All categories

3 years ago

This is a substance that cannot be broken down further by chemical means.

Physics

2 answers:

viktelen [127]3 years ago

5 0

The answer is ELEMENT

Serggg [28]3 years ago

4 0

I think its an element

You might be interested in

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

3 years ago

A 4.4 kg marble (really big heavy marble) is accelerating down an incline. When it reaches level ground it slows down to a stop

KengaRu [80]

Mass=4.4kg

acceleration=-1.74m/s^2

Use newtons second law

$\\ \rm\longmapsto Force=ma$

$\\ \rm\longmapsto Force=4.4(-1.74)$

$\\ \rm\longmapsto Force=-7.656N$

initial velocity=u

Final velocity=v=0

Acceleration=a=-1.74m/s^2

Time=t=1.27s

$\\ \rm\longmapsto a=\dfrac{v-u}{t}$

$\\ \rm\longmapsto u=v-at$

$\\ \rm\longmapsto u=0-(-1.74)(1.27)$

$\\ \rm\longmapsto u=1.74(1.27)$

$\\ \rm\longmapsto u=2.2m/s$

4 0

2 years ago

On average, both arms and hands together account for 13% of a person's mass, while the head is 7.0% and the trunk and legs accou

Feliz [49]

Answer:

Angular velocity, $N_f = 242.36 rpm$

Explanation:

The mass of the skater, M = 74.0 kg

Mass of each arm, $m_{a} = 0.13 * \frac{M}{2}$ ( since it is 13% of the whole body and each arm is considered)

$m_{a} = 0.13 * 37\\m_a = 4.81 kg$

Mass of the trunk, $m_{t} = M - 2m_{a}$

$m_t = 74 - 2(4.81)\\m_{t} = 64.38 kg$

Total moment of Inertia = (Moment of inertia of the arms) + (Moment of inertia of the trunks)

$(I_{T} )_i = 2(\frac{m_{a}L^2 }{12} + m_a(0.5L + R)^2) + 0.5 m_t R^2$

$(I_{T} )_i = 2(\frac{4.81 * 0.7^2 }{12} + 4.81(0.5*0.7 + 0.175)^2) + 0.5 *64.38* 0.175^2\\(I_{T} )_i = 3.052 + 0.986\\(I_{T} )_i = 4.038 kgm^2$

The final moment of inertia of the person:

$(I_{T} )_f = \frac{1}{2} MR^{2} \\(I_{T} )_f = \frac{1}{2} * 74*0.175^{2}\\(I_{T} )_f = 1.133 kg.m^2$

According to the principle of conservation of angular momentum:

$(I_{T} )_i w_{i} = (I_{T} )_f w_{f}\\w_{i} = 68 rpm = (2\pi * 68)/60 = 7.12 rad/s\\4.038 * 7.12 =1.133* w_{f}\\w_{f} = 25.38 rad/s\\w_{f} = \frac{2\pi N_f}{60} \\25.38 = \frac{2\pi N_f}{60}\\N_f = (25.38 * 60)/2\pi \\N_f = 242.36 rpm$

3 0

2 years ago

What is the initial position

labwork [276]

Answer:

Initial position of a body is the position of the body before accelerating or increasing its velocity the position changes and then that position is the final position.

hope it is helpful...

6 0

2 years ago

The period of the wave is

Karo-lina-s [1.5K]

Answer:

2.5 s

Explanation:

3 0

3 years ago