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Trava [24]
3 years ago
6

What is a vector quantity?​

Physics
1 answer:
Nat2105 [25]3 years ago
7 0

Answer:

A quantity that has magnitude and direction. It's usually represented by an arrow whose direction is the same direction is the same as that of the quantity and whose length is proportional to the quantity's magnitude

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Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. The initial v
wlad13 [49]
1) In a perfectly inelastic collision, the two balls stick together after the collision. In this type of collision, the total kinetic energy of the system is not conserved, while the total momentum is conserved.
If we callv_f the final velocity of the two balls that stick together, the conservation of the total momentum before and after the collision can be written as
m_a v_{Ai} + m_b v_{Bi} = (m_A+m_B)v_f (1)
where
m_A=7 kg is the mass of ball A
m_B=2 kg is the mass of ball B
v_{Ai}=6 m/s is the initial velocity of ball A
v_{Bi}=-12 m/s is the initial velocity of ball B (taken with a negative sign, since it goes in the opposite direction of ball A)

If we solve (1) to find v_f, we find that the final velocity of the balls is
v_f= \frac{m_Av_{Ai}+m_Bv_{Bi}}{m_A+m_B}= \frac{(7\cdot 6)+(2 \cdot (- 12))}{7+2}= \frac{18}{9}=2 m/s
and the positive sign means the two balls are going to the right.


2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
m_A v_{Ai}+m_B v_{Bi} = m_A v_{fA} + m_B v_{fB}
\frac{1}{2}m_A v_{Ai}^2+ \frac{1}{2}m_B v_{Bi}^2= \frac{1}{2}m_Av_{fA}^2+ \frac{1}{2}m_B v_{fB}^2
where
v_{fA} is the final velocity of ball A
v_{fB} is the final velocity of ball B

If we solve simultaneously the two equations, we find:
v_{fA}= \frac{v_{Ai}(m_A-m_B)+2m_Bv_{Bi}}{m_A+m_B} = \frac{(6)(7-2)+2(2)(-12)}{7+2}=-2 m/s
v_{fB}= \frac{v_{Bi}(m_B-m_A)+2m_Av_{Ai}}{m_A+m_B} = \frac{(-12)(2-7)+2(7)(6)}{7+2}= \frac{144}{9}=16 m/s
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.

3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
p_i = m_A v_{A} + m_B v_{fB} = (7\cdot 6)+(2 \cdot (-12))=42-24=18 m/s
and the total momentum after the collision is:
p_f = m_A v_{A} + m_B v_{fB} = (7\cdot (-2))+(2 \cdot 16)=-14+32=18 m/s

3 0
3 years ago
The acquisition of electric charge without contact between charged and/or uncharged substances.
Lera25 [3.4K]

Answer:

induction

Explanation:

I just did usatestprep

4 0
3 years ago
Read 2 more answers
2 Points<br> Which of the following are types of matter?
Hatshy [7]
What are the options?
4 0
3 years ago
A ball rolls of buildings that is 100m high calculate the time that it takes for ball to hit the ground​
LUCKY_DIMON [66]

Answer:

2as=v2-u2

2000=v2

V=44

V=u+at

44/10=t

T=4.4seconds

5 0
3 years ago
A player holds two baseballs a height h above the ground. He throws one ball vertically upward at speed v0 and the other vertica
Degger [83]

Answer:

a)  v = √(v₀² + 2g h),    b)      Δt = 2 v₀ / g

Explanation:

For this exercise we will use the mathematical expressions, where the directional towards at is considered positive.

The velocity of each ball is

ball 1. thrown upwards vo is positive

        v² = v₀² - 2 g (y-y₀)

in this case the height y is zero and the height i = h

        v = √(v₀² + 2g h)

ball 2 thrown down, in this case vo is negative

         v = √(v₀² + 2g h)

The times to get to the ground

ball 1

         v = v₀ - g t₁

         t₁ = \frac{v_{o}  - v }{ g}

ball 2

         v =  -v₀ - g t₂

         t₂ = -  \frac{v_{o}  + v }{ g}  

From the previous part, we saw that the speeds of the two balls are the same when reaching the ground, so the time difference is

       Δt = t₂ -t₁

       Δt = \frac{1}{g} \ [(v_{o} - v)  - ( - v_{o}  - v) ]

       Δt = 2 v₀ / g

6 0
3 years ago
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