Which of the following is an advantage of sexual reproduction is that

A positive point charge Q is located at x=a and a negative point charge −Q is at x=−a. A positive charge q can be placed anywher

Scilla [17]

Answer:

$\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x$

Explanation:

The Coulomb's Law gives the force by the charges:

$\vec{F} = K\frac{q_1q_2}{r^2}\^r$

Let us denote the positon of the charge q on the y-axis as 'y'.

The force between 'Q' and'q' is

$F_1 = K\frac{Qq}{x^2 + y^2}\\F_1_x = F_1\cos(\theta)$

where Θ is the angle between $F_1$ and x-axis.

$F_1_x = K\frac{Qq}{x^2 + y^2}(\frac{x}{\sqrt{x^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}$

whereas

$F_2_x = K\frac{-Qq}{a^2 + y^2}(-\frac{a}{\sqrt{a^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}$

Finally, the x-component of the net force is

$\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x$

8 0

3 years ago

Find the tension in the two ropes that are holding the 4.2 kg object in place.

Blababa [14]

Answer:

The tension in the two ropes are;

T1 = 23.37N T2 = 35.47N

Explanation:

Given mass of the object to be 4.2kg, the weight acting on the bag will be W= mass × acceleration due to gravity

W = 4.2×10 = 42N

The tension acting on the bag plus the weight are three forces acting on the bag. We need to find tension in the two ropes that will keep the object in equilibrium.

Using triangular law of force and sine rule to get the tension we have;

If rope 1 is at 57.6° with respect to the vertical and rope 2 is at 33.8° with respect to the vertical, our sine rule formula will give;

T1/sin33.8° = T2/sin57.6° = 42/sin{180-(33.8°+57.6°)}

T1/sin33.8° = T2/sin57.6° = 42/sin88.6°

From the equality;

T1/sin33.8° = 42/sin88.6°

T1 = sin33.8°×42/sin88.6°

T1 = 23.37N

To get T2,

T2/sin57.6°= 42/sin88.6°

T2 = sin57.6°×42/sin88.6°

T2 = 35.47N

Note: Check attachment for diagram.

7 0

3 years ago

Calculate the magnitude of the maximum orbital angular momentum Lmax for an electron in a hydrogen atom for states with a princi

erica [24]

Answer:

L_max= 0

Explanation:

The formula for magnitude of maximum orbital angular momentum is given by

$L_{max}= \sqrt{l(l+1)\bar{h}}$

l= orbital quantum number

l= n-1

n= shell number or principal quantum number

for n=1 , l=0

therefore, $L_{max}= \sqrt{0(0+1)\bar{h}}$

L_max= 0

3 0

3 years ago

You push a 2.3 kg block against a horizontal spring, compressing the spring by 17 cm. then you release the block, and the spring

horsena [70]

Some dogs may inherit a susceptibility to epilepsy.

6 0

3 years ago

According to the Aufbau principle,

asambeis [7]

The correct answer is A, electrons enter orbitals of lowest energy first. The Aufbau principle states that electrons orbiting atoms fill the lowest energy levels available before filling higher levels. Following this, molecules can go into the most stable electron configuration.

3 0

3 years ago