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3 years ago

14

You have a 78.7 mF capacitor initially charged to a potential difference of 11.5 V. You discharge the capacitor through a 3.03 Ω

resistor. What is the time constant?

1 answer:

Aloiza [94]3 years ago

8 0

Answer:

$\tau=0.23\;second$

Explanation:

Given,

$C=78.7\;mF\\V=11.5\;V\\R=3.03\;\Omega\\$

Time constant

$\tau=RC\\\tau=78.7\times10^{-3}\times3.03\\\tau=238.461\times10^{-3}\;second\\\tau=0.23\;second$

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3 years ago

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Please help me, this is a physics test.

sweet-ann [11.9K]

Answer:

a = 2 [m/s²]

Explanation:

To be able to solve this problem we must make it clear that the starting point when the time is equal to zero, the velocity is 5 [m/s] and when three seconds have passed the velocity is 11 [m/s], this point is the final point or the final velocity.

We can use the following equation.

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2 years ago

The 2.6-kg collar is released from rest at A and slides down the inclined fixed rod in the vertical plane. The coefficient of ki

neonofarm [45]

Answer:

The attached diagram explains the system,

Sum of Fy = 0

N=9.81

N - mgCos60 = 0

F= ukN= (0.53)(9.81) =

F= 5.12 N

So

F.d= 1/2(mv.v) - mgdsin60

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(a) v = 2.436 m/s

For deflection

-F.x = 1/2(mv.v) - mgxsin60 + 1/2 (k*x*x)

by solving for with values of v, m, g, F, k

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3 years ago

A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.

AveGali [126]

Answer:

The magnitude of electric force is $7.2\times10^{-3} N$

Explanation:

Coulomb's Law:

The force of attraction or repletion is

directly proportional to the products of charges i.e $F\propto q_1q_2$

inversely proportional to the square of distance i.e $F\propto \frac{1}{r^2}$

$\therefore F\propto \frac{q_1q_2}{r^2}$

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Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, $r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5$

we know,

$cos \theta = \frac{base }{hypotenuse}$

$\Rightarrow cos \theta = \frac{4 }{r}$

Total force $F_Q = 2.F_1 cos\theta \hat{i}$

$=2k\frac{Qq_1}{r^2} cos\theta \hat i$

$=2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i$

$=8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i$ [ r=5]

$=7.2\times10^{-3}\hat i$ N

The magnitude of electric force is $7.2\times10^{-3} N$

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3 years ago

A motorist inflates the tires of her car to a pressure of 180 kPa on a day when the temperature is -8.0° C. When she arrives at

GuDViN [60]

Apply Gay-Lussac's law:

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Final P and T values:

P = 245kPa, T = ?

Set the initial and final P/T values equal to each other and solve for the final T:

180/265.15 = 245/T

T = 361K

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3 years ago