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3 years ago

13

A 730 kg Mini runs into a stationary 2500 kg SUV. If the Mini was moving at 10. M , how s fast will it be moving after making a

completely inelastic collision with the SUV?

1 answer:

spayn [35]3 years ago

5 0

Answer:

2.26m/s

Explanation:

Since it is an inelastic collision, the mini and the Suv move together after the collision. The law of conservation of momentum states that the total momentum before and after the collision is conserved:

Using law of conservation of momentum,

$m_{1} u_{1}+m_{2} u_{2}=\left(m_{1}+m_{2}\right) v$

Given that,

$m_{1}$ = 730 kg

$u_{1}$ = 10 m/s

$u_{2}$ = 0

$m_2 =$ 2500kg

v = ?

When substituting the given values in the above equation, we get

$730(10) + 2500(0) = (730+2500)v\\\\7300 + 0 = 3230v\\\\v = \frac{7300}{3200} \\\\= 2.26m/s$

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3 years ago

An island reports that in the year 2000 there were 240 babies born. That same year, 100 individuals died.

lyudmila [28]

Answer:

+ 140

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Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

djverab [1.8K]

Answer:

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As we know from the conservation of mass, the rate at which any amount of fluid mass ( $m_{1}$ ) is entering in a system is equal to the rate at which the same amount of fluid mass ( $m_{2}$ ) is leaving the system.

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Now, according to the problem, as the density of the fluid does not change, we can write

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas through which the fluid is passing and $v_{1}$ and $v_{2}$ are the velocities of the fluid through the respective cross-sectional areas.

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3 years ago

You get hit in the head with a baseball as it is falling. What are the action and reaction forces?

german

Answer:

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Explanation:

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2 years ago

Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.

rjkz [21]

Answer:

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

Explanation:

Given

Before collision:

Momentum of block A = $p_{A1}$ = $-100\ kg\ ms^{-1}$

Momentum of block B = $p_{B1}$ = $-150\ kg\ ms^{-1}$

After collision:

Momentum of block A = $p_{A2}$ = $-200\ kg\ ms^{-1}$

Applying law of conservation of momentum to find momentum of block B after collision $p_{B2}$ .

$p_{A1}+p_{B1}=p_{A2}+p_{B2}$

Plugging in the given values and simplifying.

$-100-150=-200+p_{B2}$

$-250=-200+p_{B2}$

Adding 200 to both sides.

$200-250=-200+p_{B2}+200$

$-50=p_{B2}$

∴ $p_{B2}=-50\ kg\ ms^{-1}$

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

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3 years ago