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4 years ago

If measuring experimental results, what can you predict about the work output of a 1200 Watt hair dryer?

Physics

2 answers:

Artyom0805 [142]4 years ago

5 0

A.) The work output of the hairdryer will be less than the work input.

Elden [556K]4 years ago

3 0

I am 90% sure it’s A

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Who was known for being a pilot and an astronaut that walked on the moon?.

Mkey [24]

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2 years ago

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The kinetic theory states that the higher the temperature, the faster the

max2010maxim [7]

Answer: the higher the kinetic energy

Explanation:

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3 years ago

a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How f

harkovskaia [24]

Solution:

With reference to Fig. 1

Let 'x' be the distance from the wall

Then for $\Delta$ DAC:

$tan\theta = \frac{d}{x}$

⇒ $\theta = tan^{-1} \frac{d}{x}$

Now for the $\Delta$ BAC:

$tan\theta = \frac{d + h}{x}$

⇒ $\theta = tan^{-1} \frac{d + h}{x}$

Now, differentiating w.r.t x:

$\frac{d\theta }{dx} = \frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}]$

For maximum angle, $\frac{d\theta }{dx}$ = 0

Now,

0 = [/tex]\frac{d}{dx}[tan^{-1} \frac{d + h}{x} - tan^{-1} \frac{d}{x}][/tex]

0 = $\frac{-(d + h)}{(d + h)^{2} + x^{2}} -\frac{-d}{x^{2} + d^{2}}$

$\frac{-(d + h)}{(d + h)^{2} + x^{2}} = \frac{{d}{x^{2} + d^{2}}$

After solving the above eqn, we get

x = $\sqrt{\frac{d}{d + h}}$

The observer should stand at a distance equal to x = $\sqrt{\frac{d}{d + h}}$

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3 years ago

A pump jack scaffold must be fitted with two ______ gripping mechanisms to prevent slippage.

FromTheMoon [43]

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3 years ago

A 790kg car moving at 7 m/s takes a turn around a circle with a radius of 20 m. Determined the net force (in Newton’s) acting up

prisoha [69]

1935.5 N is the "net force" acting on a car.

<u>Explanation</u>:

Given that,

Mass of the car is 790 kg.

Velocity of the car is 7 m/s. (v)

It turned around with 20 m. (r)

We know that, Net force = m × a

$\text { Here, acceleration of the car is radial acceleration } a_{\mathrm{rad}}=\frac{v^{2}}{r}$