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3 years ago

Which is the correct answer ?

Physics

2 answers:

GarryVolchara [31]3 years ago

8 0

The answer is A, direction of flow.

hodyreva [135]3 years ago

7 0

D ............................

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Jason and Guy are throwing watermelons straight up from the ground. (They’re making fruit salad.) Jason throws his watermelon at

NISA [10]

Answer:2t

Explanation:

Given

Jason throwing watermelon at speed v and it spend t time in air

time to reach max height

$v_f=u_i+at'$

0=v-gt'

$t'=\frac{v}{g}$

and time to reach bottom is also t'

t=2t'

$t=2\frac{v}{g}$

For guy throws his watermelon at speed 2v

So total time in air will be

$t''=2\frac{2v}{g}$

$t''=4\frac{v}{g}$

t''=2t

5 0

3 years ago

A man standing on a bus remains still when the bus is at rest. When the bus moves forward and then

Hoochie [10]

This is an example of inertia - the body keeps it's energy because there is no force applied to it. When we try to stop it's motion, it resists. A man is not rigidly attached to the bus, so he keeps moving forward, at least until he hits the front window from inside. Answer is D.

6 0

3 years ago

A disoriented physics professor drives 3.22 km north, then 4.75 km west, then 1.90 km south. find the magnitude and direction of

Andre45 [30]

It can be noted that the directions west and north are perpendicular to each other. hence we can use the right angle formula for calculating total distance in these directions. Also South is just negative north. This technique is called euclidean distance norm. It is the basis of Cartesian coordinate system.

Total distance = 3.22 N + 4.75 W +1.90 S

= 1.32 N + 4.75 W

D = $\sqrt{N^2 + W^2}$

D = 4.93 Km

direction = atan(N/W) = 15.53 deg north of west

7 0

3 years ago

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.

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