**Ijmb physics questions and answers**

*IJMB PHYSICS*

physics

number 1a

Fundamental quantities, also known as base quantities or fundamental physical quantities, are the basic measurable properties in a system of measurement. These quantities cannot be defined in terms of other quantities and serve as the foundation for deriving other derived quantities.

Here are three examples of fundamental quantities along with their units in the International System of Units (SI):

1. Length: The fundamental quantity representing the extent of a one-dimensional space.

Unit: Meter (m)

2. Mass: The fundamental quantity representing the amount of matter in an object.

Unit: Kilogram (kg)

3. Time: The fundamental quantity representing the duration or sequence of events.

Unit: Second (s)

It’s worth noting that the selection of fundamental quantities may vary depending on the system of measurement being used. The SI system is the most widely used international system and forms the basis for scientific and technical measurements.

physics number 2

(a) When the lift is stationary, the reading on the spring balance will be equal to the weight of the object, which is the force exerted by gravity on the object.

The weight (force due to gravity) can be calculated using the formula:

Weight = mass * acceleration due to gravity

In SI units, the acceleration due to gravity is approximately 9.8 m/s².

Weight = 6 kg * 9.8 m/s² = 58.8 N

Therefore, the reading on the spring balance when the lift is stationary will be 58.8 Newtons.

(b) When the lift is moving with an acceleration of 0.3 m/s² upwards, we need to consider the additional force acting on the object due to the lift’s acceleration.

The net force acting on the object in this case will be the sum of its weight (force due to gravity) and the force caused by the lift’s acceleration.

Net force = Weight + Force due to acceleration

Weight = 6 kg * 9.8 m/s² = 58.8 N

Force due to acceleration = mass * acceleration

Force due to acceleration = 6 kg * 0.3 m/s² = 1.8 N

Net force = 58.8 N + 1.8 N = 60.6 N

Therefore, the reading on the spring balance when the lift is moving upwards with an acceleration of 0.3 m/s² will be 60.6 Newtons.

(c) If the lift cable breaks and the lift falls freely under gravity, the object will experience weightlessness or zero apparent weight. This is because both the object and the spring balance will fall with the same acceleration due to gravity.

In this scenario, the reading on the spring balance will be zero Newtons since there is no upward force acting on the object.

Please note that this assumes ideal conditions and neglects any air resistance or other external forces that might affect the actual motion of the falling lift.

11

(i) Vector quantities: These are quantities that have both magnitude (size or value) and direction. They are represented by vectors, which are mathematical objects with both magnitude and direction. Examples of vector quantities include displacement, velocity, acceleration, force, and momentum.

Scalar quantities: These are quantities that have only magnitude (size or value) but no direction. They are represented by scalars, which are mathematical objects with magnitude only. Examples of scalar quantities include mass, temperature, time, distance, speed, and energy.

Now let’s determine the class of each of the given quantities:

Momentum: Momentum is a vector quantity because it has both magnitude (mass times velocity) and direction.

Current: Current is a scalar quantity because it represents the flow of charge and has magnitude only.

Electrical energy: Energy is a scalar quantity as it only has magnitude and no direction.

Acceleration due to gravity: Acceleration due to gravity is a vector quantity because it has both magnitude (9.8 m/s^2) and direction (downwards towards the center of the Earth).

Mass: Mass is a scalar quantity as it represents the amount of matter and has magnitude only.

ii) The term “relative velocity” refers to the velocity of an object as observed from the perspective of another moving object. It is the difference between the velocities of the two objects relative to a common frame of reference.

(iii) Initial velocity of the truck, V_truck = 12 m/s (constant velocity)

Acceleration of the automobile, a_auto = 3 m/s^2

Initial velocity of the automobile, u_auto = 0 m/s (at rest)

To find the distance beyond the starting point where the automobile overtakes the truck, we can use the equations of motion. The equation for the distance traveled (s) by an object with initial velocity (u), acceleration (a), and time (t) is:

s = ut + (1/2)at^2

Let’s assume the automobile overtakes the truck after time t. At that point, the distance traveled by the automobile (s_auto) will be equal to the distance traveled by the truck (s_truck). Since the truck has a constant velocity, its distance traveled is given by:

s_truck = V_truck * t

For the automobile, its initial velocity is u_auto = 0 m/s, and its acceleration is a_auto = 3 m/s^2. The distance traveled by the automobile is:

s_auto = (1/2) * a_auto * t^2

Since the two distances are equal, we can equate the expressions:

V_truck * t = (1/2) * a_auto * t^2

Simplifying the equation, we have:

2 * V_truck = a_auto * t

Substituting the given values:

2 * 12 = 3 * t

t = 8 seconds

Now we can find the distance traveled by the automobile at this time:

s_auto = (1/2) * a_auto * t^2

s_auto = (1/2) * 3 * (8)^2

s_auto = 96 meters

Therefore, the automobile will overtake the truck at a distance of 96 meters beyond its starting point. To find the final velocity of the automobile at that time, we can use the equation:

v_auto = u_auto + a_auto * t

v_auto = 0 + 3 * 8

v_auto = 24 m/s

So, the automobile will be traveling at 24 m/s when it overtakes the truck.

14b

The term “specific heat capacity” refers to the amount of heat energy required to raise the temperature of a given amount of substance by a certain amount. It is the measure of how much heat energy is needed to raise the temperature of a substance per unit mass. The specific heat capacity of a substance is typically denoted by the symbol ‘c’ and is expressed in units of J/(kg·K) (joules per kilogram per Kelvin).

Specific latent heat, on the other hand, refers to the amount of heat energy required to change the state of a substance without changing its temperature. It is specifically associated with phase changes, such as melting, vaporization, or condensation. The specific latent heat of a substance is the amount of heat energy required per unit mass to undergo a phase change. It is usually denoted by the symbol ‘L’ and has units of J/kg (joules per kilogram).

Now, regarding the densities of substances ‘A’ and ‘B,’ let’s assume the density of substance ‘A’ is ‘ρA’ and the density of substance ‘B’ is ‘ρB.’ According to the given information, the ratio of the densities is 5:6. Mathematically, this can be represented as:

ρA/ρB = 5/6

This means that the density of substance ‘A’ is 5/6 times the density of substance ‘B.’ In other words, if the density of substance ‘B’ is taken as a reference, the density of substance ‘A’ is 5/6 of that reference density.

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