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k0ka [10]
3 years ago
9

a 3.46 kg box is sitting at rest on a flat floor. a.) what is the weight of the box. b.) what is the normal force on the box

Physics
1 answer:
just olya [345]3 years ago
4 0

A) The weight of the box is given by:

W = mg

W = weight, m = mass, g = gravitational acceleration

Given values:

m = 3.46kg, g = 9.81m/s²

Plug in and solve for W:

W = 3.46(9.81)

W = 33.9N

B) The box is sitting at rest on the floor. This means that the box is in equilibrium; the two forces acting on it, its weight and the normal force that the floor exerts on it, are equal in magnitude and opposite in direction. Therefore the normal force N acting on the box has a magnitude of 33.9N

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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

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Answer: The drag force goes up by a factor of 4

Explanation:

The <u>Drag Force</u> equation is:

F_{D}=\frac{1}{2}C_{D}\rho A_{D}V^{2} (1)

Where:

F_{D} is the Drag Force

C_{D} is the Drag coefficient, which depends on the material

\rho is the density of the fluid where the bicycle is moving (<u>air in this case) </u>

A_{D} is the transversal area of the body or object

V the bicycle's velocity

Now, if we assume C_{D}, \rho and A_{D} do not change, we can rewrite (1) as:

F_{D}=C.V^{2} (2)

Where C groups all these coefficients.

So, if we have a new velocity V_{n} , which is the double of the former velocity:

V_{n}=2V (3)

Equation (2) is written as:

F_{D}=C.V_{n}^{2}=C.(2V)^{2}

F_{D}=4CV^{2} (4)

Comparing (2) and (4) we can conclude<u> the Drag force is four times greater when the speed is doubled.</u>

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What is the area of the bottom of a tank 30.0 cm long and 15.0 cm wide?​
Igoryamba

Answer:

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Explanation:

If you want to get area you need the Length, width,and height.

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34kurt

Answer:

the speed is equal to 6 m/s

Explanation:

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