All categories

3 years ago

Select the statement that correctly describes how light travels.

Physics

1 answer:

Paladinen [302]3 years ago

5 0

The speed of light, traveling in a vacuum, will not change if the light source is moving.

The speed of light in vacuum does not depend on whether the light source is moving or not.

You might be interested in

If you wanted to

VLD [36.1K]

Answer:

Option C

Explanation:

According to the formula

$\\ \boxed{\sf R=\rho\dfrac{\ell}{A}}$

If we use wide wire we increase the area of cross section so resistance decreases

5 0

2 years ago

Create a model in the space below that demonstrates how action potentials ensue. A sample model can be found under the Unit Pack

Pepsi [2]

Answer:

K+NA+30

Explanation:

8 0

3 years ago

Q1) An action exerted on an object which may change the object's

Sunny_sXe [5.5K]

Answer: force...

Explanation:

4 0

3 years ago

Conduction of a nerve impulse would be the fastest in _________

Vinil7 [7]

The options of the given are:

A. A large diameter myelinated fiber

B. A small diameter myelinated fiber

C. A large unmyelinated fiber

D. A small unmyelinated fiber

E. A small fiber with multiple Schwann cells

Answer: Option A, A large diameter myelinated fiber.

Explanation:

The conduction of the nerve impulse would be greatest in the myelinated fiber because the main function of the myelin sheath is to increase the speed of the impulse at which the electrical signals propagate.

In case of the unmyelinated sheath the nerve impulse travels slowly as the conduction waves but in case of the large diameter myelinated sheath the signals travel via saltatory conduction( hop)

In this type of propagation the signals are transferred from the node of Ranvier in one neuron to next node which increases the overall velocity of the action potentials.

8 0

3 years ago

Xelga [282]

Answer:

About 7.67 m/s.

Explanation:

Mechanical energy is always conserved. Hence:

$\displaystyle \begin{aligned} E_i & = E_f \\ \\ U_i + K_i &= U_f + K_f\end{aligned}$

Where <em>U</em> is potential energy and <em>K</em> is kinetic energy.

Let the bottom of the slide be where potential energy equals zero. As a result, the final potential energy is zero. Additionally, because the child starts from rest, the initial kinetic energy is zero. Thus:

$\displaystyle U_i = K_f$

Substitute and solve for final velocity:
$\displaystyle \begin{aligned} mgh_i &= \frac{1}{2}mv_f^2 \\ \\ 2gh_i &= v^2_f \\ \\ v_f &= \sqrt{2gh_i} \\ \\ & =\sqrt{2(9.8\text{ m/s$^2$})(3.00\text{ m})} \\ \\ & \approx 7.67\text{ m/s} \end{aligned}$

In conclusion, the child's speed at the bottom of the slide is about 7.67 m/s.

8 0

2 years ago