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Ymorist [56]
3 years ago
11

Give an example in which velocity is zero yet acceleration is not.

Physics
2 answers:
Bezzdna [24]3 years ago
7 0

Answer:

Explanation:

The time rate of change of displacement is called velocity and it is a vector quantity.

The time rate of change of velocity is called acceleration, it is also a vector quantity.

A body can have zero velocity and non zero acceleration.

For example, when you throw a ball vertically upwards with some velocity, the ball reaches to maximum height. At maximum height its velocity is zero but an acceleration is acting on the ball which is the acceleration due to gravity and which acts vertically downwards always.

NeTakaya3 years ago
4 0

Answer:

(1) Oscillation of simple pendulum (2) Dropping a ball from a certain height vertically

Explanation:

Instantaneous velocity can be zero even if acceleration is non zero .

Example 1 : Consider the case of a simple pendulum which is oscillating, when bob is at the extreme position, then the bob stops for a moment ( velocity is zero there ) but acceleration is non zero due to which it reverses it’s motion .

Example 2 : When someone drop the ball from a certain height then the initial velocity of the ball is zero at that point but the acceleration is not zero it is in downward direction . This downward acceleration will increase the velocity from zero.

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In a transformer, energy is carried from the primary coil to the secondary coil by magnetic field in the iron core.

To find the answer, we have to know more about the transformer.

<h3>How transformer works?</h3>
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5 0
2 years ago
A 90 kg man stands in a very strong wind moving at 17 m/s at torso height. As you know, he will need to lean in to the wind, and
victus00 [196]

Answer:

a)  t=195.948N.m

b)  \phi=13.6 \textdegree

Explanation:

From the question we are told that:

Density \rho=1.225kg/m^2

Velocity of wind v=14m/s

Dimension of rectangle:50 cm wide and 90 cm

Drag coefficient \mu=2.05

a)

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F=\frac{1}{2}\muA\rhov^2

F=\frac{1}{2}2.05(50*90*\frac{1}{10000})*1.225*17^2

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t=163.29*1.2*sin90

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b)

Generally the equation for torque due to weight is mathematically given by

t=d*Mg*sin90

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d=sin \phi

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t=sin \phi*Mg*sin90

195.948=833sin \phi

\phi=sin^{-1}\frac{195.948}{833}

\phi=13.6 \textdegree

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