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3 years ago

14

Where in a river would you find fish that are adapted to low light and low oxygen levels?

1 answer:

snow_tiger [21]3 years ago

7 0

The dead Zone is where you were find fish that are adapted to low light and low oxygen levels.

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The difference in electrical potential energy between two places is called:

Natalka [10]

The answer is B hope this helped you !!

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3 years ago

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Something that flows easily, like water

Yanka [14]

Sweat and blood flow easily too

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4 years ago

A passenger travels from the first floor to the 60th floor, a distance of 210 m in 35 s. what is the elevators speed

Sati [7]

Distance=210m
Time=35s

$\\ \tt\bull\leadsto Speed=\dfrac{Distance}{Time}$

$\\ \tt\bull\leadsto Speed=210/35=6m/s$

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3 years ago

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We can reasonably model a 90-W incandescent lightbulb as a sphere 7.0cm in diameter. Typically, only about 5% of the energy goes

Ronch [10]

Answer:

292.3254055 W/m²

469.26267 V/m

$1.56421\times 10^{-6}\ T$

Explanation:

P = Power of bulb = 90 W

d = Diameter of bulb = 7 cm

r = Radius = $\frac{d}{2}=\frac{7}{2}=3.5\ cm$

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

c = Speed of light = $3\times 10^8\ m/s$

The intensity is given by

$I=\frac{P}{A}\\\Rightarrow I=\frac{90}{4\pi 0.035^2}\\\Rightarrow I=5846.50811\ W/m^2$

5% of this energy goes to the visible light so the intensity is

$I=0.05\times 5846.50811\\\Rightarrow I=292.3254055\ W/m^2$

The visible light intensity at the surface of the bulb is 292.3254055 W/m²

Energy density of the wave is

$u=\frac{1}{2}\epsilon_0E^2$

Energy density is also given by

$\frac{I}{c}=\frac{1}{2}\epsilon_0E^2\\\Rightarrow E=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E=\sqrt{\frac{2\times 292.3254055}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E=469.26267\ V/m$

The amplitude of the electric field at this surface is 469.26267 V/m

Amplitude of a magnetic field is given by

$B=\frac{E}{c}\\\Rightarrow B=\frac{469.26267}{3\times 10^8}\\\Rightarrow B=1.56421\times 10^{-6}\ T$

The amplitude of the magnetic field at this surface is $1.56421\times 10^{-6}\ T$

7 0

3 years ago

The block has a weight of 75 lblb and rests on the floor for which μkμk = 0.4. The motor draws in the cable at a constant rate o

WINSTONCH [101]

The given question is incomplete. The complete question is as follows.

The block has a weight of 75 lb and rests on the floor for which $\mu k$ = 0.4. The motor draws in the cable at a constant rate of 6 ft/sft/s. Neglect the mass of the cable and pulleys.

Determine the output of the motor at the instant $\theta = 30^{o}$ .

Explanation:

We will consider that equilibrium condition in vertical direction is as follows.

$\sum F_{y} = 0$

N - W = 0

N = W

or, N = 75 lb

Again, equilibrium condition in the vertical direction is as follows.

$\sum F_{x} = 0$

$T_{2} - F_{k}$ = 0

$T_{2} = \mu_{k} N$

= $0.4 \times 75 lb$

= 30 lb

Now, the equilibrium equation in the horizontal direction is as follows.

$\sum F_{x} = 0$

$T Cos (30^{o}) + T Cos (30^{o}) = T_{2}$

$2T Cos (30^{o}) = T_{2}$

or, T = $\frac{T_{2}}{2 Cos (30^{o})}$

= $\frac{30}{2 Cos (30^{o})}$

= $\frac{30}{1.732}$

= 17.32 lb

Now, we will calculate the output power of the motor as follows.

P = Tv

= $17.32 lb \times 6$

= $103.92 \times \frac{1}{550} \times \frac{hp}{ft/s}$

= 0.189 hp

or, = 0.2 hp

Thus, we can conclude that output of the given motor is 0.2 hp.

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3 years ago

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