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Montano1993 [528]
3 years ago
15

In a simple RC circuit, at t=0 the switch is closed with the capacitor uncharged. If C=30µF, =50V and R=10k, what is the poten

tial difference across the capacitor when I=2.0mA? At what rate is energy being stored in the capacitor when I = 2.0mA?
Physics
1 answer:
sergij07 [2.7K]3 years ago
3 0

Answer:

Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W

Explanation:

As we know that the current in the circuit at given instant of time is

i = 2.0 mA

R = 10 k ohm

now we know by ohm's law

V = iR

V = (2 mA)(10 kohm)

V = 20 volts

so voltage across the capacitor + voltage across resistor = V

V_c + 20 = 50

V_c = 30 V

Now we know that

U = \frac{q^2}{2C}

here rate of change in energy of the capacitor is given as

\frac{dU}{dt} = \frac{q}{C} \frac{dq}{dt}

\frac{dU}{dt} = (30)(2 mA)

\frac{dU}{dt} = 0.06 W

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scoray [572]
500 because 250 x 2 = 500
3 0
4 years ago
When jumping straight down, you can be seriously injured if you land stiff-legged. One way to avoid injury is to bend your knees
Naya [18.7K]

Answer:

a)  F = 1.26 10⁵ N,  b)  F = 2.255 10³ N, c)   F_ {soil} = 3078 N

Explanation:

For this exercise we will use the relationship between momentum and moment

          I = Δp

          F t = p_f -p₀

a) with stiff legs, final speed is zero, initial velocity is down

         Ft = 0-p₀

         F =  m v / t

let's calculate

         F = 84.0 6.82 / 4.56 10⁻³

         F = 1.26 10⁵ N

         

b) bending the legs

         

let's calculate

         F = 84.0 6.82 / 0.254

         F = 2.255 10³ N

     

c) It is requested to calculate the force of the ground on the man

            ∑ F = F_soil -W

           F_soil = F + W

           F_ {soil} = 2.255 103 + 84 9.8

           F_ {soil} = 3078 N

3 0
3 years ago
If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of laun
zvonat [6]

Answer:

H = 1/2 g t^2    where t is time to fall a height H

H = 1/8 g T^2   where T is total time in air  (2 t  = T)

R = V T cos θ       horizontal range

3/4 g T^2 = V T cos θ       6 H = R    given in problem

cos θ = 3 g T / (4 V)           (I)

Now t = V sin θ / g     time for projectile to fall from max height

T = 2 V sin θ / g

T / V = 2 sin θ / g

cos θ = 3 g / 4 (T / V)     from (I)

cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ

tan θ = 2/3      

θ = 33.7 deg

As a check- let V = 100 m/s

Vx = 100 cos 33.7 = 83,2

Vy = 100 sin 33,7 = 55.5

T = 2 * 55.5 / 9.8 = 11.3 sec

H = 1/2 * 9.8 * (11.3 / 2)^2 = 156

R = 83.2 * 11.3 = 932

R / H = 932 / 156 = 5.97        6 within rounding

3 0
3 years ago
Suppose that diameters of a new species of apple have a bell-shaped distribution with a mean of 7.2cm and a standard deviation o
Svet_ta [14]

Answer:

97.5%

Explanation:

By the empirical rule (68-95-99.7),

  1. 68% of data are within <em>μ </em>- <em>σ</em> and <em>μ </em>+ <em>σ</em>
  2. 95% of data are within <em>μ </em>- 2<em>σ</em> and <em>μ </em>+ 2<em>σ</em>
  3. 99.7% of data are within <em>μ </em>- 3<em>σ</em> and <em>μ </em>+ 2<em>σ</em>

<em>σ </em> and <em>μ</em> are the standard deviation and the mean respectively.

From the question,

<em>μ</em> = 7.2 cm

<em>σ</em> = 0.38 cm

7.96 = 7.2 + (<em>n</em> × 0.38)

<em>n</em> = 2

Hence, 7.96 represents <em>μ </em>+ 2<em>σ</em>.

P(X < <em>μ </em>+ 2<em>σ</em>) = P(X < <em>μ</em>) + P(<em>μ</em> < X < <em>μ </em>+ 2<em>σ</em>)

P(X < <em>μ</em>) is the percentage less than the mean = 50%.

P(<em>μ</em> < X < <em>μ </em>+ 2<em>σ</em>) is half of P(<em>μ </em>- 2<em>σ</em> < X < <em>μ </em>+ 2<em>σ</em>) = 95% ÷ 2 = 47.5%.

Considering this, for apples that are no more than 7.96 cm,

P(X < 7.96) = P(X < 7.2) + P(7.2 < X < 7.96) = 50% + 47.5% = 97.5%

<em />

5 0
3 years ago
An escaped convict runs 1.70 km due East of the prison. He then runs due North to a friend's house. If the magnitude of the conv
olga2289 [7]

Answer:

The right approach will be "47° north of east".

Explanation:

The given values are:

East of prison

= 1.70 km

Displacement vector

= 2.50 km

Now,

The direction will be:

⇒  Cos \ \theta =\frac{1.7}{2.5}

⇒            =0.68

⇒         \theta=47.16^{\circ}

i,e.,       \theta = 47^{\circ} (north of east)

6 0
3 years ago
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