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kotykmax [81]
3 years ago
8

The valence electron occur in what part of the atom

Physics
1 answer:
Burka [1]3 years ago
5 0
<span>The valence electrons occur in the outer shell of the atom. </span>
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An object of mass 8kg is attached to massless string of length 2m and swum with a tangential velocity of 3 what is the tension o
Paul [167]

Answer:

36 N

Explanation:

If the object of mass, m = 8 kg is swung in a horizontal circle of radius, r = 2m = length of string with tangential velocity v = 3 m/s, the tension in the string is the centripetal force which is T = mv²/r

= 8 kg × (3 m/s)²/2 m

= 4 kg × 9 m/s²

= 36 N

6 0
3 years ago
A dragster runs the quarter mile in 8.96 s. What is the car's velocity (in ft/s) at the finish line?
lara [203]
The first thing you should know to answer this question is the following conversion:
 1mi = 5280feet
 We have then that the speed is:
 v = ((1/4) * (5280)) / (8.96)
 v = 147.32 feet / s
 Answer:
 the car's velocity (in ft / s) at the finish line is 147.32 feet / s
7 0
3 years ago
Which of the following quantities is inversely proportional to the gravitational pull between two objects?
Contact [7]

Answer:

C

Explanation:

Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces

I hope this helps a little bit

7 0
2 years ago
Read 2 more answers
Please help ASAP! Willing to give brainliest.
poizon [28]

Answer: B

Explanation:the voltage is just like the force that drives the current through out the circui... When trippled, the force increases and the current increases since the resistance in the circuit remains constant.

3 0
3 years ago
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Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103
son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

PE = -\frac{GMm}{R}

As M=m, then

PE = -\frac{Gm^2}{R}

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

U = -\frac{2Gm^2}{R}

dU = -\frac{2Gm^2}{R}

dKE = -dU

Therefore replacing we have that,

\frac{1}{2}mv^2 =\frac{Gm^2}{2R}

Re-arrange to find v,

v= \sqrt{\frac{Gm}{R}}

v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}

v = 816.7m/s

Therefore the  velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

2r = 2*10^3m

Therefore

dU = Gm^2(\frac{1}{R}-\frac{1}{2r})

v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}

v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}

v = 1.83*10^7m/s

Therefore the velocity when they are about to collide is 1.83*10^7m/s

7 0
3 years ago
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