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4 years ago

What can affect gravity Helpppppp!!!

1 answer:

7 0

gravity increases with height. gravity is significantly less on high mountains or tall buildings and increases as we lose height (which is why falling objects speed up) gravity is caused by the Earth spinning. gravity affects things while they are falling but stops when they reach the ground

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Convert Gravitational constant (G) = 6.67×10^-11 Nm²kg^-2 to cm³ g ^-1 s^-2.

GaryK [48]

Answer:

6.67×10⁻⁸ cm³/g/s²

Explanation:

6.67×10⁻¹¹ Nm²/kg²

= 6.67×10⁻¹¹ (kg m/s²) m²/kg²

= 6.67×10⁻¹¹ m³/kg/s²

= 6.67×10⁻¹¹ m³/kg/s² × (100 cm/m)³ × (1 kg / 1000 g)

= 6.67×10⁻⁸ cm³/g/s²

3 0

3 years ago

A major limitation of using photovoltaic cells to generate electricity is that they

Marianna [84]

Answer:

Explained below:

Explanation:

The major limitation in the use of Photovoltaic cells to generate electricity is its unpredictability about the climate that whether it will be a shining day or a cloudy day and it does not work at night also because of no presence of sunlight as it totally depends on it. It can be used as an alternative but cannot be used as chief source of electricity.

8 0

3 years ago

A charge of −20 µC is distributed uniformly over the surface of a spherical conductor of radius 11.0 cm. Determine the electric

Alex73 [517]

Answer:

(a) $-6.76\times 10^{12}\ N/C$

(b) $-1.352\times 10^{13}\ N/C$

(c) $-7.2\times 10^{11}\ N/C$

Explanation:

(a)

Given:

Charge on sphere (Q) = $-20\ \mu C=-20\times 10^{-6}\ C$

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 5 cm = 0.05 m

Coulomb's constant (k) = $9\times 10^{9}\ Nm^2/C^2$

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:

$E=(\frac{kQ}{R^3})r$

Plug in the given values and solve for 'E'. This gives,

$E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.05\\\\E_{in}=-1.352\times 10^{14}\times 0.05\\\\E_{in}=-6.76\times 10^{12}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)$

(b)

Given:

Charge on sphere (Q) = $-20\ \mu C=-20\times 10^{-6}\ C$

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 10 cm = 0.10 m

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:

$E=(\frac{kQ}{R^3})r$

Plug in the given values and solve for 'E'. This gives,

$E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.10\\\\E_{in}=-1.352\times 10^{14}\times 0.10\\\\E_{in}=-1.352\times 10^{13}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)$