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Contact [7]
2 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]2 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]2 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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PART ONE
stepladder [879]

Answer:

1. A1, B2, C3

2. 47.1°

Explanation:

Sum of forces in the x direction:

∑Fₓ = ma

f − Fᵥᵥ = 0

f = Fᵥᵥ

Sum of forces in the y direction:

∑Fᵧ = ma

N − W = 0

N = W

Sum of moments about the base of the ladder:

∑τ = Iα

Fᵥᵥ h − W (b/2) = 0

Fᵥᵥ h = ½ W b

Fᵥᵥ (l sin θ) = ½ W (l cos θ)

l Fᵥᵥ sin θ = ½ l W cos θ

The correct set of equations is A1, B2, C3.

At the smallest angle θ, f = Nμ.  Substituting into the first equation, we get:

Nμ = Fᵥᵥ

Substituting the second equation into this equation, we get:

Wμ = Fᵥᵥ

Substituting this into the third equation, we get:

l (Wμ) sin θ = ½ l W cos θ

μ sin θ = ½ cos θ

tan θ = 1 / (2μ)

θ = atan(1 / (2μ))

θ = atan(1 / (2 × 0.464))

θ ≈ 47.1°

5 0
3 years ago
Find the magnitude of the resultant of forces 6N and 8N acting at 240° to each other
Alexxx [7]

Answer:

magnitude of the resultant of forces is 11.45 N

Explanation:

given data

force F1 = 6N

force F2 = 8N

angle = 240°

solution

we get here resultant force that is express as

F(r) = \sqrt{F_1^2+F_2^2+2F_1F_2cos\ \theta}    ..............1

put here value and we get

F(r) = \sqrt{6^2+8^2+2\times 6\times 8 \times cos240}

F(r) =  11.45 N

so magnitude of the resultant of forces is 11.45 N

6 0
3 years ago
At the local playground, a 21-kg child sits on the right end of a horizontal teeter-totter, 1.8 m from the pivot point. On the l
Ksju [112]

Answer:

By convention a negative torque leads to clockwise rotation and a positive torque leads to counterclockwise rotation.

here weight of the child =21kgx9.8m/s2 = 205.8N

the torque exerted by the child Tc = - (1.8)(205.8) = -370.44N-m ,negative sign is inserted because this torque is clockwise and is therefore negative by convention.

torque exerted by adult Ta = 3(151) = 453N , counterclockwise torque.

net torque Tnet = -370.44+453 =82.56N , which is positive means counterclockwise rotation.

b) Ta = 2.5x151 = 377.5N-m

Tnet = -370.44+377.5 = 7.06N-m , positive ,counterclockwise rotation.

c)Ta = 2x151 = 302N-m

Tnet = -370.44+302 = -68.44N-m, negative,clockwise rotation.

5 0
2 years ago
Olivia wants to find out whether a substance will fluoresce. She says she should put it in a microwave oven. Do you agree with h
Sophie [7]
I disagree.
Many substances show fluorescence under ULTRAVIOLET light, NOT microwaves. :)
5 0
3 years ago
Read 2 more answers
Question 2
Ksivusya [100]

5.77 ×10^1^4 Hz is the green photon's frequency .

The distance between similar points (adjacent crests) in adjacent cycles of a waveform signal that is propagated in space is known as the wavelength. A wave's wavelength is often measured in meters (m), centimeters (cm), or millimeters (mm) (mm). The relationship between frequency and wavelength is inverse.

<h3>Given:</h3>

Wavelength of green light = 520 nm

f = c / λ

where, f = Frequency

            c = Speed of light = 3 × 10^8 m/s

            λ = Wavelength of light

∴ f = c / λ

  f = \frac{3*10^8}{520 * 10^-^9}

    = 5.77 ×10^1^4 Hz

Therefore,  5.77 ×10^1^4 Hz is the green photon's frequency .

Learn more about wavelength here:

brainly.com/question/10728818

#SPJ1

4 0
1 year ago
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