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3 years ago

Atoms of the same element that differ only in the number of neutrons are known as.

Physics

1 answer:

True [87]3 years ago

7 0

Answer:

They are known as isotopes

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Balance the equation: ___ SiO2 + ___ HF  ___ SiF4 + ___ H2O

olga_2 [115]

Let's start from the Fluorine (F), which has 4 atoms on the right side: this means we must put a factor 4 in front of HF to balance the number of atoms of Fluorine.
Nowe we have 4 atoms of hydrogen, H, on the left side. Since we have a $H_2$ on the right side (2 atoms), we must put a factor 2 in front of $H_2 O$ , to balance the number of atoms of hydrogen.
So now the atoms of oxygen are also balanced (2 on both sides), as well as the atoms of Silicon (Si), so the balanced reaction is

$Si O_2 + 4 HF \rightarrow SiF_4 + 2 H_2 O$

3 0

3 years ago

A railroad car of mass m is moving with speed u when it collides with and connects to a second railroad car of mass 3m, initiall

Black_prince [1.1K]

Take a look at the picture. The speed of the connected cars is 1/4 of the single car’s speed, and the kinetic energy is 1/4 of the single car’s kinetic energy

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3 years ago

What is the mass of a person who weighs 653 N at earths surface

miss Akunina [59]

Answer:

66.6 kg

Explanation:

The weight of a person at Earth's surface is given by

$W=mg$

where

m is the mass of the person

$g=9.8 m/s^2$ is the acceleration of gravity

The person in the problem has a weight of

W = 653 N

Therefore, we can rearrange the equation to find his mass:

$m=\frac{W}{g}=\frac{653}{9.8}=66.6 kg$

7 0

4 years ago

A 700-kg car, driving at 29 m/s, hits a brick wall and rebounds with a speed of 4.5 m/s. what is the car's change in momentum du

Viefleur [7K]

The change in the momentum of the car due to collision between car and wall is $\fbox{\begin\ -23450\text{ kg.m}/\text{s}\end{minispace}}$ or $\fbox{\begin\\-2.3450 \times {10^4}\,\text{kg.m}/\text{s}\end{minispace}}$ .

Further Explanation:

Let us consider the car is moving towards the right direction and it collides with the wall. After the collision, the car will bounce back or car will rebound opposite to the direction of its motion i.e., towards left. Therefore, the final velocity of the car is opposite to the direction of the initial velocity of the car.

Given:

The mass of the car is $700\,{\text{kg}}$ .

The velocity of the car before collision is $29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ .

The velocity of the car after collision is $4.5\text{ m}/\text{s}$ .

Concept:

The momentum of an object is defined as the product of mass of object and the velocity with which the object is moving.

The initial momentum of the car is:

$\fbox{\begin\\{p_i}= m{v_i}\end{minispace}}$ …… (I)

Here, ${p_i}$ is the initial momentum of the car, $m$ is the mass of the car and ${v_i}$ is the initial velocity of the car.

Substitute $700\,{\text{kg}}$ for $m$ and $29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for ${v_i}$ in equation (I).

$\begin{gathered}{p_i} = \left( {700\,{\text{kg}}} \right) \cdot \left( {29\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}} \right) \\ = 20300\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\ \end{gathered}$

The final momentum of the car is defined as the product of mass of car and the velocity of the car after collision or final velocity of the car.

The final momentum of the car is:

$\fbox{\begin\\{p_f} = m{v_f}\end{minispace}}$ …… (II)

Here, ${p_f}$ is the final momentum and ${v_f}$ is the final velocity.

Substitute $700\,{\text{kg}}$ for $m$ and - $4.5\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for ${v_f}$ in equation (II).

$\begin{aligned}{p_f}&=\left( {700\,{\text{kg}}} \right)\cdot\left( {- 4.5\,{{\text{m}}\mathord{\left/{\vphantom {{\text{m}}{\text{s}}}} \right.\kern-\nulldelimiterspace}{\text{s}}}}\right)\\&=-3150\,{{{\text{kg}} \cdot {\text{m}}}\mathord{\left/{\vphantom{{{\text{kg}}\cdot {\text{m}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}$

The change in the momentum of the car after collision is the difference between the momentum of car before collision and the momentum of car after collision.

The change in momentum of the car is:

$\fbox{\begin\Delta p = {p_f} - {p_i}\end{minispace}}$ …… (III)

Here, $\Delta p$ is the change in the momentum of the car.

Substitute $- 3150\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for ${p_f}$ and $20300\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ for ${v_i}$ in equation (III).

$\begin{aligned}\Delta{p}&=-3150\text{ kg}.\text{m}/\text{s}\ -20300\text{ kg}.\text{m}/\text{s}\\&=-23450\text{ kg}.\text{m}/\text{s}\end{aligned}$

Thus, the change in the momentum of the car due to collision between car and wall is $\fbox{\begin\ -23450\text{ kg.m}/\text{s}\end{minispace}}$ or $\fbox{\begin\\-2.3450 \times {10^4}\,\text{kg.m}/{s}\end{minispace}}$ .

Learn more:

1. The motion of a body under friction brainly.com/question/4033012/

2. A ball falling under the acceleration due to gravity brainly.com/question/10934170/

3.Conservation of energy brainly.com/question/3943029/

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

Change in momentum, collision, initial velocity, final velocity, initial momentum, final momentum,-23450 kgm/s^2, -23450 kgm/s2, -2.3450*10^6 kgm/s^2, -2.3450*10^6 kgm/s2.

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3 years ago

Plot a graph based on the passage

Mnenie [13.5K]

Hence, the horizontal velocity of the rover is 1.73 m/s

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3 years ago