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Diano4ka-milaya [45]
3 years ago
15

A book is sitting on the dashboard of a car that is stopped at a traffic light as the car starts to move forward the book slides

backward off the dashboard what does the book actually do
Physics
1 answer:
Zina [86]3 years ago
3 0

Answer:

The book remained in its state of rest before the car started to move forward as no direct force acted on it.

Explanation:

According to Newton's first law of motion, a body will continue in its present state of rest, or if it is in motion, will continue to move with uniform speed in a straight line unless aced upon by an external force. This tendency of a body to remain in its state of reset or uniform motion in a straight line is known as inertia and is directly proportional to the mass of the body. The more massive a body, the more inertia it possesses. Thus Newton's first law is also known as the law of inertia.

Considering the case of the book on the dashboard of a stationary car which suddenly starts to move. While the car is stopped at the traffic light, the dashboard where the book sits and the book are both at rest. When the car begins to move forward, the dashboard moves forward with it. However as the book is not a part of the car, no force is directly acting on it, so the book so it stays at rest due to its inertia.

Therefore, as the car is moving forward, the stationary book appears to move backward from the reference point of the car, sliding off the dashboard.

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Answer:

a) 2.94 N

b) 4.90 N

Explanation:

Let us assume that the weight of the ball is 0.98 N

Solution:

a) An object’s gravitational potential energy depends on two factors which are height and its weight (or mass). The equation for gravitational potential energy (PE) is given as:  

Potential energy = weight (w) * height (h)

PE = wh

Potential energy at 2 m shelf = weight * height = 0.98 N * 2 m = 1.96 N

Potential energy at 5 m shelf = weight * height = 0.98 N * 5 m = 4.90 N

The work needed to lift the ball from the 2-m shelf to the 5-m shelf = Potential energy at 5 m shelf - Potential energy at 2 m shelf

The work needed to lift the ball from the 2-m shelf to the 5-m shelf = 4.90 N - 1.96 N = 2.94 N

b) Potential energy at 5 m shelf = weight * height = 0.98 N * 5 m = 4.90 N

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Liquid water at 120 kPa enters a 7-kW pump where its pressure is raised to 5.6 MPa. If the elevation difference between the exit
ziro4ka [17]

Answer:

The answer is given below

Explanation:

Things provided in the statement:

Pressure <em>P1</em> = 120 kPa and <em>P2</em> = 5.6 MP or 5600 kPa

Power, <em>W</em> = 7 kW

Elevation difference = ∆z = 10 m

Mass of flow = m˙

So potential energy changes may be significant

Specific volume of water V= 0.001 m³/kg

Now putting the values in the formula

Power, <em>W </em>= m˙  x V (<em>P1 - P2</em>) + m˙ x g x ∆z

             7 = m˙  x 0.001 (5600 - 120 ) + m˙ x 9.8 x 10 x (1 kJ/kg/ 1000 m^2/s^2)

             7 = m˙ x 5.48 + m˙ x 0.098

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Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

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a) The Reynold number is equal to:

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The boundary layer thickness is equal to:

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Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar

The boundary layer is equal to:

\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft

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\tau=0.332(\frac{2.359x10^{-5}*3 }{2}  )(493015.6^{0.5} )=0.0082

c) The drag coefficient is equal to:

C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019

The friction drag is equal to:

F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf

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