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Contact [7]
3 years ago
14

Where in the motion is the magnitude of the force from the spring on the object zero? Where in the motion is the magnitude of th

e force from the spring on the object a maximum? Where in the motion is the magnitude of the net force on the object zero? Where in the motion is the magnitude of the net force on the object a maximum?
Physics
1 answer:
kkurt [141]3 years ago
8 0

<em></em>

Answer:

1. The magnitude of the force from the spring on the object is zero on <em>Equilibrium.</em>

2. The magnitude of the force from the spring on the object is a maximum on <em>The top and bottom.</em>

3. The magnitude of the net force on the object is zero on <em>The Bottom.</em>

4. The magnitude of the force on the object is a maximum on <em>the Top.</em>

Explanation:

<em>1. Because the change in position delta X is zero.</em>

<em>2. Because of delta X.</em>

<em>3. Beacuse, the force of gravity and the force of the spring oppose each other to keep the block at rest, away from the equilibrium position.</em>

<em>4. Because, the force of the spring from compressiom and the force of gravity both act on the mass.</em>

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A car accelerates uniformly from rest to a speed of 6.6 m/s in 6.5 s. Find the acceleration the car presents during this time?
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1.02 m/s²

Explanation:

The following data were obtained from the question:

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Acceleration can simply be defined as the change of velocity with time. Mathematically, it can be expressed as:

a = (v – u) / t

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Initial velocity (u) = 0 m/s

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2 years ago
The volumes of a sample of lead and a sample of platinum are the same: 2 cm3. The density of lead is 11.3 g/cm3 and the density
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Read 2 more answers
One end of a string is fixed. An object attached to the other end moves on a horizontal plane with uniform circular motion of ra
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Answer:

If both the radius and frequency are doubled, then the tension is increased 8 times.

Explanation:

The radial acceleration (a_{r}), measured in meters per square second, experimented by the moving end of the string is determined by the following kinematic formula:

a_{r} = 4\pi^{2}\cdot f^{2}\cdot R (1)

Where:

f - Frequency, measured in hertz.

R - Radius of rotation, measured in meters.

From Second Newton's Law, the centripetal acceleration is due to the existence of tension (T), measured in newtons, through the string, then we derive the following model:

\Sigma F = T = m\cdot a_{r} (2)

Where m is the mass of the object, measured in kilograms.

By applying (1) in (2), we have the following formula:

T = 4\pi^{2}\cdot m\cdot f^{2}\cdot R (3)

From where we conclude that tension is directly proportional to the radius and the square of frequency. Then, if radius and frequency are doubled, then the ratio between tensions is:

\frac{T_{2}}{T_{1}} = \left(\frac{f_{2}}{f_{1}} \right)^{2}\cdot \left(\frac{R_{2}}{R_{1}} \right) (4)

\frac{T_{2}}{T_{1}} = 4\cdot 2

\frac{T_{2}}{T_{1}} = 8

If both the radius and frequency are doubled, then the tension is increased 8 times.

5 0
2 years ago
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