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3 years ago

There are some limiting factors that will completely prevent you from exercising regularly.

Physics

2 answers:

Serjik [45]3 years ago

6 0

I am pretty sure it is false

natka813 [3]3 years ago

4 0

True cause what if you have a disability

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A certain aircraft can fly 612 miles with the wind in 3 hours and travel the same distance against the wind in 6 hours. What is

Andreas93 [3]

Answer:

51 mph

Explanation:

Given: A certain aircraft can fly 612 miles with the wind in 3 hours and travel the same distance against the wind in 6 hours.

Formula: $Speed=\dfrac{Distance}{Time}$

Let speed of aircraft be x mph and speed of wind y mph

Aircraft along wind:

Actual speed of aircraft = x + y

Time taken = 3 hours

Distance = 612 miles

$\therefore x+y=\dfrac{612}{3}$

$x+y=204----------(1)$

Aircraft against wind:

Actual speed of aircraft = x - y

Time taken = 6 hours

Distance = 612 miles

$\therefore x-y=\dfrac{612}{6}$

$x-y=102----------(2)$

By solving equation (1) and equation (2) and we get

x = 153 mph

y = 51 mph

Hence, The speed of wind will be 51 mph

6 0

3 years ago

What unit is commonly used when measuring the energy rating of an electrical appliance

lapo4ka [179]

usually it is watts

8 0

3 years ago

Read 2 more answers

A speedy rabbit is hopping to the right with a velocity of 4.0 \,\dfrac{\text m}{\text s}4.0 s m 4, point, 0, start fraction,

Mrrafil [7]

Answer: 38.25 m

Explanation:

In this situation we need to find the distance $d$ between the rabbit and the carrot, and we can use the following equation, since the rabbit's acceleration is constant:

$V^{2}=V_{o}^{2} + 2ad$ (1)

Where:

$V=13 m/s$ is the rabbit's maximum velocity (final velocity)

$V_{o}=4 m/s$ is the rabbit's initial velocity

$a=2 m/s^{2}$ is the rabbit's acceleration

$d$ is the distance between the rabbit and the carrot

Isolating $d$ :

$d=\frac{V^{2}-V_{o}^{2}}{2a}$ (2)

$d=\frac{(13 m/s)^{2}-(4 m/s)^{2}}{2(2 m/s^{2})}$ (3)

Finally:

$d=38.25 m$

8 0

4 years ago

Read 2 more answers

NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA

kkurt [141]

Answer:

The answer is " $q=0.0945\,C$ ".

Explanation:

Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

$U_i=\frac{1}{4\pi\epsilon_0} \times \frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}$

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

$U_f=\frac{1}{4\pi\epsilon_0}\ \times \frac{4\times4q^2}{( \frac{5}{\sqrt{2}})}$

Potential energy shifts:

$= U_f -U_i \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+( \frac{5}{\sqrt{2})^2)}}\right ) \\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ 15 +( \frac{5}{2})}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{30+5}{2})}}\right )\\\\$

$=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{ (\frac{35}{2})}}\right )\\\\=\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{17.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 24.74- 5 }{87.5}}\right )\\\\ =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{ 19.74- 5 }{87.5}}\right )\\\\ =\frac{4q^2}{\pi\epsilon_0}\left ( 0.2256 }\right )\\\\= \frac{0.28 \times q^2}{ \epsilon_0}\\\\=q^2\times31.35 \times10^9\,J$

Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.

$=\frac{GMm}{R}-\frac{GMm}{R+h} \\\\=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right ) \\\\ =(6.67\times10^{-11}\times6.0\times10^{26})\left(\frac{1}{64\times10^{5}}-\frac{1}{67\times10^{5}} \right ) \\\\=(6.67\times6.0\times10^{15})\left(\frac{67 \times 10^{5} - 64 \times 10^{5} }{ 4,228 \times10^{5}} \right ) \\\\$