<h2>Question:</h2>
Appliances connected so that they form a single pathway for charges to flow are connected in a(n)
<h2>Answer:</h2>
<u>A</u><u>.</u><u> </u><u>Series</u><u> </u><u>circuit</u><u> </u>
<h2>
<u>#CARRYONLEARNING</u><u> </u></h2><h2>
<u>#STUDYWELL</u><u> </u></h2>
A billiard ball collides with a stationary identical billiard ball to make it move. If the collision is perfectly elastic, the first ball comes to rest after collision.
<h3>Why does the first ball comes to rest after collision ?</h3>
Let m be the mass of the two identical balls.
u1 = velocity before the collision of ball 1
u2 = 0 = velocity of second ball that is at rest
v1 and v2 are the velocities of the balls after the collision.
From the conservation of momentum,
∴ mu1 + mu2 = mv1 + mv2
∴ mu1 = mv1 + mv2
∴ u1 = v1 + v2
In an elastic collision, the kinetic energy of the system before and after collision remains same.

∴ 
∴ 
∴
₁
₂ = 0
- It is impossible for the mass to be zero.
- Because the second ball moves, velocity v2 cannot be zero.
- As a result, the velocity of the first ball, v1, is zero, indicating that it comes to rest after collision.
<h3>What is collision ?</h3>
An elastic collision is a collision between two bodies in which the total kinetic energy of the two bodies remains constant. There is no net transfer of kinetic energy into other forms such as heat, noise, or potential energy in an ideal, fully elastic collision.
Can learn more about elastic collision from brainly.com/question/12644900
#SPJ4
You do this one just like the other one that I just solved for you.
For this one ...
The density of the object is 2.5 gm/cm³.
We know that every cm³ of it we have contains 2.5 gm of mass.
We have to find out how many cm³ we have.
The question tells us: We have 2.0 cm³.
Each cm³ of space that the object occupies contains 2.5 gm of mass.
So the 2.0 cm³ that we have contains (2 x 2.5 gm) = 5 gms.
That's the mass of our object.
Answer:
The circular solar orbital speed at 4.0AU is 1/4( one fourth) that at 1.0AU
Explanation:
am = mvr= angular momentum
am4= 4mvt
am1= mvp1
Vt=1/4vp
Vp=4vt
am1= 4mvt
am1=am4
The circular solar orbital speed at 4.0AU is 1/4 (one fourth) that at 1.0AU