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3 years ago

As you sit in a chair, in what direction (or directions) do you exert an action force?

2 answers:

5 0

As we sit in a chair, Action force will be only in one direction and that direction would be downward only.

In short, Your Answer would be Option A

Hope this helps!

charle [14.2K]3 years ago

4 0

Answer:

a. downward only

Explanation:

I just took the quiz! :)

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Is kcl and kvl applicable to ac circuit?

Sedaia [141]

Yes, they both are, and you'd just be working with complex numbers instead of real number.

8 0

2 years ago

The national high magnetic field laboratory holds the world record for creating the strongest magnetic field. for brief periods

max2010maxim [7]

Newton’s 2nd law states that Force is equal to the product of mass (m) and acceleration (a):

F = m a ---> 1

While in magnetic forces, force can also be expressed as:

F = q v B ---> 2

where,

q = total charge

v = velocity = 45 cm / s = 0.45 m / s

B = the magnetic field = 85 T

First we solve for the total charge, q:

q = 3.8 × 10^-23 g (1 mol / 23 g) (6.022 × 10^23 electrons / mol) (1.602 × 10^-19 C / electron)

q = 1.594 × 10^-19 C

We equate equations 1 and 2 then solve for acceleration a:

m a = q v B

a = q v B / m

a = [1.594 × 10^-19 C * 0.45 m / s * 85 T] / 3.8 × 10-26 kg

a = 160,437,862.2 m/s^2

Therefore the maximum acceleration of Na ions is about 160 × 10^6 m/s^2.

5 0

3 years ago

Read 2 more answers

Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote

Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of $1.8 \times 10^{-8}\; \rm C$ each, the electric potential at the midpoint of segment $\rm AB$ would be approximately $8.3 \times 10^{3}\; \rm V$ .

Explanation:

Convert the unit of the length of each side of this triangle to meters: $10\; \rm cm = 0.10\; \rm m$ .

Distance between the midpoint of $\rm AB$ and each of the three charges:

$d({\rm A}) = 0.050\; \rm m$ .
$d({\rm B}) = 0.050\; \rm m$ .
$d({\rm C}) = \sqrt{3} \times (0.050\; \rm m)$ .

Let $k$ denote Coulomb's constant ( $k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}$ .)

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm A})}$ .

Electric potential due to the charge at $\rm B$ : $\displaystyle \frac{k\, q}{d({\rm B})}$ .

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm C})}$ .

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of $\rm AB$ due to all these three charges would be:

$\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}$ .

4 0

2 years ago

A lorry of mass 4000 kg is travelling at a speed of 4 m/s.

bulgar [2K]

Answer:

KE = 32,000J

v = 8m/s

Explanation:

KE = .5*m*v²

KE = .5*4000kg*(4m/s)²

KE = 32,000J

32,000J = .5*1000kg*v²

v² = 64

v = 8m/s

7 0

2 years ago

The girl makes a microscope with a 3.0 cm focal length objective and a 5.0 cm eyepiece. The microscope tube length is 10 cm. Use

saul85 [17]

To solve this problem we will use the concepts related to Magnification. Magnification is the process of enlarging the apparent size, not physical size, of something. This enlargement is quantified by a calculated number also called "magnification".

The overall magnification of microscope is

$M = \frac{Nl}{f_ef_0}$