The fastest possible conduction velocity of action potentials is observed in _______

A particle with a charge of 4.4 * 10^-5 C is released in an electric field whose magnitude is 750 N/C. What is the force that th

padilas [110]

Answer:

F = 3.3×10^ -2 N

.................

4 0

3 years ago

NEED NOW ALMOST DUE

Arte-miy333 [17]

Answer:

Asteroids are objects made of clay and silicate that orbit the sun but are too small to be considered planets.

Most asteroids revolve around the sun in an orbit between those of Mars and Jupiter.

They form a wide band called the Asteroid belt.

Other asteroids have orbits that cross Earth’s orbit. These asteroids are called Earth-crossers.

Asteroids probably consist of matter that never agglomerated into a planet when the solar system was forming.

The comet’s core is composed of ice and dust.

Comets heat up and begin to change from a solid to a gas as they approach the sun.

The matter surrounding a comet’s core is vaporized and forms a very bright halo of ice or dust not sure, and an enormous cloud of dust or gasses not sure envelopes the head of the comet.

7 0

3 years ago

A physics student of mass 43.0 kg is standing at the edge of the flat roof of a building, 12.0 m above the sidewalk. An unfriend

Dmitry_Shevchenko [17]

Answer:

The speed of the student just before she lands, v₂ is approximately 8.225 m/s

Explanation:

The given parameters are;

The mass of the physic student, m = 43.0 kg

The height at which the student is standing, h = 12.0 m

The radius of the wheel, r = 0.300 m

The moment of inertia of the wheel, I = 9.60 kg·m²

The initial potential energy of the female student, P.E.₁ = m·g·h₁

Where;

m = 43.0 kg

g = The acceleration due to gravity ≈ 9.81 m/s²

h = 12.0 h

∴ P.E.₁ = 43 kg × 9.81 m/s² × 12.0 m = 5061.96 J

The kinetic rotational energy of the wheel and kinetic energy of the student supporting herself from the rope she grabs and steps off the roof, K₁, is given as follows;

$K_1 = \dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2$

The initial kinetic energy, 1/2·m·v₁² and the initial kinetic rotational energy, 1/2·m·ω₁² are 0

∴ K₁ = 0 + 0 = 0

The final potential energy of the student when lands. P.E.₂ = m·g·h₂ = 0

Where;

h₂ = 0 m

The final kinetic energy, K₂, of the wheel and student is give as follows;

$K_2 = \dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2$

Where;

v₂ = The speed of the student just before she lands

ω₂ = The angular velocity of the wheel just before she lands

By the conservation of energy, we have;

P.E.₁ + K₁ = P.E.₂ + K₂

∴ m·g·h₁ + $\dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2$ = m·g·h₂ + $\dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2$

Where;

ω₂ = v₂/r

∴ 5061.96 J + 0 = 0 + $\dfrac{1}{2} \times 43.0 \, kg \times v_{2}^2+\dfrac{1}{2} \times 9.60 \, kg\cdot m^2 \cdot \left (\dfrac{v_2}{0.300 \, m} }\right ) ^2$