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S_A_V [24]
3 years ago
13

A light spring having a force constant of 115 N/m is used to pull a 9.00 kg sled on a horizontal frictionless ice rink. The sled

has an acceleration of 2.10 m/s2. Part A By how much does the spring stretch if it pulls on the sled horizontally
Physics
1 answer:
Ksivusya [100]3 years ago
8 0

Answer:

Stretch in the spring = 0.1643 (Approx)

Explanation:

Given:

Mass of the sled (m) = 9 kg

Acceleration of the sled (a) = 2.10 m/s ²

Spring constant (k) = 115 N/m

Computation:

Tension force in the spring  (T) = ma

Tension force in the spring  (T) = 9 × 2.10

Tension force in the spring  (T) = 18.9 N

Tension force in the spring = Spring constant (k) × Stretch in the spring

18.9 N = 115 N  × Stretch in the spring

Stretch in the spring = 18.9 / 115

Stretch in the spring = 0.1643 (Approx)

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The magnitude of the force on the charge by the influence of the magnetic field will be 6.6*10^-3 N

<h3>What is magnetic force?</h3>

Magnetic force, attraction or repulsion that arises between electrically charged particles because of their motion.The magnitude of the magnetic force acting on the charge is given by:

\rm F=qvBsin\theta

where

The magnitude of the charge   q=5.7\ \mu C =5.7\times 10^{-6} \ C

The velocity of the charge        v=4.5\times 10^5\ \frac{m}{s}

The magnitude of the magnetic field   3.2\ mT=0.0032\ T

The angle between the directions of v and B  \theta =90^o-37^o=53^o

By substituting the values we will get:

F=(5.7\times 10^}-6})(4.5\times 10^5)(0.0032)(Sin53^o)

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