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Luba_88 [7]
3 years ago
7

Which of the following is not a way in which humans directly impact algal bloom production? a. industrial pollution b. wastewate

r discharge c. fertilizer in aquatic ecosystems d. fishing Please select the best answer from the choices provided
Physics
1 answer:
Monica [59]3 years ago
6 0

d. fishing

The other options all directly affect algal bloom production because they affect the nutrients in the water, and an overabundance of certain nutrients in the water is what causes algal bloom.

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A force gives a 5.0 kg object an acceleration of 2.0 m/s 2. The same force would give a 20 kg object an acceleration of _____. 0
Oksi-84 [34.3K]

m = 5 kg

a = 2 m/s²

to find the force that accelerates the 4 kg object @ 2 m/s²

F = ma = 5 kg x 2 m/s² = 10 N

To find what acceleration 10 N would give a 20 kg object

a = F/m = 10 N/20 kg = 0.5 m/s

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3 years ago
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You are part of the designing team for a new solar thermal power plant. The plant will use thermal energy storage to produce ele
Harman [31]

Explanation:

Below is an attachment containing the solution.

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Which type of energy can be sensed by the eyes?
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Answer:

light energy..

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A toroidal coil of N turns has a central radius b and a square cross section of side a. Find its self-inductance.
Xelga [282]

Answer:

L = \frac{\mu_0 N^2 (a^2)}{2\pi b}

Explanation:

As we know that magnetic field due to torroid is given as

B = \frac{\mu_0 N i}{2\pi b}

this is approximately constant magnetic field along the axis of the torroid

now the flux linked with one coil of the torroid is given as

\phi = B.A

\phi = \frac{\mu_0 N i}{2\pi b}(a^2)

now total flux of N number of coils is given as

\phi_{total} = \frac{\mu_0 N^2 i(a^2)}{2\pi b}

now we know that self inductance is the property of coil in which flux of the coil will link with the current in the coil

So we know that

L = \frac{\phi}{i}

L = \frac{\mu_0 N^2 (a^2)}{2\pi b}

3 0
3 years ago
A flywheel with a radius of 0.300 m starts from rest and accelerates with a constant angular acceleration of 0.600 rad/s2. Compu
boyakko [2]

Answer:

0.42 m/s²

Explanation:

r = radius of the flywheel = 0.300 m

w₀ = initial angular speed = 0 rad/s

w = final angular speed = ?

θ = angular displacement = 60 deg = 1.05 rad

α = angular acceleration = 0.6 rad/s²

Using the equation

w² = w₀² + 2 α θ

w² = 0² + 2 (0.6) (1.05)

w = 1.12 rad/s

Tangential acceleration is given as

a_{t} = r α = (0.300) (0.6) = 0.18 m/s²

Radial acceleration is given as

a_{r} = r w² = (0.300) (1.12)² = 0.38 m/s²

Magnitude of resultant acceleration is given as

a = \sqrt{a_{t}^{2} + a_{r}^{2}}

a = \sqrt{0.18^{2} + 0.38^{2}}

a = 0.42 m/s²

8 0
3 years ago
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