Decreases, stays the same, increases.
The volume decreases because as air is cooled, the individual molecules collectively possess less kinetic energy and the distances between them decrease, thus leading to a decrease in the volume they occupy at a certain pressure (please note that my answer only holds under constant pressure; air, as a gas, doesn't actually have a definite volume).
The mass stays the same because physical processes do not create or destroy matter. The law of conservation of mass is obeyed. You're only cooling the air, not adding more air molecules.
The density decreases because as the volume decreases and mass stays the same, you have the same mass occupying a smaller volume. Density is mass divided by volume, so as mass is held constant and volume decreases, density increases.
Answer:
Explanation:
We know that Impulse = force x time
impulse = change in momentum
change in momentum = force x time
Force F = .285 t -.46t²
Since force is variable
change in momentum = ∫ F dt where F is force
= ∫ .285ti - .46t²j dt
= .285 t² / 2i - .46 t³ / 3 j
When t = 1.9
change in momentum = .285 x 1.9² /2 i - .46 x 1.9³ / 3 j
= .514i - 1.05 j
final momentum
= - 3.1 i + 3.9j +.514i - 1.05j
= - 2.586 i + 2.85j
x component = - 2.586
y component = 2.85
0 mark is your answer because you want to start at 0
Answer:
11.7 m/s
Explanation:
To find its speed, we first find the acceleration of the center of mass of a rolling object is given by
a = gsinθ/(1 + I/MR²) where θ = angle of slope = 4, I = moment of inertia of basketball = 2/3MR²
a = 9.8 m/s²sin4(1 + 2/3MR²/MR²)
= 9.8 m/s²sin4(1 + 2/3)
= 9.8 m/s²sin4 × (5/3)
= 1.14 m/s²
To find its speed v after rolling for 60 m, we use
v² = u² + 2as where u = initial speed = 0 (since it starts from rest), s = 60 m
v = √(u² + 2as) = √(0² + 2 × 1.14 m/s × 60 m) = √136.8 = 11.7 m/s
Answer:
Frequency, 
Explanation:
Wavelength of a light wave is 450 nm. It is required to find the frequency of this light wave. The speed of light is given by c. So,
f is the frequency of this light
So, the frequency of this light is 
.
