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3 years ago

Can you answer this math homework? Please!

Physics

1 answer:

notka56 [123]3 years ago

5 0

$\large \mathfrak{Solution : }$

Average speed :

$\boxed{ \dfrac{distance \: \: covered}{total \: \: time \: \: taken} }$

$\dfrac{100}{8}$

$12.5 \: \: m/s$

The horizontal (flat) line on a Distance - time graph shows that the body is at rest because the value of Distance (on y - axis ) is same for increasing time ( on x - axis ), it basically shows that the object is at the same position with the change in time.

therefore, correct option is :

$a. \: \: the \: \: object \: \: is \: \: not \: \: moving$

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Whats the answer<br> ----------------------------

GenaCL600 [577]

Answer:

repel each other

Explanation:

The magnitude of the charge of an electron is called... ... If a positively-charged glass rod is suspended so that it turns easily and another positively-charged glass rod is brought close to it, the two rods will... Repel each other.

8 0

2 years ago

A space station, in the form of a wheel 140 m in diameter, rotates to provide an "artificial gravity" of 3.90 m/s2 for persons w

Zigmanuir [339]

Radial acceleration is given by

$a_{rad}= \frac{v^2}{r}$
where

$v=r \omega$
then

$a_{rad}= \frac{r^2 \omega^2}{r}=r\omega^2$

Now

$70\omega^2=3.90 \frac{m}{s^2} \\ \\ \omega= \sqrt{ \frac{3.9}{70} }$

Using the relation

$\omega=2 \pi f$

$2 \pi f= \sqrt{ \frac{3.9}{70} }\\ \\ f= \frac{1}{2 \pi}\sqrt{ \frac{3.9}{70} }Hz$

Putting into rpm

$\frac{60}{2 \pi}\sqrt{ \frac{3.9}{70}} =2.254rpm$

8 0

3 years ago

What type of energy does the box have after it is done being pulled?

Ksivusya [100]

Assuming that the box is moving when it is being pulled, Work is done on the box.

So work is the Force times the distance

W=Fd

But what is work actually ? When something moves due to force over some change in distance, it have energy.

But where does this energy come from ? Does it magically appear ? The energy comes from the applied force onto the box.

So the energy have been transferred. And it’s like that throughout the universe

Now to save time, I’ll just tell you the answer: kinetic energy

:)

3 0

3 years ago

(b) The distance of mass from mass A if there is no gravitational force acted on C

shepuryov [24]

Answer:

(a) The force, acting on object 'C' is approximately 2.66972 × 10⁻¹⁰ Newtons

(b) The distance of 'C' from 'A', in the direction particle 'B' if there is no meters gravitational force acting on 'C' is appromimately 0.829 meters or 1.877 meters

Explanation:

The given parameters are;

The mass of particle, A, m₁ = 2 kg

The mass of particle, B, m₂ = 0.3 kg

The mass of particle, C, m₃ = 0.05 kg

The distance between particle 'A' and particle 'B', r₁ = 0.15 m

The distance between particle 'B' and particle 'C', r₂ = 0.05 m

(a) The gravitational force, 'F', is given as follows;

$F =G \times \dfrac{m_{1} \times m_{2}}{r^{2}}$

Where;

F = The force between the two masses

G = The gravitation constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m₁ = The mass of object 1

m₂ = The mass of object 2

If 'C' is placed at 0.05 m from 'B', we have;

F₂₃ = 6.67430 × 10⁻¹¹ × 0.05 × 0.3/(0.05²) ≈ 4.00458 × 10⁻¹⁰

The gravitational force between force between particle 'B' and particle 'C', F₂₃ = 4.00458 × 10⁻¹⁰ N (towards the right)

F₁₃ = 6.67430 × 10⁻¹¹ × 0.05 × 2/(0.1²) ≈ × 10⁻¹⁰

The gravitational force between force between particle 'A' and particle 'B', F₁₃ = 6.6743 × 10⁻¹⁰ N (towards the left)

The force, 'F', acting on object 'C' = F₁₃ - F₂₃

F = (6.6743 - 4.00458) × 10⁻¹⁰ = 2.66972 × 10⁻¹⁰ N

The force, acting on object 'C' ≈ 2.66972 × 10⁻¹⁰ N

(b), When there is no gravitational force acting on 'C', let the distance of 'C' from 'A' = x

We have;

F₂₃ = F₁₂

$F_{23} =G \times \dfrac{m_{1} \times m_{2}}{r_1^{2}} = F_{13} =G \times \dfrac{m_{1} \times m_{3}}{r_2^{2}}$

By plugging in the values and removing like terms, we get;

$\dfrac{0.3 \times 0.05}{(1.15 - x)^{2}} = \dfrac{2 \times 0.05}{x^2}$

(1.15 - x)² × 2 × 0.05 = 0.3 × 0.05 × x²

0.1·x² - 0.23·x + 1.3225 = 0.015·x²

0.1·x² - 0.23·x + 1.3225 - 0.015·x² = 0

0.085·x² - 0.23·x + 0.13225= 0

x = (0.23± √((-0.23)² - 4 × 0.085 × ( 0.13225)))/(2 × 0.085))

x ≈ 0.829, or x ≈ 1.877

Therefore, the distance of 'C' from 'A', if there is no gravitational force acting on 'C', x ≈ 0.829 m, or x = 1.877 m, in the direction of 'B'

7 0

2 years ago

Juan measured the temperature of salt water. He then added 273 to the measured value. Which conversion is Juan most likely doing

Semmy [17]

If Juan used a Celsius thermometer, it would tell him the Celsius temperature.

If he added 273 to that number, he'd have the "absolute" or Kelvin temperature.

7 0

3 years ago