pupil is circular, so the system must be solved in polar coordinates, so a numerical constant is introduced.

           θ = 1.22 λ / D

Where D is the diameter of the pupil

 Let's apply this equation to our case

        θ = 1.22 600 10⁻⁹ / 0.460 10⁻²

        θ = 1.591 10⁻² rad

This is the angle separation to solve the two light sources

6 0
3 years ago
A charge of 19 nC is uniformly distributed along a straight rod of length 15 m that is bent into a circular arc with a radius of
Salsk061 [2.6K]

Answer:

The magnitude of the electric field at the center of curvature of the arc is 3.87 N/C

Explanation:

Please see the attachments below

8 0
3 years ago
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