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3 years ago

6

Explain how an object moving at a constant speed can have a nonzero acceleration

2 answers:

SOVA2 [1]3 years ago

8 0

Its very simple if a body is moving in circle the magnitude of its velocity remain constant but its direction changes because velocity is directed towards tangent and at every point in a cirlce its direction will be different (along tangent) so velocity is not uniform .As acceleration is the rate change of velocity so it will be non zero because velocity is changing due to its direction.

laila [671]3 years ago

8 0

Acceleration depends on the change in an object’s velocity. An object moving at a constant speed can experience a nonzero acceleration if the direction of the object’s motion changes.

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Answer:

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The equation Bold r (t )equals(8 t plus 9 )Bold i plus (2 t squared minus 8 )Bold j plus (6 t )Bold k is the position of a parti

KATRIN_1 [288]

Explanation:

It is given that, the position of a particle as as function of time t is given by :

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Read 2 more answers

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is

dmitriy555 [2]

Answer:

v = 1.15*10^{7} m/s

Explanation:

given data:

charge/ unit area $= \sigma = 1.99*10^{-7} C/m^2$

plate seperation = 1.69*10^{-2} m

we know that

electric field btwn the plates is $E = \frac{\sigma}{\epsilon}$

force acting on charge is F = q E

Work done by charge q id $\Delta X =\frac{ q\sigma \Delta x}{\epsilon}$

this work done is converted into kinectic enerrgy

$\frac{1}{2}mv^2 =\frac{ q\sigma \Delta x}{\epsilon}$

solving for v

$v = \sqrt{\frac{2q\Delta x}{\epsilon m}$

$\epsilon = 8.85*10^{-12} Nm2/C2$

$v = \sqrt{\frac{2 1.6*10^{-19}1.99*10^{-7}*1.69*10^{-2}}{8.85*10^{-12} *9.1*10^{-31}}$

v = 1.15*10^{7} m/s

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3 years ago