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2 years ago

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Can the velocity of a body revese the direction when acceleration is constant?

1 answer:

TEA [102]2 years ago

7 0

Answer:

Yes, the velocity of the object can reverse direction when its acceleration is constant. For example consider that the velocity of any object at any time t is given as: ... At At t = 0 sec, the magnitude of velocity is 2m/s and is moving in the forward direction i.e.v (t) = -2.

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A floating ice block is pushed through a displacement d = (14 m) i hat - (11 m) j along a straight embankment by rushing water,

Alika [10]

Explanation:

Given that,

Displacement in ice block, $d=14i-11j$

Force exerted by water, $F=158i-179j$

To find,

Work done by the force during the displacement.

Solve,

We know that the product of force and displacement is called work done. It is also equal to the dot product of force and displacement as :

$W=F.d$

$W=(158i-179j).(14i-11j)$

We know that, i.i = j.j = k.k = 1

$W=2212+1969=4181\ J$

So, the work done by the force on the block during the displacement is 4181 Joules.

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2 years ago

__________ is/are formed between two air masses that have different temperatures. (4 points) A: Wind B:The jet stream C:Clouds D

lidiya [134]

The answer is B: the jet stream.

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2 years ago

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two strands of DNA produced during replication are copies of each other because each strand in the double helix is

Luden [163]

The correct answer you should be looking for is complementary. :)

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2 years ago

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Plzzz answer this question correctly

tekilochka [14]

Answer:

changing the direction in which a force is exerted

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3 years ago

A student fills a tank of radius r with water to a height of h1 and pokes a small, 1.0 cm diameter hole at a distance h2 from th

Alik [6]

when a hole is made at the bottom of the container then water will flow out of it

The speed of ejected water can be calculated by help of Bernuolli's equation and Equation of continuity.

By Bernoulli's equation we can write

$Po + \frac{1}{2}\rho v_1^2 + \rho g h = Po + \frac{1}{2}\rho v_2^2 + \rho g *0$

Now by equation of continuity

$A_1v_1 = A_2v_2$

$\pi (0.2)^2 v_1 = \pi (0.01)^2 v_2$

from above equation we can say that speed at the top layer is almost negligible.

$v_1 = 0$

now again by equation of continuity

$\rho g h = \frac{1}{2} \rho v^2$

$v = \sqrt{2 g h}$

here we have

$h = h_1 - h_2$

$h = 0.50 - 0.03 = 0.47m$

now speed is given by

$v = \sqrt{2* 9.8 * 0.47}$

$v = 3.03 m/s$

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3 years ago