What is the fastest motion that can be measured in any frame of reference?

Compare and contrast a series circuit with a parallel circuit. Be sure to

Flura [38]

These are my answer if you want to use them

3 0

3 years ago

Please help, i only need help with 2,5, and 7

svet-max [94.6K]

Answer:

2. Galaxy

5. Electromagnetic radiation

7. Particle Accelerator

6 0

3 years ago

A uniform line charge of density λ lies on the x axis between x = 0 and x = L. Its total charge is 7 nC. The electric field at x

DedPeter [7]

Answer:

The electric field at x = 3L is 166.67 N/C

Solution:

As per the question:

The uniform line charge density on the x-axis for x, 0< x< L is $\lambda$

Total charge, Q = 7 nC = $7\times 10^{- 9} C$

At x = 2L,

Electric field, $\vec{E_{2L}} = 500N/C$

Coulomb constant, K = $8.99\times 10^{9} N.m^{2}/C^{2}$

Now, we know that:

$\vec{E} = K\frac{Q}{x^{2}}$

Also the line charge density:

$\lambda = \frac{Q}{L}$

Thus

Q = $\lambda L$

Now, for small element:

$d\vec{E} = K\frac{dq}{x^{2}}$

$d\vec{E} = K\frac{\lambda }{x^{2}}dx$

Integrating both the sides from x = L to x = 2L

$\int_{0}^{E}d\vec{E_{2L}} = K\lambda \int_{L}^{2L}\frac{1}{x^{2}}dx$

$\vec{E_{2L}} = K\lambda[\frac{- 1}{x}]_{L}^{2L}] = K\frac{Q}{L}[frac{1}{2L}]$

$\vec{E_{2L}} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{2L}] = \frac{63}{L^{2}}$

Similarly,

For the field in between the range 2L< x < 3L:

$\int_{0}^{E}d\vec{E} = K\lambda \int_{2L}^{3L}\frac{1}{x^{2}}dx$

$\vec{E} = K\lambda[\frac{- 1}{x}]_{2L}^{3L}] = K\frac{Q}{L}[frac{1}{6L}]$

$\vec{E} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{6L}] = \frac{63}{6L^{2}}$

Now,

If at x = 2L,

$\vec{E_{2L}} = 500 N/C$

Then at x = 3L:

$\frac{\vec{E_{2L}}}{3} = \frac{500}{3} = 166.67 N/C$

4 0

4 years ago

Why did the model spacecraft go so much faster than expected on Wednesday?

Andrew [12]

Answer:

The Wednesday test launch stored more potential energy, and launched the spacecraft at a faster speed because the stronger magnetic field closer to the magnet resulted in a greater increase in potential energy.

Explanation:

6 0

3 years ago

A high-speed flywheel in a motor is spinning at 500 rpm when a power failure suddenly occurs. the flywheel has mass 40.0 kg and

tangare [24]

<span>The flywheel is solid cylindrical disc. Moment of inertial = Â˝ * mass * radius^2 Mass = 40.0 kg Radius = Â˝ * 76.0 cm = 38 cm = 0.38 meter Moment of inertial = Â˝ * 41 * 0.36^2 Convert rpm to radians/second The distance of 1 revolution = 1 circumference = 2 * Ď€ * r The number of radians/s in 1 revolution = 2 * Ď€ 1 minute = 60 seconds 1 revolution per minute = 2 * Ď€ radians / 60 seconds = Ď€/30 rad/s Initial angular velocity = 500 * Ď€/30 = 16.667 * Ď€ rad/s 170 revolutions = 170 * 2 * Ď€ = 340 * Ď€ radians The flywheelâ€™s initial angular velocity = 16.667 * Ď€ rad/s. It decelerated at the rate of 1.071 rad/s^2 for 48.89 seconds. Î¸ = Ď‰i * t + Â˝ * Î± * t^2 Î¸ = 16.667 * Ď€ * 48.89 + Â˝ * -1.071 * 48.89^2 2559.9 - 1280 Î¸ = 1280 radians</span>

4 0

3 years ago