A 4.0-kg cart moving due east at 9.0 m/s collides with a 3.0-kg cart moving due west. The carts stick together and come to rest after the collision. What was the initial speed of the 3.0-kg cart?

Answers

Answer 1

Answer:

Since the carts stick together and come to rest after the collision, the initial speed of the 3.0-kg cart is 12 m/s.

Given the following data:

Mass of cart 1 = 4.0 kg

Velocity of cart 1 = 9.0 m/s

Mass of cart 2 = 3.0 kg

To find the initial speed of the 3.0-kg cart, we would apply the law of conservation of momentum:

For a perfectly inelastic collision:

$M_1V_1 - M_2V_2 = V_f(M_1 + M_2)$

When the carts stick together and come to rest after the collision, the final velocity is zero (0).

$M_1V_1 - M_2V_2 = 0\n\nM_1V_1 = M_2V_2\n\nV_2 = (M_1V_1)/(M_2)$

Substituting the given parameters into the formula, we have;

$V_2 = (4(9))/(3)\n\nV_2 = (36)/(3)$

Initial speed = 12 m/s

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Holly plans to attend community college next year her expenses are only $1,200 because her parents are paying the rest holly earned a &750 scholarship how much will she need to pay from her saving

Answers

Answer:

$1200-$750 is $450. So Holly needs to use $450 from her savings

Step-by-step explanation:

Answer: A 450

Step-by-step explanation:

Use the quadratic formula to solve x^2 + 9x + 10 = 0. Estimate irrational solutions to the nearest tenth. A {–3.6, 16.4}
B {–7.7, –1.3}
C {–25, 16}
D {–15.4, –2.6}

Answers

Basically, just plus your numbers into quadratic formula. A= 1, B= 9, C= 10.
The answer you will get is B.

$x^(2) + 9x + 10 = 0 \nx = \frac{-(9) +/- \sqrt{(9)^(2) - 4(1)(10)}}{2(1)} \nx = (-9 +/- √(81 - 40))/(2) \nx = (-9 +/- √(41))/(2) \nx = (-9 +/- 6.4)/(2) \nx = (-9 + 6.4)/(2) \nx = (-2.6)/(2) \nx = -1.3 \n \nx = (-9 - 6.4)/(2) \nx = (-15.4)/(2) \nx = -7.7$

It is equal to B.

How does the answer for this 09:009 is 10??