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natali 33 [55]
2 years ago
5

Is position a base or derived quantity?

Physics
1 answer:
amid [387]2 years ago
8 0

Position is measured in meters (m), so it is a base quantity.

<h3>What is base quantity?</h3>

A base or fundamental  quantity is a physical quantity, in which other quantities are derived from.

Example of fundamental quantities;

  • Mass
  • Length (position)
  • Time
  • Temperature
  • Amount of substance

<h3>What is a derived quantity?</h3>

Derived quantities are those quantities obtained or expressed from fundamental quantities.

Example of derived quantities;

  • Speed
  • Acceleration
  • Volume
  • Area
  • Density, etc

Thus, we can conclude that position measured in meters (m) is a base quantity.

Learn more about base quantities here: brainly.com/question/14480063

#SPJ1

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1.Suppose someone pulls a cart up a ramp a distance of 85cm along the ramp with a force of 15N.
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1. 12.75 J

Assuming that the force applied is parallel to the ramp, so it is parallel to the displacement of the cart, the work done by the force is

W=Fd

where

F = 15 N is the magnitude of the force

d = 85 cm = 0.85 m is the displacement of the cart

Substituting in the formula, we get

W=(15 N)(0.85 m)=12.75 J


2. 10.6 N

In this part, the cart reaches the same vertical height as in part A. This means that the same work has been done (because the work done is equal to the gain in gravitational potential energy of the object: but if the vertical height reached is the same, then the gain in gravitational potential energy is the same, so the work done must be the same).

Therefore, the work done is

W=Fd=12.75 J

However, in this case the displacement is

d = 120 cm = 1.20 m

Therefore, the magnitude of the force in this case is

F=\frac{W}{d}=\frac{12.75 J}{1.20 m}=10.6 N

3 0
3 years ago
A car of mass 487 kg travels around a flat, circular race track of radius 53.3 m. The coefficient of static friction between the
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9.96 m/s

Explanation:

mass of car, m = 487 kg

radius of track, R = 53.3 m

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acceleration due to gravity, g = 9.8 m/s^2

let v be the maximum speed so that the car can go without flying off the track.

The formula for the maximum speed is given by

v_{max}=\sqrt{\mu Rg}

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vmax = 9.96 m/s

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