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LUCKY_DIMON [66]
3 years ago
5

A typical ceiling fan running at high speed has an airflow of about 1.85 ✕ 103 ft3/min, meaning that about 1.85 ✕ 103 cubic feet

of air move over the fan blades each minute. Determine the fan's airflow in m3/s.
Physics
1 answer:
densk [106]3 years ago
7 0

Answer:

0.8726  m^3/s

Explanation:

We are to convert 1.85 x 10^3 ft^3/min to m^3/s

First, let us convert the numerator from ft3 to m3

1 ft3 = 0.0283 m3

Hence,

1.85 x 10^3 ft3 = 1.85 x 10^3 x 0.0283 m3

     = 52.355 m3

Now, let us convert the denominator from minutes to seconds

1 min = 60 sec

Therefore;

1.85 x 10^3 ft^3/min = 52.355/60  m^3/s

        = 0.8726  m^3/s

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A Hooke's law spring is compressed 12.0 cm from equilibrium, and the potential energy stored is 72.0 J. What compression (as mea
Anuta_ua [19.1K]

Answer:14.14 cm

Explanation:

Given

Spring Compression x=12 cm

Potential energy Stored in spring=72 J

Suppose k is the spring constant of spring

Potential Energy of spring is given by =\frac{kx^2}{2}

\frac{k(0.12)^2}{2}=72

k(0.12)^2=144

k=10,000 N/m

k=10 kN/m

for 100 J energy

\frac{k(x_0)^2}{2}=100

10\times 10^3\cdot (x_0)^2=200

(x_0)^2=2\times 10^{-2}

x_0=0.1414

x_0=14.14 cm

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3 years ago
A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the
andreyandreev [35.5K]

Answer:

t = 1.42 s and d = 35.5 m

Explanation:

Given that,

Velocity of a roadrunner is 25 m/s

A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.

We need to find the time before the roadrunner is under the overpass and  how far away from the overpass is the roadrunner when the coyote drops the net.

d=ut+\dfrac{1}{2}at^2\\\\\text{Here, u = 0 and a = g}\\\\d=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2d}{g}} \\\\t=\sqrt{\dfrac{2\times 10}{9.8}} \\\\t=1.42\ s

Let d is the distance traveled. So,

d = vt

d = 25 m/s × 1.42 s

d = 35.5 m

5 0
3 years ago
A sample of an unknown substance has a mass of 89.5 g. If 345.2 J of heat are required to heat the substance from 285 K to 305 K
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Heat gained in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:<span>

Heat = mC(T2-T1)
345.2 = 89.5(C)(305 - 285)
C = 0.1928 </span>J/g•K

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