Answer:
To calculate the tension on a rope holding 1 object, multiply the mass and gravitational acceleration of the object. If the object is experiencing any other acceleration, multiply that acceleration by the mass and add it to your first total.
Explanation:
The tension in a given strand of string or rope is a result of the forces pulling on the rope from either end. As a reminder, force = mass × acceleration. Assuming the rope is stretched tightly, any change in acceleration or mass in objects the rope is supporting will cause a change in tension in the rope. Don't forget the constant acceleration due to gravity - even if a system is at rest, its components are subject to this force. We can think of a tension in a given rope as T = (m × g) + (m × a), where "g" is the acceleration due to gravity of any objects the rope is supporting and "a" is any other acceleration on any objects the rope is supporting.[2]
For the purposes of most physics problems, we assume ideal strings - in other words, that our rope, cable, etc. is thin, massless, and can't be stretched or broken.
As an example, let's consider a system where a weight hangs from a wooden beam via a single rope (see picture). Neither the weight nor the rope are moving - the entire system is at rest. Because of this, we know that, for the weight to be held in equilibrium, the tension force must equal the force of gravity on the weight. In other words, Tension (Ft) = Force of gravity (Fg) = m × g.
Assuming a 10 kg weight, then, the tension force is 10 kg × 9.8 m/s2 = 98 Newtons.
I think its Coulomb's law<span>
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<h2><u>Question</u><u>:</u><u>-</u></h2>
Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the work done by Ryan?
<h2><u>Answer:</u><u>-</u></h2>
<h3>Given,</h3>
=> Force applied by Ryan = 10N
=> Distance covered by the book after applying force = 30 cm
<h3>And,</h3>
30 cm = 0.3 m (distance)
<h3>So,</h3>
=> Work done = Force × Distance
=> 10 × 0.3
=> 3 Joules
The deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h is 
.
The acceleration in opposite direction is known as the deceleration. Basically the deceleration is negative value of the acceleration since the negative sign depicts its opposite in direction.
The given data:
time, t = 1.1 s
initial speed, u = 1000 km/h = 
final speed, v = 0 m/s
So we will be using the equation of motion, that is,
v = u + at
Hence , the deceleration of the rocket is .
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