RANGKAIAN RLC SERI PADA TEGANGAN BOLAK BALIK 220V R=80 OHM XL= 90 OHM XC = 30 OHM MAKA TEGANGAN PADDA INDUKTOR ADALAH???

The unit of energy is a derived unit why

olga2289 [7]

Answer:

The unit of energy is joule which depends upon the fundamental unit kg, m and sec. So, the unit of energy is a derived unit.

Explanation:

7 0

2 years ago

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A piece of wire 29 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria

skelet666 [1.2K]

Answer:

Explanation:

Total length of the wire is 29 m.

Let the length of one piece is d and of another piece is 29 - d.

Let d is used to make a square.

And 29 - d is used to make an equilateral triangle.

(a)

Area of square = d²

Area of equilateral triangle = √3(29 - d)²/4

Total area,

$A = d^{2}+\frac{\sqrt3}{4}\left ( 29-d \right )^{2}$

Differentiate both sides with respect to d.

$\frac{dA}{dt}=2d- \frac{\sqrt3}{4}\times 2(29-d)$

For maxima and minima, dA/dt = 0

d = 8.76 m

Differentiate again we get the

$\frac{d^{2}A}{dt^{2}}= + ve$

(a) So, the area is maximum when the side of square is 29 m

(b) so, the area is minimum when the side of square is 8.76 m

8 0

3 years ago

an ice skater starts with a velocity of 2.25 m/s in a 50.0 direction. after 8.33 m/s in a 120 direction. what is the y-component

Nata [24]

The y-component of the acceleration is $0.28 m/s^2$

Explanation:

The y-component of the ice skater acceleration can be calculated with the equation

$a_y = \frac{v_y-u_y}{t}$

where

$v_y$ is the y-component of the final velocity

$u_y$ is the y-component of the initial velocity

t is the time elapsed

Here we have:

Initial velocity is $u=2.25 m/s$ at $\theta_1=50.0^{\circ}$ , so its y-component is $u_y = u sin \theta_1 = (2.25)(sin 50.0^{\circ})=1.72 m/s$

Final velocity is $v=4.65 m/s$ at $\theta_2=120.0^{\circ}$ , so its y-component is $v_y = v sin \theta_2 = (4.65)(sin 120.0^{\circ})=4.03 m/s$

The time elapsed is

t = 8.33 s

Therefore, the y-component of the acceleration is

$a_y = \frac{4.03-1.72}{8.33}=0.28 m/s^2$

Learn more about acceleration:

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#LearnwithBrainly

7 0

3 years ago

CHEGG You stretch a spring with spring constant k = 1.2x104 N/m to extend 6.0 cm away from its equilibrium position. How much do

lubasha [3.4K]

Answer:

The elastic potential energy of the spring change during this process is 21.6 J.

Explanation:

Given that,

Spring constant of the spring, $k=1.2\times 10^4\ N/m$

It extends 6 cm away from its equilibrium position.

We need to find the elastic potential energy of the spring change during this process. The elastic potential energy of the spring is given by the formula as follows :

$E=\dfrac{1}{2}kx^2\\\\E=\dfrac{1}{2}\times 1.2\times 10^4\times (0.06)^2\\\\E=21.6\ J$

So, the elastic potential energy of the spring change during this process is 21.6 J.

4 0

3 years ago

A spring has a spring constant of 90N/m.How much potential energy does it store when stretched by 2 cm?

il63 [147K]

Answer:

The potential energy stored in the spring is 0.018 J.

Explanation:

Given;

spring constant, k = 90 N/m

extension of the spring, x = 2 cm = 0.02 m

The potential energy stored in the spring is calculated as;

U = ¹/₂kx²

where;

U is the potential energy stored in the spring

Substitute the given values in the equation above;

U = ¹/₂ x 90 N/m x (0.02 m)²

U = 0.018 J

Therefore, the potential energy stored in the spring is 0.018 J.

5 0

3 years ago