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PIT_PIT [208]
4 years ago
13

A 350 kg mass, constrained to move only vertically, is supported by two springs, each having a spring constant of 250 kN/m. A pe

riodic force with a maximum value of 100 N is applied to the mass with a frequency of 2.5 rad/s. Given a damping factor of 0.125, the amplitude of the vibration is
Physics
1 answer:
Arlecino [84]4 years ago
3 0

Answer:

Explanation:

Expression for amplitude of forced damped oscillation is as follows

A = \frac{F_0}{\sqrt{m(\omega^2-\omega_0^2)^2+b^2\omega^2} }

where

ω₀ =\sqrt{\frac{k}{m} }

ω₀ = \sqrt{\frac{500000}{350} }

=37.8

b = .125 ,

ω = 2.5

m = 350  

A = \frac{100}{\sqrt{350(37.8^2-2.5^2)^2+.125^2\times2.5^2} }

A = 3.75 mm . Ans

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A real heat engine operates between temperatures tc and th. during a certain time, an amount qc of heat is released to the cold
tino4ka555 [31]

q_{c} = Heat released to cold reservoir

q_{h} = Heat released to hot reservoir

W_{max} = maximum amount of work

t_{c} = temperature of cold reservoir

t_{h} = temperature of hot reservoir

we know that

\frac{q_{c}}{q_{h}}=\frac{t_{c}}{t_{h}}

q_{h} = (\frac{t_{h}}{t_{c}})q_{c}                                eq-1

maximum work is given as

W_{max} = q_{h} - q_{c}

using eq-1

W_{max} =  (\frac{t_{h}}{t_{c}})q_{c} - q_{c}



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4 years ago
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asambeis [7]
Recall the equation for magnetic force:

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7 0
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How come walking converts chemical energy into mechanical energy
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3 0
3 years ago
Read 2 more answers
A horizontal clothesline is tied between 2 poles, 12 meters apart. When a mass of 4 kilograms is tied to the middle of the cloth
Olegator [25]

Answer:

T=26.03 N

Explanation:

Given that

Distance between poles = 12 m

Mass of block m= 4 kg

Sag distance = 5 m

Lets take tension in the clothesline is T.

The component of tension in vertical direction will be T cosθ.

By force balancing

2 T cosθ = 40

here tan\theta =\dfrac{5}{6}

θ=39.80°

2 T cos39.8 = 40

T=26.03 N

6 0
3 years ago
Estimate the volume of a typical house (2050 feet squared in size and 10 feet tall) answer in units of meters squared
Aloiza [94]

Answer:

Volume = 6248.48 m^{3}

Explanation:

Given:

The area of the house A = 2050\ ft^{2}

The height of the house h=10\ ft

We need to find the volume of a typical house.

Solution:

We find the volume of the house by multiplying the area of the house and height of the house.

Volume = Area\times height

Volume = A\times h

Area and height of the house are known, so we substitute these value in above equation.

Volume = 2050\times 10

Volume = 20500\ ft^{3}

Now we convert the unit from feet to meter.

Divide the volume by 3.2808 for m^{3}

Volume = \frac{20500}{3.2808}

Volume = 6248.48\ m^{3}

Therefore, the volume of the house is 6248.48 m^{3}

8 0
4 years ago
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