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5 days ago

An object moving at a constant velocity travels 274 m in 23 s. what is its velocity?

Physics

2 answers:

Softa [2K]5 days ago

6 0

V= 274 meters / 23 sec

V= 11.91 meters per sec

ValentinkaMS [2.4K]5 days ago

4 0

Solution:

Velocity, v = 11.91 m/s

Justification:

We have the following details:

Distance traveled by the object, d = 274 m

Time taken, t = 23 s

We are tasked with determining the object's velocity. It can be calculated using the formula of distance divided by time, that is,[[(TAG_19]]

$v=\dfrac{d}{t}$

$v=\dfrac{274\ m}{23\ s}$

v = 11.91 m/s

Thus, the object's velocity is 11.91 m/s. This is the solution we seek.

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Sav [2226]

Answer:

Please refer to the explanation

Explanation:

Race distance is 5km

Top speed = 45 yards

Converting yards to kilometers:

1km equals 1093.613 yards

x = 45 yards

(1093.613 * x) = 45

x = 45 / 1093.613

x = 0.0411480 km

Where x indicates the maximum distance he can sustain his highest speed in kilometers.

Thus, from the data available, we can determine that Lamar will not be able to maintain his maximum speed for the full 5km race, as he can only sustain it for 0.0411 kilometers.

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11 days ago

Calculate the weight of a 4.5 kg rabbit.

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6 days ago

a. For a spring-mass oscillator, if you double the mass but keep the stiffness the same, by what numerical factor does the perio

kicyunya [2264]

Answer:

a) factor $b=\sqrt{2}$

b) factor $b=\frac{1}{2}$

c) factor $b=1$

d) factor $b=1$

Explanation:

For an oscillating spring-mass system, the time period is expressed as:

$T=\frac{1}{f}$

$T={2\pi} \sqrt{\frac{m}{k} }$

where:

$f=$ represents the frequency of oscillation

$m=$ signifies the mass linked to the spring

$k=$ is the spring's stiffness constant

a) If the mass is doubled:

New mass, $m'=2m$

Thus, the new time period:

$T'=2\pi\sqrt{\frac{m'}{k} }$

$T'=2\pi\sqrt{\frac{2m}{k} }$

$T'=\sqrt{2}\times 2\pi\sqrt{\frac{m}{k} } }$

$T'=\sqrt{2} \times T$

this leads to factor $b=\sqrt{2}$ as per the question.

b) When the stiffness constant is quadrupled, holding other factors constant:

New stiffness constant, $k'=4k$

Thus, the new time period:

$T'=2\pi\sqrt{\frac{m}{k'} }\\\\T'=2\pi\sqrt{\frac{m}{4k} }\\\\T'=\frac{1}{2} \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=\frac{1}{2} \times T$

this results in factor $b=\frac{1}{2}$ as required.

c) When both mass and stiffness constant are quadrupled:

New stiffness, $k'=4k$

New mass, $m'=4m$

Thus, the new time period:

$T'=2\pi\sqrt{\frac{m'}{k'} }\\\\T'=2\pi\sqrt{\frac{4m}{4k} }\\\\T'=1 \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=1 \times T$

which leads to factor $b=1$ as stated in the question.

d) If amplitude is quadrupled, the time period remains unaffected because T does not depend on amplitude as demonstrated by the equation.

Thus, factor $b=1$

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Answer:

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According to Newton's third law, for every action, there is an equal and opposite reaction.

Refer to the documents for a thorough solution and clarification of the issue.

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