8. slit-like opening to the larynx

Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin.

Sergio039 [100]

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge $q_{1}= 3.8\times10^{-6}\ C$

Second charge $q_{2}=3.2\times10^{-6}\ C$

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

$E_{1}=E_{2}$

$\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}$

$\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}$

$\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}$

$x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}$

Put the value into the formula

$x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}$

$x=1.69\ m$

Hence, The distance is 1.69 m.

5 0

3 years ago

A student pulls a block over a rough surface with a constant force FP that is at an angle θ above the horizontal, as shown above

gizmo_the_mogwai [7]

Answer:

B.The force of friction between the block and surface will decrease.

Explanation:

The force of friction is given by

$F_s = \mu N$

where $\mu$ is the coefficient of friction and $N$ is the normal force.

When the student pulls on the block with force $F_p$ at an angle $\theta$ , the normal force on the block becomes

$N = Mg- F_psin(\theta)$

and hence the frictional force becomes

$F_s = \mu (Mg- F_psin(\theta))$ .

Now, as we increase $\theta$ , $sin(\theta)$ increases which as a result decreases the normal force $Mg- F_psin(\theta)$ , which also means the frictional force decreases; Hence choice B stands true.

<em>P.S: Choice D is tempting but incorrect since the weight </em> $W=mg$ <em> is independent of the external forces on the block. </em>

6 0

3 years ago

Read 2 more answers

Phyllis is calculating the path of a space probe that will travel to Pluto. On the way to Pluto, the probe will pass Mars, Satur

inysia [295]

Answer:

Jupiter

Explanation:

4 0

3 years ago

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What includes the hippocampus, amygdala and hypothalamus?

marysya [2.9K]

Answer:

the limbic system has its input and processing side (the limbic cortex, amygdala and hippocampus) and an output side (the septal nuclei and hypothalamus).

Explanation:

hope it helps

6 0

3 years ago

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If 90 grams of gold has a volume of 5 cm3, calculate the density (include the units and show your work or you get no credit

ehidna [41]

Answer:

The density of gold is of 18 grams per cm3.

Explanation:

The mass density of a homogeneous material expresses how much mass of that material is present in a given volume. Since the density of an object is obtained by dividing its mass by its volume, to obtain the density of gold, its 90 grams of mass must be divided by its 5 cm3 volume, performing the following calculation:

90/5 = X

18 = X

Thus, the density of gold is 18 grams per cm3.

6 0

3 years ago