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4 years ago

? is the change in position relative to an initial location

Physics

2 answers:

umka21 [38]4 years ago

7 0

Answer: c. displacement

Explanation:

Let's begin by stating clear that motion is the change of position of a body at a certain time. So, during this motion, the body will have a trajectory and a displacement.

In the specific case of the displacement, it is defined as the distance in a straight line between the initial and final position (is a vector magnitude).

Therefore:

Displacement s the the change in position relative to an initial location.

Elodia [21]4 years ago

6 0

Answer:

c: displacement

Explanation:

When a person moves from one point to another, the shortest distance between the two points in a straight line is what we call as the displacement. It is a vector quantity. Which means the change in position with respect to an initial position will be displacement.

Speed is a measure of change of distance relative to time. Velocity is a measure of change of displacement relative to time and acceleration is a measure of change of velocity relative to time.

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The biodiversity of cultivated crops is currently under threat

Angelina_Jolie [31]

C. Establishing a seed bank with a variety of seeds of

different crop species

Explanation:

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3 years ago

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3 years ago

The maximum speed of a mass m on an oscillating spring is vmax . what is the speed of the mass at the instant when the kinetic a

umka21 [38]

Let
A = the amplitude of vibration
k = the spring constant
m = the mass of the object

The displacement at time, t, is of the form
x(t) = A cos(ωt)
where
ω = the circular frequency.

The velocity is
v(t) = -ωA sin(ωt)

The maximum velocity occurs when the sin function is either 1 or -1.
Therefore
$v_{max} = \omega A$
Therefore
$v(t) = -V_{max} sin(\omega t)$

The KE (kinetic energy) is given by
$KE = \frac{m}{2}v^{2} = \frac{m}{2} V_{max}^{2} sin^{2} (\omega t)$

The PE (potential energy) is given by
$PE = \frac{k}{2} x^{2} = \frac{k}{2} A^{2} cos^{2} (\omega t)$

When the KE and PE are equal, then
$v^{2} = \frac{k}{m} A^{2} cos^{2} (\omega t)$

For the oscillating spring,
$\omega ^{2} = \frac{k}{m} \\ V_{max} = \omega A = \sqrt{ \frac{k}{m} } A$
Therefore
$v^{2} = \frac{k}{m} \frac{m}{k} V_{max}^{2} cos^{2} ( \sqrt{ \frac{k}{m} t} ) \\ v = V_{max} \,cos( \sqrt{ \frac{k}{m} t} )$

Answer: $v(t) = V_{max} cos( \sqrt{ \frac{k}{m} t} )$

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4 years ago

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Dafna1 [17]

Answer:

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