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jolli1 [7]
3 years ago
10

A 40 kg skier starts at the top of a 12-meter high slope at the bottom she is traveling 10 m/s how much energy does she lose fri

ction?
A. she doesn't lose any because mechanical energy is always conserved
B. 4,704 J
C. 2,000 J
D. 2,704 J
Physics
2 answers:
Brrunno [24]3 years ago
8 0

Answer:

D. 2,704 J

Explanation:

I just took the quiz and got it right.

Please say thanks by putting a heart, so I know that I helped you

Iteru [2.4K]3 years ago
7 0
Potential energy at top:
PE = mgh
PE = 40 x 9.81 x 12
P.E = 4,708.8 J

Kinetic energy at bottom:
KE = 1/2 mv²
KE = 1/2 x 40 x 10²
K.E = 2,000 J

P.E = K.E + Frictional losses
Frictional losses = 4708 - 2000
Frictional losses = 2708 J

The answer is D.
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Which object represents a negatively charged particle? which object represents a positively charged molecule? which object repre
kari74 [83]

The answers to your questions are as written below:

  • The objects that represents a negatively charged particle is : Image B
  • The object that represents a positively charged molecule is : Image A
  • The object that represents an uncharged molecule is : Image C
  • The object the will not move when in an electric fied is : Image C

<h3>Different types of charges molecules</h3>

A negatively charged molecule move inwards when placed in an electric field while positively charged molecule placed in a electric field will move outwards the electric field.

A neutral/uncharged molecule will remains still when placd in an elctric field due to the absence of charges.

Hence we can concude that the answers to your questions are as listed above.

Learn more about electric charges :brainly.com/question/857179

#SPJ4

attached below is the missing image

8 0
2 years ago
Water is flowing through a channel that is 12m wide with a
Alexus [3.1K]

Answer:

Velocity from second channel will be 1.6875 m/sec

Explanation:

We have given width of the channel , that is diameter of the channel 1 d_1 = 12 m

So radius r_1=\frac{d_1}{2}=\frac{12}{2}=6m

Speed through the channel 1 v_1=0.75m/sec

Width , that is diameter of the channel 2 d_2=4m

So r_2=2m

From continuity equation

A_1v_1=A_2v_2

\pi \times 12^2\times 0.75=4\times \pi\times  4^2\times v_2

v_2=1.6875m/sec

So velocity from smaller channel will be 1.6875 m /sec

6 0
3 years ago
Please help on this one?
Setler79 [48]

I may be wrong but I think the answer is open. Hope this helps.

8 0
4 years ago
PLEASE HELP ON QUESTION
mina [271]

Answer:

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Explanation:

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2 years ago
PLS HELP!! IN QUIZ RN!
kakasveta [241]

Answer:

D Energy Transformation

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3 years ago
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