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3 years ago

5

Two charged objects are separated by some distance. the charge on the first object is greater than the charge on the second obje

ct. how does the force between the two objects compare?

2 answers:

oksian1 [2.3K]3 years ago

8 0

Force will be compared by coulomb law

M12 = M21

Svetlanka [38]3 years ago

7 0

The forces in both directions are equal. Their strength depends on the PRODUCT of the charges, not on their individual values or the comparison between them. Same goes for gravity. Your weight on Earth is exactly the same as the Earth's weight on YOU.

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Which of these statements is correct about photosynthesis and decomposition?

Ede4ka [16]

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3 years ago

A charge q produces an electric field of strength 4E at a distance of d away. Determine the electric field strength at a distanc

gizmo_the_mogwai [7]

Answer:

c.) 36E

Explanation:

The magnitude of the electric field is given by the expression

$E=k \frac{q}{d^2}$ (1)

where k is the Coulomb's constant, q is the charge that generates the field, and d is the distance from the charge.

In this problem, we have that the magnitude of the field at a distance d is 4E, so we can rewrite the previous equation as

$4E = k\frac{q}{d^2}$

Now we want to determine the electric field at a distance of $d'=\frac{1}{3}d$ away. Substituting into (1), we find

$E' = k \frac{q}{d'^2}=k \frac{q}{(\frac{1}{3}d)^2}=9 k \frac{q}{d^2}$ (2)

We also know that

$4E = k\frac{q}{d^2}$ (3)

So combining (2) with (3), we find a relationship between the original field and the new field:

$E' = 9 \cdot (4E) = 36E$

7 0

3 years ago

What obstacles or challenges are encountered in magnetic levitation?

olasank [31]

Mike Hawk Moe Lester Hugh Janice

5 0

3 years ago

A 80 kg skier grips a moving rope that is powered by an engine and is pulled at constant speed to the top of a 25 degrees hill.

zloy xaker [14]

Answer:

$P=28.085\,hp$

Explanation:

Given that:

mass of 1 skier, $m=80kg$

inclination of hill, $\theta=25^{\circ}$

length of inclined slope, $l=220m$

time taken to reach the top of hill, $t=2.3 min= 138 s$

coefficient of friction, $\mu=0.15$

<em>Now, force normal to the inclined plane:</em>

$F_N=m.g.cos\theta$

$F_N=80\times 9.8\times cos25^{\circ}$

$F_N=710.54\,N$

<em>Frictional force:</em>

$f=\mu.F_N$

$f=0.15\times 710.54$

$f=106.58\,N$

<em>The component of weight along the inclined plane:</em>

$W_l=m.g.sin\theta$

$W_l=80\times 9.8\times sin25^{\circ}$

$W_l=331.33\,N$

<em>Now the total force required along the inclination to move at the top of hill:</em>

$F=f+W_l$

$F=106.58+331.33$

$F=437.91\,N$

<em>Hence the work done:</em>

$W=F.l$

$W=437.91\times 220$

$W=96340.80\,J$

<em>Now power:</em>

$P=\frac{W}{t}$

$P=\frac{96340.80}{138}$

$P=698.12\,W$

<u>So, power required for 30 such bodies:</u>

$P=30\times 698.12$

$P=20943.65\,W$

$P=\frac{20943.65}{745.7}$

$P=28.085\,hp$

8 0

3 years ago

Which sentence describe newton's third law?

strojnjashka [21]

C. Forces are always in pairs

3 0

3 years ago