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12345 [234]
3 years ago
11

Why is the outer surface of a teapot slightly polished

Physics
1 answer:
evablogger [386]3 years ago
6 0

Answer:

Well teapots are metal. If it were plastic, putting it on top of an oven it would melt. And metals are usually shiny.

Explanation:

If you have any questions feel free to ask in the comments - Mark

Also when you have the chance please mark me brainliest.

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A box is sliding down an incline tilted at a 11.1° angle above horizontal. The box is initially sliding down the incline at a sp
raketka [301]

Answer:s=0.68 m

Explanation:

Given

Inclination \theta =11.1^{\circ}

Speed of block(u)=1.6 m/s

Coefficient of kinetic Friction \mu _k=0.39

deceleration provided by friction=g\sin \theta -\mu _kg\cos \theta [/tex]

Using v^2-u^2=2as

Final velocity v=0

0-1.6^2=2(g\sin \theta -\mu _kg\cos \theta )s

s=\frac{-1.6^2}{2\cdot (9.8\sin 11.1-0.39\times 9.8\times \cos 11.1)}

s=0.68 m

5 0
3 years ago
What planet are people working on so they can move there
Nata [24]

Answer:

mars

Explanation:

just search it up

8 0
3 years ago
Read 2 more answers
A projectile is launched at an angle of 60° from the horizontal and at a velocity of
gayaneshka [121]

Answer:

60*12.0= 720 = v/60 * 12.0 squared which is 1,728

Explanation:

Horizontal velocity component: Vx = V * cos(α)

5 0
3 years ago
A person drives 70 km/h in 1 hour to the east, then 80 km/h for another hour to the east. What
andre [41]

Answer: The average velocity is 150 km/h

Explanation: 70+80=150

6 0
3 years ago
From mechanics, you may recall that when the acceleration of an object is proportional to its coordinate, d2xdt2=−kmx=−ω2x , suc
marysya [2.9K]

Answer:

    q = q₀ sin (wt)

Explanation:

In your statement it is not clear the type of circuit you are referring to, there are two possibilities.

1) The circuit of this problem is a system formed by an Ac voltage source and a capacitor, in this case all the voltage of the source is equal to the voltage at the terminals of the capacitor

                    ΔV = ΔV_{C}

we assume that the source has a voltage of the form

                    ΔV = ΔV₀o sin wt

The capacitance of a capacitor is

                   C = q / ΔV

                  q = C ΔV sin wt

the current in the circuit is

                    i = dq / dt

                    i = c ΔV₀ w cos wt

if we use

                  cos wt = sin (wt + π / 2)

we make this change by being a resonant oscillation

we substitute

                  i = w C ΔV₀ sin (wt + π/2)

With this answer we see that the current in capacitor has a phase factor of π/2 with respect to the current

2) Another possible circuit is an LC circuit.

In this case the voltage alternates between the inductor and the capacitor

                     V_{L} + V_{C} = 0

                      L di / dt + q / C = 0

the current is

                      i = dq / dt

                       

they ask us for a solution so that

                    L d²q / dt² + 1 / C q = 0

                     d²q / dt² + 1 / LC q = 0

this is a quadratic differential equation with solution of the form

                    q = A sin (wt + Ф)

to find the constant we derive the proposed solution and enter it into the equation

                di / dt = Aw cos (wt + Ф)

                d²i / dt²= - A w² sin (wt + Ф)

                 - A w² + 1 /LC  A = 0

                  w = √ (1 / LC)

To find the phase factor, for this we use the initial conditions for t = 0

in the case of condensate for t = or the charge is zero

                 0 = A sin Ф

                  Ф = 0

             

                  q = q₀ sin (wt)

6 0
3 years ago
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