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3 years ago

A car traveling 20 m/s to the north collides with a car traveling 10 m/s to the

Physics

2 answers:

Mrac [35]3 years ago

6 0

Answer:

0 kgm/s

Explanation:

a p e x

sergiy2304 [10]3 years ago

5 0

Answer: C

Explanation:

In collision, whether elastic or inelastic collisions, momentum is always conserved. That is, the momentum before collision will be equal to the momentum after collision.

Change in momentum of the system will be momentum after collision minus total momentum before collision.

Since momentum is a vector quantity, the direction will also be considered.

Momentum = MV - mU

Let

M = 800 kg is going north

at V = 20 m/s and the other car

m= 800 kg is going south

at U = 10m/s.

Substitute all the parameters into the formula

Momentum = (800 × 20) - (800 × 10)

= 8000 kgm/s

The final momentum after collision will also be equal to 8000 kgm/s

Change in momentum = 8000 - 8000

Change in momentum = 0

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3 years ago

A car passes point “A” and then 120 meters later. It’s velocity was measured 21 m/s. If it’s acceleration was constant at 0.853

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Recall that

${v_f}^2-{v_i}^2=2a\Delta x$

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So we have

$\left(21\dfrac{\rm m}{\rm s}\right)^2-{v_i}^2=2\left(0.853\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m)$

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3 years ago

A roller-coaster car has a mass of 1040 kg when fully loaded with passengers. As the car passes over the top of a circular hill

rusak2 [61]

Answer:

a.6373.5 N

b.-3837.6 N

Explanation:

Mass of roller coaster=m=1040 kg

Radius=r=24 ,

a. $v=9.4m/s$

Normal force= $F_N$

According to question

$mg-F_N=\frac{mv^2}{r}$

Where $g=9.81 m/s^2$

Substitute the values

$1040\times 9.81-F_N=\frac{1040\times (9.4)^2}{24}$

$10202.4-F_N=3828.9$

$F_N=10202.4-3828.9=6373.5 N$

b. $v=18m/s$

$g=9.81 m/s^2$

$1040\times 9.81-F_N=\frac{1040\times (18)^2}{24}$

$10202.4-F_N=14040$

$F_N=10202.4-14040=-3837.6 N$

8 0

4 years ago