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2 years ago

5

Letters A B, C and D represent locations on a bar magnet. Which location has the greatest magnetic force? (l) A (3) C ets (2) B

(4) D

1 answer:

Elena L [17]2 years ago

6 0

Answer: The magnetic field of a bar magnet is strongest at either pole of the magnet. It is equally strong at the north pole compared with the south pole. The force is weaker in the middle of the magnet and halfway between the pole and the center. So it would be D.

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The following are secondary sources of energy except a.coal b crude oil c nuclear d wind e wood

monitta

Answer:

d) Wind

Explanation:

Secondary energy is energy produced by converting energy available in its natural state in the environment. Hence Wind is a primary source not a secondary source

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2 years ago

An 18.7 g sample of platinum metal increases

vredina [299]

Answer:

0.092

Explanation:

because i said sooooo

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2 years ago

A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri

GREYUIT [131]

Answer:

$|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))$

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

$\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r$

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)$

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

$\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}$

Now, substitute this into 'dF':

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)$

Now we can integrate dF over the rod.

$\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)$

4 0

2 years ago

With 51 gallons of fuel in its tank, the airplane has a weight of 2390.7 pounds. What is the weight of the plane with 81 gallons

Shtirlitz [24]

Answer: 2561.7 pounds

Explanation:

If we assume the total weight of an airplane (in pounds units) as a <u>linear function</u> of the amount of fuel in its tank (in gallons) and we make a Weight vs amount of fuel graph, which resulting slope is 5.7, we can use the slope equation of the line:

$m=\frac{Y-Y_{1}}{X-X_{1}}$ (1)

Where:

$m=5.7$ is the slope of the line

$Y_{1}=2390.7pounds$ is the airplane weight with 51 gallons of fuel in its tank (assuming we chose the Y axis for the airplane weight in the graph)

$X_{1}=51gallons$ is the fuel in airplane's tank for a total weigth of 2390.7 pounds (assuming we chose the X axis for the a,ount of fuel in the tank in the graph)

This means we already have one point of the graph, which coordinate is:

$(X_{1},Y_{1})=(51,2390.7)$

Rewritting (1):

$Y=m(X-X_{1})+Y_{1}$ (2)

As Y is a function of X:

$Y=f_{(X)}=m(X-X_{1})+Y_{1}$ (3)

Substituting the known values:

$f_{(X)}=5.7(X-51)+2390.7$ (4)

$f_{(X)}=5.7X-290.7+2390.7$ (5)

$f_{(X)}=5.7X+2100$ (6)

Now, evaluating this function when X=81 (talking about the 81 gallons of fuel in the tank):

$f_{(81)}=5.7(81)+2100$ (7)

$f_{(81)}=2561.7$ (8) This means the weight of the plane when it has 81 gallons of fuel in its tank is 2561.7 pounds.

3 0

2 years ago

An electron with a charge of -1.6 × 10-19 coulombs experiences a field of 1.4 × 105 newtons/coulomb. What is the magnitude of th

neonofarm [45]

Answer:

Electric force, $F=2.24\times 10^{-14}\ N$

Explanation:

It is given that,

Charge on an electron is $-1.6\times 10^{-19}\ C$

Electric field, $E=1.4\times 10^5\ N/m$

We need to find the magnitude of the electric force on this electron due to this field. The electric force is given by :

$F=qE\\\\F=1.6\times 10^{-19}\times 1.4\times 10^5\\\\F=2.24\times 10^{-14}\ N$

So, the electric force is $2.24\times 10^{-14}\ N$ .

6 0

2 years ago