A 5,400 W motor is used to do work. If the motor is used for 640 s, about how much work could it do?

I started to solve this problem but I’m not sure on what to do next.

snow_tiger [21]

ANSWER:

3408.81 kg

STEP-BY-STEP EXPLANATION:

Given:

v = 111 m/s

Ek = 21000000 J

We have that the formula for kinetic energy is as follows:

$E_k=\frac{1}{2}\cdot m\cdot v^2$

We substitute the values given in the exercise and solve for m (mass)

$\begin{gathered} 21000000=\frac{1}{2}\cdot m\cdot111^2 \\ m=\frac{21000000\cdot2}{111^2} \\ m=3408.81\text{ kg} \end{gathered}$

The mass of the helicopter is 3408.81 kilograms.

7 0

1 year ago

I will give Brainliest to whoever can answer!!!!

Alexus [3.1K]

Answer:

The nodes and anti nodes would reverse roles.

Explanation:

I believe it has to do with the path differences. If waves are in phase, then the path differences are such that the waves reach the screen with crests superimposing crests and troughs superimposing troughs. This happens when the periods of each wave are equal or the paths themselves differ by a whole number multiple of the wavelength (λ, 2λ, 3λ, ...).

Now make these waves out of phase. Then half of the waves will travel half a wavelength farther than the rest. So the path difference will be 0.5λ, 1.5λ, 2.5λ, ....

4 0

3 years ago

Mars has two moons, Phobos and Deimos. Phobos orbits Mars at a distance of 9380 km from Mars's center, while Deimos orbits at 23

Sloan [31]

Answer:

The ratio is $\frac{T_1}{T_2} = 3.965$

Explanation:

From the question we are told that

The radius of Phobos orbit is R_2 = 9380 km

The radius of Deimos orbit is $R_1 = 23500 \ km$

Generally from Kepler's third law

$T^2 = \frac{ 4 * \pi^2 * R^3}{G * M }$

Here M is the mass of Mars which is constant

G is the gravitational constant

So we see that $\frac{ 4 * \pi^2 }{G * M } = constant$

$T^2 = R^3 * constant$

=> $[\frac{T_1}{T_2} ]^2 = [\frac{R_1}{R_2} ]^3$

Here $T_1$ is the period of Deimos

and $T_1$ is the period of Phobos

So

$[\frac{T_1}{T_2} ] = [\frac{R_1}{R_2} ]^{\frac{3}{2}}$

=> $\frac{T_1}{T_2} = [\frac{23500 }{9380} ]^{\frac{3}{2}}]$

=> $\frac{T_1}{T_2} = 3.965$

8 0

3 years ago

Why do noble gasses rarely react with other elements?

ollegr [7]

Noble gasses have an outer shell full of electrons. A full outer energy level is the most stable arrangement of electrons. As a result, noble gases cannot become more stable by reacting with other elements and gaining or losing valence electrons. Therefore, noble gases are rarely involved in chemical reactions and almost never form compounds with other elements.

3 0

2 years ago

If element x has 44 protons how many electrons does it have?

Jlenok [28]

The correct answer is '44'.

In fact, the atom of an element has an equal number of protons and electrons: therefore, if element X has 44 protons, it should have 44 electrons as well.

For curiosity: the chemical element with 44 protons and 33 electrons is Ruthenium.

5 0

3 years ago