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Flauer [41]
3 years ago
11

A skier starts from rest down a slope 500.0 M long, the skier accelerates at a constant rate of 2.00 m/s/s, what's the velocity

of the skier at the bottom of the hill
Physics
1 answer:
nevsk [136]3 years ago
8 0
We can use the kinematic equation
(v_f)^2 = (v_i)^2 + 2*a*d
where Vf is what we are looking for
Vi is 0 since we start from rest
a is acceleration
and d is the distance

we get
(Vf)^2 = (0)^2 + 2*(2)*(500)
(Vf)^2 = 2000
Vf = about 44.721
or 44.7 m/s   [if you are rounding this by significant figures]
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stephen buys a new moped . he travels 4km south and then 6km east. how far does he need to go to get back where he started??
stiks02 [169]

Answer:

My answer is 7.2 km

Explanation:

When Stephen goes to the south and then to the east, he is drawing a right triangle, where the 4 km and 6 km sides are the cathetus of a right triangle.

Then we use the Pithagorean theorem to solve this problem. We need to find the hypotenuse.

c² = a² + b²

c² = 4² + 6²

c² = 16 + 36

c² = 52

c = 7.2 km

7 0
3 years ago
Is Time and Light both connected in anyway?​
MissTica

Answer:

It is not connected by any way

8 0
2 years ago
If a 42kg rolling object slows from 11.5m/s to 3.33m/s how much work did friction do
geniusboy [140]

Answer: 1608.39 J

Explanation: Given that the

mass M = 42kg

U = 11.5m/s

V = 3.33m/s

how much work did friction do

Work done = Force × distance

Work done = Ma × distance

But acceleration a = V/t

Work done = M × V/t × d

Work done = M × V × d/t

Where d/t = velocity

Therefore,

Work done = M × U × V

Work done = 42 × 11.5 × 3.33

Work done = 1608.39 J

8 0
3 years ago
An observation which is descriptive is considered
oksano4ka [1.4K]
A descriptive observation may very well be a mixture of both quantitative and qualitative as it can utilize elements of both types. Qualitative deals with the kinds of observations that cannot be measured in numerical form. Quantitative data is just that.
6 0
3 years ago
Read 2 more answers
Two moles of an ideal monatomic gas go through the cycle abc. For the complete cycle, 800 J of heat flows out of the gas. Proces
KiRa [710]

a) Sketches of all possible pv-diagrams for the cycle are attached below

b) The work W_{ac} for the process Ca is : 2462.8  J

<u>Given data :</u>

Amount of heat flowing out = 800 J

Ta = 200 K

Tb = 300 K

R = 800

<u>B) Determine the </u><u>work W </u><u>for the process</u><u> Ca</u><u> </u>

Wₐs = -pdv

       = - [ pVb - pVa ] ---- ( 1 )

note : pVb = nRTb ,  pVa = nRTa

Equation ( 1 ) becomes

= -nR [ Tb - Ta ]

= - 2(8.314 ) [ 300 - 200 ]

= - 1662.87

given that W_{bs} = 0 which is isochonic

dv = 0 ( cyclic process ) = d∅ - dw

∴ 0 = 800 - ( Wₐs  +  W_{ac} )

Therefore : W_{ac} = 800 + 1662.8 = 2462.8  J

Hence we can conclude that the work W for the process Ca = 2462.8  J

Learn more about Pv diagrams : brainly.com/question/25401637

Download pdf
3 0
2 years ago
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