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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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Waves on a pond are an example of which kind of wave?
KATRIN_1 [288]
<span>
</span><span>Waves on a pond are an example of which kind of wave? 

</span>B. surface waves
6 0
4 years ago
Planetary orbits... are spaced more closely together as they get further from the Sun. are evenly spaced throughout the solar sy
BaLLatris [955]

Answer:

E) are almost circular, with low eccentricities.

Explanation:

Kepler's laws establish that:

All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).

A planet describes equal areas in equal times (Kepler's second law).

The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).

T^{2} = a^{3}

Where T is the period of revolution and a is the semi-major axis.

Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.

However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)

In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).

Therefore, option A and B can not be true.

In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.  

4 0
3 years ago
A body with initial velocity 8.0 m/s moves along a straight line with constant acceleration and travels
Aleksandr [31]

Answer:

<em>(a) The average velocity is 16 m/s</em>

<em>(b) The acceleration is 0.4 m/s^2</em>

<em>(c) The final velocity is 24 m/s</em>

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity (or the speed) of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, final speed is calculated as follows:

v_f=v_o+at\qquad\qquad [1]

The distance traveled by the object is given by:

\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]

(a) The average velocity is defined as the total distance traveled divided by the time taken to travel that distance.

We know the distance is x=640 m and the time taken t= 40 s, thus:

\displaystyle \bar v=\frac{x}{t}=\frac{640}{40}=16

The average velocity is 16 m/s

Using the equation [1] we can solve for a:

\displaystyle a=\frac{v_f-v_o}{t}

(c) From [2] we can solve for a:

\displaystyle a= 2\frac{x-v_ot}{t^2}

Since vo=8 m/s, x=640 m, t=40 s:

\displaystyle a= 2\frac{640-8\cdot 40}{40^2}=0.4

The acceleration is 0.4 m/s^2

(b) The final velocity is calculated by [1]:

v_f=8+0.4\cdot 40

v_f=8+16=24

The final velocity is 24 m/s

3 0
3 years ago
What is a procedure that could be used to separate a mixture f sugar, black pepper, and pebbles?
Virty [35]
Dissolving can separate the three. Because pepper doesn't mix with water and neither do pebbles.
3 0
3 years ago
It's a snowy day and you're pulling a friend along a
NeX [460]

The coefficient of friction between the sled and the snow is 0.119.

To find the answer, we need to know about the friction.

<h3>How to find the coefficient of friction between the sled and the snow?</h3>
  • Whenever a body moves over the surface of another body, a force come into play, which acts parallel to the surface of contact and oppose the relative motion. This opposing force is called friction.
  • To solve the problem, we have to draw the free body diagram of the given system.
  • We have given with the following values,

                                     a=0\\\alpha =35^0\\T=75N\\m=57kg

Here, acceleration will be equal to zero, because the velocity is given as constant.

  • Thus, from the diagram, we can write the balancing equations as follows,

                                      ma=Tcos\alpha -f\\\where\\f=kN\\\N+Tsin\alpha=mg\\Thus,\\N=mg-Tsin\alpha

  • Substituting N in f and f in the equation of ma, then we get,

                   ma= Tcos\alpha -k(mg-Tsin\alpha )

  • Substituting values, we get the coefficient of friction as,

                    0=(75*cos35)-k((57*9.8)-(75sin35))\\\\k((57*9.8)-(75sin35))=(75*cos35)\\\\515.6k=61.44\\\\k=\frac{61.44}{515.6}=0.119

Thus, we can conclude that, the coefficient of friction between the sled and the snow is 0.119.

Learn more about the friction here:

brainly.com/question/28107059

#SPJ1

6 0
2 years ago
Read 2 more answers
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