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charle [14.2K]
3 years ago
12

What surface features does mars have that are also common on earth?

Physics
2 answers:
Eva8 [605]3 years ago
6 0
Volcanoes, Sand Dunes, and Large Canyons
dedylja [7]3 years ago
4 0
<span>Both planets, Mars and the Earth are terrestrial planets. This means that their surface is rocky, they have Mountains, canyons, valleys, ice caps..</span>



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State a situation in which force is applied on a body, but no work is done​
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Why do streets and highways have speed limits rather than velocity limits?
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3 years ago
A circular disc of mass 20kg and radius 15cm is mounted in an horizontal cylindrical axle of radius
disa [49]

Using the concepts of energy, rotational Newton's second law and rotational kinematics we can find the kinematic energy of the system formed by the disk and the cylindrical axis

          KE = 0.23 J

given parameters

  • Disk radius R = 15 cm = 0.15 m
  • Cylinder radius r = 1.5 cm = 0.0015 m
  • Disk mass M = 20 kg
  • Time t = 1.2 s
  • Force F = 12 N

to find

  • Kinetic energy (KE)

This exercise must be solved in parts:

1st part. Endowment kinetic energy is the energy due to the circular motion of an object and is described by the equation

         KE = ½ I w²

Where KE is the kinetic energy, I the moment of inertia and w the angular velocity

The moment of inertia is a magnitude that measures the inertia for rotational movement, it is a scalar quantity, therefore it is additive. In this system it is composed of two bodies, the disk and the cylindrical axis, for which the total moment of inertia it is

         I_{ total} = I_{ disk} + I_{ cylinder}

the moments of inertia with respect to an axis passing through the center of mass are tabulated

disk          I_{disk} = ½ M R²

cylinder   I_{cylinder} = ½ m r²

where M and m are the masses of the disk and cylinder respectively, R and r their radii

         I_{total} = ½ (M R² + m r²) = ½ M R² ( 1 + \frac{m}{M} \ (\frac{r}{R})^2 )

         I_{total} = ½ M R² ( 1+ \frac{m}{20}  (\frac{0.015}{0.15} )^2 ) = \frac{1}{2} M R² (1 + 0.005 m)

As the shaft mass  is much lighter than the disk mass , the last term is very small, which is why we despise it.

         I_{total} = ½ M R²

2nd part. Let's use Newton's second law for endowment motion

        τ = I α

        α = \frac{\tau }{I_{total}}l

        τ = F R

        α = \frac{F \ R}{I_{total}}

With the rotational kinematics expressions, we assume that the system starts from rest (w₀ = 0)

        w = w₀ + α  t

where w is the angular velocity, alpha is the angular acceleration and t is the time

        w = 0 + \frac{\tau }{I_{total}} \ t

we substitute in the kinetic energy equation

        KE = ½ I_{total}  ( \frac{ \tau }{I_{total}} \ t )²

        KE = ½ \frac{ \tau^2 }{I_{total}} \ t^2

let's substitute

        KE = \frac{F^2 \ R^4}{M \ R^2 } \ t^2

        KE = F² R² t² / M

let's calculate

        KE = 12² 0.15² 1.2² / 20

        KE = 0.23 J

With the concepts of energy and rotational kinematics we can find the kinetic energy of the system is

       KE = 0.23 j

learn more about rotational kinetic energy here:

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Halogens therefore react most vigorously with Group 1 and Group 2 metals of all main group elements.
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