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frez [133]
2 years ago
11

1. As the angle of the ramp is increased, the normal force increases /decreases / remains the same and the friction-force increa

ses /decreases / remains the same. [1 Point] 2. As the angle of the ramp is increased, the force parallel increases /decreases / remains the same. [1 Point] 3. The angle at which the force down the plane was equal to the force of friction (for the cabinet) was _____________. [1 Point] 4. Consider a very low (~ zero) friction, 5.0 kg skateboard on a ramp at an angle of 15o to the horizontal. What would be the net force that would cause acceleration when the skateboard is allowed to move
Physics
1 answer:
ser-zykov [4K]2 years ago
7 0

(1) As the angle of the ramp is increased, the normal force decreases.

(2) As the angle of the ramp is increased, the parallel force increases.

(3) The angle at which the force down the plane was equal to the force of friction is zero degree.

(4) The net force that would cause acceleration is 47.33 N.

<em>Let the </em><em>angle</em><em> of inclination of the </em><em>ramp</em><em> = θ</em>

(1)

The normal force on an object on the ramp inclined to the ramp is calculated as follows;

F_n = mgcos (\theta)

when θ is 0;

F_n = mgcos (0)\\\\F_n = mg

when θ is 90;

F_n = mgcos(90)\\\\F_n = 0

Thus, as the angle of the ramp is increased, the normal force decreases.

(2)

The parallel force on an object on the ramp inclined to the ramp is calculated as follows;

F_x = mgsin(\theta)\\\\

when θ is 0;

F_x = mgsin(\theta)\\\\F_x = mgsin(0) \\\\F_x = 0

when θ is 90;

F_x = mgsin(90)\\\\F_x = mg

Thus, as the angle of the ramp is increased, the parallel force increases.

(3)

The force of friction is calculated as follows;

F_n = \mu F_n

F_k = \mu mgcos(\theta)

F_k = \mu mg cos(0)\\\\F_k = \mu mg

Thus, the angle is zero degree

(4)

The net force that would cause acceleration is calculated as follows;

F_k = Fn\\\\F_k = mg cos(\theta)\\\\F_k = 5 \times 9.8 \times cos(15)\\\\F_k = 47.33 \ N

Learn more here: brainly.com/question/14121363

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Answer:

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1. momentum is the product of velocity and mass.

2.Momentum is a vector quantity.

3.Momentum have kg.m/s unit.

So the following option are correct.

c. Momentum is the product of mass and velocity

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g. Momentum has unit of kgm/s.

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Bob and John are pulling in different directions. If Bob is pulling to the right with a force of 10N, and John is pulling to the
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Answer:

The net force on the box is 2 N to the left.

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Bob: +10 N

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Students perform a set of experiments by placing a block of mass m against a spring, compressing the spring a distance x along a
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Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.

  • The statement that indicates how the relationship between <em>v</em> and <em>x</em> changes is;<u> As </u><u><em>x</em></u><u> increases, </u><u><em>v</em></u><u> increases, but the relationship is no longer linear and the values of </u><u><em>v</em></u><u> will be less for the same value of </u><u><em>x</em></u><u>.</u>

Reasons:

The energy given  to the block by the spring = \mathbf{0.5  \cdot k  \cdot x^2}

According to the principle of conservation of energy, we have;

On a flat plane, energy given to the block = 0.5  \cdot k  \cdot x^2 = kinetic energy of

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Therefore;

0.5·k·x² = 0.5·m·v²

Which gives;

x² ∝ v²

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On a plane inclined at an angle θ, we have;

The energy of the spring = \mathbf{0.5  \cdot k  \cdot x^2}

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The energy given to the block = 0.5 \cdot k \cdot x^2 - m \cdot g  \cdot sin(\theta) = The kinetic energy of block as it leaves the spring = \mathbf{0.5  \cdot m  \cdot v^2}

Which gives;

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Which is of the form;

a·x² - b = c·v²

a·x² + c·v² = b

Where;

a, b, and <em>c</em> are constants

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  • As <em>x</em> increases, <em>v</em> increases, however, the value of <em>v</em> obtained will be lesser than the same value of <em>x</em> as when the block is on a flat plane.

<em>Please find attached a drawing related to the question obtained from a similar question online</em>

<em>The possible question options are;</em>

  • <em>As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x</em>
  • <em>The relationship is no longer linear and v will be more for the same value of x</em>
  • <em>The relationship is still linear, with lesser value of v</em>
  • <em>The relationship is still linear, with higher value of v</em>
  • <em>The relationship is still linear, but vary inversely, such that as x increases, v decreases</em>

<em />

Learn more here:

brainly.com/question/9134528

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