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zzz [600]
3 years ago
11

Ethel throws a rock downward, and air resistance is negligible. Compared to a rock that is dropped, the acceleration of the thro

wn rock (after it is thrown) is
Physics
1 answer:
Scorpion4ik [409]3 years ago
3 0

The accelerations will be the same because the acceleration of each rock is caused only by gravity, which accelerates objects the same amount no matter what the velocity is.

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If a refrigerator draws 2.2 A at 120 V, What is its resistance?
Ierofanga [76]
Note that Ohm's law is given by
V = IR

Hence, V is 120V and I is 2.2A

Substitute the values:
120 = 2.2R

Therefore, R = 120/2.2
R = 54.5 Ohms
3 0
3 years ago
the radius of planet is drecresing by 50%without changing its mass . the acceleration due to gravity will be?​
tigry1 [53]

4 times more then real

Explanation:

if the radius of the earth decrease by 50% then the acceleration due to gravity increases by 4 times.

4 0
3 years ago
Technician A says that a proper U-joint inspection can be done with the driveshaft in place on the vehicle. Technician B says th
Julli [10]

Answer:

TECHNICIAN B

Explanation: Drive shaft is an elongated part of the drive system made of steel of a vehicle that transmits the <em>ENGINE TORQUE to the WHEELS</em> of a vehicle. For proper inspection a technician is expected to completely remove the driveshaft to check all its compactment. It is  Cylindrical and usually used for <em>REAR-WHEEL-VEHICLES</em>. Drive shafts are known by different names like propeller shaft,Cardan shaft etc it has different types which range from

<em>One-piece drive shaft ,Two-piece drive shaft  to Slip-in-tube drive shaft .</em>

6 0
3 years ago
What is the momentum of a 3 kg bowling ball moving at 3 m/s?
Nataly [62]

Explanation:

<h3>p = mv</h3>

  • <em>p</em> denotes momentum
  • <em>m</em> denotes mass
  • <em>v</em> denotes velocity

→ p = 3 kg × 3 m/s

→ <u>p</u><u> </u><u>=</u><u> </u><u>9</u><u> </u><u>kg</u><u>.</u><u>m</u><u>/</u><u>s</u>

<u>Option</u><u> </u><u>D</u><u> </u><u>is</u><u> </u><u>corre</u><u>ct</u><u>.</u>

5 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 ) \\\\

=( 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
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