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Tasya [4]
2 years ago
4

Now, suppose that Zak's younger cousin, Greta, sees him sliding and takes off her shoes so that she can slide as well (assume he

r socks have the same coefficient of kinetic friction as Zak's). Instead of getting a running start, she asks Zak to give her a push. So, Zak pushes her with a force of 125 N over a distance of 1.00 m. If her mass is 20.0 kg, what distance d2 does she slide after Zak's push ends? Remember that the frictional force acts on Greta during Zak's push and while she is sliding after the push.
Physics
1 answer:
N76 [4]2 years ago
7 0

Answer:  1.55 m

Explanation:

from the question we were given

force = 125 N

distance = 1 m

mass = 20 kg

first we find the work done in pushing her = force x distance

   work done = 125 x 1 = 125 j

now we find the frictional force = (  coefficient of friction x mass x acceleration due to gravity ), taking the coefficient of friction as o.25 and acceleration due to gravity as 9.8 m/s

frictional force = 0.25 x 20 x 9.8 = 49 N

the frictional force must do the same amount of work Zak did in pushing his sister so as to bring her to a stop. now we can get the distance she slides by applying the formula work done = frictional force x distance

  125 = 49 x distance

distance = 125 / 45 = 2.55 m

to get how far she slides after Zak's push stops we have to subtract the distance she was pushed for from the total distance she slid

= 2.55 - 1 = 1.55 m

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A particle decelerates uniformly from a speed of 30 cm/s to rest in a time interval of 5.0 s. It then has a uniform acceleration
wel

Answer:

V=20cm/s

Explanation:

The average speed is the distance total divided the time total:

V=X/T

First stage:

T1=5s

v_{f}  =v_{o} - at

But, v_{f}  =0   (decelerates to rest)

then: a =v_{o} /t=0.3/5=0.06m/s^{2}

on the other hand:

x =v_{o}*t - 1/2*at^{2}=0.3*5-1/2*0.06*5^{2}=0.75m

X1=75cm

Second stage:

T2=5s

x =v_{o}*t + 1/2*at^{2}=0+1/2*0.1*5^{2}=1.25m

X2=125cm

Finally:

X=X1+X2=200cm

T=T1+T2=10s

V=X/T=20cm/s

8 0
2 years ago
According to the nebular theory of solar system formation, what key difference in their early formation explains why the jovian
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Answer:

The Jovian planets formed beyond the Frostline while the terrestrial planets formed in the Frostline in the solar nebular

Explanation:

The Jovian planets are the large planets namely Saturn, Jupiter, Uranus, and Neptune. The terrestrial planets include the Earth, Mercury, Mars, and Venus. According to the nebular theory of solar system formation, the terrestrial planets were formed from silicates and metals. They also had high boiling points which made it possible for them to be located very close to the sun.

The Jovian planets formed beyond the Frostline. This is an area that can support the planets that were made up of icy elements. The large size of the Jovian planets is as a result of the fact that the icy elements were more in number than the metal components of the terrestrial planets.

3 0
2 years ago
(a) A 15.0 kg block is released from rest at point A in the figure below. The track is frictionless except for the portion betwe
castortr0y [4]

Answer:

(a) coefficient of friction = 0.451

This was calculated by the application of energy conservation principle (the total sum of energy in a closed system is conserved)

(b) No, it comes to a stop 5.35m short of point B. This is so because the spring on expanding only does a work of 43 J on the block which is not enough to meet up the workdone of 398 J against friction.

Explanation:

The detailed step by step solution to this problems can be found in the attachment below. The solution for part (a) was divided into two: the motion of the body from point A to point B and from point B to point C. The total energy in the system is gotten from the initial gravitational potential energy. This energy becomes transformed into the work done against friction and the work done in compression the spring. A work of 398J was done in overcoming friction over a distance of 6.00m. The energy used in doing so is lost as friction is not a conservative force. This leaves only 43J of energy which compresses the spring. On expansion the spring does a work of 43J back on the block is only enough to push it over a distance of 0.65m stopping short of 5.35m from point B.

Thank you for reading and I hope this is helpful to you.

4 0
2 years ago
7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle
jasenka [17]

Weight of the carriage =(m+M)g =142.1\ N

Normal force =Fsin(\theta) + W = 197.1\ N

Frictional force =\mu N=27.59\ N

Acceleration =4.66\ m\ s^{-2}

Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

  • So Weight(W) =(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with F_x, force of 110\ N acting vertically downward.Both are downward and Normal is upward so Normal force =Summation\ of\ both\ forces

  • Normal force (N) = Fsin(\theta)+W=110sin(30) + 142.1 =197.1\ N
  • Frictional force (f) =\mu N=0.14\times 197.1 =27.59\ N

To calculate acceleration we will use Newtons second law.

That is Force is product of mass and acceleration.

We can see in the diagram that F_y=Horizontal and F_x=Vertical component of forces.

So Fnet = Fy(Horizontal) - f(friction) = m\times a

  • Acceleration (a) =\frac{Fcos(\theta)-\mu N}{mass(m)} =\frac{(95.26-27.59)}{14.5}= 4.66\ m\ s{^2 }

So we have the weight of the carriage, normal force,frictional force and acceleration.

3 0
2 years ago
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P = 68.125 kW

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Explanation:

4 0
2 years ago
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