All categories

3 years ago

The earliest known records describing electric charge involve the use of _____.

Physics

2 answers:

Anika [276]3 years ago

8 0

Yes..the answer to this question is "Amber" .... hope this helps!!

blondinia [14]3 years ago

5 0

Amber is the answer to this question

You might be interested in

In major league baseball, the pitcher's mound is 60 feet from the batter. If a pitcher throws a 85 mph fastball, how much time e

Nostrana [21]

The given from your problem are the following:
V = 85mph (This is miles per hour)
d = 60 feet

If you notice the units do not match. Before we can do anything else, we need to make the figures match.

In this case, we will convert 85miles per hour to feet per hour. There are 5,280 feet in 1 mile.
$\frac{85miles }{hr}$ x $\frac{5,280feet}{1miles} = \frac{448,800feet}{hr}$

But wait! If you think about the scenario, you are looking for how long it will take for the ball to reach the bat. The most applicable unit of time to use here is second. It would be very hard to really measure a short and instantaneous event in hours. So we convert it into feet per second:

There are 3,600 seconds in 1 hour.

$\frac{448,800feet }{hour}$ x $\frac{1hour}{3,600seconds} = \frac{448,800feet}{3,600 seconds} = 124.67ft/s$

So now we have our new given as:

v = 124.67ft/s
d = 60 ft

The formula for time can be derived from the formula from velocity, which is:
$velocity = \frac{distance}{time}$

The formula of time will then be:
$time= \frac{distance}{velocity}$

All you need to do is plug in what you know and solve for what you don't know.

$time= \frac{60feet}{124.67ft/s}$

$time= 0.48s$

The answer then is 0.48s.

If you want this in hours, just divide the value in seconds by 3,600. The answer would then be 0.00013hr. (See how small it is? This is why seconds would be a more appropriate measure.)

8 0

3 years ago

In terms of the sequence of the scientific method, what is the immediate purpose of doing an experiment? A. asking the question.

densk [106]

In terms of the scientific method, the immediate purpose of doing an experiment is gathering data. You cannot draw conclusions or form a proper hypothesis without some sort of basis and data.

4 0

3 years ago

Read 2 more answers

What is the force per unit area at this point acting normal to the surface with unit nor- Side View √√ mal vector n = (1/ 2)ex +

Mumz [18]

Complete Question:

Given $\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$ at a point. What is the force per unit area at this point acting normal to the surface with $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$ ? Are there any shear stresses acting on this surface?

Answer:

Force per unit area, $\sigma_n = 28 MPa$

There are shear stresses acting on the surface since $\tau \neq 0$

Explanation:

$\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$

equation of the normal, $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$

$\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

Traction vector on n, $T_n = \sigma \b n$

$T_n = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

$T_n = \left[\begin{array}{ccc}\frac{23}{\sqrt{2} }\\0\\\frac{27}{\sqrt{33} }\end{array}\right]$

$T_n = \frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z$

To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.

$\sigma_n = T_n . \b n$

$\sigma \b n = (\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z) . ((1/ \sqrt{2} ) \b e_x + 0 \b e_y +(1/ \sqrt{2}) \b e_z)\\\\\sigma \b n = 28 MPa$

If the shear stress, $\tau$ , is calculated and it is not equal to zero, this means there are shear stresses.

$\tau = T_n - \sigma_n \b n$

$\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau = \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z$

$\tau = \sqrt{(-5/\sqrt{2})^2 + (27/\sqrt{2})^2 + (5/\sqrt{2})^2} \\\\ \tau = 19.74 MPa$

Since $\tau \neq 0$ , there are shear stresses acting on the surface.

3 0

3 years ago

A certain superconducting magnet in the form of a solenoid of length 0.72 m can generate a magnetic field of 3.5 T in its core w

ira [324]

Answer:

55407

Explanation:

we have given that magnetic field B=3.5 T

current through the coil=90 A

Length of solenoid =0.72 m

we know the formula of magnetic field

$B=\frac{\mu _0NI}{l}$

so $N=\frac{Bl}{\mu _0I}=\frac{3.5\times 0.72}{4\pi \times 10^{-6}\times 90}=55407.277=55407\ turns$

so the number of turn in solenoid will be 55407

3 0

2 years ago

A system absorbs 194 kj of heat and the surroundings do 120 kj of work on the system. internal eneergy change

notsponge [240]

We can solve the problem by using the first law of thermodynamics, which states that:
$\Delta U = Q-W$
where
$\Delta U$ is the change in internal energy of the system
Q is the heat absorbed by the system
W is the work done by the system

In our problem, the heat absorbed by the system is Q=+194 kJ, while the work done is W=-120 kJ, where the negative sign means the work is done by the surroundings on the system. Therefore, the variation of internal energy is
$\Delta U= Q-W=+194 kJ - (-120 kJ)=+314 kJ$

6 0

3 years ago