Identify and explain the role of all the parts of the female reproductive system.

Which is the relationship between algae and fungus?

aleksklad [387]

the correct answer is mutualism

6 0

3 years ago

My Notes An electron is released from rest on the axis of a uniform positively charged ring, 0.500 m from the ring's center. If

dsp73

Answer:

Velocity of the electron = v = 1.2\times 10^8\ m/s.

Explanation:

Given,

Mass of the electron = $m_e\ =\ 9\times 10^{-31}\ kg$
Charge on the electron = $q_e\ =\ 1.62\times 10^{-19}\ C$
Charge density of the ring = $\rho\ =\ +1.00\times 10^{-6}\ C/m$
Radius of the ring = R = 0.70 m
Distance between the electron ant the center or the ring = x = 0.5 m

Now total charge on the ring = $Q\ =\ \rho\times 2\pi R$

Potential energy due to the charged ring to the point on the x-axis is

$P.E.\ =\ \dfrac{KQq_e}{\sqrt{R^2\ +\ x^2}}\\$

Let v be the velocity of the electron at the center of the ring.

Total kinetic energy of the electron = $\dfrac{1}{2}m_ev^2\\$

Now, From the conservation of energy,

the total potential energy of the electron at initially is converted to the total kinetic energy of the electron at the center of the ring,

$\therefore P.E.\ =\ K.E.\\\Rightarrow \dfrac{KQ}{\sqrt{R^2\ +\ x^2}}\ =\ \dfrac{1}{2}m_ev^2\\\Rightarrow v\ =\ sqrt{\dfrac{2 KQq_e}{m_e\sqrt{R^2\ +\ x^2}}}\\\Rightarrow v\ =\ \sqrt{\dfrac{2Kq_e\rho \times 2\pi R}{m_e\sqrt{R^2\ +\ x^2}}}\\\Rightarrow v\ =\ \sqrt{\dfrac{2\times 9\times 10^9\times 1.0\times 10^{-6}\times 2\times 3.14\times 0.7\times 1.6\times 10^{-19}}{9\times 10^{-31}\times \sqrt{0.7^2\ +\ 0.5^2}}}\\\Rightarrow v\ =\ 1.2\times 10^8\ m/s.$

Hence the velocity of the electron on the center of the charged ring is $1.2\times 10^8\ m/s.$

5 0

3 years ago

Physics 1 Course. please help.

Aleks [24]

Answer:

a. 1.64 m/s²

Explanation:

Centripetal acceleration is the square of tangential velocity divided by the radius.

a = v²/r

First, convert km/h to m/s.

30.0 km/h (1000 m/km) (1 h / 3600 s) = 8.33 m/s

Find the acceleration.

a = (8.33 m/s)² / (42.4 m)

a = 1.64 m/s²

6 0

3 years ago

Frame S' passes frame S in the usual way. Two events are simultaneous in S'.

yuradex [85]

Answer:

c)They can also be simultaneous in S if their separation is zero.

Explanation:

By relativity theory, we can say two events when seen from two different reference frames can only be simultaneous when they are at the same space location and occur simultaneously in at least one reference frame, therefore when Frame S′ usually passes Frame S. Two occurrences in S′ are simultaneous, therefore these occurrences can be simultaneous in S when their separation is 0 (that is they are at the same location)

And therefore option c. If their separation is zero, they can also be simultaneous in S.

8 0

3 years ago

Calculate (A⃗ ×B⃗ )⋅C⃗ for the three vectors A⃗ with magnitude A = 4.86 and angle θA = 23.5 ∘ measured in the sense from the +x

trasher [3.6K]

Answer:

(A⃗ ×B⃗ )⋅C⃗ = 69.868

Explanation:

We simplify the cross product first, thereafter the solution of the cross product is now simplified with the dot product as shown in the step by step calculation in the attachment

4 0

3 years ago