Ian at 11:18, 20 May 2010

2010-05-20T11:18:29Z

←Older revision		Revision as of 11:18, 20 May 2010
Line 3:		Line 3:
	Here <math>T</math> is the tension in the cable and <math>a</math> is the acceleration of the packet.		Here <math>T</math> is the tension in the cable and <math>a</math> is the acceleration of the packet.

-	The acceleration is in the opposite direction to the motion.This means mathematically the acceleration is negative. <math></math>	+	The acceleration is in the opposite direction to the motion.This means mathematically the acceleration is negative, <math>a=-0 \textrm{.}1\text{ m}{{\text{s}}^{\text{-2}}}</math>

	Applying Newton's second law		Applying Newton's second law

Ian: New page: Image:10.1.gif Here $T$ is the tension in the cable and $a$ is the acceleration of the packet. The acceleration is in the opposite direction to the motion.This ...

2010-05-20T11:15:14Z

New page: Image:10.1.gif Here <math>T</math> is the tension in the cable and <math>a</math> is the acceleration of the packet. The acceleration is in the opposite direction to the motion.This ...

New page

[[Image:10.1.gif]]

Here <math>T</math> is the tension in the cable and <math>a</math> is the acceleration of the packet.

The acceleration is in the opposite direction to the motion.This means mathematically the acceleration is negative. <math></math>

Applying Newton's second law

<math>\begin{align}
& \uparrow :\ \text{Resultant}\ \text{Force}=T-300\times 9 \textrm{.}81=
ma = 300\times (-0 \textrm{.}1) \\
& \\
& T=-30+2943\approx 2910\ \text{N} \\
\end{align}</math>

Solution 10.1c - Revision history

Ian at 11:18, 20 May 2010

Ian: New page: Image:10.1.gif Here T is the tension in the cable and a is the acceleration of the packet. The acceleration is in the opposite direction to the motion.This ...

Ian: New page: Image:10.1.gif Here $T$ is the tension in the cable and $a$ is the acceleration of the packet. The acceleration is in the opposite direction to the motion.This ...