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Phantasy [73]
3 years ago
8

What does the force diagram look like for a book at rest on a desk?

Physics
1 answer:
Sever21 [200]3 years ago
7 0
An arrow representing the weight of the book acts downwards on the table. An arrow of equal magnitude, but opposite direction represents the normal reaction force of the table on the book.
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Match the following list of key words with their definitions
Korvikt [17]

Question 1 :

1.  VELOCITY

2.  SPEED

3.  NEWTON

4.  MOMENTUM

5.  MASS

Question 2

1.     FORCE

2.    FRICTION

3.    GRAVITY

4.    ELECTROAGNETIC FORCE

5.    ACCELERATION

6 0
2 years ago
A 17.6-kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally
leonid [27]

In order get the block up to a speed of 3.58 m/s in 1.77 s, it must undergo an acceleration <em>a</em> of

<em>a</em> = (3.58 m/s) / (1.77 s) ≈ 2.02 m/s²

When the spring is getting pulled, Newton's second law tells us

• the net vertical force is

∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0

where ∑ <em>F</em> is the net force, <em>n</em> is the magnitude of the normal force, and <em>mg</em> is the weight of the block - it follows that <em>n</em> = <em>mg</em> ; and

• the net horizontal force is

∑ <em>F</em> = <em>F</em> - <em>f</em> = <em>ma</em>

where <em>F</em> is the applied force, <em>f</em> is kinetic friction, <em>m</em> is the block's mass, and <em>a</em> is the acceleration found earlier. <em>F</em> stretches the spring by <em>x</em> = 0.250 m, so we have

<em>F</em> - <em>f</em> = <em>kx</em> - <em>µn</em> = <em>kx</em> - <em>µmg</em> = <em>ma</em>

where <em>k</em> is the spring constant and <em>µ</em> is the coefficient of kinetic friction.

When the block is being pulled at a constant speed, Newton's second law says

• the net vertical force is still

∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0

so that <em>n</em> = <em>mg</em> again; and

• the net horizontal force is

∑ <em>F</em> = <em>F</em> - <em>f</em> = 0

This time, <em>F</em> stretches the spring by <em>y</em> = 0.0544 m, so we have

<em>F</em> - <em>f</em> = <em>ky</em> - <em>µmg</em> = 0

Solve the equations in boldface for <em>k</em> and <em>µ</em> :

<em>kx</em> - <em>µmg</em> = <em>ma</em>

<em>ky</em> - <em>µmg</em> = 0

==>   <em>k</em> (<em>x</em> - <em>y</em>) = <em>ma</em>

==>   <em>k</em> = <em>ma</em> / (<em>x</em> - <em>y</em>)

==>   <em>k</em> = (17.6 kg) (2.02 m/s²) / (0.250 m - 0.0544 m) ≈ 182 N/m

Then

<em>ky</em> - <em>µmg</em> = 0

==>   <em>µ</em> = <em>ky </em>/ (<em>mg</em>)

==>   <em>µ</em> = (182 N/m) (0.0544 m) / ((17.6 kg) <em>g</em>) ≈ 0.0574

3 0
3 years ago
. The naturally occurring charge on the ground on a fine day out in the open country is –1.00nC/m2 . (a) What is the electric fi
JulijaS [17]

Answer:

a) -113 N/C

b) -423.71 V

Explanation:

The Step by step explanation and solution has been attached in the attachment section.

4 0
3 years ago
How do you know if you have all the forces needed for a FBD?
Korolek [52]
It’s so many FBD meaning be more specific so we can help you
5 0
3 years ago
A swimming pool, 10.0 m by 4.0 m, is filled with water to a depth of 3.0 m at a temperature of 20.2°c. if the energy needed to r
Tomtit [17]

You need to find the mass of water in the pool.
Find the volume (10 x 4 x 3) = 120 m3

Water has a density of 1000g/m3,so 120 m3 = 120 x 1000 = 120 000 kg

[delta]H = 4.187 x 120 000 x 3.4 (and the units will be kJ)

You then use the heat of combustion knowing that each mole of methane releases 891 kJ of heat so if you divide 891 into the previous answer, you will get the number of moles of CH4


8 0
3 years ago
Read 2 more answers
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