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2 years ago

Which statement best describes the electrons in this illustration ?

Physics

2 answers:

Nonamiya [84]2 years ago

7 0

A. electrons are negatively charged outside nucleus

koban [17]2 years ago

5 0

A is obviously true
the electrons are located around the nucleus (nucleus is made up of protons and neutrons.)
electrons are negatively charge particles
protons are positively charged particles
neutrons are neutral
hope this helps

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1. Applied research observational evidence 2. Basic research the experimental factor that changes in response to a change in the

lesya [120]

Answer:

This question is about matching each definition with its correct term. Please find the term matched with their appropriate definition below.

Explanation:

1. Empirical evidence: An empirical evidence is an observational evidence i.e an evidence gathered by observation or use of senses.

2. Dependent variable: Dependent variable is an experimental factor that changes in response to a change in the independent variable. In other words, it is dependent on the independent variable.

3. Applied research: Applied research is a type of research oriented at solving a present problem or need. It encompasses the production of products that can be sold for profit.

4. Hypothesis: A hypothesis in an experiment is a proposed explanation for a scientific problem that itself can be tested by experimentation. A hypothesis aims at providing a testable explanation to an observed problem.

5. Control: A control is a quantity in an experiment that remains unchanged or constant. It is kept the same by the experimenter for all groups in the experiment in order not to influence the outcome.

6. Basic research: Basic research is the research that expands knowledge in a particular area. It is the kind of research that aims at filling a knowledge void or satiating curiosity.

7. Independent variable: The independent variable is the experimental factor that is changed or manipulated deliberately by the scientist.

8 0

3 years ago

What is the density of a piece of wood with a mass of 25 kg<br> and a volume of 0.0385 m³?

Orlov [11]

Answer:

649kg/m^3

Explanation:

Let p be the density of this particular object.

Formula for density:

$p = \frac{mass \: (in \: kg)}{volume \: (in \: {m}^{3}) }$

We can substitute the givenmass and volume to find density of the object.

$p = \frac{25kg}{0.0385 {m}^{3} } \\ = 649kg \: per \: {m}^{3}$

Therefore the density of this object is 649kg/m^3.

7 0

1 year ago

Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce

lutik1710 [3]

Answer:

speed of electrons = 3.25 × $10^{7}$ m/s

acceleration in term g is 3.9 × $10^{17}$ g.

radius of circular orbit is 2.76 × $10^{-4}$ m

Explanation:

given data

voltage = 3 kV

magnetic field = 0.66 T

solution

law of conservation of energy

PE = KE

qV = 0.5 × m × v²

v = $\sqrt{\frac{2qV}{m}}$

v = $\sqrt{\frac{2\times 1.6 \times 10^{-19}\times 3}{9.1\times 10^{-31}}$

v = 3.25 × $10^{7}$ m/s

and

magnetic force on particle movie in magnetic field

F = Bqv

ma = Bqv

a = $\frac{Bqv}{m}$

a = $\frac{0.67\times 1.6\times 10^{-19}\times 3.25\times 10^7}{9.1\times 10^{-31}}$

a = 3.82 × $10^{18}$ m/s²

and acceleration in term g

a = $\frac{3.82\times 10^{18}}{9.81}$

a = 3.9 × $10^{17}$ g

acceleration in term g is 3.9 × $10^{17}$ g.

and

electron moving in circular orbit has centripetal force

F = $\frac{mv^2}{r}$

Bqv = $\frac{mv^2}{r}$

r = $\frac{mv}{Bq}$

r = $\frac{9.1\times 10^{-31}\times 3.25\times 10^7}{0.67\times 1.6\times 10^{-19}}$

r = 2.76 × $10^{-4}$ m

radius of circular orbit is 2.76 × $10^{-4}$ m

8 0

3 years ago

At t = 0 the components of a radio-controlled car's velocity are vx = 0.5 m/s and vy = 1.2 m/s. Find the components of the car's

FinnZ [79.3K]

Answer:

x-component of velocity = 5.7 m/s

y-component of velocity = -1.4 m/s

Explanation:

Use first equation of motion to find components of velocity at a given time:

$v = u + at$

where, $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration and $t$ is the time.

Given:

$u_x= 0.5 m/s\\u_y=1.2 m/s\\a_x=2 m/s^2\\a_y=-1m/s^2\\t = (2.6-0)s =2.6 s$

$v_x=u_x+a_xt\\\Rightarrow v_x=0.5+2\times 2.6\\\Rightarrow v_x=5.7 m/s$

$v_y=u_y+a_yt\\\Rightarrow v_y=1.2-1\times 2.6\\\Rightarrow v_y=-1.4 m/s$

5 0

2 years ago

Select the statement that is not true about the Hubble Telescope:

xxMikexx [17]

Answer:

The option is B is not true for Hubble telescope.

4 0

2 years ago