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Mkey [24]
2 years ago
7

Calculate the power output of a 1.5 g fly as it walks straight up a windowpane at 2.4 cm/s .

Physics
1 answer:
Pepsi [2]2 years ago
6 0

Answer:

Power output, P = 0.35 watts

Explanation:

It is given that,

Mass of the fly, m = 1.5 g = 0.0015 kg

Speed of the fly, v = 2.4 cm/s = 0.024 m/s

We have to find the power output of the fly. Power output is given by :

P=\dfrac{W}{t}

And W = F.d

d = distance traveled

P=\dfrac{F.d}{t}

Since, v = d /t

So, P = F . v

P=mg\times v

P=0.0015\ kg\times 9.8\ m/s^2\times 0.024\ m/s

P = 0.00035 watts

or P = 0.35 watts

Hence, this is the required solution.

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<h3>pressure = force / area</h3>

<h3>force = 84 N</h3><h3>pressure = 6 × 10 - 5 = 55 m2</h3>

<h3>pressure = 84 / 55</h3>

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hope that helps and please tell me if i am wrong :)

8 0
2 years ago
Which formula is used to find fluctuation of the shape of body
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Answer:

varn=n1+1ehvkT–1

Explanation:

This is Einstein's equation.

5 0
2 years ago
A windowpane is half a centimeter thick and has an area of 1.0 m2. The temperature difference between the inside and outside sur
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To solve this problem it is necessary to apply the concepts related to the heat flux rate expressed in energetic terms. The rate of heat flow is the amount of heat that is transferred per unit of time in some material. Mathematically it can be expressed as:

\frac{Q}{t} = \frac{kA}{L} (T_H - T_C)

Where

k = 0.84 J/s⋅m⋅°C (The thermal conductivity of the material)

A = 1m^2 Area

L = 5*10^{-3}m Length

T_H= Temperature of the "hot"reservoir

T_C= Temperature of the "cold"reservoir

Replacing with our values we have that,

\frac{Q}{t} = \frac{kA}{L} (T_H - T_C)

\frac{Q}{t} = \frac{(0.84)(1)}{0.005} (15)

\frac{Q}{t} = 2520J/s

Therefore the correct answer is B.

3 0
2 years ago
The PVT behavior of a certain gas is described by the equation of state: P(V − b) = RT where b is a constant. If in addition CV
alexdok [17]

Answer:

shown in the attachment

Explanation:

The detailed step by step and necessary mathematical application is as shown in the attachment.

6 0
2 years ago
A chair of mass 30.0 kg is at rest on a horizontal floor. The floor is not frictionless. You push on the chair with a force of 8
miv72 [106K]
First make sure you draw a force diagram. You should have Fn going up, Fg going down, Ff going left and another Fn going diagonally down to the right. The angle of the diagonal Fn (we'll call it Fn2) is 35° and Fn2 itself is 80N. Fn2 can be divided into two forces: Fn2x which is horizontal, and Fn2y which is vertical. Right now we only care about Fn2y.

To solve for Fn2y we use what we're given and some trig. Drawing out the actual force of Fn2 along with Fn2x and Fn2y we can see it makes a right triangle, with 80 as the hypotenuse. We want to solve for Fn2y which is the opposite side, so Sin(35)=y/80. Fn2y= 80sin35 = 45.89N

Next we solve for Fg. To do this we use Fg= 9.8 * m. Mass = 30kg, so Fg = 9.8 * 30 = 294N.

Since the chair isn't moving up or down, we can set our equation equal to zero. The net force equation in the vertical direction will be Fn + Fn2y -Fg = 0. If we plug in what we know, we get Fn + 45.89 -294 = 0. Then solve this algebraically.

Fn +45.89 -294 = 0
Fn +45.89 = 294
Fn = 248.11 N

You'll get a more accurate answer if you don't round Fn2y when solving for it, it would be something along the lines of 45.88611 etc
7 0
2 years ago
Read 2 more answers
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