What is imperceptible human motion?

Two spherical conductors are separated by a distance much larger than either of their radii. Sphere A has a radius of 11.5 cm an

bonufazy [111]

Explanation:

As the given spheres are connected by a thin wire so, the potential on the spheres are the same.

$\frac{q_{1}}{r_{1}} = \frac{q_{2}}{r_{2}}$ ......... (1)

Hence, total charge will be as follows.

$q_{1} + q_{2}$ = Q = -95.5 nC .......... (2)

Using the above two equations, the final equation will be as follows.

$q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}$

and, $q_{1} = \frac{Qr_{1}}{r_{1} + r_{2}}$

Hence, we will calculate the charge on sphere B after the equilibrium is reached as follows.

$q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}$

= $\frac{-95.5 \times 74.4 cm}{(11.5 + 74.4) cm}$

= 82.714 nC

Thus, we can conclude that the charge on sphere B after equilibrium has been reached is 82.714 nC.

5 0

3 years ago

A 120-kg roller coaster cart is being tested on a new track, and a crash-test dummy is loaded into it. The roller coaster starts

avanturin [10]

Answer:

vb = 22.13 m/s

So, the only thing that was measured here was the height of point A relative to point B. And the Law of Conservation of Energy was used.

Explanation:

In order to find the speed of roller coaster at Point B, we will use the law of conservation of Energy. In this situation, the law of conservation of energy states that:

K.E at A + P.E at A = K.E at B + P.E at B

(1/2)mvₐ² + mghₐ = (1/2)m(vb)² + mg(hb)

(1/2)vₙ² + ghₐ = (1/2)(vb)² + g(hb)

where,

vₙ = velocity of roller coaster at point a = 0 m/s

hₙ = height of roller coaster at point a = 25 m

g = 9.8 m/s²

vb = velocity of roller coaster at point B = ?

hb = Height of Point B = 0 m (since, point is the reference point)

Therefore,

(1/2)(0 m/s)² + (9.8 m/s²)(25 m) = (1/2)(vb)² + (9.8 m/s²)(0 m)

245 m²/s² * 2 = vb²

vb = √(490 m²/s²)

vb = 22.13 m/s

So, the only thing that was measured here was the height of point A relative to point B. And the Law of Conservation of Energy was used.

5 0

3 years ago

Our eyes are typically 6 cm apart. Suppose you are somewhat unique, and yours are 9.50 cm apart. You see an object jump from sid

Serhud [2]

Answer: 12.67 cm, 8 cm

Explanation:

Given

Normal distance of separation of eyes, d(n) = 6 cm

Distance of separation is your eyes, d(y) = 9.5 cm

Angle created during the jump, θ = 0.75°

To solve this, we use the formula,

θ = d/r, where

θ = angle created during the jump

d = separation between the eyes

r = distance from the object

θ = d/r

0.75 = 9.5 / r

r = 9.5 / 0.75

r = 12.67 cm

θ = d/r

0.75 = 6 / r

r = 6 / 0.75

r = 8 cm

Thus, the object is 12.67 cm far away in your own "unique" eyes, and just 8 cm further away to the normal person eye

8 0

3 years ago

If two forces are in the same direction then do they cancel each other out

stellarik [79]

Im pretty sure that they do

3 0

3 years ago

A children's roller coaster is released from the top of a track. If its maximum speed at ground level is 3 m/s, find the height

san4es73 [151]

Answer:

h = 0.46 m

Explanation:

According to the law of conservation of energy:

Potential Energy Lost by Roller Coaster = Kinetic Energy Gained by Roller Coaster

$mgh = \frac{1}{2}mv^2\\\\2gh = v^2\\\\h = \frac{v^2}{2g}$

where,

h = height = ?

v = speed at bottom = 3 m/s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

$h = \frac{(3\ m/s)^2}{(2)(9.81\ m/s^2)}$

h = 0.46 m

4 0

3 years ago