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3 years ago

Identify three situations in which convection occurs.

2 answers:

3 0

A boiling pot of water (the water travels in a current throughout the pot), a hot air balloon (hot air rises, making the balloon rise) , and cup of a steaming, hot liquid (hot air rises, creating steam) are all situations where convection occurs.

serg [7]3 years ago

3 0

Sample Response: Convection occurs in a pot of boiling water and in a lava lamp. It also occurs in Earth’s interior and in the Sun’s interior. Convection also takes place in the atmosphere, where it drives air circulation, and in the ocean, where it is partially responsible for ocean circulation.

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Samantha is investigating the densities of several liquids. Which action is the result of her creative thinking.

Yuki888 [10]

It’s B my friending

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2 years ago

Find the time t2 that it would take the charge of the capacitor to reach 99.99% of its maximum value given that r=12.0ω and c=50

defon

Answer:

Explanation:

Given that, .

R = 12 ohms

C = 500μf.

Time t =? When the charge reaches 99.99% of maximum

The charge on a RC circuit is given as

A discharging circuit

Q = Qo•exp(-t/RC)

Where RC is the time constant

τ = RC = 12 × 500 ×10^-6

τ = 0.006 sec

The maximum charge is Qo,

Therefore Q = 99.99% of Qo

Then, Q = 99.99/100 × Qo

Q = 0.9999Qo

So, substituting this into the equation above

Q = Qo•exp(-t/RC)

0.9999Qo = Qo•exp(-t / 0.006)

Divide both side by Qo

0.9999 = exp(-t / 0.006)

Take In of both sodes

In(0.9999) = In(exp(-t / 0.006))

-1 × 10^-4 = -t / 0.006

t = -1 × 10^-4 × - 0.006

t = 6 × 10^-7 second

So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge

8 0

3 years ago

Circular motion occurs on earth true or false

Bingel [31]

Circular Motion does occur on earth, so yes, true

8 0

3 years ago

Read 2 more answers

Two forces F1 and F2 of equal magnitude are applied to a brick of mass 20kg lying on the floor as shown in the figure above. If

jeka57 [31]

Let $F_{1}=F_{2}=F$ .

Normal force equals (using Newton's third law) $N=mg+F\sin30^{o}-F\sin37^{o}$ .

$F_{f}\leq \mu N = \mu(mg+F\sin30^{o}-F\sin37^{o})$ , but $F_{f}\leq F(\cos30^o+\cos37^o)$ for all $F_{f}$ (in order to start moving the break). Therefore $F(\cos 30^o+\cos37^o)\geq \mu(mg+F\sin30^{o}-F\sin37^{o})$ , solving for $F$ : $F\geq \frac{\mu mg}{\cos30^o+\cos37^o-\mu\sin30^o+\mu\sin37^o}\approx 46,91\; \textbf{N}$