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Contact [7]
3 years ago
6

A force of gravity pulls downward on a book on a table. What force prevents the book from accelerating downward?

Physics
2 answers:
Nadusha1986 [10]3 years ago
3 0

Answer: The Normal Force

Explanation:

According to Third Newton's Law,  also konwn as The principle of action and reaction :

<em>"If a body A exerts an action on another body B, it performs on A another action that is the same and in the opposite direction."</em>

In this case this is what happens with the book on a table,  gravity force pulls it downward, but the Normal Force (perpendicular to the surface of the table) acts in the opposite direction with the same magnitude, preventing the book from accelerating downward.

Gnoma [55]3 years ago
3 0

Answer: The support / normal force

Explanation: The force that prevents downward acceleration is the support force.

The force that acts on the book at rest balancing up the gravitational force is the normal force.

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A horizontal uniform bar of mass 3 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the
kirill115 [55]

Answer:

T₁ = 2.8125 N

Explanation:

The equilibrium equation of the moments at the point where string 2 is located on the bar is like this:

∑M₂ = 0

M₂ = F*d

Where:

∑M₂  : Algebraic sum of moments in the the point (2) of the bar

M₂ : moment in the point 2 ( N*m)

F  : Force ( N)

d  : Horizontal distance of the force to the point 2 ( N*m

Data

mb = 3 kg : mass of the  bar

mm = 1.5 kg :  mass of the  monkey

L = 3m : lengt of the bar

g = 9.8 m/s²: acceleration due to gravity

Forces acting on the bar

T₁ : Tension in string 1 (vertical upward)

T₂ : Tension in string 2 (vertical upward)

Wb :Weihgt of the bar (vertical downward)

Wm: Weihgt of the monkey  (vertical downward)

Calculation of the weight of the bar (Wb) and of the monkey(Wm)

Wb = m*g = 3 kg*9.8 m/s² = 29.4 N

Wm = m*g = 1.5 kg*9.8 m/s² = 14.7 N

Calculation of the distances  from forces the point 2

d₁₂ = (3-0.6) m = 2.4m  : Distance from T1 to the point 2

db₂ = (3÷2) m = 1.5 m : Distance from Wb to the point 2

dm₂ = (3÷2) m = 1.5 m : Distance from Wm to the point 2

Equilibrium  of moments at the point  2 on the bar

∑M₂ = 0

T₁(d₁₂) - Wb(db₂) - Wm(dm₂) = 0

T₁(2.4) -3*(1.5) - 1.5*(1.5) = 0

T₁(2.4) =3*(1.5) + 1.5*(1.5)

T₁(2.4) =6.75

T₁ = 6.75 / (2.4)

T₁ = 2.8125 N

5 0
3 years ago
What occurs when two Stars collide into each other?
ddd [48]

Answer:

A stellar collision.

Explanation:

A stellar collision is the coming together of two stars caused by stellar dynamics within a star cluster, or by the orbital decay of a binary star due to stellar mass loss or gravitational radiation, or by other mechanisms not yet well understood.

5 0
3 years ago
Read 2 more answers
If an object is being pulled by two forces, one 4 N to the left and the other 2 N to the right, what is the net
nika2105 [10]

Answer:

2N

Explanation:

subtract  rthe two forces to see which is greater

4-2=2

6 0
3 years ago
The magnetic field at point P due to a 2.0-A current flowing in a long, straight, thin wire is 8.0 μT. How far is point P from t
Tanzania [10]

Answer:

r = 0.05 m = 5 cm

Explanation:

Applying ampere's law to the wire, we get:

B = \frac{\mu_oI}{2\pi r}\\\\r =  \frac{\mu_oI}{2\pi B}

where,

r = distance of point P from wire = ?

μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²

I = current = 2 A

B = Magnetic Field = 8 μT = 8 x 10⁻⁶ T

Therefore,

r = \frac{(4\pi\ x\ 10^{-7}\ N/A^2)(2\ A)}{2\pi(8\ x\ 10^{-6}\ T)}\\\\

<u>r = 0.05 m = 5 cm</u>

8 0
3 years ago
A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and t
Katen [24]

Answer:

0.358g

Explanation:

Density of Helium = 0.179g/L

ρ=m/v

m=ρv

when the volume was 2L

m1= 0.179*2

m1=0.358g

when the volume increased to 4L

m2= 0.179*4

m2=0.716g

gram of helium added = 0.716g-0.358g

=0.358g

5 0
3 years ago
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