Balance in the water cycle means what two are equal

Gravity is the force that keeps us on the Earth. It pulls us towards the center of the Earth. If you were to move from the surfa

Nikitich [7]

<h2>Answer: B. Gravitational potential energy </h2>

Explanation:

<em>The gravitational potential energy is the energy that a body or object possesses, due to its position in a gravitational field. </em>

That is why this energy depends on the relative height of an object with respect to some point of reference and associated with the gravitational force.

In the case of the <u>Earth</u>, in which <u>the gravitational field is considered constant</u>, the value of the gravitational potential energy $U_{p}$ will be:

$U_{p}=mgh$

Where $m$ is the mass of the object, $g$ the acceleration due gravity and $h$ the height of the object.

As we can see, the value of $U_{p}$ is directly proportional to the height.

6 0

3 years ago

The force F⃗ pulling the string is constant; therefore the magnitude of the angular acceleration α of the wheel is constant for

abruzzese [7]

Answer:

The answer is " $\boxed{\boxed{\omega = \sqrt{\frac{2fd}{kmr^2}}}}$ "

Explanation:

$\to d= r \theta \\\\ \to \theta =\frac{d}{r}\\\\\to \omega^{r} - \omega_{0}^{r} = 2 \alpha \theta\\\\\to \omega^{r} = 2 \alpha \theta - \omega_{0}^{r} \\\\\to \omega^{r} = 2 (\frac{F}{Kmr}) \frac{d}{r}\\\\\to \omega = \sqrt{\frac{2fd}{kmr^2}}$

5 0

3 years ago

Your answer should be precise to 0.1 m/s. Use a gravitational acceleration of 10 m/s/s. At it lowest point, a pendulum is moving

saw5 [17]

Explanation:

It is given that,

Speed, v₁ = 7.7 m/s

We need to find the velocity after it has risen 1 meter above the lowest point. Let it is given by v₂. Using the conservation of energy as :

$\dfrac{1}{2}mv_1^2=\dfrac{1}{2}mv_2^2+mgh$

$v_2^2=v_1^2-2gh$

$v_2^2=(7.7)^2-2\times 10\times 1$

$v_2=6.26\ m/s$

So, the velocity after it has risen 1 meter above the lowest point is 6.26 m/s. Hence, this is the required solution.

4 0

3 years ago

Read 2 more answers

А A pool of water of refractive index

babymother [125]

Answer:

Apparent depth = 45 cm

Explanation:

The refractive index of water in a pool, n = 4/3

Real depth, d = 60 cm

We need to find its apparent depth when viewed vertically through air. The ratio of real depth to the apparent depth is equal to the refractive index of the material. Let the apparent depth is d'. So,

$n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{60}{\dfrac{4}{3}}\\\\d'=45\ cm$

So, the apparent depth is 45 cm.

3 0

2 years ago

An unstable atomic nucleus has a mass of 17.010-27kg, and starts out at rest. When it decays, it the original nucleus disintegra

slega [8]

Answer:

Part a)

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

$E = 4.4 \times 10^{-13} J$

Explanation:

As per momentum conservation we know that there is no external force on this system so initial and final momentum must be same

So we will have

$m_1v_1 + m_2v_2 + m_3v_3 = 0$

$(5 \times 10^{-27})(6 \times 10^6\hat j) + (8.4 \times 10^{-27})(4 \times 10^6\hat i) + (3.6 \times 10^{-27}) v = 0$

$(30\hat j + 33.6\hat i)\times 10^6 + 3.6 v = 0$

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

By equation of kinetic energy we have

$E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2$

$E = \frac{1}{2}(5 \times 10^{-27})(6\times 10^6)^2 + \frac{1}{2}(8.4 \times 10^{-27})(4 \times 10^6)^2 + \frac{1}{2}(3.6 \times 10^{-27})(8.33^2 + 9.33^2) \times 10^{12}$

$E = 9\times 10^{-14} + 6.72 \times 10^{-14} + 2.82\times 10^{-13}$

$E = 4.4 \times 10^{-13} J$

8 0

2 years ago