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Sholpan [36]
3 years ago
10

A 4.0-m-long, 500 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building

under construction. A 70 kg construction worker stands at the far end of the beam.
What is the magnitude of the torque about the bolt due to the worker and the weight of the beam?
Physics
1 answer:
Andre45 [30]3 years ago
5 0

Answer:

T=12544 N*m

Explanation:

Given

L=4.0m

ms=500kg

mw=70kg

Torque is the force in a distance the relation is proportional so the torque of weight first is:

Ts = Fs*d

Ts = ms*g*L

Ts = 500kg*9.8m/s^2*2m

Ts = 9800 N*m

now torque of the worker

Tw = Fw*d

Tw = 70kg*9.8m/s^2*4m

Tw = 2744 N*m

Torque net is

Tnet = Tw+Ts

Tnet= 2744 + 9800 =12544 N*m

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You are a member of an alpine rescue team and must project a box of supplies, with mass m, up an incline of constant slope angle
AlekseyPX

Answer:

v = \sqrt{2gh + \frac{2\mu_kgh}{tan\theta}}

Explanation:

When we push the box from the bottom of the incline towards the top then by work energy theorem we can say that

Work done by all the forces = change in kinetic energy of the system

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here we know that

F_f = \mu_k mg cos\theta

also we know that the length of the incline is given as

s = \frac{h}{sin\theta}

now we have

- mgh - \mu_k mgcos\theta(\frac{h}{sin\theta}) = -\frac{1}{2}mv^2

so we have

v = \sqrt{2gh + \frac{2\mu_kgh}{tan\theta}}

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3 years ago
Carlos uses a rope to pull his car 30 m to a parking lot because it ran out of gas. If Carlos exerts 2,000 N of force to pull th
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<span>5.8 × 104 J
i already checked it on edge

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6 0
3 years ago
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A particle executes simple harmonic motion with an amplitude of 1.69 cm. At what positive displacement from the midpoint of its
m_a_m_a [10]

Answer: 0.0146m

Explanation: The formula that defines the velocity of a simple harmonic motion is given as

v = ω√A² - x²

Where v = linear velocity, A = amplitude = 1.69cm = 0.0169m, x = displacement.

The maximum speed of a simple harmonic motion is derived when x = A, hence v = ωA

One half of maximum speed = speed of motion

3ωA/2 = ω√A² - x²

ω cancels out on both sides of the equation, hence we have that

A/2 = √A² - x²

(0.0169)/2 = √(0.0169² - x²)

0.00845 = √(0.0169² - x²)

By squaring both sides, we have that

0.00845² = 0.0169² - x²

x² = 0.0169² - 0.00845²

x² = 0.0002142

x = √0.0002142

x = 0.0146m

5 0
4 years ago
Which of the following is NOT true of fusion?
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C., used in power plants I think. 
6 0
4 years ago
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An emission ray has a frequency of 5.10 × 10¹Ê Hz. Given that the speed of light is 2.998 × 10§ m/s, what is the wavelength of t
scoundrel [369]
The answer is 5.88 · 10⁻⁷<span> m.</span>

To calculate this we will use the light equation:

v = λ · f,
where:
v - the speed of light (units: m/s)
<span>λ - the wavelength of the ray (units: m)
</span>f - the frequency of the ray (units: Hz = 1/s <span>since Hz means cycles per second (f=1/T))
</span>
It is given:
f = 5.10 · 10¹⁴ Hz = 5.10 · 10¹⁴<span> 1/s
v = 2.998 </span>· 10⁸<span> m/s
</span><span>λ = ?
</span>
If v = λ · f, then λ = v ÷ f:
λ = 2.998 · 10⁸ m/s ÷ 5.10 · 10¹⁴ 1/s
   = 0.588 · 10⁸⁻¹⁴ · m
   = 0.588 · 10⁻⁶ m
   = 5.88 · 10⁻⁷ m
6 0
4 years ago
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