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3 years ago

(eopt)s. Find the power dissipated in a 25 r heaterconnected to a 220V outer.

Physics

1 answer:

Alex Ar [27]3 years ago

6 0

Answer:

P = 1936 W

Explanation:

Given that,

Voltage, V = 220 V

The resistance of the heater, R = 25 ohms

We need to find the power dissipated in the heater. We know that the power dissipated by the electric appliances is given by :

$P=\dfrac{V^2}{R}\\\\P=\dfrac{(220)^2}{25}\\\\P=1936\ W$

So, the required power is equal to 1936 W.

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KE = 1/2 * m * v^2
KE = 1/2 * 0.135 * 40^2
KE = 1/2 * 0.135 * 1600
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3 years ago

Which law is associated with inertia

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3 years ago

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3 years ago

Una partícula se mueve en el plano XY efectúa un desplazamiento mientras actúa sobre ella una fuerza constante. X= (4i + 3j) m,

dsp73

Answer:

a) La magnitud del desplazamiento es de 5 m

La magnitud de la fuerza es 20 N

b) El trabajo realizado por la fuerza es de 100 J

c) El ángulo entre la fuerza y el plano es 0 °

Explanation:

a) La magnitud del desplazamiento se encuentra por la relación;

$\left | X \right | = \sqrt{X_{x}^{2}+X_{y}^{2}}$

Lo que da;

$\left | X \right | = \sqrt{4^{2}+3^{2}} = 5 \ m$

De manera similar, la magnitud de la fuerza, F, se encuentra como sigue;

$\left | F \right | = \sqrt{F_{x}^{2}+F_{y}^{2}}$

Lo que da;

$\left | F \right | = \sqrt{16^{2}+12^{2}} = 20 \ N$

b) El trabajo, W, realizado por la fuerza = Fuerza, F × Distancia, X

∴ Ancho = 20 N × 5 m = 100 N · m = 100 J

c) La dirección de la fuerza viene dada por la siguiente fórmula;

$tan^{-1} \left (\dfrac{F_y}{F_x} \right ) = tan^{-1} \left (\dfrac{12}{16} \right ) = 38.9^{\circ}$

La dirección del plano viene dada por la siguiente fórmula;

$tan^{-1} \left (\dfrac{X_y}{X_x} \right ) = tan^{-1} \left (\dfrac{3}{4} \right ) = 38.9^{\circ}$

Por tanto, el ángulo entre la fuerza y el plano = 0 °

La fuerza actúa a lo largo del plano.

6 0

3 years ago

A 5.50-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical

iris [78.8K]

Answer:

17.71N/m

Explanation:

The period of the spring is expressed according to the expression;

$T = 2 \pi \sqrt{\frac{m}{k} } \\$

m is the mass of the object

k is the force constant

Given

m = 5.50kg

T = 3.50s

Substitute into the formula;

$T = 2 \pi \sqrt{\frac{m}{k} } \\3.5 = 2 (3.14) \sqrt{\frac{5.5}{k} } \\3.5 = 6.28 \sqrt{\frac{5.5}{k} } \\\frac{3.5}{6.28} = \sqrt{\frac{5.5}{k} } \\0.557 = \sqrt{\frac{5.5}{k} } \\square \ both \ sides\\0.557^2 = (\sqrt{\frac{5.5}{k} })^2 \\0.3106 = \frac{5,5}{k}\\k = \frac{5.5}{0.3106}\\k = 17.71N/m$

Hence the force constant of the spring is 17.71N/m

4 0

3 years ago

(eopt)s. Find the power dissipated in a 25 r heaterconnected to a 220V outer.​

(eopt)s. Find the power dissipated in a 25 r heaterconnected to a 220V outer.