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3 years ago

Select all that apply.

2 answers:

8 0

The true statements are

"Its temperature will rise until reaching 0°C."
And
"Its temperature will remain at 0°C until all the ice melts."

So, basically once the ice has reached 0°C we can conclude that the ice has probably fully melted. However the temperature has to stay there if we want the ice to melt.

We must remember that the melting point of ice is 0°C, and that Ice is completely melted at approximately 0°C so we must have those two statements true

klemol [59]3 years ago

4 0

Temperature will rise until reaching 0°C.

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2 years ago

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3 years ago

For the following elementary reaction 2br• -> br2-. The rate of consumption of the reaction and the rate of formation of prod

Scorpion4ik [409]

Answer: $-\frac{1}{2}\times \frac{d[Br^.]}{dt}=+\frac{d[Br_2]}{dt}$

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Rate of a reaction is defined as the rate of change of concentration per unit time.

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$2Br^.\rightarrow Br_2$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

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2 years ago

What is the total resistance in this circuit?

saul85 [17]

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2 years ago

A tiny 0.0250 -microgram oil drop containing 15 excess electrons is suspended between to horizontally closely-spaced metal plate

qaws [65]

Answer:

(a) 12 × 10⁻³ C = 12 mC (b) The lower plate

Explanation:

Given

mass of oil drop, m = 0.025 μg = 0.025 × 10⁻⁶

radius of plates, r = 6.50 cm = 6.5 × 10⁻² m

k = 1/4πε₀ = 9.0 × 10⁹ Nm²/C²

electric charge, e= 1.6 × 10⁻¹⁹ C

charge on oil drop, q = 15e

charge on plates, Q = ?

First, we find the charge density of the plates, D = Q/A where Q = charge on plates and A = area of plates. Since the plates are circular, the area is given by A=πr² where r = radius of plates. D=Q/πr²

Also, the electric field, E between the plates is given by E = D/ε =Q/Aε = Q/ε₀πr².

The force on the oil drop due to the electric field between the plates is given by F = qE = qQ/ε₀πr².

Since the oil drop is suspended between the plates, it means that the electric force due to the field on the oil drop balances the weight of the oil drop. So, since weight of oil drop W = mg where g = 9.8 m/s². F =W (for oil drop suspension).

So, qQ/ε₀πr²=mg

So, Q=mgε₀πr²/q

From k = 1/4πε₀, ε₀=1/4πk

So, Q = mgπr²/4πkq = mgr²/4kq = (0.025 × 10⁻⁶ × 9.8 × (6.5 × 10⁻²)²)÷(4 × 9 × 10⁹ × 15 × 1.6 × 10⁻¹⁹)= 0.012 C = 12 mC

(b) The lower plate must be positive because, the direction of the electric field must be upwards, so as to balance out the weight of the oil drop so as to suspend it.

6 0

2 years ago