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iren2701 [21]
3 years ago
15

A small bolt with a mass of 33.0 g sits on top of a piston. The piston is undergoing simple harmonic motion in the vertical dire

ction with a frequency of 3.05 Hz. What is the maximum amplitude that the piston can oscillate without the bolt losing contact with the piston's surface?
Physics
1 answer:
densk [106]3 years ago
5 0

Answer:

0.027m

Explanation:

the bolt loses contact with the piston only when acceleration due to gravity equals acceleration of piston

ω² * A = g where ω is angular velocity, A amplitude, g acceleration due to gravity

ω is given by 2πf, ω² is 4π²f²

A= g/4π²f² depending on the value of g used either 10m/s² or 9.8m/s²,

i used 10m/s² in this answer

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A 410-kg piano is being unloaded from a truck by rolling it down a ramp inclined at 25°. There is negligible friction and the ra
SpyIntel [72]

Answer:BOONK GANG

Explanation:

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5 0
3 years ago
A car covers 120 km in 3 hours calculate its speed in m/s​
8_murik_8 [283]

Answer:

11.1 m/s

Explanation:

120 km = 120 x 1,000 = 120,000 m

3 hours = 3 x 60 x 60 = 3600 x 3 = 10,800 s

speed = 120,000 / 10,800

= 1200/108

= 11.1 m/s

6 0
2 years ago
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ikadub [295]
I think the answer is A?
7 0
2 years ago
the frequency of a beam of uv light is 1.0 ×10 ^15hz what is the energy in one quantum of this light express it in ev? ​
kap26 [50]

Answer:

4.14 eV

Explanation:

f = 1.0 ×10^15 Hz

h= 6.63×10^-34 J s (  this is called PLANCK 'S CONSTANT)

ENEGY = E = ?

E = hf  ( THIS IS FORMULA FOR ENERGY OF ONE QUANTA OR ONE PHOTON )

E= 6.63×10^-34×1.0 ×10^15

E = 6.63×10^-19 J

As 1eV = 1.6×10^-19 J so changing energy in eV from joules we will divide energy by 1.6×10^-19

hence E in eV = 6.63×10^-19/(1.6×10^-19)

          E = 4.14 eV

7 0
3 years ago
An inventor claims to have invented a heat engine that receives 750kJ of heat from a source at 400K and produces 250kJ of net wo
IRISSAK [1]

Answer:

the claim is not valid or reasonable.

Explanation:

In order to test the claim we will find the maximum and actual efficiencies. maximum efficiency of a heat engine can be found as:

η(max) = 1 - T₁/T₂

where,

η(max) = maximum efficiency = ?

T₁ = Sink Temperature = 300 K

T₂ = Source Temperature = 400 K

Therefore,

η(max) = 1 - 300 K/400 K

η(max) = 0.25 = 25%

Now, we calculate the actual frequency of the engine:

η = W/Q

where,

W = Net Work = 250 KJ

Q = Heat Received = 750 KJ

Therefore,

η = 250 KJ/750 KJ

η = 0.333 = 33.3 %

η > η(max)

The actual efficiency of a heat engine can never be greater than its Carnot efficiency or the maximum efficiency.

<u>Therefore, the claim is not valid or reasonable.</u>

3 0
3 years ago
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