Answer:
67 kPa
Explanation:
Given that,
Initial volume, V₁ = 25 cm³
Initial pressure, P₁ = 100 kPa
Final volume, V₂ = 15 cm³
We need to find the change in pressure of the gas. The relation between the volume and pressure of a gas is given by :

or
= 167 kPa
The change in pressure,
= P₂ - P₁
= 167 kPa - 100 kPa
= 67 kPa
Hence, the correct option is (a).
Answer:
Weight of body = 490 Newton
Explanation:
Mass of the body is the total amount of matter contained in the body but Weight of the body is the force exerted by gravity on the body. Weight of the body depends upon acceleration due to gravity of the planet.
Thus,
Weight = mass × acceleration due to gravity
W = mg
W = weight of the body
m = mass of the body
g = acceleration due to gravity
W = 50 × 9.8
W = 490 
W = 490 Newton
Hence, weight of the body is 490 Newton.
Answer:
The intensity of sound at rock concert is 10¹⁰ greater than that of a whisper.
Explanation:
The intensity of sound is given by;

where;
I is the intensity of the sound
I₀ is the threshold of sound intensity = 1 x 10⁻¹² W/m²
The intensity of sound at a rock concert

The intensity of sound of a whisper

Thus, the intensity of sound at rock concert is 10¹⁰ greater than that of a whisper.
Answer:
488.6KN
Explanation:
Hello!
the first step to solve this problem we must find the pressure exerted at the bottom of the tank (P) which is the sum of the external air pressure (P1 = 92kPa), the pressure inside the tank (P2 = 100kPa) and the pressure due to the weight of the water (P3), taking into account the above we have the following equation
P=P1+P2+P3
to find the pressure at the bottom of the tank due to the weight of the water we use the following equation

where
α=density=1 g/cm^3=1000kg/M^3
H=height=14.1m
g=gravity=3.71m/s^2
solving
P3=(1000)(14.1)(3.71)=52311Pa=52.3kPa
P=P1+P2+P3
P=100kPa+92kPa+52.3kPa=244.3kPa
finally to solve the problem we remember that the pressure is the force exerted on the area
