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dimaraw [331]
3 years ago
11

Does the distance between two charged objects affect the force of attraction?

Physics
1 answer:
miv72 [106K]3 years ago
5 0

Answer:

Both forces act along the line joining the objects like masses or charges. And both forces are inversely proportional to the square of the distance between the objects, this is known as the inverse-square law.

Explanation:

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I really don’t get this one
worty [1.4K]

Compounds are molecules with 2 or more elements

So the answer would be the third one

CO2;H2O

4 0
3 years ago
Two wheel gears are connected by a chain. The larger gear has a radius of 8 centimeters and the smaller gear has a radius of 3 c
dezoksy [38]

Answer:

B.The linear velocity of the gears is the same. The linear velocity is 432π centimeters per minute.

Explanation:

As we know that the small gear completes 24 revolutions in 20 seconds

so the angular speed of the smaller gear is given as

\omega = 2\pi\frac{24}{20}

\omega = 2.4\pi rad/s

Now we know that the tangential speed of the chain is given as

v = r \omega

so we have

v = (3 cm)(2.4\pi)

v = 7.2 \pi cm/s

v = 432\pi cm/min

Since both gears are connected by same chain so both have same linear speed and hence correct answer will be

B.The linear velocity of the gears is the same. The linear velocity is 432π centimeters per minute.

8 0
3 years ago
According to research, college graduates are more likely to engage in all of the following EXCEPT:
Ksivusya [100]
Strange question. I’d say B)
7 0
3 years ago
Read 2 more answers
If a 2 cm vector represents speed, and 1 cm equals 5 m/s, how fast is the represented
Ne4ueva [31]

Answer: 10m/s

Explanation: Since a single vector with length of 1cm expresses 5m/s, the vector which has a length doubled from the original vector should have the speed which is also doubled.

6 0
3 years ago
A typical neutron star may have a mass equal to that of the Sun but a radius of only 20 km.(a) What is the gravitational acceler
Pepsi [2]

Answer:

331665750000\ m/s^2

3257806.62409 m/s

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of Sun = 1.989\times 10^{30}\ kg

r = Radius of Star = 20 km

u = Initial velocity = 0

v = Final velocity

s = Displacement = 16 m

a = Acceleration

Gravitational acceleration is given by

g=\dfrac{GM}{r^2}\\\Rightarrow g=\dfrac{6.67\times 10^{-11}\times 1.989\times 10^{30}}{20000^2}\\\Rightarrow g=331665750000\ m/s^2

The gravitational acceleration at the surface of such a star is 331665750000\ m/s^2

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 331665750000\times 16+0^2}\\\Rightarrow v=3257806.62409\ m/s

The velocity of the object would be 3257806.62409 m/s

8 0
3 years ago
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