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3 years ago

What is atomic composition

Physics

1 answer:

Mazyrski [523]3 years ago

4 0

Answer:

The atom consists of a tiny nucleus surrounded by moving electrons. The nucleus contains protons, which have a positive charge equal in magnitude to the electron's negative charge. The nucleus may also contain neutrons, which have virtually the same mass but no charge.

Explanation:

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How does the law of the conservation of mass relate to "You can't get something for nothing

ivolga24 [154]

You can't get something from nothing is practically stating the law. The law means states that energy can't be created or destroyed, in other words, it's constant. So in saying so, you truly can't get something from nothing

8 0

4 years ago

Q3. A bridge is built without expansion gaps.

snow_tiger [21]

The answer is A ) much hotter

6 0

3 years ago

On the same spring day, a station near the equator has a surface temperature of 25°C, 15°C higher than the middle-latitude city

DedPeter [7]

Answer:

The air temperature at the tropopause is - 79 °C

Explanation:

We know that a station near the equator has a surface temperature of 25°C

Vertical soundings reveal an environmental lapse rate of 6.5 °C per kilometer.

The tropopause is encountered at 16 km.

In order to find the air temperature at the tropopause we are going to deduce a linear function for the temperature at the tropopause.

This linear function will have the following structure :

$f(x)=ax+b$

Where '' $a$ '' and '' $b$ '' are real numbers.

Let's write $T(x)$ to denote the temperature '' T '' in function of the distance

'' x '' ⇒

$T(x)=ax+b$

We can reorder the function as :

$T(x)=b+ax$ (I)

Now, at the surface the value of '' $x$ '' is 0 km and the temperature is 25°C so in the function (I) we write :

$T(0)=25=b+a(0)$ ⇒ $b=25$ ⇒

$T(x)=25+ax$ (II)

In (II) the value of '' $a$ '' represents the change in temperature per kilometer.

Because the temperature decrease with the height this number will be negative and also a data from the question ⇒

$T(x)=25-(6.5)x$ (III)

In (III) we deduced the linear equation. The last step is to replace by $x=16$ in (III) ⇒

$T(16)=25-(6.5)(16)=-79$

The air temperature at the tropopause is - 79 °C

6 0

3 years ago

Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the

Over [174]

Answer:

Part a)

$a= 0.32 m/s^2$

Part b)

$F_c = 3.6 N$

Part c)

$F_c = 5.5 N$

Explanation:

Part a)

As we know that the friction force on two boxes is given as

$F_f = \mu m_a g + \mu m_b g$

$F_f = 0.02(10.6 + 7)9.81$

$F_f = 3.45 N$

Now we know by Newton's II law

$F_{net} = ma$

so we have

$F_p - F_f = (m_a + m_b) a$

$9.1 - 3.45 = (10.6 + 7) a$

$a = \frac{5.65}{17.6}$

$a= 0.32 m/s^2$

Part b)

For block B we know that net force on it will push it forward with same acceleration so we have

$F_c - F_f = m_b a$

$F_c = \mu m_b g + m_b a$

$F_c = 0.02(7)(9.8) + 7(0.32)$

$F_c = 3.6 N$

Part c)

If Alex push from other side then also the acceleration will be same

So for box B we can say that Net force is given as

$F_p - F_f - F_c = m_b a$

$9.1 - 0.02(7)(9.8) - F_c = 7(0.32)$

$F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)$

$F_c = 5.5 N$

3 0

3 years ago

Vector A has a magnitude 5 m, vector B has a magnitude 7 m. Which of the following cannot be the magnitude of the vector A + B?

bulgar [2K]

I think it should be 7m

7 0

4 years ago