Describe your acceleration if you ride your bike up a hill,then ride down the other side.

A simple pendulum consists of a point mass suspended by a weightless, rigid wire in a uniform gravitation field. Check all that

9966 [12]

Answer:

f.The period is independent of the suspended mass.

Explanation:

The period of a pendulum is given by

$T=2 \pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration due to gravity

From the formula, we see that:

1) the period of the pendulum depends only on its length, L, and it is proportional to the square root of the length

2) the period does not depend neither on the mass of the pendulum, nor on its amplitude of oscillation

So, the only correct statements are

f.The period is independent of the suspended mass.

Note: statement "e.The period is proportional to the length of the wire" is also wrong, because the period is NOT proportional to the length of the wire, but it is proportional to the square root of it.

3 0

2 years ago

What is the required force to bring a 1550 kg vehicle moving at 32 m/s to a stop in a distance of 45 m?

Vilka [71]

The answer is 9 i think.

8 0

3 years ago

Velocity vector and acceleration vector in a uniform circular motion are related as.

mr_godi [17]

They are related as $\bold{\underline {v}\,.\,\underline a }= \bold{0}$

In a uniform circular motion, the magnitude of the speed does not change during the travel and only the instantaneous direction changes.
This speed is always directed along the tangent to the circle at a given point. (refer to the figure attached)
For any circular motion, the must-have acceleration is the centripetal acceleration that is directed towards the centre of the circular locus (if the motion has a tangential acceleration, it has a tangential acceleration additionally).
Therefore, both the directions of the tangential speed and the centripetal acceleration are orthogonal to each other (perpendicular: one is 90 degrees apart from the other).
In mathematics, 2 vectors ( $\underline p$ , $\underline q$ ) that are perpendicular to each other have a quality that their dot product ( $\underline p\,.\, \underline q$ ) equal to zero vector ( $\bold 0$ ) which is written as $\undeline p\,.\, \underline q = \bold 0$ .
This quality can be considered when dealing with the velocity vector and the acceleration vector in a manner $\underline v\,.\, \underline a =\bold 0$ .