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3 years ago

What is an RC circuit?

1 answer:

8 0

Is a circuit with both a resistor (R) and a capacitor (C). RC circuits are frequent element in electronic devices. They also play an important role in the transmission of electrical signals in nerve cells.

Hope this helps!

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what is the position of centre of curvature for concave and convex mirror show with the help of diagram if you can

Anon25 [30]

it is the point at infinity where it is at a distance from the curve equal to the radius of curvature lying on the normal vector. Sorry no diagram

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3 years ago

Help me please........................

stepan [7]

Answer:

Left

Explanation:

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3 years ago

At a particular instant the magnitude of the momentum of a planet is 2.05 × 10^29 kg·m/s, and the force exerted on it by the sta

Evgesh-ka [11]

727.5256266 AWNSERrRrrRr

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3 years ago

A net force, the magnitude of which is 3800 N, accelerates a 1260-kg vehicle for 10.0 s. The vehicle travels 50.0 m during this

Novay_Z [31]

Answer:

SEE EXPLANATION

Explanation:

$p = \frac{fd}{t} \\ where \: \\p = power \\ f = force \\ d = distance \\ and \: t = time \\ \\ p = \frac{3800 \times 50}{10} \\ p = \frac{190000}{10} \\ p = 19000w$

7 0

3 years ago

A 0.58 kg mass is moving horizontally with a speed of 6.0 m/s when it strikes a vertical wall. The mass rebounds with a speed of

Sunny_sXe [5.5K]

Answer:

$5.8\; {\rm kg\cdot m \cdot s^{-1}}$ .

Explanation:

If the mass of an object is $m$ and the velocity of that object is $v$ , the linear momentum of that object would be $m\, v$ .

Assume that the initial velocity of the mass is positive ( $6.0\; {\rm m\cdot s^{-1}}$ .) However, the direction of the velocity is reversed after the impact. Thus, the sign of the new velocity of the object would be negative- the opposite of that of the initial velocity. The new velocity would be $(-4.0\; {\rm m\cdot s^{-1}})$ .

Thus, the change in the velocity of the mass would be:

$\begin{aligned}& (\text{Change in Velocity}) \\ =\; & (\text{Final Velocity}) - (\text{Initial Velocity}) \\ =\; & (-4.0\; {\rm m\cdot s^{-1}}) - (6.0\; {\rm m\cdot s^{-1}}) \\ =\; & (-10\; {\rm m\cdot s^{-1})\end{aligned}$ .

The change in the linear momentum of the mass would be:

$\begin{aligned} & \text{change in momentum} \\ =\; & (\text{mass}) \times (\text{change in velocity}) \\ =\; & 0.58\; {\rm kg} \times (-10\; {\rm m\cdot s^{-1}}) \\ =\; & (-5.8\; {\rm kg \cdot m \cdot s^{-1}})\end{aligned}$ .

Thus, the magnitude of the change of the linear momentum would be $5.8\; {\rm kg \cdot m \cdot s^{-1}}$ .

7 0

3 years ago