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Anna71 [15]
3 years ago
10

A log splitting machine was invented to split logs quickly. A motor powers a metal piece that is pushed through the middle of a

log. The metal piece used to split a log is an example of what simple machine?a. wedgeb. pulleyc. screwd. lever
Physics
1 answer:
Alexandra [31]3 years ago
8 0

Answer:

a. wedge

Explanation:

The wedge machine is used to separate, lift or split

objects. It is triangular in shape with a tapered formation thick and blunt on one end, thin on the other end (the edge)

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Coulomb is a very large unit for practical use. Justify your answer if 10^10 electrons are transferred from a body/second
zlopas [31]
Given:

10^10 electrons per second

To justify that coulomb is a very large unit for practical use, we need to convert the quantity of electron given to Coulombs:

From literature, 

1 Coulomb is equivalent to 6.242×10^18 electrons<span>.

So,

= 10^10 electrons * (1 coulomb/</span><span>6.242×10^18</span> electrons) / second
<span>= 1.602 x 10^-9 coulumbs

This value is too small to be used in an actual setting. 

</span><span>
</span>
3 0
3 years ago
A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the
Rudiy27

The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

E=\dfrac{Q}{4\pi\epsilon_0 r^2}

Substitute numerical values:

E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

Learn more about Gaussian sphere here:

brainly.com/question/2004529

#SPJ4

6 0
2 years ago
An uniform electric field of magnitude E = 100 N/C is oriented along the positive y-axis. What is the magnitude of the flux of t
Ede4ka [16]

Answer:

The magnitude of the flux of electric field through a square of surface area is zero.

Explanation:

E=100 NC^{-1}\\\\A=2 m^2\\\\Electic\,\,flux\,\,flux\,\,is\,\,given\,\,as:\\\\\phi_E=E.A\,cos\,\theta

It is given that square box is parallel to yz-plane which has normal vector perpendicular to plane in x-direction. Angle between normal vector of area and electric field is 90°. Substituting in (1)

\phi_E=E.A\,cos\,(90^o)\\\\\phi_E=0

4 0
3 years ago
Fill-in-the-Blank
padilas [110]

Answer:

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YES

6 0
3 years ago
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Answer for a thanks
mariarad [96]
Well, if your question is how light affects plants,

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