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Scorpion4ik [409]
2 years ago
10

Which of the following is not a form of an inclined plane?

Physics
2 answers:
disa [49]2 years ago
8 0

Answer: piece of paper cut into a triangle  

Explanation:

A plane surface lifted at an angle to horizontal from one edge is said to be an inclined plane. A wedge, a ramp or a screw are examples of a forms of inclined plane. A screw is like a inclined plane wound on a cylinder. These are simple machines. On the other hand, a piece of paper cut into a triangle is not an inclined plane as it does rise at an angle from horizontal. It is not a machine.

Len [333]2 years ago
7 0

A piece of paper cut into a triangle

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Determine the amount of potential energy of a 5.0Kg book that is moved to three different shelves on a bookcase. The height of e
rjkz [21]
Formula to find gravitational potential energy:
mgh
m: mass
g: gravitational acceleration
h: height (relative to reference level)

so the P.E. at 1.0.m is (5x9.8x1)= 49J
P.E. at 1.5m is (5x9.8x1.5) =73.5J
P.E. at 2.0m is (5x9.8x2)=98J
8 0
3 years ago
a child pulls on a string that is attached to a car. if the child does 80.2 J of work while pulling the car 25.0 m, with what fo
12345 [234]

Answer:

F = 3.20 N

Explanation:

Given:

Work done by child = 80.2 j

Distance that the car moves = 25.0 m

We need to find the force acting on the car.

Solution:

Using work done formula as.

W = F\times d

Where:

W = Work done by any object.

F = Force (push or pull)

d = distance that the object moves.

Substitute W = 80.2\ J\ and\ d =25.0\ m in work done formula.

80.2 = F\times 25

F=\frac{80.2}{25}

F = 3.20 N

Therefore, force acting on the car F = 3.20 N

3 0
3 years ago
A body of mass 500kg moving at a speed of 10m/s reaches the speed of 50m/s in 20s.The force exerted is
Leona [35]

Answer:

answer is 1000 N

formula used-

<em><u>F= m x (v-u/t)</u></em>

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6 0
3 years ago
Block A of mass M is at rest and attached to the top of a spring. The block compresses the spring a distance d from its uncompre
Anni [7]

Answer:

a)  k = Mg / d , b)   v = √2gh , c)  v_{f} = \frac{2}{3} \ \sqrt{2gh},  d)   x² + 6d x - \frac{8}{3} dh = 0

e)the spring must compress a greater distance.

Explanation:

a) when the block of mass M is placed on the spring, we have an equilibrium condition,

             ∑ F  = 0

             F_{e}- W = 0

             k d = Mg

             k = Mg / d

b) let's use the concepts of energy to find the velocity of the block just before the collision

starting point. Position when released

          Em₀ = U = m g h

lowest point. Right at the point of shock

          Em_{f} = K = ½ m v²2

as there is no friction, energy is conserved

          Em₀ = Em_{f}

          mg h = ½ m v²

          v = √2gh

         

c) The velocity of the two blocks after the collision, we define a system formed by the two blocks, in such a way that the forces during the collision are internal and the moment is conserved

initial instant. Just before the crash

          p₀ = 2M v + M 0

final instant. Just after the shock, before the spring compression begins

         p_{f} = (2M + M) v_{f}

 the moment is preserved

          p₀ = p_{f}

          2M v = 3M v_{f}

          v_{f} = ⅔ v

          v_{f} = \frac{2}{3} \ \sqrt{2gh}

d) now we work with the joined system after the collision, let's use the concepts of energy

starting point. After shock, before beginning spring compression

        Em₀ = K = ½ (3M) v_{f}^2

        Em₀ = 3/2 M (\frac{2}{3} \ \sqrt{2gh})²

        Em₀ = 4/3 M gh

final point. With the spring fully compressed

       Em_f = K_e + U = ½ k x² + (3M) g x

in this case we have taken the zero of gravitational potential energy at the point where the blocks collide, as there is no friction, the energy is conserved

         Em₀ = Em_f

        4/3 M g h = ½ k x² + 3M g x

        ½ k x² + 3Mg x - 4/3 Mgh = 0

we substitute the expression for k

         \frac{1}{2} (\frac{Mg}{d}) x² + 3Mg x - \frac{4}{3} Mgh = 0

          \frac{x^{2} }{2d} + 3 x - \frac{4}{3}h = 0

to find the value of the spring compression, the second degree equation must be solved

          x² + 6d x - \frac{8}{3} dh = 0

         x = [-6d ±\sqrt{(36 d^{2} - 4 \frac{8}{3} dh)  } ] / 2

         x = [-6d ± 6d \sqrt{ 1 -  \frac{32}{3 \ 36}  \ \frac{h}{d}    }  ]/2

         x = 3d ( -1±  \sqrt{ 1 - 0.296 \frac{h}{d}   }  )

e) If the collision elastic force would not lose any part of the kinetic energy during the collision, therefore the speed of the block of mass M would be much higher and therefore the spring must compress a greater distance.

8 0
3 years ago
Which force is always attractive and has a very small range?
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Answer:

Strong Nuclear

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