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photoshop1234 [79]
3 years ago
13

A 2.0   kg 2.0kg2, point, 0, start text, k, g, end text cart moving right at 5.0   m s 5.0 s m ​ 5, point, 0, start fraction, st

art text, m, end text, divided by, start text, s, end text, end fraction on a frictionless track collides with a 3.0   kg 3.0kg3, point, 0, start text, k, g, end text cart initially at rest. The 2.0   kg 2.0kg2, point, 0, start text, k, g, end text cart has a final speed of 1.0   m s 1.0 s m ​ 1, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction to the left. What is the final speed of the 3.0   kg 3.0kg3, point, 0, start text, k, g, end text cart?
Physics
2 answers:
butalik [34]3 years ago
7 0
Yes así le tienes que hacer para aprender
Elena L [17]3 years ago
3 0

Answer:  The answer is 4.0 m/s.

Explanation: khan acadmey.

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kirza4 [7]

Answer:

yes

Explanation:

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3 years ago
The scientist who suggested that energy can be created under certain conditions was
tester [92]
Einstein hope this helped
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4 years ago
Read 2 more answers
An electric car is being designed to have an average power output of 4,600 W for 2 h before needing to be recharged. (Assume the
MAVERICK [17]

Answer:

a)3312 x 10⁴ J

b)I = 57.5 A

c)9200 W

Explanation:

Given that

P =4600 W

Time t= 2 h = 2 x 3600 s= 7200 s

We know that

1 W = 1 J/s

a)

Energy stored in the battery = P .t

                                              =4600 x 7200 J

                                            =3312 x 10⁴ J

b)

We know that power P given as

P = V .I

V=Voltage  ,I =Current

4600 = 80 x I

I = 57.5 A

c)

The energy supplied = 4600 x 2 = 9200 W

7 0
3 years ago
Calculate the total energy of 4.0 kg object moving horizontally at 20 m/s 50 meters above the surface.
Serhud [2]

Answer:

Correct answer:  E total = 2,800 J

Explanation:

Given:

m = 4 kg   the mass of the object

V = 20 m/s  the speed (velocity) of the object

H = 50 m the height of the object above the surface

E total = ? J

The total energy of an object is equal to the sum of potential and kinetic energy

E total = Ep + Ek

Ep = m g H   we take g = 10 m/s²

Ep = 4 · 10 · 50 = 2,000 J

Ek = m V² / 2

Ek = 4 · 20² / 2 = 2 · 400 = 800 J

E total = 2,000 + 800 = 2,800 J

E total = 2,800 J

God is with you!!!

4 0
3 years ago
Mercury is characterized by
cupoosta [38]
 <span>It's close to the sun without much atmosphere, so it's characterized by </span><span>very extreme temperatures.

Happy studying ^_^</span>
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