Problem: An ice hockey player hits a puck of mass 0.15 kilograms with a force of 100 newtons in the horizontal direction. What i

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:

The tension in the string when the speed increased is 134.53 N

Explanation:

Given;

Tension in the string, T = 120 N

initial speed of the transverse wave, v₁ = 170 m/s

final speed of the transverse wave, v₂ = 180 m/s

The wave speed is expressed as;

where;

μ represents mass per unit length

The new tension T₂ will be computed as;

Consequently, the tension in the string when the speed was increased is 134.53 N

The string vibrates in its third harmonic, where n = 3. The length of the string, l, measures 0.36 m. The frequency of the sound produced is f = 500 Hz. The speed of sound in air is 344 m/s. To find the speed of sound generated by the string in the third harmonic, we can apply the appropriate formula for frequency.

B) 14.0 N

To address this inquiry, we need to evaluate the kinetic energy of the box before and after crossing the rough section. The kinetic energy is given by the formula:

E = 0.5 M V^2

where

E = Energy

M = Mass

V = velocity

Now, utilizing the known data, we compute the energy prior and post.

Before:

E = 0.5 M V^2

E = 0.5 * 13.5kg * (2.25 m/s)^2

E = 6.75 kg * 5.0625 m^2/s^2

E = 34.17188 kg*m^2/s^2 = 34.17188 joules

After:

E = 0.5 M V^2

E = 0.5 * 13.5kg * (1.2 m/s)^2

E = 6.75 kg * 1.44 m^2/s^2

E = 9.72 kg*m^2/s^2 = 9.72 Joules

Hence, the box consumed energy equal to 34.17188 J - 9.72 J = 24.451875 J over a length of 1.75 meters. Next, we will calculate the loss per meter by dividing the energy loss by the distance traversed.
24.451875 J / 1.75 m = 13.9725 J/m = 13.9725 N

When we round to one decimal point, we arrive at 14.0 N, which corresponds with option “B.”

Answer:

a) W =400 kJ

b) W = 0 kJ

c) W =-160.944 KJ

Explanation:

Given

Process 1 ---> 2

Pressure at point (1) P1 = 10 bar = P2

Volume at point (1) V1 = 1 m^3

Volume at point (2) V2 =4 m^3

For Process 2 ---> 3, where V = constant

Pressure at point (3) P3 = 10 bar

Volume at point (3) V3 = 4 m^3

Process 3 ---> 1 defined as PV = constant.

Required

Sketch the processes on the PV coordinates To calculate the work in kJ

Work is calculated by W=

a=V2

b=V1

x=Pdx=dV

For Process 1 ---> 2 where P3 = P4 = 5 bar  

W=

a=V3

b=V2

x=4dx=dV

substituting the values here into the integral gives

W=400 kJFor Process 2 ---> 3

As V = constant in this case, the volume remains unchanged, resulting in W = 0 kJ  

For Process 3 ---> 1  By applying point (1) --> 5 x.2 = C ---> C = 1 P = 5V^-1  

 W=

a=V1

b=V3

x=1V^-1dx=dVsubstituting values into this integral results in

W=| ln V | limits a and b

  = -160.944 KJ