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JulijaS [17]
2 years ago
7

Look at the variables for this lab. Which variable would we have to change in order to change the amount of current flowing thro

ugh the wire?
Physics
1 answer:
____ [38]2 years ago
3 0
Where are the variables from the lab?
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A billiard ball with a mass of 1.5kg is moving at 25m/s and strikes a second ball with a mass of 2.3 kg that is motionless. Find
aliya0001 [1]

Answer:

The velocity of the second ball after the collision will be 16.3 m/s.

Step-by-Step Explanation:

Law of conservation of momentum states that the total momentum of an isolated remains the same before and after the collision.

Let: mass of the first ball = m1 = 1.5 kg

mass of the second ball = m2 =  2.3 kg

Momentum before the collision:

velocity of the first ball = v1 = 25 m/s

velocity of the second ball = v2 = 0 m/s

[tex] Initial momentum = m1v1 + m2v2 = 1.5*25 + 2.3*0 = 37.5 kgm/s [\tex]

Momentum after the collision:

velocity of the first ball = v1 = 0 m/s

velocity of the second ball = v2 = x

[tex]momentum after the collision= m1v1 + m2v2 = 1.5*0 + 2.3*x = 2.3*x[\tex]

According to law of conservation of momentum:

Initial momentum = Momentum after collision

37.5 = 2.3*x

⇒ x = 37.5/2.3

x = 16.3 m/s

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3 years ago
I have my finals exams and I have an activate exam paper, I wanna find it on the internet for free, does anyone know a website t
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3 years ago
Which feature of a human community is similar to a niche in a biological community?
pogonyaev

Answer: C. occupation

7 0
4 years ago
Which of these is emitted during beta decay ?
satela [25.4K]

Answer:

C. a small charged particle.

Explanation:

typically beta radiation emits an electron which is a small negativity charged particle.

hope it helps. :)

4 0
3 years ago
Read 2 more answers
A car starts from rest at a stop sign. It accelerates at 4.6 m/s^2 for 6.2 s , coasts for 2.1s , and then slows down at a rate o
ki77a [65]

Answer:

D = 271.54 m

Explanation:

given,

1. car accelerates at 4.6 m/s² for 6.2 s

2. constant speed for 2.1 s

3. slows down at 3.3 m/s²

distance travel for case 1

using equation of motion

 d_1 = u t +\dfrac{1}{2}at^2

 d_1 =\dfrac{1}{2}\times 4.6\times 6.2^2

      d₁ = 88.41 m

case 2

constant speed for 2.1 s now, we have to find velocity

v = u  + at

v = 0 + 4.6 x 6.2

v = 28.52 m/s

distance travel in case 2

d₂ = v x t

d₂ = 28.52 x 2.1 = 59.89 m

for case 3

distance travel by the car

v² = u² + 2 a s

final velocity if the car is zero

0² = 28.52² + 2 x (-3.3) x d₃

6.6 d₃ = 813.39

 d₃ = 123.24 m

total distance travel by the car

D = d₁ + d₂ + d₃

D = 88.41 + 59.89 + 123.24

D = 271.54 m

5 0
3 years ago
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