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3 years ago

11

One solution to minimize resonance with buildings is to ______ the width to span ratio.

1 answer:

Shtirlitz [24]3 years ago

3 0

Answer:

Increase

Explanation:

Resonance is a phenomenon which occurs when a body A in motion set another body B into motion of it own natural frequency. So for resonance to be minimize in a body is to increase the width to span ratio. So as to reduce the overall vibration which affects directly building resonance, the stiffness or trusses and girders should be increase. The increase in this aspect helps to reinforce building structure and support.

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Proved that<br>V = u+at<br>

kondaur [170]

Answer:

$\sf Proof \ below$

Explanation:

We know that acceleration is change in velocity over time.

$\sf a=\frac{\triangle v}{t}$

$\sf a=\frac{v-u}{t}$

v is the final velocity and u is the initial velocity.

Solve for v.

Multiply both sides by t.

$\sf at=v-u$

Add u to both sides.

$\sf at + u=v$

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3 years ago

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A mass of 25kg is moving and is stopped in 1.2 seconds with an acceleration of 1.0 m/s squared what is the force on the mass?

Phantasy [73]

F=ma therefore 25kg*1.0m/s^2=25N force on the mass

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3 years ago

A uniform, solid, 2000.0 kgkg sphere has a radius of 5.00 mm. Find the gravitational force this sphere exerts on a 2.10 kgkg poi

mars1129 [50]

Answer:

$0.0110284391534\ N$

$0.0653784219002\ N$

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

$m_1$ = Mass of sphere = 2000 kg

$m_2$ = Mass of other sphere = 2.1 kg

r = Distance between spheres

Force of gravity is given by

$F=\dfrac{Gm_1m_2}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 2000\times 2.1}{(5.04\times 10^{-3})^2}\\\Rightarrow F=0.0110284391534\ N$

The gravitational force is $0.0110284391534\ N$

$F=\dfrac{Gm_1m_2}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 2000\times 2.1}{(2.07\times 10^{-3})^2}\\\Rightarrow F=0.0653784219002\ N$

The gravitational force is $0.0653784219002\ N$

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3 years ago

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ehidna [41]

The earth is 4.54 billion years old.

7 0

3 years ago

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How many air molecules are in a 13.0×12.0×10.0 ft room (28.2 L=1 ft3)? Assume atmospheric pressure of 1.00 atm, a room temperatu

aksik [14]

Answer:

1963.93 Moles

Explanation:

-We know the standard conversion ratio for the volume of a mole is $1 Mole=22.4L$

Given volume of rooms as $13.0ft\times12.0ft\times10ft=1560 ft^3$

Convert the volume into liters: $1560\times28.2l=43992l$

#From our conversion ratio above, we get the volume of air molecules in moles as:

$V_m=\frac{43992L}{22.4L}\times 1\ mole\\\\=1963.93\ Moles$

Hence, the volume of air molecules is 1963.93 Moles

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3 years ago