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djyliett [7]
3 years ago
11

How much gravitational force do two lead balls with a mass of 8 kilograms, the centers of mass of which are 17 cm apart, affect

each other?
Physics
1 answer:
Arisa [49]3 years ago
7 0

Answer:

1.48×10⁻⁷ Newtons

Explanation:

From the question,

According to newton's law of universal gravitation.

F = Gmm'/r²........................ Equation 1

F = gravitational force, G = gravitational constant, m = mass of the first ball, m' = mass of the second ball, r = distance between the balls.

Given: m = m' = 8 kg, r = 17 cm = 0.17 m,

Constant : G = 6.67×10⁻¹¹ Nm²/kg²

Substitute these values into equation 1

F =  (6.67×10⁻¹¹×8×8)/(0.17²)

F = 1.48×10⁻⁷ N

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Natali5045456 [20]

Answer:

Radio stations have dipole type antennas

this field increases in intensity and propagates outwards,

Explanation:

Radio stations have dipole type antennas, that is, all sides are isolated from each other, when the AC signal from the radio station arrives, the lcharge begins at times and by the Lens law a field appears that opposes this movement, this field increases in intensity and propagates outwards, when the voltage reaches a maximum, the generated wave also reaches the maximum, now the incident wave begins to decrease, an electric hand appears to oppose this prisoner, and in this way a cap is created. electric .

5 0
4 years ago
A boat sails along the shore. To an observer, the boat appears to move at a speed of 22 m/s, and a man on the boat walking forwa
seropon [69]
The boat is moving at 22 m/s while the man is moving at 23.1 m/s.

That means the man, relative to the boat, is moving at 23.1-22 = 1.1 m/s.

v =d/t, so t = d/v --> t = 3/1.1 = 2.7 s
7 0
3 years ago
What is the frequency of an x- ray if the wavelength is 4.5 E - 10m?
AnnZ [28]
V (speed) = F (frequency) x Wavelength
If we rearrange the formula, making frequency the subject;
F (frequency) = Speed ÷ Wavelength
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4 0
3 years ago
When all group iia elements lose their valence electrons, the remaining electron configurations are the same as for what family
Orlov [11]
Noble Gases

I hope I helped

ΩΩΩΩΩΩΩΩΩΩ

8 0
3 years ago
Read 2 more answers
A very long straight wire has charge per unit length 1.44×10-10C/m.
4vir4ik [10]

Answer:

Distance of the point where electric filed is 2.45 N/C is 1.06 m            

Explanation:

We have given charge per unit length, that is liner charge density \lambda =1.44\times 10^{-10}C/m

Electric field E = 2.45 N/C

We have to find the distance at which electric field is 2.45 N/C

We know that electric field due to linear charge is equal to

E=\frac{\lambda }{2\pi \epsilon _0r}, here \lambda is linear charge density and r is distance of the point where we have to find the electric field

So 2.45=\frac{1.44\times 10^{-10} }{2\times 3.14\times 8.85\times 10^{-12}\times r}

r = 1.06 m

So distance of the point where electric filed is 2.45 N/C is 1.06 m

3 0
3 years ago
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