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Citrus2011 [14]
3 years ago
11

What is the instantaneous speed at 5 seconds? (remember s=d/t)

Physics
2 answers:
Sliva [168]3 years ago
5 0

Answer:

5 seconds is a poor time to ask about, because the speed abruptly changes at exactly 5 seconds.

Up until that time, the speed has been 1 m/s. And then, at exactly 5 seconds, it becomes zero.

_________

It's also a poor question because speed is calculated from the distance covered, but the graph shows displacement, not distance. You can't really tell the distance covered from a displacement graph. For example, if an object happens to be moving in a circle around the place where it started, then the total distance covered keeps increasing, but its displacement is constant.

Read more on Brainly.com - brainly.com/question/6618489#readmore

Explanation:

TEA [102]3 years ago
4 0
5 seconds is a poor time to ask about, because the speed abruptly changes at exactly 5 seconds.

Up until that time, the speed has been 1 m/s. And then, at exactly 5 seconds, it becomes zero.
_________

It's also a poor question because speed is calculated from the distance covered, but the graph shows displacement, not distance. You can't really tell the distance covered from a displacement graph. For example, if an object happens to be moving in a circle around the place where it started, then the total distance covered keeps increasing, but its displacement is constant.
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An open cart is rolling to the left on a horizontal surface. A package slides down a chute and lands in the cart. Which quantity
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Explanation:

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A long, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive charge density λ = 1 nC/m . A se
zheka24 [161]

Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, \lambda = 1\ nC = 1\times 10^{- 9}\ C

Charge density of rod 2, \lambda = - 1\ nC = - 1\times 10^{- 9}\ C

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}

Also,

\vec{E} = \frac{2K\lambda }{R}                  (1)

where

K = electrostatic constant = \frac{1}{4\pi \epsilon_{o} R}

R = Distance

\lambda = linear charge density

Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C

\vec{E} = 1800\ N/C     (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C

\vec{E'} = 1800\ N/C     (towards)

Now, the total field at the origin is the sum of both the fields:

\vec{E_{net}} = 1800 + 1800 = 3600\ N/C

7 0
3 years ago
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