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ANEK [815]
3 years ago
10

A large crane consists of a 20 m, 3,000 kg arm that extends horizontally on top of a vertical tower. The arm extends 15 m toward

s the lifting end and 5 m towards the counterweight. If the crane is to lift a 5,000 kg load, what must the weight of the counterweight be in order to maintain static equilibrium
Physics
1 answer:
Anni [7]3 years ago
7 0

Answer:

m=18000kg

Explanation:

From the question we are told that:

Crane Length l=20m

Crane Mass m_a=3000kg

Arm extension at lifting end l_l=15m

Arm extension at counter weight end l_c=5m

Load M_l=5000kg

Generally the equation for Torque Balance is mathematically given by

T_1 *l_c-(m_a*g) *l_c-(T_2)*l_l=0

mg*5 *-(3000*9.8) *5-(5000*9.8)*15=0

m=18000kg

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5 0
3 years ago
Which of the following best describes what causes the pases of the moon.?
Alborosie
The correct answer is<span> The earth casts a shadow on the moon

The Earth casts a shadow on the moon and that's why we can't see the remaining parts. The parts that we can see are what we call the lunar phases.</span>
4 0
3 years ago
A heavy anvil is suspended by a 0.75 m long steel wire that has a mass of 12 g. When the wire is plucked, it hums at its fundame
Dima020 [189]

Explanation:

It is given that,

length of steel wire, l = 0.75 m

Mass of the wire, m = 12 g = 0.012 kg

Fundamental frequency, f = 120 Hz

We need to find the mass of the anvil (m'). The fundamental frequency is given by :

f=\dfrac{v}{2l}

v is the speed of the mass

Speed is given by :

v=\sqrt{\dfrac{T}{\mu}}

\mu is the mass per unit length,\mu=\dfrac{m}{l}

f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}

T is the tension in the wire,

f=\dfrac{1}{2l}\sqrt{\dfrac{Tl}{m}}

T=4f^2lm

T=4(120)^2\times 0.75\times 0.012

T = 518.4 N

Tension in the wire, T = m' g

m'=\dfrac{T}{g}

m'=\dfrac{518.4}{9.8}

m' = 52.89 kg

So, the mass of the anvil is 52.89 kg. Hence, this is the required solution.

6 0
4 years ago
Unpolarized light of intensity I0 = 950 W/m2 is incident upon two polarizers. The first has its polarizing axis vertical, and th
Ket [755]

Answer:

Intensity of the light (first polarizer) (I₁) = 425 W/m²

Intensity of the light (second polarizer) (I₂) = 75.905 W/m²

Explanation:

Given:

Unpolarized light of intensity (I₀) = 950 W/m²

θ = 65°

Find:

a. Intensity of the light (first polarizer)

b. Intensity of the light (second polarizer)

Computation:

a. Intensity of the light (first polarizer)

Intensity of the light (first polarizer) (I₁) = I₀ / 2

Intensity of the light (first polarizer) (I₁) = 950 / 2

Intensity of the light (first polarizer) (I₁) = 425 W/m²

b. Intensity of the light (second polarizer)

Intensity of the light (second polarizer) (I₂) = (I₁)cos²θ

Intensity of the light (second polarizer) (I₂) = (425)(0.1786)

Intensity of the light (second polarizer) (I₂) = 75.905 W/m²

5 0
4 years ago
The force that opposes motion to moving parts is _____
Colt1911 [192]
The force that opposes motion to moving parts is F<span>riction</span><span>

Hope this helped!
</span>
6 0
4 years ago
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