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tatiyna
3 years ago
11

A motorcycle of mass 100 kilograms slowly rolls off the edge of a cliff and falls for three seconds before reaching the bottom o

f a gully. Its momentum upon hitting the ground is
Physics
1 answer:
Digiron [165]3 years ago
6 0

Answer:

2940 kg m/s

Explanation:

First of all, we need to find the final velocity of the motorcycle when it hits the ground. The horizontal component of the velocity is zero, so we can just find the vertical component, which is given by:

v_y = v_0+gt

where

v0 =0 is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration due to gravity

t = 3 s is the time of the fall

Substituting into the formula, we find

v_y=0+(9.8 m/s^2)(3 s)=29.4 m/s

And so, we can now find the momentum of the motorcycle upon hitting the ground, which is given by the product between the mass and the velocity:

p=mv=(100 kg)(29.4 m/s)=2940 kg m/s

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The first thing you should know to solve this problem is the conversion of pounds to kilograms:
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 1lb ---> 0.45Kg
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 Clearing x we have:
 x = ((125) / (1)) * (0.45) = 56.25 Kg.
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 her mass expressed in kilograms is 56.25 Kg.
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3 years ago
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Answer:

es un motor de combustión interna con encendido por chispa.

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3 years ago
If the magnetic flux through a loop increases according to the relation; where , is in milliweber (mWb) and t is in seconds. Cal
Alekssandra [29.7K]

The magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.

<h3>emf induced in the loop</h3>

The magnitude of e.m.f induced in the loop is calculated as follows;

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dФ/dt = 12t + 7

at t = 2 seconds

emf = dФ/dt = 12(2) + 7 = 31 V

Thus,  the magnitude of e.m.f induced in the loop when t = 2 s is 31 Volts.

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2 years ago
A 6 kilogram block in outer space is moving at -100 m/s (to the left). It suddenly experiences three forces as shown below.
Alika [10]

Newton's second law and the kinematic relations allow to find the results for the questions about forces and the movement of the block are:

    B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º

    C) The maximum force is: F = 24 N

    D) The time to stop the block is: t = 25 s

 

Newton's second law establishes a relationship between the net force, the mass, and the acceleration of the body. In the special case that the acceleration is zero it is called the equilibrium condition.

B) They indicate a diagram of forces on the block, let's look for the components of the force that the block maintains with zero acceleration, in the attached we have a free-body diagram including the force applied to keep the system in equilibrium.

x-axis

      -10 + 12 sin 60 + Fₓ = 0

        Fₓ = 10- 12 sin 60 = -0.39 N

y-axis

       12 cos 60 - 6 + F_y = 0

        F_y = 6 - 12 cos 60 = 0 N

We can give the result of the force in two ways:

  • Form of coordinates F = -0.39 i ^ N
  • Form of module and angle.

Let's use Pythagoras' theorem to find the modulus.

       F = \sqrt{F_x^2 + F_y^2 } \\F = \sqrt{0.39^2 +0^2}  

       F = 0.39N

We use trigonometry for the angle.

       tan \theta = \frac{F_y}{F_x}

       tan θ=  0º

The component of the force is negative therefore this angle is in the second quadrant, to measure the angle from the positive side of the x axis in a counterclockwise direction.

        θ = 180 + θ'

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C) if the three forces can be moved and the maximum force occurs when they are all linear.

          10+ 6 + 6 + F = 0

          F = -24 N

D) if we maintain this force and eliminate the other three, the block stops, let's look for its acceleration.

          a = \frac{F}{m}  

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The acceleration is in the opposite direction of the initial velocity of the block v₀ = -100 m / s

If we use kinematic relations.

        v = v₀ - a t

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         t = \frac{0-v_o}{a}

         t = 100/4

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    B) the force applied to maintain the system is equilibrium is: F = 0.39 N with an angle of tea = 180º

    C) The maximum force is: F = 24 N

    D) The time to stop the block is: t = 25 s

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