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3 years ago

Scientists use tables and graphs to chart and analyze _________.

Physics

1 answer:

MissTica3 years ago

6 0

Explanation:

Tables and graphs are visual representations. They are used to organise information to show patterns and relationships. A graph shows this information by representing it as a shape. Researchers and scientists often use tables and graphs to report findings from their research.

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A piece of purple plastic is charged with 3. 13×106 extra electrons compared to its neutral state. what is its net electric char

horsena [70]

A piece of purple plastic is charged with 3. 13×106 extra electrons compared to its neutral state, then the net electric charge in coulomb would be - 5.008×10⁻¹³ coulombs.

<h3>What is an electric charge?</h3>

Charged material experiences a force when it is exposed to an electromagnetic field due to the physical property of electric charge. You can have a positive or negative electric charge (commonly carried by protons and electrons respectively). Unlike charges attract one another while like charges repel one another. We refer to an object as neutral if it has no net charge.

The charge on one electron is -1.6 ×10⁻¹⁹ coulomb.

Then the charge on the 3.13×10⁶ extra electrons compared to its neutral state

=-1.6×10⁻¹⁹ ×(3.13×10⁶)

As given in the problem A piece of purple plastic is charged with 3.13×10⁶ extra electrons compared to its neutral state then the net electric charge in coulombs would be - 5.008×10⁻¹³ coulombs.

Learn more about an electric charge from here

brainly.com/question/8163163

#SPJ4

4 0

2 years ago

On earth, the solar constant is approximately I = 1367 W/m2 . This corresponds to the power per unit area that a detector at the

d1i1m1o1n [39]

Answer:

Explanation:

for earth -sun system

E = σ T⁴ , E is radiation emitted by sun at temperature T

E = 4π R₁² x solar constant on earth , R₁ is distance between earth and sun.

for marsh -sun system

E = σ T⁴ , E is radiation emitted by sun at temperature T

E = 4π R₂² x solar constant on Mars , R₂ is distance between earth and sun.

from two equation ,

4π R₂² x solar constant on Mars = 4π R₁² x solar constant on earth

solar constant on Mars = (R₁ / R₂ )² x solar constant on earth

6 0

3 years ago

A 50.0-g Super Ball traveling at 29.5 m/s bounces off a brick wall and rebounds at 19.0 m/s. A high-speed camera records this ev

Dima020 [189]

Answer:

$a=-10210.52\ m/s^2$

Explanation:

Given that,

Mass of a ball, m = 50 g

It is traveling at 29.5 m/s bounces off a brick wall and rebounds at 19.0 m/s.

Initial velocity, u = 29.5 m/s

Finl velocity, v =-10 m/s (as it rebounds)

We need to find the magnitude of the average acceleration of the ball during this time interval.

Acceleration = rate of change of velocity

$a=\dfrac{v-u}{t}\\\\a=\dfrac{(-19)-29.5}{4.75\times 10^{-3}}\\\\=-10210.52\ m/s^2$

So, the required acceleration is $10210.52\ m/s^2$ .

8 0

3 years ago

One particle has a mass of 3.12 x 10-3 kg and a charge of +8.8 C. A second particle has a mass of 7.1 x 10-3 kg and the same cha

never [62]

Answer: $r_i=0.016119\ m\approx 16.119\ mm$

Explanation:

Given

mass of first particle is $m_1=3.12\times 10^{-3}\ kg$

mass of second particle is $m_2=7.1\times 10^{-3}\ kg$

Charge on both the particle $q=8.8\times 10^{-6}\ C$

Now final speed of first particle is $v_1=131\ m/s$

Final separation between particles is $r=0.15\ m$

As there is no external force therefore linear momentum is conserved

$0+0=m_1v_1+m_2v_2$

$0=3.12\times 10^{-3}\times 131+7.1\times 10^{-3}\times v_2$

$v_2=-\dfrac{3.12\times 10^{-3}}{7.1\times 10^{-3}}\times 131$

$v_2=-57.56\ m/s$

Conserving total energy

Initial Kinetic energy +Initial Potential energy=Final Kinetic energy +Final Potential energy

$\Rightarrow 0+\frac{kq^2}{r_i}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+\frac{kq^2}{r_f}$

$\Rightarrow \frac{9\times 10^9\times 8.8^2\times 10^{-12}}{r_i}=\frac{1}{2}\times 3.12\times 10^{-3}\times 131^2+\frac{1}{2}7.1\times 10^{-3}\times (57.56)^2+\frac{9\times 10^9\times 8.8^2\times 10^{-12}}{0.15}$

$\Rightarrow \frac{0.696}{r_i}=26.771+11.76+4.646$

$\Rightarrow \frac{0.696}{r_i}=43.177$

$r_i=0.016119\ m\approx 16.119\ mm$

7 0

3 years ago

Calculate the thermal energy in a cubic meter of air at room temperature and one atmosphere. Is it possible to extract this ener

Ksivusya [100]

Answer:

312 kJ

Explanation:

The thermal energy contained in a cubic meter of air at room temperature (20 C) and one atmosphere (101 kPa) would be.

H = Cv * T + p * V

Being

Cv: specific heat of air at constant volme (0.72 kJ/(kg*K))

T: temperature in absolute scale (20C is 293 K)

Then:

H = 0.72 * 293 + 101 * 1 = 312 kJ

The only way to extract thermal energy from a body and turn it into useful work (such as charging a phone) is to exchange heat with it and a another colder body. So, with just the air at room temperature it is impossible to extract its heat energy.

7 0

3 years ago