Any ideas on what I should write???

How much work does the charge escalator do to move 2.30 μC of charge from the negative terminal to the positive terminal of a 3.

loris [4]

Answer:

$6.9\times 10^{-6} J$

Explanation:

We are given that

Charge=q= $2.3\mu C=2.3\times 10^{-6} C$

$1\mu C=10^{-6} C$

Potential difference=V=3 V

We know that

Work done= $V\times q$

Using the formula

Work done by charge to move from the negative terminal to the positive terminal of battery= $2.3\times 10^{-6}\times 3$ J

Work done by charge to move from the negative terminal to the positive terminal of battery= $6.9\times 10^{-6} J$

6 0

3 years ago

A thin, uniform metal bar, 2.00 m long and weighing 90.0 N, is hanging vertically from the ceiling by a frictionless pivot. Sudd

matrenka [14]

Answer:

(a) The "angular speed" is 5.88 rad/s.

Explanation:

Given values,

The length of the bar is L = 2m

The weight of the bar is w = 90 N

The metal bar is hanging vertically from the ceiling by a frictionless pivot

The mass of the ball is m = 3kg

The distance between the ceiling and the ball is d = 1.5m

$\text { The "initial speed of the ball" is } V_{i}=10 \mathrm{m} / \mathrm{s}$

$\text { The "final speed of the ball" is } V_{f}=6 \mathrm{m} / \mathrm{s}$

(a) Calculating the angular speed:

$W_{\mathrm{f}}=\left(3 \mathrm{mg} \mathrm{d} \frac{V_{i}+V_{f}}{w l^{2}}\right)$

$W_{f}=\left(3 \times 3 \times 9.8 \times 1.5 \times \frac{10+6}{90 \times 2^{2}}\right)$

$W_{f}=\frac{2116.8}{360}$

$\mathrm{W}_{\mathrm{f}}=5.88 \mathrm{rad} / \mathrm{s}$

The angular speed is 5.88 rad/s.

(b) The "angular momentum" is conserved because the torque is not exerted by "the pivot" on the system about the "axis of rotation" but the "linear momentum" is not conserved because "the pivot" exerts a "vertical" and a "horizontal force" on the system during the collision.

8 0

4 years ago

A light spring of constant 179 N/m rests vertically on the bottom of a large beaker of water. A 5.32 kg block of wood of density

Digiron [165]

Answer:

Compression of the spring: 0.18 m (downward)

Explanation:

The forces acting on the block of wood are:

- The force of gravity, acting downward, of magnitude $mg$ , where m = 5.32 kg is the mass of the block and $g=9.8 m/s^2$ is the acceleration due to gravity