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leonid [27]
3 years ago
14

A bottle of milk has a mass of 1.3 kg what is the weight of the bottle of milk

Physics
2 answers:
tigry1 [53]3 years ago
8 0
"weight is a measurement of force, and force is measure in newtons. the equation for this is F=ma where m = 1.3 and a = 9.81 (assuming you're on Earth) F = 1.3*9.81 F = 12.753 N of force."

Got this answer from 
http://openstudy.com/updates/553aafc8e4b0a876f243a798

;)
erik [133]3 years ago
5 0

Answer:

Weight. W = 12.74 N

Explanation:

Given that,

Mass of bottle of milk, m = 1.3 kg

We have to find the weight of the bottle of milk. The weight of an object is given by the product of its mass and acceleration due to gravity i.e.

F = m g

g = acceleration due to gravity on Earth's surface

F = 1.3 kg × 9.8 m/s²

F = 12.74 N

Hence, the weight of the bottle of milk is 12.74 N.

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Explanation:

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Thus the number of atoms of each element must be same on both sides of the equation so as to keep the mass same and thus balanced chemical equations are written.

K exists as atoms and oxygen exist as molecule which consists of 2 atoms. The ratio of number of atoms on both sides of the reaction are same and thus the ratio of atoms of potassium to ratio of atoms of oxygen is 4:2.

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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

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Answer:

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Explanation:

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