A book is pushed with an initial horizontal velocity of 5.0 meters per second off the top of a 1.19 meter high desk. How far awa

#1

The volume of lead measures 100 cm^3

with a density of lead at 11.34 g/cm^3

. Thus, the mass of the lead block equals density multiplied by volume

Therefore, its weight in air is noted as

Next, the buoyant force acting on the lead is defined as

We know that

After solving, we find

V = 11.22 cm^3

(ii) This corresponding volume of water exerts the same weight as the buoyant force, resulting in 0.11 N

(iii) The buoyant force measures 0.11 N

(iv) The lead block sinks in water due to its density being greater than that of water.

#2

The buoyant force acting on the lead block counterbalances its weight

(ii) This volume of mercury corresponds to the buoyant force weight, confirming that the block floats within mercury, resulting in 11.11 N as its weight.

(iii) The buoyant force is recorded as 11.11 N

(iv) Given that lead's density is less than mercury's, the lead will float in the mercury medium.

#3

Indeed, an object that has lesser density than a liquid will float; otherwise, it will sink in the liquid.

Answer:

a)106.48 x 10⁵ kg.m²

b)144.97 x 10⁵ kgm² s⁻¹

Explanation:

a)Given

m = 5500 kg

l = 44 m

The moment of inertia for one blade

= 1/3 x m l²

where m denotes the mass of the blade

l represents the length of each blade.

Substituting the necessary values, the moment of inertia for one blade is

= 1/3 x 5500 x 44²

= 35.49 x 10⁵ kg.m²

Total moment of inertia for 3 blades

= 3 x 35.49 x 10⁵ kg.m²

= 106.48 x 10⁵ kg.m²

b) The angular momentum 'L' is calculated using

L = x ω

where,

= the moment of inertia of the turbine i.e 106.48 x 10⁵ kg.m²

ω= angular velocity =2π f

f represents the frequency of rotation of the blade i.e 13 rpm

f = 13 rpm=>= 13 / 60 revolutions per second

ω = 2π f => 2π  x  13 / 60 rad / s

L= x ω =>106.48 x 10⁵ x   2π  x  13 / 60

  = 144.97 x 10⁵  kgm² s⁻¹    

Answer: SG = 2.67

The specific gravity for the sand is 2.67

Explanation:

Specific gravity is determined by the formula: density of the substance/density of water

Provided information;

Mass of sand m = 100g

The volume of sand equals the volume of water it displaces

Vs = 537.5cm^3 - 500 cm^3

Vs = 37.5cm^3

Calculating density of sand = m/Vs = 100g/37.5 cm^3

Ds = 2.67g/cm^3

Density of water Dw = 1.00 g/cm^3

Thus, the specific gravity of the sand can be expressed as

SG = Ds/Dw

SG = (2.67g/cm^3)/(1.00g/cm^3)

SG = 2.67

The specific gravity of the sand stands at 2.67

Answer:

1. The force applied by the shelf supporting the book.

Explanation:

The free body diagram for the book is represented as follows:

1 - The weight of the book acting downward

2 - The normal force exerted by the shelf upward on the book.

As the book remains stationary, these two forces balance each other, and in accordance with Newton's Third Law, the reactive force equivalent to gravity is opposite and equal to the weight of the book. This reaction force prevents the book from falling off the shelf.

Response:

1 slit width = 0.158 mm, slit separation = 0.633 mm, distance between diffraction maxima = 12.7 mm

Explanation:

This involves defining several terms:

λ, the wavelength of the beam

D, the distance from the plane to the slit

x, the distance between minima in the diffraction pattern (in single slit setups)

w, the fringe width for double slit setups

1. In a double slit experiment, the fringe width is also recognized as the distance between Maxima.

Thus, w=λD/d, leading to d=λD/w when rearranged.

So, d= (632.8 x 10^-9 x 1 x 10^3)/1

which equals d= 0.633mm.

For single slit diffraction, minima are defined by a*sinΘ= m*x

where a is the slit width and m is an integer.

For small angles, it simplifies to Θ= (x/D) = (λ/a), allowing us to solve for a:

a = λD/a

a = (632.8x10^-9 x 1)/4

yielding a= 0.158mm.

2. A grating with 20 slits/mm gives d= 1/20mm= 0.05x10^-3.

To find y (the distance between maxima), apply y= λL/d:

Finally, y= (632.8 x 10^-9 x 1)/0.05x10^-3, which results in y= 12.7mm.