Which of the following are Electromagnetic Waves?

What kind of elements tend to have higher ionization energies.

algol13

Non metals tend to have higher ionization energies

5 0

2 years ago

What type of waves moves energy forward, but the source moves up and down?

stich3 [128]

I'm pretty sure its transverse waves

4 0

3 years ago

The law of conservation of momentum states that the total momentum of interacting objects does not change . This means the total

pickupchik [31]

Answer:

The momentum of an object is equal to the product of its mass and its velocity.

Explanation:

Consider an object of mass $m$ travelling at a velocity $\vec{v}$ . The momentum $\vec{p}$ of this object would be:

$\vec{p} = m \cdot \vec{v}$ .

For the law of conservation of momentum, consider two objects: object $\rm a$ and object $\rm b$ . Assume that these two objects collided with each other.

Let $m_{\rm a}$ and $m_{\rm b}$ denote the mass of the two objects.
Let $\vec{v}_{\rm a}(\text{initial})$ and $\vec{v}_{\rm b}(\text{initial})$ denote the velocity of the two object right before the interaction.
Let $\vec{v}_{\rm a}(\text{final})$ and $\vec{v}_{\rm b}(\text{final})$ denote the velocity of the two objects right after the interaction.

The momentum of the two objects right before the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ , respectively.
The momentum of the two objects right after the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ , respectively.

The sum of the momentum of the two objects would be:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ right before the collision, and
$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ right after the collision.

Assume that the system of these two objects is isolated. By the law of conservation of momentum, the sum of the momentum of these two objects should be the same before and after the collision. That is:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) = m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ .

4 0

4 years ago

The bones of the forearm (radius and ulna) are hinged to the humerus at the elbow. The biceps muscle connects to the bones of th

sertanlavr [38]

Answer:

The force exerted by the biceps is 143.8 kgf.

Explanation:

To calculate the force exerted by the biceps, we calculate the momentum in the elbow.

This momentum has to be zero so that her forearm remains motionless.

Being:

W: mass weight (6.15 kg)

d_W= distance to the mass weight (0.425 m)

A: weight of the forearm (2.25 kg)

d_A: distance to the center of mass of the forearm (0.425/2=0.2125 m)

H: force exerted by the biceps

d_H: distance to the point of connection of the biceps (0.0215 m)

The momemtum is:

$H*d_H-A*d_A-W*d_W=0\\\\H=(A*d_A+W*d_W)/d_H\\\\H=(2.25*0.2125+6.15*0.425)/0.0215\\\\H=(0.478125+2.61375)/0.0215\\\\H=3.091875/0.0215=143.8$

The force exerted by the biceps is 143.8 kgf.

8 0

4 years ago

What are some cunductors

sdas [7]

Answer:

copper, silver, aluminum, brass

Explanation:

you could have looked that up

4 0

3 years ago