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aev [14]
3 years ago
10

A red cart has a mass of 4 kg and a velocity of 5 m/s. There is a 2-kg blue cart that is parked and not moving, thus its velocit

y is 0 m/s.
The red cart hits the blue cart.


The blue cart starts to move forward with a velocity of 6 m/s. The red cart bounces off of the blue cart and moves backwards, at a velocity of 2 m/s.


Name the TYPE of collision that occurred.

Calculate the BEFORE momentum of the red cart. Use correct units.

Calculate the BEFORE momentum of the blue cart. Use correct units.

Calculate the AFTER momentum of the red cart. Use correct units.

Calculate the AFTER momentum of the blue cart. Use correct units.

Calculate the SYSTEM’S TOTAL MOMENTUM before the collision. Use correct units.

Calculate the SYSTEM’S TOTAL MOMENTUM after the collision. Use correct units.

EXPLAIN how this example supports the Law of Conservation of Momentum. Use the evidence to support your answer of how this example supports the Law of Conservation of Momentum. Use correct units. Be specific.
Physics
1 answer:
IRISSAK [1]3 years ago
5 0

Answer:dam hold up

Explanation:

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When a boat is placed in liquid, two forces act on the boat. Gravity pulls the boat down with a force equal to the weight of the
wel

Answer:

the fraction of submerged volume is equal to the ratio of the densities of the body between the density of the fluid.

Explanation:

This is a fluid mechanics problem, where as the boat is in equilibrium with the pushing force we can write Newton's second law

                  B- W = 0

                  B = W

           

the thrust force is equal to the weight of the liquid that is dislodged

                  B = ρ g V

we substitute

             ρ g V = m g

             V = m /ρ_fluid          1

we can write the mass of the pot as a function of its density

             ρ_body = m / V_body

            m = ρ_body  V_body

             V_fluid / V_body = ρ_body / ρ _fluid         2

Equations 1 and 2 are similar, although 2 is easier to analyze, the fraction of submerged volume is equal to the ratio of the densities of the body between the density of the fluid.

The effect appears the pot as if it had a lower apparent weight

3 0
4 years ago
A man is at a car dealership, looking for a car to buy. He looks at the sticker on the driver’s window of a car and sees that th
maxonik [38]

Answer:

he will see the sticker because its behind a window bruh and thats a big daddy stack of greens

Explanation:

3 0
3 years ago
A ring of diameter 7.70 cm is fixed in place and carries a charge of 5.00 mC uniformly spread over its circumference. (a) How mu
MatroZZZ [7]

Answer:

3.974 Joule

Explanation:

Diameter of ring = 7.7 cm

a = Distance from the center = d/2 = 3.85 cm = 0.0385 m

Q = Charge = 5 mC

q = Charge to move = 3.4 mC

k = Coulomb constant = 9×10⁹ Nm²/C²

Work done will be equal to Potential energy when mass is at center

U=\frac{kQq}{a}\\\Rightarrow U=\frac{9\times 10^9\times 5\times 10^{-6}\times 3.4\times 10^{-6}}{0.0385}=3.974\ J

∴ Work to move a tiny 3.4 mC charge from very far away to the center of the ring is 3.974 Joule

4 0
3 years ago
Read 2 more answers
A dwarf planet discovered out beyond the orbit of Pluto is known to have an orbital period of 619.36 years. What is its average
Maksim231197 [3]

Answer: 72.66 AU=1.089(10)^{10} km

Explanation:

Let's begin by explaining that according to Kepler’s Third Law of Planetary motion “The square of the orbital period T of a planet is proportional to the cube of the semi-major axis a of its orbit”:

T^{2}\propto a^{3} (1)  

Now, if T is measured in years (Earth years), and a is measured in astronomical units (equivalent to the distance between the Sun and the Earth: 1AU=1.5(10)^{8}km), equation (1) becomes:  

T^{2}=a^{3} (2)  

So, knowing T=619.36 years and isolating a from (2) we have:  

a=\sqrt[3]{T^{2}} (3)  

a=\sqrt[3]{(619.36 years)^{2}} (4)  

Finally:

a=72.66 AU T his is the distance between the dwarf planet and the Sun in astronomical units

Converting this to kilometers, we have:

a=72.66 AU \frac{1.5(10)^{8}km}{1 AU}=1.089(10)^{10} km

4 0
3 years ago
A boy throws a ball into the air at 10.2 m/s. Assuming that only gravity acts on the ball, how high does it rise, in m?
ozzi
Answer: 5.31 meters

Explanation: Use conservation of energy. Initial energy equals final energy. Initially, there is only kinetic energy (because height = 0 initially). At the end, kinetic energy equals 0 because at max height, there is max potential energy and the ball stops moving for a split second.

mgh = .5mv^2
Masses cancel out
gh = .5v^2
(9.8)(h) = .5(10.2^2)
Solve for h. h = 5.31 meters
5 0
3 years ago
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