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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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A 5 kg ball of clay is moving with a speed of 25 m/s directly toward a 10 kg ball of clay which is at rest. The two clay balls c
Dafna11 [192]

Answer:

8.3m/s

Explanation:

Given parameters:

mass of clay ball  = 5kg

Speed of clay ball  = 25m/s

mass of clay ball at rest  = 10kg

speed of clay ball at rest  = 0m/s

Unknown:

Velocity after collision  = ?

Solution:

 Since the balls stick together, this is an inelastic collision:

   m1v1 + m2v2  = v(m1 + m2)

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7 0
2 years ago
An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is
Andreas93 [3]

Answer:

v=1.08\times 10^7\ m/s

Explanation:

Initial speed of the electron, u = 0

The charge per unit area of each plate, \dfrac{Q}{A}=1.69\times 10^{-7}\ C/m^2

Separation between the plates, d=1.75\times 10^{-2}\ m

An electron is released from rest, u = 0

Using equation of kinematics,

v^2-u^2=2ad..........(1)

Acceleration of the electron in electric field, a=\dfrac{qE}{m}............(2)

Electric field, E=\dfrac{\sigma}{\epsilon_o}............(3)

From equation (1), (2) and (3) :

v=\sqrt{\dfrac{2q\sigma d}{m\epsilon_o}}

v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 1.69\times 10^{-7}\times 1.75\times 10^{-2}}{9.1\times 10^{-31}\times 8.85\times 10^{-12}}}

v = 10840393.1799 m/s

or

v=1.08\times 10^7\ m/s

So, the electron is moving with a speed of 1.08\times 10^7\ m/s before it reaches the positive plate. Hence, this is the required solution.

3 0
3 years ago
What is the kinetic energy of a 0.15 kg baseball that is moving with a velocity of 50 m/s?
goblinko [34]
Using the equation:

ke \:  =  \frac{1}{2} m{v}^{2}
ke  =   \frac{1}{2} (0.15) {(50)}^{2}
ke  = 188  \: (3sf)


Ans: 188 J
7 0
2 years ago
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