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Paha777 [63]
2 years ago
5

What are the cues for a forehand stroke

Physics
1 answer:
MaRussiya [10]2 years ago
8 0

Answer:

huh

Explanation:

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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
2 years ago
Why isn't direct current used in transformers​
Furkat [3]

Answer:

No, it will not and this has a historical importance. The reason is that transformers work via induction of electrical forces by changes in magnetic fields, so the constat fields produced by dc currents won't work at all

Explanation:

4 0
2 years ago
Read 2 more answers
A plant dying after being exposed to poison represents a physical change.<br><br> True<br> False
nasty-shy [4]

Answer:

It does not represent a physical change

5 0
3 years ago
Read 2 more answers
How much heat must be removed from 200 pounds of blanched vegetables at 189 degrees F to freeze them to a temperature of 22 degr
Virty [35]

To solve this problem it is necessary to apply the concepts related to heat exchange in the vegetable and water.

By definition the exchange of heat is given by

Q =mc\Delta T

where,

m = mass

c = specific heat

\Delta T = Change in temperature

Therefore the total heat exchange is given as

\Delta Q = Q_w+Q_v

\Delta Q = m_wc_w(T_i-T_f)+m_vc_v(T_i-T_f)

Our values are given as,

Total mass is M_T = 200lb ,however the mass of solid vegetable and water is given as,

m_v= 0.4*200lb = 80lb

m_w=0.6*200lb=120lb

T_i = 183\°F\\T_f = 22\°F\\c_w = 1Btu/lbF\\c_v = 0.24Btu/lbF

Replacing at our equation we have,

\Delta Q = m_wc_w(T_i-T_f)+m_vc_v(T_i-T_f)

\Delta Q = (120)(1)(183-22)+(80)(0.24)(183-22)

\Delta Q = 22411.2Btu

Therefore the heat removed is 22411.2 Btu

6 0
2 years ago
A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow
Oksana_A [137]

Answer:

a)\omega=1.36rad/s

b)\omega=12.99rpm

c)F=705.6N

Explanation:

a) The angular velocity is related to the centripetal acceleration by the formula a_{cp}=\omega^2r, which for our purposes we will write as:

\omega=\sqrt{\frac{a_{cp}}{r}}

Since <em>we want this acceleration to be 1.5 times that due to gravity</em>, for our values we will have:

\omega=\sqrt{\frac{1.5g}{r}}=\sqrt{\frac{(1.5)(9.8m/s^2)}{(8m)}}=1.36rad/s

b) 1 rpm (revolution per minute) is equivalent to an angle of 2\pi radians in 60 seconds:

1\ rpm=\frac{2\pi rad}{60s} =\frac{\pi}{30}rad/s

Which means <em>we can use the conversion factor</em>:

\frac{1\ rpm}{\frac{\pi}{30}rad/s}=1

So we have (multiplying by the conversion factor, which is 1, not affecting anything but transforming our units):

\omega=1.36rad/s=1.36rad/s(\frac{1\ rpm}{\frac{\pi}{30}rad/s})=12.99rpm

c) The centripetal force will be given by Newton's 2nd Law F=ma, so on the centripetal direction for our values we have:

F=ma=(48kg)(1.5)(9.8m/s^2)=705.6N

8 0
2 years ago
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