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Nikolay [14]
3 years ago
11

Jamie pushes a book off a table. The push is an example of a contact force because A. Jamie used energy. B. Jamie had to touch t

he book to move it. C. the book fell to the floor. D. the book was on a flat surface.
Physics
2 answers:
gizmo_the_mogwai [7]3 years ago
6 0
From the word "contact" clearly defines it as touch. three types of forces: contact -  field -  electro-static. thus the answer is C : Jamie hd to touch the book to move it'
 
olasank [31]3 years ago
3 0

Explanation :

Contact forces are the forces that act between the objects that are in contact.

When James pushes a book off a table, the push is an example of a contact force. This is because Jamie had to touch the book to move it. For experiencing a contact force, there should be contact between two objects.

The examples of contact forces are the frictional force, tension force, normal force etc.

While gravitational force, electric force and the magnetic force are non contact force.

So, the correct option is (B) " Jamie had to touch the book to move it ".

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A ball starts at rest and rolls down an inclined plane. The ball reaches 7.5 m/s in 3 seconds. What is the acceleration?
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Answer:

a=2.5\ m/s^2

Explanation:

<u>Motion With Constant Acceleration </u>

It's a type of motion in which the velocity of an object changes uniformly over time.

The equation that describes the change of velocities is:

v_f=v_o+at

Where:

a   = acceleration

vo = initial speed

vf  = final speed

t    = time

Solving the equation for a:

\displaystyle a=\frac{v_f-v_o}{t}

The ball starts at rest (vo=0) and rolls down an inclined plane that makes it reach a speed of vf=7.5 m/s in t=3 seconds.

The acceleration is:

\displaystyle a=\frac{7.5-0}{3}

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Sammy squirrel is steering his boat at a heading of 327 degree at 18mph. The current is flowing at 4mph at a heading of 60 degre
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Answer:

  • 59.97 º at 18.23 mph

Explanation:

To find Sammy's course you have to add the two velocities (vectors), 18 mph 327º and 4 mph 60º.

To add the two vectors analytically you decompose each vector into their vertical and horizontal components.

<u>1. 18 mph 327º</u>

  • Horizontal component: 18 mph × cos (327º) = 15.10 mph

  • Vertical component: 18 mph × sin (327º) = - 9.80 mph

  • Vector notation:

       15.10\hat i-9.80\hat j

<u>2. 4 mph 60º</u>

  • Horizontal component: 4 mph × cos (60º) = 2.00 mph

  • Vertical component: 4 mph × sin (60º) = 3.46 mph

  • Vector notation:

       2.00\hat i+3.46\hat j

<u>3. Addition:</u>

You add the corresponding components:

15.10\hat i-9.80\hat j+2.00\hat i+3.46\hat j\\ \\ 17.10\hat i-6.34\hat j

To find the magnitude use Pythagorean theorem:

  • \sqrt{17.1^2+6.34^2}= 18.23

<u>4. Direction:</u>

Use the tangent ratio:

  • tan(\alpha )=opposite/adjacent=3.46/2.00=1.73

Find the inverse:

  • arctan (1.73) ≈ 59.97º
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