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3 years ago

Ingrid kicks a football with an initial velocity at 12 m/s at an angle of 45 relative to the ground

1 answer:

5 0

Equation of movement is

y(x) = x * tg (a) - $\frac{g}{2 * u^{2} * cos (a)^{2} } * x^{2}$

where y is y-coordinate, x - x coordinate, u - your initial speed, a = angle above horizon.

differentiate it to get dy/dx (speed)

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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.

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

$=( 40.02\times10^{15})\left(\frac{3 \times 10^{5}}{ 4,228 \times10^{5}} \right ) \\\\ =40.02 \times10^{15} \times 0.0007 \\\\$

$\\\\ =0.02799\times10^{10}\,J \\\\= q^2\times31.35\times10^{9} \\\\ =0.02799\times10^{10} \\\\q=0.0945\,C$

This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.

6 0

3 years ago

Which transition represents a time when water molecules are moving closer together?

slava [35]

A. Is very attractive. If it's sublimation directly from water vapor in the air to ice on the glass, then yes. But from liquid water mist to water ice, no. Ice is less dense than water. That's why cubes float in your soda. Better leave 'A' alone. . . . D. Ice pellets turn to liquid. That one's good.

8 0

3 years ago

Read 2 more answers

To determine the height of a tall building such as Sears Tower in Chicago, Illinois a ball was dropped from the top of the build

Darya [45]

Answer:

The height of Sears Tower is 1448.5 feet.

Explanation:

<h3>We apply the free fall formula to the ball: </h3><h3> $y=v_{o} *t+\frac{1}{2} *g*t^{2}$ </h3><h3>y: The vertical distance the ball moves at time t </h3><h3> $v_{o}$ i: Initial speed </h3><h3>g=Gravity acceleration= $9.8*(\frac{\frac{1ft}{0.305m} }{s^{2} } )$ </h3>

Known information

We know that the vertical distance (y) that the ball moves in 9,5s is equal to height of Sears Tower (h).

Too we know that the ball is released from rest, then, $v_{0}$ =0

Height of Sears Tower calculation:

We replace in the equation 1 the data following;

$y=h$

$v_{o} =0$

$g=32,1\frac{ft}{s^{2} }$

$t= 9,5s$

$h=0*9.5+\frac{1}{2} *32.1*9.5^{2}$

$h=1448.5 ft$

Answer: The height of Sears Tower is 1448.5 ft

6 0

3 years ago

How does your power output in climbing the stairs compare to the power output of a 100-watt light bulb? if your power could have

cricket20 [7]

1) Assuming an adult person has an average mass of m=80 kg, and assuming it takes about 30 seconds to climb 5 meters of stairs, the energy used by the person is
$E=mgh=(80 kg)(9.81 m/s^2)(5 m)=3924 J$
So the power output is
$P= \frac{E}{t}= \frac{3924 J}{30 s} \sim 130 W$

And since the estimate we made is very rough, we can say that the power output of the person is comparable to the power output of the light bulb of 100 W.

2) Based on the results we found in the previous part of the exercise, since the power output of the person is comparable to the power output of 1 light bulb of 100 W, we can say that the person could have kept burning only one 100-W light bulb during the climb.

6 0

3 years ago

Read 2 more answers

An airplane is flying in the direction 10° east of south at 701 km/hr. Find the component form ofthe velocity of the airplane, a

solniwko [45]

Answer:

The component form will be;

In the x-axis = 121.73 due west

In the y-axis = 690.35 due south

Explanation:

An image of the calculation has been attached

7 0

2 years ago