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nirvana33 [79]
3 years ago
10

On an essentially frictionless, horizontal ice rink, a skater moving at 5.0 m/s encounters a rough patch that reduces her speed

by 40 % due to a friction force that is 22 % of her weight. Use the work-energy theorem to find the length of this rough patch. Express your answer using two significant figures.
Physics
1 answer:
Leviafan [203]3 years ago
3 0

Answer:  the length of the rough patch is 4.9 m.

Explanation:

The work - energy theorem, in simple words, expresses that the change in the kinetic energy in an object, is equal to the work done on the object by non-conservative forces, like frictional ones.

In the question, we know that the skater is moving at a given speed, and due to the effects of friction, her speed is reduced to 40%.

Added to this, we are told that the friction force is equal to 22% of her weight so we can write the following:

Ff = 0.22 m. g.

Now the work done by this force, is equal to the product of the force times the distance during which the force acted, i.e,. the distance that we are looking for.

So, we can write the following in a brief:

ΔK = Wf ⇒ 1/2 m (vf² - v₀²) = -Ff . x

(The negative sign explains that the friction force always opposes to the relative movement between the two surfaces in contact).

Replacing by the values, and solving for x, we get:

1/2 ( 2² - 5²) = -0.22. 9.8. x   ⇒  x = 4.9 m

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sun, earth, capitol building, human, atom

Explanation:

Gravity is directly proportional to the size of an object. Therefore, the object with the most mass will have the greatest force of gravity. We know this from the equation. F = G\frac{m_1m_2}{r^2}

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An object of mass kg is released from rest m above the ground and allowed to fall under the influence of gravity. Assuming the f
IgorLugansk [536]

Answer:

Explanation:

From, the given information: we are not given any value for the mass, the proportionality constant and the distance

Assuming that:

the mass = 5 kg and the proportionality constant = 50 kg

the distance of the mass above the ground x(t) = 1000 m

Let's recall that:

v(t) = \dfrac{mg}{b}+ (v_o - \dfrac{mg}{b})^e^{-bt/m}

Similarly, The equation of mption:

x(t) = \dfrac{mg}{b}t+\dfrac{m}{b} (v_o - \dfrac{mg}{b}) (1-e^{-bt/m})

replacing our assumed values:

where v_=0 \ and \ g= 9.81

x(t) = \dfrac{5 \times 9.81}{50}t+\dfrac{5}{50} (0 - \dfrac{(5)(9.81)}{50}) (1-e^{-(50)t/5})

x(t) = 0.981t+0.1 (0 - 0.981) (1-e^{-(10)t}) \ m

\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}

So, when the object hits the ground when x(t) = 1000

Then from above derived equation:

\mathbf{x(t) = 0.981t-0.981(1-e^{-(10)t}) \ m}

1000= 0.981t-0.981(1-e^{-(10)t}) \ m

By diregarding e^{-(10)t} \ m

1000= 0.981t-0.981

1000 + 0.981 = 0.981 t

1000.981 = 0.981 t

t = 1000.981/0.981

t = 1020.36 sec

7 0
3 years ago
A stone is dropped from a height of 49m and simultaneously another ball is thrown upward from the ground with a speed of 40m/s.
Lilit [14]

Answer:

S1 = 1/2 g t^2     distance stone falls in time t

S2 = Vy t - 1/2 g t^2   distance thrown stone rises in time t

H = 49 = S1 + S2 = Vy t

t = 49 / 40 sec   time when stones meet

Check:

Stone 1 falls:      1/2 g t^2 = 1/2 * 9.8 * (49 / 40)^2 = 7.35 m

Stone 2 rises :  40 * (49 / 40) - 1/2 * 9.8 (49 / 40)^2 = 41.65 m

5 0
3 years ago
A crew on a spacecraft watches a movie that is two hours long. The spacecraft is moving at high speed through space. An Earth-ba
laiz [17]

Answer:

The duration of the movie is longer than 2 hrs.

Explanation:

Given:

The duration of the movie observed by the crew on the spacecraft is 2 hrs.

According to time-dilation formula:

t = \dfrac{t_{0}}{\sqrt{1 - \dfrac{v^{2}}{c^{2}}}}

Here, t is the required time, t_{0} is the original time, v is the velocity of the spacecraft and c is the velocity of light.

Since v, so t_{0} > t.

So the time required will be large.

4 0
3 years ago
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