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3 years ago

True or false ??

Physics

2 answers:

galben [10]3 years ago

6 0

Well, gravity's acceleration is -9.81 m/s^2 so false.

Vinvika [58]3 years ago

3 0

False, because objects have different masses. Remember, mass determines acceleration

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An object ends up at a final position of x=-55.25 meters after a displacement of -189.34 meters after a displacement of -189.34

Travka [436]

The initial position of the object was found to be 134.09 m.

Explanation:

As displacement is the measure of difference between the final and initial points. In other words, we can say that displacement can be termed as the change in the position of the object irrespective of the path followed by the object to change the path. So

Displacement = Final position - Initial position.

As the final position is stated as -55.25 meters and the displacement is also stated as -189.34 meters. So the initial position will be

Initial position of the object = Final position-Displacement

Initial position = -55.25 m - (-189.34 m) = -55.25 m + 189.34 m = 134.09 m.

Thus, the initial position for the object having a displacement of -189.34 m is determined as 134.09 m.

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3 years ago

Suppose you have two identical sheets of paper and you crumple one of the sheets of paper into a ball. If you drop the crumpled

sammy [17]

Answer:

Because it can easily resist air resistance.

Explanation:

Since air resistance is not negligible, the crumpled paper will reach the ground first because it can easily resist air resistance surrounding it compare to the un-crumpled one that will be influenced by the air thereby causing the un-crumpled paper to spend more time in the air

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3 years ago

When a wave is bent by traveling from one medium to another

Natalka [10]

Answer:

B. Refraction

Explanation:

Jope this helps

6 0

3 years ago

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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.

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