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Dennis_Churaev [7]
3 years ago
13

What is motion define three types of motion with examples

Physics
1 answer:
Lana71 [14]3 years ago
3 0

Explanation:

the push or pull of an object is called motion,

There are not three type of motion but there are three law of motion

1.V=u+at

2.V²-U²=2as

3.s=ut+1/2at²

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Woman has a near point of 75.0 cm. What focal length, f, is necessary for the corrective lens so that he can hold his physics bo
Komok [63]

Answer:

The focal length of the lens must be 27.3cm

Explanation:

Image distance= -75 cm

Object distance= 20.0cm

From:

1/v + 1/u = 1/f

-1/75+ 1/20 = 1/f

11/300= 1/f

f= 300/11

f= 27.3cm

The focal length of the lens must be 27.3cm

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How long will it take for a car at rest to accelerate at 7m/s^2 to a speed of 45 m/s
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What type of properties would a scientist need to know to correctly place in new element into a group?
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3 0
3 years ago
In case A below, a 1 kg solid sphere is released from rest at point S. It rolls without slipping down the ramp shown, and is lau
mestny [16]

Answer:

the block reaches higher than the sphere

\frac{y_{sphere}} {y_block} = 5/7

Explanation:

We are going to solve this interesting problem

A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp

Let's use the concept of conservation of energy

starting point. At the top of the ramp

         Em₀ = U = m g y₁

final point. At the exit of the ramp

         Em_f = K + U = ½ m v² + ½ I w² + m g y₂

notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂

energy is conserved

          Em₀ = Em_f

         m g y₁ = ½ m v² + ½ I w² + m g y₂

angular and linear velocity are related

        v = w r

the moment of inertia of a sphere is

         I = \frac{2}{5} m r²

we substitute

         m g (y₁ - y₂) = ½ m v² + ½ (\frac{2}{5} m r²) (\frac{v}{r})²

         m g h = ½ m v² (1 + \frac{2}{5})

where h is the difference in height between the two sides of the ramp

h = y₂ -y₁

         mg h = \frac{7}{5} (\frac{1}{2} m v²)

         v = √5/7  √2gh

This is the exit velocity of the vertical movement of the sphere

         v_sphere = 0.845 √2gh

B) is the same case, but for a box without friction

   starting point

          Em₀ = U = mg y₁

   final point

          Em_f = K + U = ½ m v² + m g y₂

          Em₀ = Em_f

          mg y₁ = ½ m v² + m g y₂

          m g (y₁ -y₂) = ½ m v²

          v = √2gh

this is the speed of the box

          v_box = √2gh

to know which body reaches higher in the air we can use the kinematic relations

          v² = v₀² - 2 g y

at the highest point v = 0

           y = vo₀²/ 2g

for the sphere

           y_sphere = 5/7 2gh / 2g

           y_esfera = 5/7 h

for the block

           y_block = 2gh / 2g

            y_block = h

       

therefore the block reaches higher than the sphere

         \frac{y_{sphere}} {y_bolck} = 5/7

3 0
3 years ago
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