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Arisa [49]
3 years ago
8

I need help on this question !

Physics
1 answer:
Stella [2.4K]3 years ago
4 0

Answer:

A. Create habitats for coral, but not fish

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2 years ago
What must the charge (sign and magnitude) of a particle of mass 1.41 gg be for it to remain stationary when placed in a downward
Yuri [45]

Answer:

q = 2.067 \times 10^{-5}\ C

Explanation:

Given,

mass = 1.41 g = 0.00141 Kg

Electric field,E = 670 N/C.

We know,

Force in charge due to Electric field.

F = E q

And also we know

F = m g

Equating both the equation of motion

m g = E q

q =\dfrac{mg}{E}

q =\dfrac{0.00141 \times 9.81}{670}

q = 2.067 \times 10^{-5}\ C

Charge of the particle is equal to q = 2.067 \times 10^{-5}\ C

6 0
3 years ago
HELLLLLP PLEASE, thank u :)
MrRa [10]

Answer:

Option B. 3.0×10¯¹¹ F.

Explanation:

The following data were obtained from the question:

Potential difference (V) = 100 V.

Charge (Q) = 3.0×10¯⁹ C.

Capacitance (C) =..?

The capacitance, C of a capacitor is simply defined as the ratio of charge, Q on either plates to the potential difference, V between them. Mathematically, it is expressed as:

Capacitance (C) = Charge (Q) / Potential difference (V)

C = Q/V

With the above formula, we can obtain the capacitance of the parallel plate capacitor as follow:

Potential difference (V) = 100 V.

Charge (Q) = 3.0×10¯⁹ C.

Capacitance (C) =..?

C = Q/V

C = 3.0×10¯⁹ / 100

C = 3.0×10¯¹¹ F.

Therefore, the capacitance of the parallel plate capacitor is 3.0×10¯¹¹ F.

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Mixing chemicals in a cold pack creates a reaction that:
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