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Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

5 0

3 years ago

A block with mass m = 0.450 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The oth

svetoff [14.1K]

Answer:

k = 26.25 N/m

Explanation:

given,

mass of the block= 0.450

distance of the block = + 0.240

acceleration = a_x = -14.0 m/s²

velocity = v_x = + 4 m/s

spring force constant (k) = ?

we know,

x = A cos (ωt - ∅).....(1)

v = - ω A cos (ωt - ∅)....(2)

a = ω²A cos (ωt - ∅).........(3)

$\omega = \sqrt{\dfrac{k}{m}}$

now from equation (3)

$a_x = \dfrac{k}{m}x$

$k = \dfrac{m a_x}{x}$

$k = \dfrac{0.45 \times (-14)}{0.24}$

k = 26.25 N/m

hence, spring force constant is equal to k = 26.25 N/m

8 0

3 years ago

Read 2 more answers

A neutron consists of one "up" quark of charge +2e/3 and two "down" quarks each having charge -e/3. If we assume that the down q

Anon25 [30]

Answer:

The magnitude of the electrostatic force is 120.85 N

Explanation:

We can use Coulomb's law to find the electrostatic force between the down quarks.

In scalar form, Coulomb's law states that for charges $q_1$ and $q_2$ separated by a distance d, the magnitude of the electrostatic force F between them is:

$F = k \frac{|q_1q_2|}{d^2}$

where $k$ is Coulomb's constant.

Taking the values:

$d = 4.6 \ 10^{-15} m$

$q_1 = q_2 = - \frac{e}{3} = - \frac{1.6 \ 10^{-19} \ C}{3}$

and knowing the value of the Coulomb's constant:

$k = 8.99 \ 10 ^{9} \frac{N m^2}{C^2}$

Taking all this in consideration:

$F = 8.99 \ 10 ^{9} \frac{N m^2}{C^2} \frac{ (- \frac{1.6 \ 10^{-19} \ C}{3} ) ^2}{(4.6 \ 10^{-15} m)^2}$

$F = 120.85 \ N$

8 0

3 years ago

Examine the cross-sectional slice of a stem of a mint plant. Which statement most specifically describes mint?

Maslowich

The available options are:

Mint is a dicot.

Mint is a monocot.

Mint is an angiosperm.

Mint is a bulb plant.

Answer:

Mint is a dicot.

Explanation:

Given the fact that Mint is considered to be a member of Lamiaceae, an angiosperm plant which is characterized by typically having leaves that consist of reticulate vacation and appears like veins in structure. It also has a seed that contains two cotyledons.

Hence, it is considered a DICOT PLANT due to these characteristics. The botanical name of Mint is referred to as Mentha arvensis.

7 0

2 years ago

If 12V battery maintains a 4.5A current through a resistor what is the resistance of the resistor?

Dimas [21]

Answer:

2.7ohms

Explanation:

Given parameters:

Voltage of the battery = 12V

Current = 4.5A

Unknown:

Resistance of the resistor = ?

Solution:

From Ohm's law, we know that;

V = IR

V is the voltage

I is the current

R is the resistance

So;

R = $\frac{V}{I}$ = $\frac{12}{4.5}$ = 2.7ohms

5 0

2 years ago