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Contact [7]
2 years ago
14

15. If friction was completely removed from the track, what affect would this have on the roller coaster ride?

Physics
1 answer:
finlep [7]2 years ago
4 0
Answer:
A

Explanation:
The coaster wouldn’t eventually stop because there’s no fiction to stop it. In order for a object to stop, there must be a opposing force ( in this case, fiction).
It is true that the coaster could use hills to stop, but the coaster can’t use flat section to stop.
So what I think is that the answer is A.

Fun Fact:
It would be really tough just to stand up without friction.

I hope this help and have a nice day :)
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A pendulum clock is taken to the moon (g = 1.6 N/kg). How long must the pendulum be in order for the clock to
mamaluj [8]

Answer:

0.041 m

Explanation:

T = 2π√(L/g)

1 s = 2π√(L / (1.6 N/kg))

L = 0.041 m

4 0
3 years ago
What is Albert Einstein’s contribution to the understanding of nuclear energy?
Phoenix [80]
Einstein's equations showed that matter could be converted into energy; and vice-versa
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7 0
2 years ago
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Suppose we have a 600 kilogram great "yellow" shark swimming to the right at a speed of 3 meters traveled each second as it trie
sammy [17]

Answer:

2.64 m/s

Explanation:

Given that a 600 kilogram great "yellow" shark swimming to the right at a speed of 3 meters traveled each second as it tries to get lunch. An unsuspecting 100 kilogram blue fin tuna is minding its own business swimming to the left at a speed of 0.5 meters traveled each second. GULP! After the great "yellow" shark "collides" with the blue fin tuna

Momentum = MV

Momentum of the yellow shark before collision = 600 × 3 = 1800 kgm/s

Momentum of the tun final before collision = 100 × 0.5 = 50 kgm/s

Total momentum before collision = 1800 + 50 = 1850 kgm/s

Let's assume that they move together after collision. Then,

1850 = ( 600 + 100 ) V

1850 = 700V

V = 1850 / 700

V = 2.64285 m/s

Therefore, the momentum of the shark after collision is 2.64 m/ s approximately

6 0
3 years ago
Two 1.9 kg masses are 1.1 m apart (center to center) on a frictionless table. Each has + 9.6 μC of charge.
Anni [7]

Answer:

F= 0.6 N

Explanation:

Fe(electrical force)=k q1q2/r^2

    k=9*10^9\\q1=-9.6 *10^-6 C\\q2= -9.6*10^-6 C\\r= 1.1m

So,    

        F=\frac{9*10^9*-9.6 *10^-6 C* -9.6*10^-6 C}{1.1^2}

         F= 0.6 N

5 0
3 years ago
A square plate of copper with 55.0 cm sides has no net charge and is placed in a region of uniform electric field of 82.0 kN/C d
Alex73 [517]

Answer

given,

Side of copper plate, L = 55 cm

Electric field, E = 82 kN/C

a) Charge density,σ = ?

  using expression of charge density

 σ = E x ε₀

ε₀ is Permittivity of free space = 8.85 x 10⁻¹² C²/Nm²

now,

 σ = 82 x 10³ x 8.85 x 10⁻¹²

 σ = 725.7 x 10⁻⁹ C/m²

 σ = 725.7 nC/m²

change density on the plates are 725.7 nC/m² and -725.7 nC/m²

b) Total change on each faces

   Q = σ  A

   Q = 725.7 x 10⁻⁹ x 0.55²

   Q = 219.52 nC

Hence, charges on the faces of the plate are 219.52 nC and -219.52 nC

7 0
3 years ago
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