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butalik [34]
3 years ago
6

A 5.00 kg object oscillates back and forth at the end ofa spring whose spring constant is 49.3 N/m. An obersever istraveling at

a speed of 2.80 *10^8 m/s relative to the fixed end ofthe spring. What does this observer measure for the period ofoscillation.
The answer in the back of the book is 5.57 seconds, but I keepgetting 30 seconds. Can someone help tell me what I am doing wronghere is my work:
t= period observer sees
to=real period
to=2π*√(m/k)=2π*√(5.00kg/49.3N/m)=2.00sec
t=to/(1-√(v^2/c^2)), where v=2.80*10 ^8 m/s andc=3.0*10^8 m/s
t=2.00/(1-√((2.80*10^8)^2/(3.00*10^8)^2))=30seconds
Physics
1 answer:
defon3 years ago
8 0

Answer:

5.571 sec

Explanation:

angular frequency = √ (k/m) = √ (49.3 / 5) = 3.14 rad/s

Period To = 2π / angular frequency

Period To = 2π/3.14 = 2 × 3.14 / 3.142 = 2.00 sec which you got

T measured by the observer = To / (√ (1 - (v²/c²))) = 2 / √( 1 - 0.871111) = 2 / 0.35901 = 5.571 sec

t=2.00/(1-√((2.80*10^8)^2/(3.00*10^8)^2))= should have been  ( To / (√ (1 - (v²/c²))).  where To = 2.00 sec

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Rufina [12.5K]

Answer:

The  angular speed is w = 5.89 \ rad/s

Explanation:

From the question we are told that

    The time taken is  t = 1.6 s

    The number of somersaults  is n  =  1.5

The total angular displacement during the somersault is mathematically represented as

         \theta  =  n *  2 *  \pi

substituting values

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       \theta  = 9.426 \ rad

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3 0
3 years ago
A ball is thrown up into the air with an initial velocity of 18 m/s. A) How high does the ball go? B) Calculate the time needed
kaheart [24]

Answer:

B) t = 1.83 [s]

A) y = 16.51 [m]

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f} =v_{o} -g*t

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Vf = final velocity = 0

Vo = initial velocity = 18 [m/s]

g = gravity acceleration = 9.81 [m/s²]

t = time [s]

Note: the negative sign in the above equation means that the acceleration of gravity is acting in the opposite direction to the motion.

A) The maximum height is reached when the final velocity of the ball is zero.

0 = 18 - (9.81*t)

9.81*t = 18

t = 18/9.81

t = 1.83 [s], we found the answer for B.

Now using the following equation.

y = y_{o} + v_{o}*t - 0.5*g*t^{2}\\

where:

y = elevation [m]

Yo = initial elevation = 0

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y = 16.51 [m]

7 0
3 years ago
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lys-0071 [83]
The answer to that would be that 

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8 0
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Angelina_Jolie [31]

Answer:

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As water is falling from 60m high, its potential energy from 60m high would convert to power. So the rate of change in potential energy is

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3 0
3 years ago
Water initially at 200 kPa and 300°C is contained in a piston–cylinder device fitted with stops. The water is allowed to cool at
netineya [11]

Answer:

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We replace the values in equation

Δu=u₁-u₃

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3 0
3 years ago
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