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geniusboy [140]
3 years ago
6

Mary slides down a snow-covered hill on a large piece of cardboard and then slides across a frozen pond at a constant velocity o

f 2.40 m/s. After Mary has reached the bottom of the hill and is sliding across the ice, Sue runs after her at a velocity of 4.40 m/s and hops on the cardboard. How fast do the two of them slide across the ice together on the cardboard? Mary's mass is 69.0 kg and Sue's is 56.0 kg. Ignore the mass of the cardboard and any friction between the cardboard and the snow and/or ice. (Indicate the direction with the sign of your answer.) m/s
Physics
1 answer:
Tresset [83]3 years ago
3 0

Answer: 3.288 m/s

Explanation:

Given

Mass of Mary, m1 = 69 kg

Mass of Sue, m2 = 56 kg

Speed of Mary, v1 = 2.4 m/s

Speed of Sue, v2 = 4.4 m/s

Speed of the 2 of them, v = ?

We solve this using the principle of conservation of linear momentum

m1.v1 + m2.v2 = m1.v + m2.v

m1.v1 + m2.v2 = (m1 + m2) v

v = [(m1.v1) + (m2.v2)] / (m1 + m2)

v = [(69 * 2.4) + (56 * 4.4)] / (69 + 56)

v = (164.6 + 246.4) / 125

v = 411 / 125

v = 3.288 m/s

Thus, the speed at which both Mary and Sue slide together across the ice is 3.288 m/s

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2 years ago
A car slows down at -5.00 m/s² until it comes to a stop after travelling 15.0 m. What was the initial speed of the car?
lapo4ka [179]
<h2>Answer: 12.24m/s</h2>

According to <u>kinematics</u> this situation is described as a uniformly accelerated rectilinear motion. This means the acceleration while the car is in motion is constant.

Now, among the equations related to this type of motion we have the following that relates the velocity with the acceleration and the distance traveled:

V_{f}^{2}-V_{o}^{2}=2.a.d   (1)

Where:

V_{f} is the Final Velocity of the car. We are told "the car comes to a stop after travelling", this means it is 0.

V_{o} is the Initial Velocity, the value we want to find

a is the constant acceleration of the car (the negative sign means the car is decelerating)

d is the distance traveled by the car

Now, let's substitute the known values in equation (1) and find V_{o}:

0-V_{o}^{2}=2(-5m/{s}^{2})(15m)    (2)

-V_{o}^{2}=-150{m}^{2}/{s}^{2}    (3)

Multiplying by -1 on both sides of the equation:

V_{o}^{2}=150{m}^{2}/{s}^{2}    (4)

V_{o}=\sqrt{150{m}^{2}/{s}^{2}}    (5)

Finally:

V_{o}=12.24m/s >>>This is the Initial velocity of the car

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<u>O</u><u>p</u><u>t</u><u>i</u><u>o</u><u>n</u><u> </u><u>C</u><u> </u><u>i</u><u>s</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>a</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u>

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The reactants are charcoal that is unlit + oxygen and the products are the burnt charcoal + energy.

(Explanation with formula and reason attached. Check it.)

> Benjemin360

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jeka94
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