What is the importance of star ranting labelilling of an electrical appliance?

Physics

1 answer:

Misha Larkins [42]2 years ago

5 0

Answer:

The standard of efficiency varies by product category

these are called BEE star label and they show how much electricity the appliance consumes in year each appliance gets btw one and five stars with five stars meaning that it's extremely efficienct andis likely to keep your electricity bills in check

 hope it's helpful

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A sphere has a radius of 3.9cm and a density of 7.58 g/cm cubed. What is the mass?

snow_lady [41]

Answer:

m=ρV

V=4/3 * pi * r3

V=1.3 * 3.14 * 3.9^3

V=242.14 cm^3

m=7.58 * 242.14

m=1.8 kG

Explanation:

1. We calculate volume for sphere.

2. Then we calculate mass of sphere.

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2 years ago

A right triangle ABC has sides /AB/ =9cm, /BC/=12cm find /AC/ and angles ACB and ABC if angle BAC= 90°

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3 years ago

You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo

jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$