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barxatty [35]
3 years ago
15

The mountains most commonly found at divergent plate boundaries are

Physics
1 answer:
dybincka [34]3 years ago
8 0
fault-block mountains. hope this helps
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At the train station, you notice a large horizontal spring at the end of the track where the train comes in. This is a safety de
xxMikexx [17]

Answer:

  v₀ = 9,798 ft / s

Explanation:

We can solve this problem with kinematics in one dimension, when the train stops the speed is zero, the acceleration is negative so that the train stops. Let's use the equation

         v² = v₀² - 2 a d

         v = 0

         v₀ = √2 a d

In the problem it indicates that the acceleration is g / 2, we substitute

         v₀ = √2 (g / 2) d

Let's calculate

         v₀ = √ 2 32/2 3 = √32 3

         v₀ = 9,798 ft / s

8 0
3 years ago
An object accelerates from rest to 85m/s over a distance of 36m. What acceleration did it experience?
Marizza181 [45]
Suvat
we have s, u, v and we want a
the suvat equation with these values in is: v^2 = u^2 - 2as
so a = (-v^2 + u^2)/-2s 
plug numbers in
a = (-85^2 + 0^2)/-2*36 = 7225/72 = 100.3... ms^-2
6 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
At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.5 m/s2. At the same insta
Licemer1 [7]

Answer:

6.96 s

Explanation:

<u>Given:</u>

  • u = initial speed of the automobile = 0 m/s
  • a = constant acceleration of the automobile = 2.5\ m/s^2
  • v = constant speed of the truck = 8.7 m/s

<u>Assume:</u>

  • t = time instant at which the automobile overtakes the truck.

At the moment the automobile and the truck both meat each other the distance travel by both vehicles must be the same.

\therefore \textrm{Distance traveled by the automobile }=\textrm{Distance traveled by the truck}\\\Rightarrow ut+\dfrac{1}{2}at^2=vt\\\Rightarrow (0)t+\dfrac{1}{2}\times 2.5\times t^2=8.7t\\\Rightarrow 1.25t^2=8.7t\\\Rightarrow 1.25t^2-8.7t=0\\\Rightarrow t(1.25t-8.7)=0\\\Rightarrow t = 0\,\,\,or\,\,\, t = \dfrac{8.7}{1.25}\\\Rightarrow t = 0\,\,\,or\,\,\, t = 6.96\\

Since t = 0 s is the initial condition. So, they both meet again at t = 6.96 s such that the automobile overtakes the truck.

6 0
3 years ago
How is capacitance related to the distance between the plates of a capacitor?
svetlana [45]

Answer:C

Explanation: I studied, and C is correct

5 0
3 years ago
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