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laila [671]
3 years ago
7

Un lucru care are o gaura in mijloc

Physics
1 answer:
UkoKoshka [18]3 years ago
5 0

Answer:

gogoașă ?

Explanation:

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todos los cuerpos pueden estar sobre la tierra por la fuerza de gravedad .cuando nos caemos ,saltamos y realizamos otras activid
const2013 [10]
¿Es esta una pregunta verdadera o falsa? Por cierto, no hay mucha gente que hable español en esta aplicación, así que buena suerte.
4 0
3 years ago
An uncharged 30.0-µF capacitor is connected in series with a 25.0-Ω resistor, a DC battery, and an open switch. The battery has
kolezko [41]

Answer:

(a) Maximum current through resistor is 1.43 A

(b) Maximum charge capacitor receives is 1.50\times 10^{-3}\text{ C}.

Explanation:

(a)

In an RC (resistor-capacitor) DC circuit, when charging, the current at any time, <em>t</em>, is given by

I(t) = I_0e^{-t/\tau}

Here, I_0 is the maximum current and <em>τ</em> represents time constant which is given by RC (the product of the resistance and capacitance).

The maximum current is given by

I = \dfrac{V}{R_\text{eff}}

<em>V</em> is the emf of the battery and R_\text{eff} is the effective resistance.

In this question, R_\text{eff} = 10.0 Ω + 25.0 Ω = 35.0 Ω

I = \dfrac{50.0\text{ V}}{35.0\ \Omega} = 1.43\text{ A}

(b) The maximum charge is given

<em>Q</em> = <em>CV</em>

where <em>C</em> is the capacitance of the capacitor

Q = (30.0\times10^{-6}\text{ F})(50.0\text{ V}) = 1.50\times 10^{-3}\text{ C}

3 0
3 years ago
Read 2 more answers
An athlete can exercise by making mechanical waves in ropes. What is the
saveliy_v [14]

Answer:

c

Explanation:

3 0
3 years ago
On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a const
IceJOKER [234]

Answer:

a)

Explanation:

  • Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:

        x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}   (1)

  • Since the car starts from rest, v₀ =0.
  • We know the value of t = 5 sec., but we need to find the value of a.
  • Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:

       a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2  (2)

  • Replacing a and t in (1):

       x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}  = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m.  (3)

  • Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
  • Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:

       a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2  (4)

  • Replacing v₀, at and t in (1), we have:

       x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m   (5)

  • Therefore, as the truck travels twice as far as the car, the right answer is a).
7 0
3 years ago
(4.3 x 10-2 )4.950 x 105)<br> O 21.285 x 103<br> O 2.13 x 104<br> O 21 x 103<br> O 2.1 x 104
Natalka [10]

Answer:

The answer is:

It's 21.285 × 103

7 0
3 years ago
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