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3 years ago

How much heat energy is produced by 0.5 Wh of electrical energy ?

Physics

2 answers:

Fittoniya [83]3 years ago

8 0

The answer is B

Hope it helps

<3

babymother [125]3 years ago

6 0

1.7 Btu

1 watt = 3.41214 Btu/h

1watt * 1h = 3.41214 Btu/h * h

1 = 3.41214 Btu/ (watt*h)/

0.5 watt * h = 0.5 watt*h * 3.41214 Btu/(watt*h) = 1.706 Btu

You might be interested in

How much force is needed to accelerate a 68 kilogram-skier at a rate of 1.2 m/sec^2?

DiKsa [7]

<span>Answer: Force = 81.6 N

Explanation:
According to Newton's Second law:
F = ma --- (1)

Where F = Force = ?
m = Mass = 68 kg
a = Acceleration = 1.2 m/s^2

Plug in the values in (1):
(1) => F = 68 * 1.2
F = 81.6 N (The force needed to accelerate the skier at a rate of 1.2 m/s^2)</span>

4 0

3 years ago

Read 2 more answers

When describing the interaction between magnetic poles we could say that

kifflom [539]

The Answer Is A
Hope This Helps !

8 0

3 years ago

Read 2 more answers

What happened to the balloon when it was placed on the bottle with the baking soda and vinegar and why

Veseljchak [2.6K]

Answer:

it creates a gas called carbon dioxide. The gas begins to expand in the bottle and starts to inflate the balloon

Explanation:

Why does this happen? well, The faster-moving particles inside the bottle start to move faster and faster and soon they expand to fill the balloon.

8 0

2 years ago

A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$

Now, we solve (1) for $x_f_g$

$x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m$

5 0

3 years ago

Quiero conoce amiga que tenga correo

34kurt

Hola , soy daniela .

4 0

3 years ago