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4 years ago

What happens to the momentum if you decrease an object's velocity?

1 answer:

4 0

The momentum of an object can be expressed as: p = mv, so if you were to decrease an objects velocity, it would decrease the momentum of that object as well.

Hope this helps!

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Arrange the balls in order from greatest amount of gravitational potential energy to least

Aloiza [94]

F e d b c a
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3 years ago

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What’s the answer to this problem ? with work pls !!

mars1129 [50]

This is the answer to Question 5

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3 years ago

A 0.144-kg baseball is moving toward home plate with a speed of 43 m/s when

algol [13]

I would say 648858. bc yes

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3 years ago

A 23.5 g piece of aluminum metal is initially at 100.0°C. It is dropped into a coffee cup-calorimeter containing 130.0 g of wate

vivado [14]

Answer: The molar heat capacity of aluminum is $25.3J/mol^0C$

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of water = 130.0 g

$m_2$ = mass of aluminiunm = 23.5 g

$T_{final}$ = final temperature = $26.0^oC=(273+26)K=299K$

$T_1$ = temperature of water = $23^oC=(273+23)K=296K$

$T_2$ = temperature of aluminium = $100^oC=273+100=373K$

$c_1$ = specific heat of water= $4.184J/g^0C$

$c_2$ = specific heat of aluminium= ?

Now put all the given values in equation (1), we get

$130.0\times 4.184\times (299-296)=-[23.5\times c_2\times (299-373)]$

$c_2=0.938J/g^0C$

Molar mass of Aluminium = 27 g/mol

Thus molar heat capacity = $0.938J/g^0C\times 27g/mol=25.3J/mol^0C$

5 0

4 years ago

A bullet with a mass of 5 gramsand speed of 560 m/sis fired horizontally atablock of wood with a mass of 2 kg. The block rests o

Anni [7]

Momentum is conserved if and only if sum of all forces which are exserted on system equals zero. In our situation there are only internal forces, so by Newton's third law their vector sum is 0.

So $mv=(m+M)v' \Leftrightarrow v'=\frac{mv}{m+M}$ .

Kinetic energy of system at first: $\frac{mv^2}{2}=784\;\textbf{J}$ . After: $\frac{(m+M)\frac{m^2v^2}{(m+M)^2} }{2}=\frac{m^2v^2}{2(m+M)}\approx 1,96\; \textbf{J}$ . The secret is that other energy is in work of deformation forces (they in turn heat a bullet and a block).

Answer is A)

5 0

3 years ago