What do the planets Saturn, Neptune, and Uranus have in common?

What parameters affects the inductance of a coil

Kazeer [188]

Answer:

Number of Wire Turns in the Coil.

Explanation:

The greater the number of turns of wire in the coil, the greater the inductance. Fewer turns of wire in the coil results in lesser inductance. More coils of wires indicate a greater amount of magnetic field force for a given amount of coil current.

4 0

2 years ago

Which substance is a greenhouse gas that is emitted by human activities, is the most widespread, and remains in the air the long

e-lub [12.9K]

Answer:

I am not really sure, but it is probably Carbon Dioxide

Explanation:

3 0

4 years ago

Read 2 more answers

3) How far will 20 N of force stretch a spring with a spring constant of 140 N/m?

muminat

Answer:

7 meters, 2.8 meters

Explanation:

work done (nm) = force (n) * distance (m)

140= 20 * m

140/20 = m

m=7 meters

140= 50 * m

140/50 = m

m= 2.8 meters

4 0

2 years ago

Give an example of a situation in which you would describe an

Viktor [21]

It would be A because it would make sense

5 0

3 years ago

A baseball is batted from a height of 1.09 m with a speed of

kobusy [5.1K]

(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.

(b) The maximum height above the ground reached by the ball is 8.6 m.

(c) The distance off course the ball would be carried is 0.38 m.

(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

<h3>Horizontal and vertical components of the ball's velocity</h3>

Vx = Vcosθ

Vx = 39.7 x cos(17.8)

Vx = 37.8 m/s

Vy = Vsin(θ)

Vy = 39.7 x sin(17.8)

Vy = 12.14 m/s

<h3>Maximum height reached by the ball</h3>

$H = \frac{v^2 sin^2(\theta)}{2g} \\\\H = \frac{(39.7)^2 \times (sin17.8)^2}{2(9.8)} \\\\H = 7.51 \ m$

Maximum height above ground = 7.51 + 1.09 = 8.6 m

<h3>Distance off course after 2 second </h3>

Upward speed of the ball after 2 seconds, V = V₀y - gt

Vy = 12.14 - (2x 9.8)

Vy = - 7.46 m/s

Horizontal velocity will be constant = 37.8 m/s

Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)

$V = \sqrt{(-7.46)^2 + (37.8)^2} \\\\V = 38.53 \ m/s$

<h3>Resultant speed of the ball and crosswind</h3>

$V = \sqrt{38.52^2 + 4^2} \\\\V = 38.72 \ m/s$

<h3>Distance off course the ball would be carried</h3>

d = Δvt = (38.72 - 38.53) x 2

d = 0.38 m

The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.

Learn more about projectiles here: brainly.com/question/11049671

5 0

2 years ago