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SVEN [57.7K]
3 years ago
13

A machine part is undergoing SHM with a frequency of 5.00Hz and amplitude 1.80cm . How long does it take the part to gofrom x=0

to x= -1.80 cm
well i know that x=Acos(ωt)
and

ω=2πf so we have -1.8cm= 1.8cm * cos(2πft)

we end up with -1=cos(2πft) I know that cos(π)=-1so ft =1/2 but if that is true

then I have 5s^-1*t= 1/2 gives t= .1s

not .05s like the book says can anyone help me with the problem I'mhaving.
Physics
1 answer:
kifflom [539]3 years ago
5 0

Answer:

0.05 second

Explanation:

frequency, f = 5 H

time period is defined as the reciprocal of the frequency.

T = 1/f = 1/5 = 0.2 s

Amplitude, A = 1.80 cm

Time period is also defined as the time taken to complete one oscillation.

As the particle moves from one extreme position to one extreme position is one forth the time period of the oscillation.

As the particle goes from x = 0 to x = - 1.80 cm, so the time taken by the particle is one forth the time period of the oscillation.

t = t / 4 = 0.2 / 4 = 0.05 second.

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A tiger travels 3m/s^2 for 4.1s,what was its initial speed if it's final speed was 55k/h?
zaharov [31]
Acceleration a=3m/s^2
time t= 4.1seconds
Final velocity V= 55km/h
initial velocity U= ?
First convert V to m/s
36km/h=10m/s
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Using the formula V= U+at
U= V-at
U= 15.28-3*4.1=15.28-12.3=2.98m/s
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8 0
3 years ago
A sample of an ideal gas has a volume of 2.37 L at 2.80×102 K and 1.15 atm. Calculate the pressure when the volume is 1.68 L and
iogann1982 [59]

Answer:

p_2 = 1.76 atm

Explanation:

given data:

v_1 = 2.37 L

v_2 = 1.68 L

p_1 =1.15 atm

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t_1 = 280 K

t_2 = 304 K

from Gas Law Equation

, WE HAVE

\frac{p_1 v_1}{t_1} =\frac{p_2 v_2}{t_2}

Putting the values

\frac{1.15*2.37}{280}  =\frac{p_2 *1.68}{304}

9.733*10^{-3} = \frac{p_2 *1.68}{304}

9.733*10^{-3}*304 = p_2*1.68

\frac{9.733*10^{-3}*304}{1.68} =p_2

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7 0
4 years ago
A 1.00 kg block of ice, at -25.0°C, is warmed by 35 kJ of energy. What is the final temperature of the ice?
ahrayia [7]

Answer:

-8.4°C

Explanation:

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The heat sustain by an object is given as;

H = m× c× (T2-T1)

Where H is heat transferred

m is mass of substance

T2-T1 is the temperature change from starting to final temperature T2.

c- is the specific heat capacity of ice .

Note : specific heat capacity is an intrinsic capacity of a substance which is the energy substained on a unit mass of a substance on a unit temperature change.

Hence ; 35= 1× c× ( T2-(-25))

35= c× ( T2+25)

35 =2.108×( T2+25)

( T2+25)= 35/2.108= 16.60°{ approximated to 2 decimal place}

T2= 16.60-25= -8.40°C

C, specific heat capacity of ice is =2.108 kJ/kgK{you can google that}

6 0
3 years ago
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Answer:

Paying for employees seminars and workshops related to their careers

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8 0
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