If a body starts moving from rest its initial velocity is ų V zero m/s

Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges

Nostrana [21]

Answer:

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

$E_1 = E_2$

Let the position where net field is zero will lie at distance "r" from q1

$\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}$

now we will have

$\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}$

now square root both sides

$\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}$

now we have

$\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1$

so we have

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

8 0

3 years ago

Semi-trailer trucks have an odometer on one hub of a trailer wheel. The hub is weighted so that it does not rotate, but it conta

shtirl [24]

Answer:

1020 km

Explanation:

A complete rotation of the wheel equals a distance of 1 circumference.

The circumference is

$C = \pi d$

where d is the diameter of the wheel.

300,000 rotations = $300000\pi d = 300000\times\pi\times1.08\text{ m} = 1017876.0\ldots\text{ m}$

In kilometers, this is = 1017876/1000 km = 1020 km

6 0

3 years ago

A 20 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at

stellarik [79]

Answer: 0.516 ft/s

Explanation:

Given

Length of ladder L=20 ft

The speed at which the ladder moving away is v=2 ft/s

after 1 sec, the ladder is 5 ft away from the wall

So, the other end of the ladder is at

$\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft$

Also, at any instant t

$\Rightarrow l^2=x^2+y^2$

differentiate w.r.t.

$\Rightarrow 0=2xv+2yv_y\\\\\Rightarrow v_y=-\dfrac{x}{y}\times v\\\\\Rightarrow v_y=-\dfrac{5}{19.36}\times 2=0.516\ ft/s$

5 0

3 years ago

A shovel is the third class lever

Nookie1986 [14]

Answer:

Yes

Explanation:

In a third-class lever, the effort force lies between the resistance force and the fulcrum. Some kinds of garden tools are examples of third-class levers. When you use a shovel, for example, you hold one end steady to act as the fulcrum, and you use your other hand to pull up on a load of dirt.

8 0

3 years ago

Two equal charges with magnitude Q and Q experience a force of 12.3442 when held at a distance r. What is the force between two

andre [41]

Answer:

197.5072.

Explanation:

According to the Coulomb's law, the magnitude of the electrostatic force of interaction between two charges $\rm q_1$ and $\rm q_2$ which are separated by the distance $\rm d$ is given by

$\rm F = \dfrac{kq_1q_2}{d^2}.$

where, k is the Coulomb's constant.

For the case, when,

$\rm q_1 = Q.$
$\rm q_2 = Q.$
$\rm d=r.$
$\rm F=12.3442.$

Then, using Coulomb's law,

$\rm 12.3442 = \dfrac{kQQ}{r^2}=\dfrac{kQ^2}{r^2}\ \ \ \ .......\ (1).$

For the case, when,

$\rm q_1 = 2Q.$
$\rm q_2 = 2Q.$
$\rm d=\dfrac r2.$

Then, using Coulomb's law, the new electric force between the charges is given by,

$\rm F' = \dfrac{k(2Q)(2Q)}{\left (\dfrac r2\right )^2}\\=\dfrac{k\ 4Q^2}{\dfrac{r^2}{4}}\\=4\times 4 \times \dfrac{kQ^2}{r^2}\\=16\ \dfrac{kQ^2}{r^2}\\=16\times 12.3442\ \ \ \ \ \ \ \ (Using\ (1))\\=197.5072.$

8 0

2 years ago

If a body starts moving from rest itsinitial velocity isų Vzerom/s​

If a body starts moving from rest itsinitial velocity isų Vzerom/s